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This watermark does not appear in the registered version - http://www.clicktoconvert.com 1 UNIT-I LESSON: 1 – SYMMETRY ELEMENTS AND SYMMETRY OPERATIONS CONTENTS 1.0. AIMS AND OBJECTIVES 1.1. INTRODUCTION 1.2. IDENTICAL CONFIGURATION 1.3. EQUIVALENT CONFIGURATION 1.4. SYMMETRY OPERATION 1.5. SYMMETRY ELEMENT 1.6. ROTATION AXIS OF SYMMETRY (Cn ) 1.7. PLANE OF SYMMETRY OR MIRROR PLANE (σ) 1.8. CENTER OF SYMMETRY OR INVERSION CENTER (i) 1.9. ROTATION-REFLECTION AXIS N-FOLD (Sn ) 1.10. IDENTITY (E) 1.11. INVERSE OPERATIONS 1.11.1. INVERSE OF s And i 1.11.2. INVERSE OF ROTATION [Cn -1 ] 1.11.3. INVERSE OF Sn 1.12. LET US SUM UP 1.13. CHECK YOUR PROGRESS 1.14. LESSON - END ACTIVITIES 1.15. REFERENCES 1.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on symmetry elements and symmetry operations to the students. On successful completion of this lesson the student should have: * Understand the symmetry elements and symmetry operations. 1.1. INTRODUCTION Symmetry is a very fascinating phenomenon in nature. It is found in geometrical figures such as a cube, a sphere, an equilateral triangle, a rectangle, a square, a regular pentagon, a regular hexagon etc. Its importance was recognized by eminent Greek philosophers Pythagoras and Plato. 1.2. IDENTICAL CONFIGURATION An identical configuration is the one which is not only indistinguishable from the original one but also identical with it. This watermark does not appear in the registered version - http://www.clicktoconvert.com 2 1.3. EQUIVALENT CONFIGURATION An equivalent configuration is the one which cannot be distinguished from the original one but need not be identical with it. 1.4. SYMMETRY OPERATION A symmetry operation is a movement of the molecule such that the resulting configuration of the molecule is indistinguishable from the original. 1.5. SYMMETRY ELEMENT A symmetry element is a geometrical entity such as a line or a space or a point about which an operation of rotation or reflection or inversion is done. 1.6. ROTATION AXIS OF SYMMETRY (Cn) It is also called rotational axis, if a rotation around an axis by 360°/n results in a molecule indistinguishable from the original. Examples are water (C2 ) and ammonia (C3 ). A molecule can have more than one symmetry axis and the one with the highest number of n is called the principal axis and takes the z-axis in a Cartesian coordinate system. This axis of symmetry can be explained by taking the example of triangular planar boron trichloride molecule. In boron trichloride molecule an axis of symmetry is located perpendicular to the plane containing all the atoms. This is known as the C3 axis of symmetry. In general the symbol for proper axis of symmetry is Cn where n is known as the order of the axis. The order of the axis is given by the number of rotations by q, to get the identical configuration. n i s æ 2p ö alternatively given by the formula n = ç ÷ , where q is the minimum angle of rotation to obtain è q ø the equivalent configuration. n has non- zero positive integral values. Boron trichloride molecule has three C2 axes of symmetry in addition to the C3 axis (Fig.1.1). The C3 axis in this molecule is known as principal axis. In general, if there are Cn axes of different orders in a molecule, the axis with the highest order is referred to as the principle axis. C2 Cl B Cl Cl (a) This watermark does not appear in the registered version - http://www.clicktoconvert.com 3 C3 Cl Cl B Cl (b) Fig.1.1. (a). The C2 axis of symmetry in BCl3 molecule. (b). The C3 principal axis in BCl3 molecule. 1.7. PLANE OF SYMMETRY OR MIRROR PLANE (σ) If reflection through a plane leaves an identical copy of the original molecule it has a plane of symmetry. Water has two of them: one in the plane of the molecule itself and one perpendicular to it. A symmetry plane parallel with the principal axis is dubbed vertical (σv ) and one per perpendicular to it horizontal (σh ). A third plane exists: if a symmetry plane bisects the angle between two n-fold axes that are perpendicular to the principal axis the plane is dihedral (δd). A plane can also be identified by its plane (xz),(yz) in the Cartesian coordination plane. C2 C2 sv sv O O H H H H Fig. 1.2. The reflection planes in water molecule. 1.8. CENTER OF SYMMETRY OR INVERSION CENTER (i) A molecule has a center of symmetry when for any atom in the molecule an identical atom can be found when it moves in a straight through this center an equal distance on the other side. An example is xenon tetrafluoride but not cisplatin even though both molecules are square planar. This is a point such that any line drawn through it meets the same atom at equal distances in opposite directions. All homo-nuclear diatomic molecules possess the centre of symmetry. This watermark does not appear in the registered version - http://www.clicktoconvert.com 4 Fig.1.3 lists the molecules with centre of symmetry. This element of symmetry is also called S2 axis. A particular symmetry element generates many symmetry operations. A Cn axis generates a set of operations C n1 , C n2 , C n3 ,....., C nn . The C nn operation is equivalent to the identity operation. H H O H H C C O F C N N H H F Fig.1. 3. Diagram showing molecules with centre of symmetry. 1.9. ROTATION-REFLECTION AXIS N-FOLD (Sn ) It is also called improper rotational axis. Molecules with this symmetry element can have a 360°/n rotation around an axis followed by a reflection in a plane perpendicular to it without a net change. An example is tetrahedral silicon tetrafluoride with three S3 axes and the staggered conformation of ethane with S6 symmetry. It is the line about which a rotation by a specific angle followed by reflection in a plane perpendicular to the rotation axis is performed. Fig.1.4 shows the S6 axis in staggered from of ethane. H H H C H H H C H H C H s C6 H H C H H H C C H H Fig.1.4. The S6 axis in staggered form of ethane. 1.10. IDENTITY (E) H H This watermark does not appear in the registered version - http://www.clicktoconvert.com 5 This is a default symmetry element and every molecule has one. 1.11. INVERSE OPERATIONS Suppose for a molecule we carry out an operation P followed by Q such that Q returns all the atoms of the molecule to their original position then, Q is said to be the inverse of P. In such cases, QP = E =PQ Algebraically we can express Q = P-1 , thus we can write P-1 P = E = PP-1 because an operation and its inverse always commute. 1.11.1. INVERSE OF s And i In the case of inversion and reflection, the carrying out of these operations in succession leads to identity E, i.e. s.s = s2 = E and i.i= i2 = E. Hence in these cases, these operations themselves are their own inverse that i = i-1 and s = s-1 . 1.11.2. INVERSE OF ROTATION [Cn-1 ] In the case of rotation, simply carrying out an operation for the second time does not give the original configuration. In Cn is the clockwise rotation by (2p/n)0 then Cn -1 is an anticlockwise rotation by (2p/n)0 about Cn . Then, Cn -1 Cn = E. For example, in NH3 molecule C 3-1C 3 = E . C3 C3 C3-1 C3+ N H1 H2 H3 Clockwise rotation Anti-clockwise rotation N H2 H3 N H1 H2 H3 H1 At the same time, for NH3 , C32 C31 = E The above way of carrying out the symmetry operations successively is algebraically represented as multiplication. If P and Q are two symmetry operations, PQ is a combined operation of carrying out Q first and then P. By conversion, first operation is written at right. For example, in the second case shown above for NH3 molecule, carrying out C2 first and then sv is represented as sv C2 . 1.11.3. INVERSE OF Sn As with Cn -1 , Sn-1 can be defined as rotation anticlockwise by (2p/n)0 followed by reflection with perpendicular plane. We can prove for improper axis. This watermark does not appear in the registered version - http://www.clicktoconvert.com 6 ( )( ) S n-1 S n1 = C n-1s h C n1s h = E C n1s = s C n1 Since ( )( ) ( )( S n-1S n = Cn-1s h Cn1s h = Cn-1s h s hCn1 ( ) ) = C n-1 s h s h C n1 = C n-1 EC n = C n-1C n = E However, we can express Sn -1 interms of clockwise rotation about the same axis. We know that Sn n= E when n is odd. Then S nn -1 S n1 = E (n-even) Thus, S n-1 = S nn -1 S n2 n-1 S n1 = E (n-odd) Thus, S n-1 = S n2 n -1 1.12. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Identical configuration Equivalent configuration Symmetry operation Symmetry element Rotation axis of symmetry (Cn ) Plane of symmetry or mirror plane (σ) Center of symmetry or inversion center (i) Rotation-reflection axis n-fold (Sn ) Identity (E) Inverse operations Inverse of s and i Inverse of rotation [Cn -1 ] Inverse of Sn 1.13. CHECK YOUR PROGRESS 1. 2. What is n improper axis of rotation? What are the operations generated byS5 ? How many of these are the distinct operations of S5 ? Prove the following: a) S2 =I b) S 62 = C 3 and c) S nn = E This watermark does not appear in the registered version - http://www.clicktoconvert.com 7 1.14. LESSON - END ACTIVITIES 1. 2. (a) Distinguish between (1). Symmetry element and symmetry operations. (2). Proper and improper rotation. (b) Show that C2 (z) and s(xy) commute. What is an inverse operation? Is this equivalent to any other combination of operations? Give an example. 1.15. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications. This watermark does not appear in the registered version - http://www.clicktoconvert.com 8 LESSON: 2 – GROUPS AND THEIR BASIC PROPERTIES CONTENTS 2.0. AIMS AND OBJECTIVES 2.1. INTRODUCTION 2.2. GROUP 2.2.1. BASIC PROPERTIES OF A GROUP 2.2.2. ORDER OF GROUP 2.2.3. ABELIAN GROUP 2.2.4. NON-ABELIAN GROUP 2.2.5. ISOMORPHISM 2.3. SIMILARITY TRANSFORMATION AND CLASSES 2.3.1. SIMILARITY TRANSFORMATION 2.3.2. CLASS 2.4. GROUP MULTIPLICATION TABLE 2.4.1. IMPORTANT CHARACTERISTICS OF A GROUP MULTIPLICATION TABLE 2.5. SYMMETRY CLASSIFICATION OF MOLECULES INTO POINT GROUPS: 2.6. DIFFERENCE BETWEEN POINT GROUP AND SPACE GROUP 2.7. LET US SUM UP 2.8. CHECK YOUR PROGRESS 2.9. LESSON - END ACTIVITIES 2.10. REFERENCES 2.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on groups and their basic properties to the students. On successful completion of this lesson the student should have: * Understand the groups and their basic properties. 2.1. INTRODUCTION Having defined various symmetry operations in Lesson 1 we may now ask ourselves whether it is possible to classify the molecules into ‘Groups’ on the basis of the symmetry elements they posses. Is it possible to define certain symmetry groups so that all molecules belonging to a certain group have the same type of symmetry operations? Luckily, the answer is ‘Yes’. This means that we can give an accurate description of the symmetry of any molecule by knowing to which group it belongs. 2.2. GROUP A group is a complete set of members which are related to each other by certain rules. Each member may be called an element. This watermark does not appear in the registered version - http://www.clicktoconvert.com 9 2.2.1. BASIC PROPERTIES OF A GROUP Certain rules have to be satisfied by the elements so that they form a group. These rules are the following: 1. 2. The product of any two elements and the square of any element must be elements of group (closure property). There must be one element in the group which commutes with everyone of the elements and leaves it unchanged. The associative law of multiplication should be valid. For every element there must be a reciprocal (inverse) and this reciprocal is also an element of the group. 3. 4. RULE 1 If A and B are the element of the group and if AB = C, C must be a member of the group. Product AB means that we perform the operation B first and then operation A i.e., the sequence of operations is from right to left. It should be noted that the other product BA need not be same as AB. BA means doing A first and then performing the operation B later. Let BA = D. D must be a member of the group by rule 1. Usually AB BA and so C D. However, there may be some special elements A and B each that AB = BA. Then A and B are said to commute with each other or the multiplication of A and B is commutative. Such a group where any two elements commute is called an abelian group. H2 O belongs to an abelian group. The four symmetry operations for H2 O are E, C2z, sv(xz) and s’v(yz).The inter-relationships between these operations are given in the group multiplication table (Table 2.1). Table 2.1. E E E C2z, C2z sv(xz) sv(xz) s’v(yz) s’v(yz) C2z C2z E s’v(yz) sv(xz) sv(xz) sv(xz) s’v(yz) E C2z sv(xz) C2z E s’v(yz) s’v(yz) Note that each member is its own inverse. The product of any two operations is found among the four members. ( ) E 2 = C 22 = s v' 2 =E s v s v' = s v' s v = C2 ; C2 s v = s v C2 = s v' etc All these are noted from the table. This watermark does not appear in the registered version - http://www.clicktoconvert.com 10 RULE 2 Each group must necessarily have an element which commutes with every other element of the group and leaves it unchanged. Let A and B be the elements of the group. Let X be the element satisfying rule 2. i.e. XA = AX = A and also XB = BX = B. BA = BX2 A; BX2 = B = BE, where we have set X2 = E (identity) It is clear BEn = B, n being any integer. This kind of element E which does not effect any change when multiplied with any element is a unique element and is called an identity operation E. For water, E, the identity operation satisfies this rule. It is so for all molecules. RULE 3 Associative law of multiplication must be valid. This means ABCD is the same as (AB) (CD), (A) (BCD) or (ABC) (D). ABC is the same as A(BC) or (AB)C. For example, we have for water ( ) C 2s v Es v' = (C 2s v ) Es v' = s v' s v' = E ( ) C 2s v' Es v = C 2 s v' E s v = C 2s v s v' = C 2 Multiplication simply means successive application of the symmetry operations in the order right to left. RULE 4 Inverse of an element A is denoted by A-1 (this does not mean 1/A). It is simply an element of the group such that A-1 A=E. In case of symmetry groups, A-1 is that element which undoes or annuls the effect of A. For H2 O we have, for example, C2 C2 = E. Therefore C 2-1 = C 2 i.e., C2 is its own inverse. This is true of all other elements for H2 O. But this is not general. For example, C 62 = C3 ¹ E . Therefore C 6-1 is not C6 . 2.2.2. ORDER OF GROUP The total number of elements present in a group is known as the order of the group. It is denoted by n. Example: 1. 2. Water molecule belongs to C2v group of order 4 because it contains 4 elements namely E, C2z, sv and s’v. Ammonia belongs to C3v group of order 6 as it contains 6 elements namely E, C3 1 , C32 , sv(1), sv(2) and sv(3). This watermark does not appear in the registered version - http://www.clicktoconvert.com 11 2.2.3. ABELIAN GROUP A group is said to be abelian if for all pairs of elements of the group, the binary combination is commutative. That is AB = BC; BC = CB – and so on. Example: The elements of C2v point group E, C2z, sv and s’v form an abelian group as all the elements of this group commute with each other. 2.2.4. NON-ABELIAN GROUP A group is said to be non-abelian if the commutative law does not hold for the binary combinations of the elements of the group, i.e., AB BA. Example: The elements of C3v point group E, C 3 1 , C3 2 , sv(1), sv(2) and sv(3) donot consecutive an abelian group since the elements donot follow commutative law. 2.2.5. ISOMORPHISM Two groups are supposed to be isomorphic if they obey the following rules. 1. 2. 3. Both have same order and structure. There is a one-to-one correspondence in all respects between the members of the two groups. If A1 , B 1 , C 1 , D 1 and A2 , B 2 , C 2 , D2 are the members of the two isomorphic groups, A1 corresponds to A2 , B1 corresponds to B2 and so on. The relationship between the any two members of a group is exactly the same as the relationship between the corresponding members of the other group. Let us take the three groups listed below: (i) E, C2 (ii) E, i (iii) E, sh All the three are isomorphic groups. E = C 22 ; C 2 E = C 2 E = i 2 ; iE = i E = s h2 ; s h E = s h etc. 2.3. SIMILARITY TRANSFORMATION AND CLASSES 2.3.1. SIMILARITY TRANSFORMATION Let A and X be the elements of a group and let us define B such that B = X-1 AX This watermark does not appear in the registered version - http://www.clicktoconvert.com 12 B is called the similarity transform of A by X, or A is said to be subjected to similarity transformation with respect to X. If A and B are related by a similarity transformation they are called conjugate elements. Take the NH3 molecule, for instance. Z axis is the C3 axis. Z Z s "' N Hc 120° C3Z N Hc Ha N Ha Hb Hb Hc Hb Ha s"' Z N Hc Ha C32Z N Ha Hb Hb Hc Fig. 2.1. Illustration of similarity transformation on NH3 . 1. 2. 3. There are three reflection planes. These are usually designated as follows: Plane formed by z-axis and NHa bond: s’ or sa or sv’. Plane formed by z-axis and NHb bond: s’’ or sb or sv’. Plane formed by z-axis and NHc bond: s’’’ or sc or sv’. Let us prefer the designation s’, s’’ and s’’’. Let us perform a reflection (s’’’) with respect to the plane formed by NHc and z-axis. Let us perform s again. s’’’2 = E. Now let us find the similarity transform of C3 w.r.t. s’’’, i.e., (s’’’)-1 C3 s’’’ = ? It is seen from the Fig. that (s’’’)-1 C3 s’’’ = C 32 . Remember (s’’’) = (s’’’)-1 . Thus C3 and C 32 are conjugate elements. The following rules about conjugate elements are notable: 1. 2. Every element is conjugate of itself because every element is the similarity transforms of itself w.r.t. identity (E): E = E-1 and A = E-1 AE. If A is the conjugate of B then B is the conjugate of A. This means that if A is the similarity transform of B by X, B is the similarity transform of A by X-1 . We have A = X-1BX This watermark does not appear in the registered version - http://www.clicktoconvert.com 13 But (X-1 )-1 AX-1 = XA X-1 = X (X-1 BX) X-1 = (XX-1 ) B (XX-1 ) = B (associative law) 3. If A is the conjugate of B and B is the conjugate of C, then A, B and C are mutually conjugate. 2.3.2. CLASSES A complete set of elements which are conjugate to one another is called a class of the group. Let us consider NH3 . Set us the coordinate system in such a manner that ZNHa is in the yz plane. (Fig. 1) s’ is then syz. Without disturbing the NH3 molecule rotate the coordinate system by 120° w.r.t z axis, i.e., subject the coordinate system to C3 . Now yz plane is ZNHb . syz is s’, s’’ and s’’’ are equivalent. s’ becomes same as that of s’’ if we change the coordinate system by a symmetry operation (C3 ) of the point group. s’ and s’’ are therefore in the same class. Example: Show that the three reflections of NH3 constitute a class. It is not difficult to show -1 that C .C 3 = E . Hence, C32 = (C3 ) . Let as perform the similarity transformation of s’ by C3 in NH3 . 2 3 C3-1s 'C3 = C32s 'C3 = s " Thus s’ and s’’ are conjugate. Similarly we can show that s’, s’’ and s’’’ are mutually conjugate. Therefore s’, s’’ and s’’’ form a class. 2.4. GROUP MULTIPLICATION TABLE Every group is characterized by a multiplication table. The relationship between the elements of the binary combinations is reflected in the multiplication table. Consider a water molecule. It has four symmetry elements, viz., E, C2 (z), sv (xz) and sv ’(yz) (Fig.2.2). This watermark does not appear in the registered version - http://www.clicktoconvert.com 14 sv (xz) z C2 y O H H sv '(yz) x Fig. 2.2. The Four symmetry elements of H2 O molecules We can easily show that the product of any two symmetry elements is one of the four elements of the group. Thus, for instance, C2 (z)sv (xz) = sv ’(yz). Proceeding this way the symmetry operations of H2 O can be listed in a group multiplication table (GMT) (Table 2.2). E C2 (z) sv (xz) sv ’(yz) E C2 (z) sv (xz) sv ’(yz) E C2 (z) sv (xz) sv ’(yz) C2 (z) E sv ’(yz) sv (xz) sv (xz) sv ’(yz) E C2 (z) sv ’(yz) sv (xz) C2 (z) E Table 2.2. Group multiplication table of the symmetry operations of H2 O molecule 2.4.1. IMPORTANT CHARACTERISTICS OF A GROUP MULTIPLICATION TABLE 1. 2. 3. 4. 5. It consists of h rows and h columns where h is the order of the group. Each column and row is labeled with group element. The entry in the table under a given column and along given row is the product of the elements which head that column and the row (multiplication rule is strictly followed). At the intersection of the column labeled by Y and the row labeled by X, we found the element which is the product XY. The following rearrangement theorem holds good for every ‘Group Multiplication Table’. “Each row and each column in the table lists each of the group elements once and only once. No two rows may be identical nor any two columns be identical. Thus each row and each column is a rearranged list of the group elements”. 2.5. SYMMETRY CLASSIFICATION OF MOLECULES INTO POINT GROUPS Molecules can be classified into point groups depending on the characteristic set of symmetry elements possessed by them. A molecular group is called a point group since all the elements of symmetry present in the molecule intersect at a common point and this point remains fixed under all the symmetry operations of the molecule. The symmetry groups of the molecules are denoted by specific symbols known as Schoenflies notations. This watermark does not appear in the registered version - http://www.clicktoconvert.com 15 Table 2.3. Some Molecular Point Groups Point Group C1 C2 C3 Cs C2v C3v C¥v C2h D2h D3h D4h Symmetry Elements Examples D6h Td Oh C 6 H6 E,2C6 ,6C2 (^ to C6 ),3sv , 3sd, sh , C2 ,2C3 ,2S6 , 2S3 ,i CH4 E, 4C3 ,3C2 ,3S4 (coincidence with C2 ),6sd E,3C4 ,4C3 ,3S4 and 3C2 (both coincident with the C4 SF6 axes), 6C2 ,4S6 , 3sh , 6sd E CHFClBr E,C2 H2 O 2 E,C3 C 2 H6 NOCl E,sv H2 O,CH2 =O, pyridine E,C2 ,2sv NH3 ,CHCl3 ,PH3 E,C3 ,3sv HCl, NO,CO E,C¥,¥sv trans CHCl=CHCl E,C2 , sh , i CH2 =CH2 ,naphthalene E,3C2 , 3s, i BF3 (trigonal planar) E,2C3 , 3C2 (^ to C3 ),3sv , sh , 2S3 E,C4 , 4C2 (^ to C4 ),2sv , 2sd, sh , C2 , S4 (coincidence with [PtCl4 ]2- ( s q u a r e planar) C4 ), i 2.6. DIFFERENCE BETWEEN POINT GROUP AND SPACE GROUP Symmetry operations do not alter the energy of the molecule. Further in all the above operations the centre of the molecule is not altered as none of the operations involve a total translational movement of the molecule. Whatever happens to the molecule, the centre (point) is not changed. At least one point is fixed. Hence these are classified as ‘point group’ operations. In case of crystals operations such as ‘screw rotations’ and glide plane reflections can be additionally specified. Screw rotation involves a rotation with respect to an axis and then a translation in the direction of the same axis. Glide plane reflection is a reflection in a plane followed by a translation along a line in that plane. These are particular to crystals and the classification comes under what is known as space group. Note that here even the centre changes. Thus in short, in point group, there is at least one point (centre) which is not altered after all operations while in space group it is not possible to identify such a stationary point. 2.7. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Group Basic properties of a group Order of group Abelian group Non-abelian group This watermark does not appear in the registered version - http://www.clicktoconvert.com 16 Ø Ø Ø Ø Ø Isomorphism Similarity transformation and classes Group multiplication table Symmetry classification of molecules into point groups Difference between point group and space group 2.8. CHECK YOUR PROGRESS 1. 2. Explain why a set of numbers cannot form a group by the process of division. Explain why the set of integers between 0 and µ do not form a group under the process of multiplication. 2.9. LESSON - END ACTIVITIES 1. 2. Construct the multiplication table for the C3v point group to which NH3 molecule belongs. Draw the structure of three distinct isomers of C2 H2 Cl2 and determine their point groups. Which of them is polar? 2.10. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications. This watermark does not appear in the registered version - http://www.clicktoconvert.com 17 UNIT-II LESSON 3: REDUCIBLE AND IRREDUCIBLE REPRESENTATIONS CONTENTS 3.0. AIMS AND OBJECTIVES 3.1. INTRODUCTION 3.2. REDUCIBLE REPRESENTATION 3.3. IRREDUCIBLE REPRESENTATION 3.4. GRAND/GREAT ORTHOGONALITY THEOREM (G.O.T.) 3.5. CHARACTER TABLES FOR POINT GROUPS 3.6. CALCULATION OF CHARACTER VALUES OF REDUCIBLE REPRESENTATION PER UNSHIFTED ATOM FOR EACH TYPE OF SYMMETRY OPERATION 3.6.1. IDENTITY (E) 3.6.2. INVERSION AT THE CENTRE OF SYMMETRY (i) 3.6.3. REFLECTION IN A SYMMETRY PLANE (s) 3.6.4. PROPER ROTATION Cn ’ 3.7. DETERMINATION OF TOTAL CARTESIAN REPRESENTATION T3N 3.8. DETERMINATION OF DIRECT SUM FROM TOTAL CARTESIAN REPRESENTATION 3.9. LET US SUM UP 3.10. CHECK YOUR PROGRESS 3.11. LESSON - END ACTIVITIES 3.12. REFERENCES 3.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on reducible and irreducible representations to the students. On successful completion of this lesson the student should have: * Understand the reducible and irreducible representations. 3.1. INTRODUCTION The set of matrices corresponding to the symmetry operations of a group is called its representation. Representations can be classified into (a) Reducible representations (reps) and (b) irreducible representations (irreps). 3.2. REDUCIBLE REPRESENTATION Let A, B, C… Be the matrices which form the representation of a group and let X be the similarity transformation matrix of this group such that This watermark does not appear in the registered version - http://www.clicktoconvert.com 18 X-1 AX = A’ ----- (1) X-1 BX = B’ ----- (2) X-1 CX = C’ ----- (3) Then, if X is the proper transformation matrix, we have a1' 0 a2' X-1AX = A' = a3 ' 0 a4' ----- (4) The new matrix A’ is now blocked out along the diagonal into smaller matrices a1 ’, a2 ’, a3’, a4’, etc., with the off-diagonal elements equal to zero. Similarly, we have b1' 0 b2' X-1BX = B' = b3' 0 b4' ----- (5) This is expressed by saying that the given sets of matrices form a reducible representation (rep). 3.3. IRREDUCIBLE REPRESENTATION If it is not possible to find a similarity transformation which will reduce the matrices A, B, C ... to block-diagonalized form, the representation is called an irreducible representation (irrep). 3.4. GRAND/GREAT ORTHOGONALITY THEOREM (G.O.T.) This is the most important theorem of group theory. It concerns the matrix elements which constitute the irreps of a point group. Mathematically it is stated as follows: å G (R ) i R G j (R ) m ' n ' = * mn h li l j d ij d mm 'd nn ' ----- (6) This watermark does not appear in the registered version - http://www.clicktoconvert.com 19 Here Gi and Gj are the ith and jth irreps of a point group of order h with dimensions li and lj, respectively; Gi (R)mn is the mnth matrix element corresponding to the symmetry operation R belonging to the ith irrep and Gj (R)mn, is the complex conjugate of the m’n’th matrix element corresponding to the symmetry operation R belonging to the jth irrep. The ds are the well known ‘Kronecker deltas’ which have the following property: ì1, i = j ì1, m = m' ì1, n = n' d ij = í ; d mm ' = í ; d nn ' = í î0, i ¹ j î0, m ¹ m' î0, n ¹ n' ----- (7) The summation is performed operations R of the molecule. If the matrix elements are real, then Gi (R) m’n’ = Gj (R)m’n’ ----- (8) The following three cases arise for the G.O.T. assuming that the matrix elements are real: 1. For two different irreps, i å G (R ) i mn j, m=m’ and n=n’, G j (R )mn = 0 ----- (9) R 2. For the same irrep i=j, m m’ and n n’, å Gi (R )mn G j (R )m'n' = 0 ----- (10) R 3. For an irrep I and for m=m’, n=n’, å [G (R ) ] 2 i mn ----- (11) = h li R In practice, we do not use the G.O.T. in the form given above but in a slightly different form involving the characters of the irreps. 3.5. CHARACTER TABLES FOR POINT GROUPS For practical purposes, it is sufficient to know only the characters of each symmetry class of a point group to which a molecule belongs. A character table lists the characters of all the symmetry classes for all the irreps of a group. Character tables for the C2v and C3v point groups are given in Tables 3.1 and 3.2. Table 3.1. Character table for C2v point group C2v I A1 A2 B1 B2 E 1 1 1 1 C2 (z) 1 1 -1 -1 sv (xz) II 1 -1 1 -1 Table 3.2. Character table for C3v point group sv ’(yz) 1 -1 -1 1 III z Rz x, Ry y, Rx IV x2 ,y2 ,z2 xy xz yz This watermark does not appear in the registered version - http://www.clicktoconvert.com 20 C3v I A1 A2 E E 1 1 2 2C3 II 1 1 -1 3sv 1 -1 0 III IV z x2 +y2 ,z2 Rz (x,y)(Rx ,Ry ) (x2 - 2 ,xy)(xz,yz) The character tables can be obtained from the properties of the irreps given above. We can explain the character table by dividing it into four sections I, II, III, IV. Section I. In the top row, the Schoenflies symbol for the point group is given. This section also lists the Mulliken symbols for the different irreps. Symbols A and B are used to label one dimensional irreps and E and T to label two dimensional and three-dimensional irreps, respectively. The nomenclature E should not be confused with the identity operation. A is used when the character for the rotation about the principal axis is +1 and B when it is -1. In other words, A stands for symmetric and B for antisymmetric to such rotation. For a molecule having a centre of symmetry, the subscripts g and u are used to label the irreps that are respectively symmetric and antisymmetric to inversion through the centre of symmetry. Subscripts 1 and 2 are used to label the irreps that are resoectuvekt symmetric and antisymmetric to reflection in a vertical plane s v . The superscripts’ “are used it denote the irreps that are respectively symmetric and antisymmetric to reflection in a horizontal plane s h . Section II. This section gives the characters for all the symmetry operations of different irreps. The characters of the identity operation for the one-dimensional, two-dimensional and three- dimensional irreps are, respectively, 1, 2 and 3. Section III. It gives the transformation properties of the Cartesian coordinates x,y,z and rotations Rx , Ry , Rx about these axes. Section IV. It gives the transformation properties of the binary products of Cartesian coordinates xy,yz,zx etc. and the squares of the coordinates x2 , y2 , z2 , x2 + y2 , x2 + y2 +z2 etc. The Cartesian coordinates, their squares and binary products, etc., listed in sections III and IV are referred to as the basis functions on which the symmetry operations operate. 3.6. CALCULATION OF CHARACTER VALUES OF REDUCIBLE REPRESENTATION PER UNSHIFTED ATOM FOR EACH TYPE OF SYMMETRY OPERATION The contribution to c (R ) per unshifted atom for all symmetry operations R can be worked out in the following manner. 3.6.1. IDENTITY (E) In this case, all three vectors remain unchanged for every unshifted atom as shown in Fig 3.1 where x’=x, y’=y and z’=z. The transformation matrix therefore, includes diagonal elements: This watermark does not appear in the registered version - http://www.clicktoconvert.com 21 +1 0 0 0 +1 0 0 0 +1 Then c (E ) per unshifted atom is +3. z' z E y' y x x' Fig. 3.1 3.6.2. INVERSION AT THE CENTRE OF SYMMETRY (i) Fig. 3.2 shows the effect for each unshifted atom, where x’ = - x, y’ = -y, and z’ = -z. Therefore, the matrix contains the following diagonal elements: -1 0 0 0 -1 0 0 0 -1 Thus c (i ) per unshifted atom is -3. z x' i y y' x z' Fig. 3.2 This watermark does not appear in the registered version - http://www.clicktoconvert.com 22 3.6.3. REFLECTION IN A SYMMETRY PLANE (s) The effect of any s on an unshifted atom is typically shown in Fig. 3.3, where x’ = x, y’ = y and z’ = z. The transformation matrix, therefore, +1 0 0 0 -1 0 0 0 +1 Thus c (s ) per unshifted atom is +4. z' z s y y' x x' Fig. 3.3 3.6.4. PROPER ROTATION Cn ’ Rotation is by (360/n)0 , usually about z axis. For the unshifted atom, the result is as shown in Fig 3.4, where q = (360/n)0 . x’ =z, contributing +1 to c (C n' ) and x.y go to x’, y’ respectively. z' y Cn z x x' y' Fig. 3.4 3.7. DETERMINATION OF TOTAL CARTESIAN REPRESENTATION T3N The total Cartesian representation T3N can be derived from the contribution c (R ) per unshifted atom by simple arithmetic multiplication of c (R ) with the number of unshifted atoms for every symmetry operation. Hence the calculation of T3N involves two steps (i) to count the number of unshifted atoms for every symmetry operation. (ii) to calculate the contribution to This watermark does not appear in the registered version - http://www.clicktoconvert.com 23 c (R ) for every unshifted atom in every type of symmetry operation. T 3N is also called the reducible representation of the group for a particular transformation. ILLUSTRATIONS (i) For H2 0 molecule This molecule belongs to C2v group. The number of unshifted atom for each symmetry operation and the resultant T3N can be given as C2v sxz syz 1 1 3 -1 1 3 E C2 unshifted atoms 3 T 3N 9 O H H (ii) For POCl3 molecule This molecule belongs to C3v point group. Since it is made of five atom, it will give 15 ´ 15 matrices. Using the method of unshifted atoms, the reducible representations can be worked out as given below C2v E 2C3 3sv O unshifted atoms 5 2 3 P T 3N 15 0 3 Cl Cl Cl Rotation by C3 1 or C32 leaves P and O unshifted. Reflection by any sv leaves P, O and one Cl unshifted. 3.8. DETERMINATION OF DIRECT SUM FROM TOTAL CARTESIAN REPRESENTATION The direct sum of total Cartesian representation T3N can be determined using the reduction formula. This watermark does not appear in the registered version - http://www.clicktoconvert.com 24 ILLUSTRATION (1) POCl3 molecule The unshifted atoms and total Cartesian representation for this molecule which belongs to C3v point group is given by C 3v unshifted atoms T 3N E 2C 3 5 2 15 0 C3v 3sv A1 3 3 E 2C3 1 1 1 A2 1 1 -1 E 2 -1 0 By applying reduction formula 1 [(1 ´ 15 ´ 1) + (2 ´ 0 ´ 1) + (3 ´ 3 ´ 1)] = 4 6 1 a( A2 ) = [(1 ´ 15 ´ 1) + (2 ´ 0 ´ 1) + (3 ´ 3 ´ -1)] = 1 6 1 a(E ) = [(1 ´ 15 ´ 2 ) + (2 ´ 0 ´ -1) + (3 ´ 3 ´ 0 )] = 5 6 a( A1 ) = Therefore, the direct sum for total Cartesian representation is T3N = 4A1 + A2 + 5E (2) Reducible representation and direct sum for T3N for [PtCl4 ] 2[PtCl4 ] 2- belongs to D4h point group Cl Cl Pt Cl Cl D4h E 2C4 C2 unshifted atoms 5 1 1 3 T 3N 15 1 -1 -3 2C2' i 2S4 sh 2sv 2sd 1 1 1 5 3 1 -1 -3 -1 5 3 1 2C2'' 3sv Using the character table for D4h point group and reduction formula, it can be shown that This watermark does not appear in the registered version - http://www.clicktoconvert.com 25 T3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu 3.9. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Reducible Representation Irreducible representation Grand/great Orthogonality theorem (G.O.T.) Character tables for point groups Calculation of character values of reducible representation per unshifted atom for each type of symmetry operation Identity (e) Inversion at the centre of symmetry (i) Reflection in a symmetry plane (s) Proper rotation Cn Determination of total cartesian representation T3n Determination of direct sum from total cartesian representation 3.10. CHECK YOUR PROGRESS 1. 2. Define reducible and irreducible representation. Construct the C2v character table. 3.11. LESSON – END ACTIVITIES 1. 2. State and explain the Great Orthogonality Theorem. Use the conclusions obtained from the Orthogonality theorem to construct the character table for C2v group. Using the Great Orthogonality theorem to construct the character table for C3v point group. 3.12. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications. This watermark does not appear in the registered version - http://www.clicktoconvert.com 26 LESSON 4: GROUP THEORY AND VIBRATIONAL SPECTROSCOPY CONTENTS 4.0. AIMS AND OBJECTIVES 4.1. INTRODUCTION 4.2. GROUP THEORY AND NORMAL MODES OF VIBRATION OF POLYATOMIC MOLECULES 4.3. INFRA-RED ABSORPTION AND RAMAN SCATTERING SPECTROSCOPY 4.4. DETERMINATION OF SYMMETRY PROPERTIES OF VIBRATIONAL MODES 4.5. SYMMETRY SELECTION RULES FOR INFRA – RED RAMAN SPECTRA 4.6. MUTUAL EXCLUSION RULE 4.7. LET US SUM UP 4.8. CHECK YOUR PROGRESS 4.9. LESSON - END ACTIVITIES 4.10. REFERENCES 4.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on group theory and vibrational spectroscopy to the students. On successful completion of this lesson the student should have: * Understand the group theory and vibrational spectroscopy. 4.1. INTRODUCTION Group theory is a very powerful tool at the hands of a chemist, a theorist and a spectroscopist. It finds many applications the details of which are beyond the scope of this lesson. Some important applications of group theory are listed below. 1. Construction of hybrid orbitals. 2. Construction of SALCs (symmetry adapted linear combinations of atomic orbitals). SALCs are used in molecular orbital theory (MOT) of chemical bonding. 3. Determination of the irreducible to which the vibrational modes of molecules belong. 4. Determining which spectral transitions in infrared and Raman spectra are allowed or forbidden. 5. Determining the selection rules for transitions in carbonyl compounds and other chromophores. It is found that the former transitions are forbidden whereas the latter are allowed. 6. Determining which molecules are polar or nonpolar. 4.2. GROUP THEORY AND NORMAL MODES OF VIBRATION OF POLYATOMIC MOLECULES Group theory helps in two aspects of vibrational spectroscopy. This watermark does not appear in the registered version - http://www.clicktoconvert.com 27 (i) (ii) Firstly helps in the classification of the normal modes of vibrations according to the irreducible representations of the point group of the molecule. Secondly, it helps in qualitatively finding out Raman and IR spectral activity of the fundamentals as well as overtone and combination bands. The number of fundamental modes of vibrations can be worked out in the following manner: Let us consider a molecule having N atoms. If we specify the position of each atom in space with the coordinates x, y and z, then there will be 3N coordinates (degrees of freedom) for the entire system. Since the molecule has translational, rotational and vibrational motion, these 3N coordinates (degrees of freedom) can be assigned to each type of motion as given below. Motion Degrees of freedom to describe the motion Linear 3 2 3N-5 Translation Rotation Vibration Non linear 3 3 3N-6 These degrees of freedom for vibrational motion are called the normal or fundamental modes of vibration. Depending on the type of molecule, these normal modes may be active either in IR or Raman or both. Example, the normal modes of vibration in H2 O and CO2 molecules. O H O H C O O O H O H H C O O H C O O C O A knowledge of the symmetry of vibrational modes in molecules will be helpful in predicting whether these modes of vibrations will give rise to infrared or Raman spectrum or both these spectra. 4.3. INFRA-RED ABSORPTION AND RAMAN SCATTERING SPECTROSCOPY This watermark does not appear in the registered version - http://www.clicktoconvert.com 28 In infra-red spectrum, the sample is irradiated with infra-red radiation leading to absorption of the radiation at frequencies corresponding to the absorptions give the vibrational frequencies of that molecule. Thus, infrared technique is a direct measurement of the vibrational frequencies which lie in infrared region. In Raman spectroscopy, the molecules are irradiated with uv or visible radiation causing perturbation of the molecule which in turn induces vibrational transitions. As a result, energy is taken up from or given out to the incident radiation, which is scattered at a shifted frequency. The differences in frequency between incident light and Raman scattered light correspond to vibrational frequencies. Further in Raman spectroscopy, plane-polarized light is used as incident light. The scattered light may be still polarized or depolarized. Hence, certain frequencies and hence certain vibrational modes may be found to give polarized Raman light and others give depolarized scattered light. Since in Raman spectroscopy a higher energy radiation than IR is used, the resulting data would consist of a series of infra-red absorption lines and a series of Raman scattered lines. These lines could be assigned to different vibrational modes only if the symmetry properties of these modes are clearly understood. 4.4. DETERMINATION OF SYMMETRY PROPERTIES OF VIBRATIONAL MODES The simple procedure for obtaining the representations of vibrational modes and hence understanding their symmetry properties involve the following steps: (i) Assign point group to the given molecule. (ii) Deduce the reducible representation T3N for all the symmetry operations of the point group using the relation c xyz (R ) = U R c i (R ) ------ (1) Where c xyz (R ) = The character for the operation R in the reducible representation U R = The number of unshifted atoms for the operation R c i (R ) = The character for operation R per unshifted atom. (iii) The reducible representation T 3N for the molecule is split into the various irreducible representations of the point group by using the standard reduction formula ( h )å g c (R )c (R ) ai = 1 p p i p ----- (2) Rp (iv) The irreducible representations T3N thus obtained correspond to the translational, rotational and vibrational degrees of freedom. Thus T3N can be written as T3N = Tx + Ty + Tz + Rx + Ry + Rz +Tvib Tvib = T3N- [Tx + Ty + Tz + Rx + Ry + Rz] This watermark does not appear in the registered version - http://www.clicktoconvert.com 29 ILLUSTRATIONS (a) Water molecule: (i) Water molecule belongs to C2v point group. (ii) The symmetry elements of this point group are E C2v s xz s yz C2 O syz H H sxz (iii) The reducible representation T3N for this point group is C2v E C2 sxz syz T 3N 9 -1 1 3 (iv) This reducible representation is decomposed into various irreducible representations using the standard reduction formula and by using the character table for this group. C2v E C2 sxz syz A1 1 1 1 1 Tz A2 1 1 -1 -1 Rz B1 1 -1 1 -1 T x ,R y B2 1 -1 -1 1 T y,R x 1 [1.9.1 + 1.(- 1).1 + 1.1.1 + 1.3.1] = 3 4 1 a A 2 = [1.9.1 + 1.(- 1).1 + 1.1.(- 1) + 1.3.(- 1)] = 1 4 a A1 = This watermark does not appear in the registered version - http://www.clicktoconvert.com 30 1 [1.9.1 + 1.(- 1)(. - 1) + 1.1.1 + 1.3.(- 1)] = 2 4 1 = [1.9.1 + 1.(- 1)( . - 1) + 1.1.(- 1) + 1.3.1] = 3 4 a B1 = aB2 Thus T3N is given by T3N = 3A1 + A2 + 2B1 + 3B2 (v) The sum of the irreducible representations of vibrational modes Tvib is related to T3N by the relation Tvib = T3N- [Tx + Ty + Tz + Rx + Ry + Rz] Using the values of Tx , Ty , Tz , Rx , Ry and Rz in the character table for this groups , we get Tvib = (3A1 + A2 + 2B1 + 3B2 ) – (B1 + B2 + A1 + B2 + B1 + A1 ) Tvib = 2A1 + B2 Thus the three normal modes of vibration of water molecule belong to A1 and B2 representations. Of these, those vibrations which belong to A1 symmetry are called totally symmetric vibrations. SO2 molecule also has the same symmetry properties as water molecule and hence its sum of representations of vibrational modes Tvib is also Tvib = 2A1 + B2 (b) Ammonia Molecule (i) (ii) This belongs to C3v point group The various symmetry elements of this group are E C2v s xz s yz C2 N H (iii) The reducible representation T3N using reduction formula, we get T 3N (iv) H H E 2C3 12 0 3sv By decomposing T3N using reduction formula, we get 2 This watermark does not appear in the registered version - http://www.clicktoconvert.com 31 T3N = 3A1 + A2 + 4E (v) A reference to the character for C3v group gives Tx + Ty + Tz = A1 + E Rx + Ry + Rz = A2 + E Thus Tvib is obtained as Tvib = 2A1 + 2E Thus ammonia has four vibrational modes. (c ) BF3 molecule (i) (ii) This molecule belongs to D3h point group. The various symmetry operations present are E 2C3 3C2 s h 2S3 3sv F F B F (iii) The reducible representation T3N for BF3 molecule is T 3N (iv) E 2C3 3C2 sh 2S3 3sv 12 0 -2 4 -2 2 This representation is decomposed by reduction formula to give T3N = A1 ’ + A2 ’ + 3E’+ 2A2 ”+E”. The symmetries of the translations (Tx, Ty, Tz) and rotations (Rx, Ry and Rz) as given by the character table for D3h are Translation Rotation (T x ,T y) : E" (R x ,R y) : E" T z : A 2" R z : A 2" This watermark does not appear in the registered version - http://www.clicktoconvert.com 32 Thus Tvib is given by Tvib = 2A1 ’+ 2E’+A2 ” (d) By a similar treatment the Tvib for the following molecules are obtained as Molecule SO2 Symmetry T3N C2v 3A1 + A2 + 2B1 + 3B2 Tvib 2A1 + B2 POCl3 C3v 4A1 + A2 +5E 3A1 + 3E PtCl4 2- D4h A1g + A2g + B1g + B2g+Eg +2A2u+B2u+3Eu A1g+ B1g+ B2g+A2u+B2u+2Eu RuO4 Td A1 + E +T1 + 3T2 A1 + E +2T2 Knowing the symmetries of all the vibrational modes of a molecule, it is possible to predict which of them will be active in the infra-red and Raman spectra. To do this we must have knowledge of symmetry selection rules for two effects. 4.5. SYMMETRY SELECTION RULES FOR INFRA – RED RAMAN SPECTRA (A) SELECTION RULE FOR IR Consider a transition from the vibrational ground state of a molecule with wave function y 0 to an excited vibrational state with wave function y i . The probability of such a transition Pi occurring is given by Pi = òy 0 my i dt Where m = dipole moment of the molecule (a vector) dT =implies integration carried over all possible variables of the wave functions. The vector m can be split into three components mx , my , mz along the three Cartesian coordinates and one of the three integrals is given by Pik = òy 0 m ky i dt (k = x, y, z ) Symmetry selection rules: (i) If one of the three integrals òy 0 m ky i dt is non-zero, then that vibrational mode is inactive. This occurs when the product y 0 my 1 is totally symmetric, u, the character of this direct product function is +1 for all symmetry opera5tions of the relevant point group. This watermark does not appear in the registered version - http://www.clicktoconvert.com 33 (ii) If the integral òy 0 m ky i dt is zero, the probability of that transition is zero. Then it is said to be forbidden in infra-red. How to find out symmetry of a particular vibration The integral consists of three product functions. Of these (i) y o , the ground state vibrational wave function, is always totally symmetric. (ii) The symmetry properties of mk and those of a translational vector along the same axis Tk are the same. (iii) The symmetry properties of y i are the same as those of vibrational mode i. Thus, if a vibrational mode has the same symmetry property as one of the translation vectors, Tx , Ty , Tz for that point group, then a transition from the ground state to that mode (in excited state) will be infra-red active. (B) RAMAN SELECTION RULE The probability of a vibrational transition occurring in Raman scattering is given by Pi = òy 0ay i dt Where a = Polarizability of the molecule (a tenser) The Raman effect depends upon a molecular dipole induced in the molecule by the electromagnetic field of incident radiation. The induced dipole is proportional to the polarizability (a) of the molecule, which a measure of the ease with which the molecular electron distribution could be disterted. Since a, is a tensor, there are only six distinct components, viz, [For vibrational transitions ajk = akj (where j,k = x,y,z )] SELECTION RULE For a vibrational mode to be Raman active one of the six integral of the form òy 0 a jky i dt ¹ 0 ( j , k = x, y , z ) As in the case of IR, the above integral will be non-zero only if a jk has the same symmetry as the mode described by the wave function yi. It is found that a jk has the same symmetry properties as x2 , a xy as xy and so on) Thus, a transition from the ground state to that mode would be Raman active, if a normal mode has the same symmetry as one of these binary combinations of x,y and z. Since the selection rules for infra – red and Raman spectra have different physical bases; there is no relationship between them. Thus an infra-red active mode might or might not be Raman active or conversely. This watermark does not appear in the registered version - http://www.clicktoconvert.com 34 4.6. MUTUAL EXCLUSION RULE Consider a molecule which has a centre of symmetry. Point groups of molecules with this element of symmetry have two sets of irreducible representations. The representations which are symmetric with respect to inversion are called g representations. The representations which are antisymmetric to inversion are called u representations. Let us consider the inversion of a Cartesian coordinate x through the centre of inversion. The coordinate x becomes –x. Therefore all representations generated by x,y, or z must belong to a u representation. On the other hand, the product of two coordinates x and y does not change sign on inversion (-x.- y = xy). The product xy generates a g representation. All the other quadratic or binary coordinates also generate the g representation. From these rules we can conclude that in centrosymmetric molecules, the vibrational modes belonging to g symmetry species are Raman active and the modes belonging to u symmetry species are IR active. This rule is called the mutual exclusion rule. Another way of stating this rule is as follows: If a molecule has a centre symmetry, then any vibration that is active in the IR is inactive in the Raman and vice versa. Therefore, we can infer that a molecule has no centre of symmetry if the same vibration appears in both IR and Raman. Table1 lists the IR active and Raman active vibrational modes in some centrosymmetric and noncentrosymmetric molecules. CO2 , C2 H2 and N2 F2 possess centre of symmetry. It is seen from Table1 that the IR active modes in these molecules are Raman inactive and vice versa. H2 O, NH3 , HCN and BF3 do not have centre of symmetry. In BF3 the vibrational mode with symmetry species E’ is IR active and Raman active. In H2 O, NH3 and HCN, all the modes are IR active and Raman active. Table1 Molecule Point Group Symmetry Species IR Active Species Raman Active Species CO2 C 2 H2 N2 F2 H2 O NH3 BF3 HCN D¥h D¥h C2h C2v C3v D3h C¥v A1g,A1u,E1u 2A1g,E1g,E1u, A1u 3Ag,Au,2Bu 2A1 ,B2 2A1 ,2E A1 ’, 2E ’, A2 ” 2A1 ,E1 A1u,E1u E1u, A1u Au,Bu A1 ,B2 A1 ,E A2 ”, E ’ A1 ,E1 A1g A1g,E1g Ag A1 ,B2 A1 ,E A1 ’, E ’,E” A1 ,E1 4.7. LET US SUM UP In this lesson, we: Pointed out Ø Group theory and normal modes of vibration of polyatomic molecules Ø Infra-red absorption and Raman scattering spectroscopy Ø Determination of symmetry properties of vibrational modes This watermark does not appear in the registered version - http://www.clicktoconvert.com 35 Ø Symmetry selection rules for infra – red Raman spectra Ø Mutual exclusion rule 4.8. CHECK YOUR PROGRESS 1. 2. How is mutual exclusion principle explained on the basis of symmetry of vibrational modes? Taking water as example, illustrate how the various vibrational modes form basis for group representation? 4.9. LESSON – END ACTIVITIES 1. 2. Explain the selection rules in IR and Raman spectroscopy from symmetry point of view. Show that the normal modes form the basis for irreducible representations with a suitable example. 4.10. REFERENCES 1. K.V. Raman, Group Theory and its applications to Chemistry, Tata McGraw-Hill Publishing Company limited, New Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. 3. V. Ramakrishnan, M.S Gopinathan, Group Theory in Chemistry, Vishal Publications. This watermark does not appear in the registered version - http://www.clicktoconvert.com 36 UNIT-III LESSON 5: THE TIME-DEPENDENT AND TIME-INDEPENDENT SCHRÖDINGER EQUATIONS CONTENTS 5.0. AIMS AND OBJECTIVES 5.1. INTRODUCTION 5.2. THE TIME-DEPENDENT SCHRÖDINGER EQUATION 5.2.1. ONE-DIMENSIONAL EQUATION FOR A FREE PARTICLE 5.2.2. OPERATORS FOR MOMENTUM AND ENERGY 5.2.3. EXTENSION TO THREE DIMENSIONS 5.2.4. INCLUSION OF FORCES 5.3. TIME-INDEPENDENT SCHRODINGER EQUATION 5.4. REQUIREMENTS OF THE ACCEPTABLE WAVE FUNCTION 5.5. BORN’S INTERPRETATION OF THE WAVE FUNCTION 5.6. LET US SUM UP 5.7. CHECK YOUR PROGRESS 5.8. LESSON - END ACTIVITIES 5.9. REFERENCES 5.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on the time-dependent and time- independent Schrödinger equations to the students. On successful completion of this lesson the student should have: * Understand the time-dependent and time- independent Schrödinger equations. 5.1. INTRODUCTION Classical mechanics applies only to macroscopic particles. For microscopic “particles” we require a new form of mechanics, which we will call quantum mechanics. Schrodinger formulated an important fundamental equation in 1926. Schrodinger argued that if micro-particles like electrons could behave like waves, the equation of wave motion could be successfully applied to them. 5.2. THE TIME-DEPENDENT SCHRÖDINGER EQUATION 5.2.1. ONE-DIMENSIONAL EQUATION FOR A FREE PARTICLE The wave function of a localized free particle is the one given in eqn. This watermark does not appear in the registered version - http://www.clicktoconvert.com 37 ¥ Y ( x, t ) = ò A(k )exp[ikx - iw (k )t ]dk ----- (1) -¥ For a free particle, the classical expression for energy is 2 p E= x 2m 2 2 é 1 mv 2 = 1 m v ù E = ê 2 2 m ú ê ú ê px 2 = m2v 2 ú ê ú ê p = mv ú êë úû ----- (2) Replacing px by kh and E by hw we get w= hk 2 2m ----- (3) 2p h h é Q k h = ´ = = ê l 2p l ê êhw = h ´ 2pu = hu êë 2p ù pú ú ú úû Substituting this value of w in Eq. (1) ¥ Y ( x, t ) = éæ hk 2 ç ( ) A k exp i kx êç ò 2m -¥ ëè öù t ÷÷údk øû ----- (4) Differentiating Y ( x, t ) with respect to t, we get ¥ éæ ¶Y ih hk 2 2 ç ( ) = k A k exp i kx êç ¶t 2m -ò¥ 2m ëè öù t ÷÷ú dk øû ----- (5) Differentiating Y ( x, t ) twice with respect to x, we get ¥ éæ ¶ 2Y hk 2 2 ç ( ) = k A k exp i kx êç ò 2m ¶x 2 -¥ ëè öù t ÷÷ú dk øû ----- (6) Combining Eqs. (5) and (6), we have ih ¶Y ( x, t ) h2 ¶ 2Y =¶t 2m ¶x 2 ----- (7) This watermark does not appear in the registered version - http://www.clicktoconvert.com 38 which is the one-dimensional Schrodinger equation for a free particle. 5.2.2. OPERATORS FOR MOMENTUM AND ENERGY To obtain the operators for momentum and energy, Eq. (7) may be written as 1 æ ¶ öæ ¶ ö æ ¶ö ç ih ÷ Y ( x , t ) = ç - ih ÷ç - ih ÷Y ( x, t ) 2m è ¶x øè ¶x ø è ¶t ø ----- (8) From a comparison of Eqs. (2) and (8), it may be concluded that the energy ‘E’ and momentum ‘P’ can be considered as the differential operators. E ® ih ¶ ¶ and Px ® ih ¶t ¶x ----- (9) operating on the wave function Y ( x, t ) . Eq. (7) is obtained even if the operator for ‘P’ is taken as ¶ ¶ in place of - ih . ih ¶x ¶x 5.2.3. EXTENSION TO THREE DIMENSIONS The one-dimensional treatment given above can easily be extended to three dimensions. The three-dimensional wave packet can be written as ¥ Y (r, t ) = ò A(k )exp[i(k.r - wt )]dk x dk y dk z ----- (10) -¥ Proceeding on similar lines as in the on-dimensional case we get the three-dimensional Schrodinger equation for a free particle as ih ¶Y (r, t ) h2 2 =Ñ Y (r, t ) ¶t 2m ----- (11) An analysis similar to the one made for one-dimensional system leads to the following operators for energy and momentum E ® ih ¶ and P ® ihÑ ¶t ----- (12) 5.2.4. INCLUSION OF FORCES Modification of the free particle equation to a system moving in a potential V(r,t) can easily be done. The classical energy expression for such a system is given by p2 E= + V (r , t ) 2m ----- (13) This watermark does not appear in the registered version - http://www.clicktoconvert.com 39 Schrodinger then made the right guess regarding the operators for r and t as r ® r and t ® t ----- (14) Replacing E,p,r and t in Eq. (13) by their operators and allowing the operator equation to operate on the wave function Y (r, t ) , we get ù ¶Y (r, t ) é h 2 2 ih = êÑ + V (r, t )ú Y (r, t ) ¶t ë 2m û ----- (15) which is the time-dependent Schrodinger equation for a particle of mass ‘m’ moving in a potential V(r,t). The quantity in the square bracket in Eq. (15) is the operator for the Hamiltonian of the system. In general, it’s solution will be complex because of the presence of i in the equation. The equation cannot be relativistically invariant as it contains first derivative in time and second derivative in space coordinates. 5.3. TIME-INDEPENDENT SCHRODINGER EQUATION The time-dependent Schrodinger equation (15) describes the evolution of quantum systems using time-dependent wave function Y (r, t ) .It completely neglects the time dependence of the operators. If the Hamiltonian operator does not depend on time, the variables r and t of the wave function Y (r, t ) can be separated into two functions y (r ) and f (t ) . Y (r, t ) = y (r )f (t ) ----- (1) Substituting this value of Y (r, t ) in Eq. (15) ih ù ¶Y (r, t ) é h 2 2 = êÑ + V (r, t )ú Y (r, t ) ¶t ë 2m û and dividing throughout by y (r ) and f (t ) we get ù 1 df (t ) 1 é h2 2 ih = Ñ + V (r )ú Y (r ) êf (t ) dt y (r ) ë 2m û ----- (2) The left side of this equation is a function of time and right side function of space coordinates. Since‘t’ and ‘r’ are independent variables, each side must be equal to a constant, say E. This gives rise to the equations 1 df (t ) iE =f (t ) dt h ----- (3) This watermark does not appear in the registered version - http://www.clicktoconvert.com 40 and é h2 2 ù Ñ + V (r )úy (r ) = Ey (r ) êë 2m û ----- (4) Solution of Eq. (3) is straightforward and is given by æ iEt ö f (t ) = C expç ÷ è h ø ----- (5) where ‘c’ is a constant. The equation for y (r ) , Eq. (4), is the time- independent Schrodinger equation or simply Schrodinger equation. Since y (r ) determines the amplitude of the wave functiony (r, t ) , it is called the amplitude equation. Equation (1) now takes the form æ iEt ö Y (r, t ) = y (r ) expç ÷ è h ø ----- (6) The constant C is included in the normalization constant fory (r ) . Significance of the separation constant, E, can be understood by differentiating y (r, t ) in Eq. (1) with respect to time and multiplying by ih . Then ih ¶Y (r, t ) = EY (r, t ) ¶t ----- (7) Multiplying both sides of Eq. (7) by Y * from left and integrating over the space coordinates from - ¥ to ¥ . We get ¥ æ ¶ö ò Y * çè ih ¶t ÷øY (r, t ) = E ----- (8) -¥ As the left side Eq. (8) is the expectation value of the energy operator, the constant ‘E’ is the energy of the system. The same can be understood from Eq. (4) as - h2 2 Ñ + V (r ) 2m is the operator associated with the Hamiltonian of the system. 5.4. REQUIREMENTS OF THE ACCEPTABLE WAVE FUNCTION The interpretation given to Y or Y 2 imposes certain restrictions on acceptable values of Y . This watermark does not appear in the registered version - http://www.clicktoconvert.com 41 2 A physical system is described by the probability density Y (r , t ) and the normalization integral equation. 2 ¥ ò Y(r , t ) dt = 1 -¥ For the probability density to be unique and the total probability to be unity, the wave function must be finite and single valued at every point in space. The probability current density j, equation j (r , t ) = ih Y ÑY * - Y * ÑY 2m ( ) Another important parameter of the probability interpretation contains Y and ÑY . Hence Y has to be continuous and ÑY must be finite. The Schrodinger equation has the term Ñ 2 Y .For Ñ 2 Y to exist ÑY must be continuous. For a wave function Y (r, t ) to be acceptable, Y (r, t ) and ÑY must be finite, single valued and continuous at all points in space. Any function that is finite, single valued and continuous can be taken as an acceptable (well-behaved) wave function. 5.5. BORN’S INTERPRETATION OF THE WAVE FUNCTION The wave function Y (r, t ) has no physical existence, since it can be complex. Also it cannot be taken as a direct measurement of the probability as (r,t) since the probability is real and non-negative. However, Y (r, t ) must in someway be an index of the presence of the particle at (r,t). A universally accepted statistical interpretation was suggested by Born in 1926.He interpreted the product of Y (r, t ) and its complex conjugate Y * as the position probability density P(r,t). P (r , t ) = Y * (r , t )Y (r , t ) = Y (r , t ) 2 ----- (1) 2 The quantity y (r , t ) dt is the probability of finding the system at time‘t’ in the small 2 volume of element dt surrounding the point r. When Y (r , t ) dt is integrated over the entire space one should get the total probability, which is unity. Therefore ¥ ò Y(r , t ) 2 dt = 1 ----- (2) -¥ For Eq. (2) to be finite, Y (r, t ) must tend to zero sufficiently rapidly as r ® ±¥ .Hence, one can multiply Y (r, t ) by a constant, say N, so that NY satisfies the condition in Eq, (2).Then This watermark does not appear in the registered version - http://www.clicktoconvert.com 42 ¥ N 2 ò 2 Y (r , t ) dt = 1 ----- (3) -¥ The constant ‘N’ is called the normalization constant and Eq. (3) the normalization condition. Since the Schrodinger equation is a linear differential equation, NY is a solution of it. The wave functions for which the integral in Eq. (2) does not converge will be treated depending on the nature of the functions. 5.6. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø The time-dependent Schrödinger equation One-dimensional equation for a free particle Operators for momentum and energy Extension to three dimensions Inclusion of forces Time- independent Schrodinger equation Requirements of the acceptable wave function Born’s interpretation of the wave function 5.7. CHECK YOUR PROGRESS 1. 2. What are the requirements of an acceptable wave function? Outline the Born’s interpretation of the wave function. 5.8. LESSON – END ACTIVITIES 1. 2. Is the time dependent Schrodinger equation relativistically invariant? Explain. Derive the time independent Schrodinger Equation. 5.9. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006- 2007. This watermark does not appear in the registered version - http://www.clicktoconvert.com 43 LESSON 6: OPERATORS CONTENTS 6.0. AIMS AND OBJECTIVES 6.1. INTRODUCTION 6.2. ALGEBRA OF OPERATORS 6.2.1. ADDITION AND SUBTRACTION 6.2.2. MULTIPLICATION 6.3. COMMUTATOR OPERATOR 6.4. LINEAR OPERATOR 6.5. EIGENVALUES AND EIGENFUNCTIONS 6.6. BASIC POSTULATES OF QUANTUM MECHANICS 6.6.1. POSTULATE-I 6.6.1.1. RULES FOR SETTING UP A QUANTUM MECHANICAL OPERATORS 6.6.1.2. SOME OPERATORS OF INTEREST 6.6.1.2.1. MOMENTUM OPERATOR 6.6.1.2.2. HAMILTONIAN OPERATOR 6.6.1.2.3. ANGULAR MOMENTUM OPERATOR 6.6.2. POSTULATE-II 6.6.2.1. SCHRODINGER EQUATION AS AN EIGENVALUE EQUATION 6.6.3. POSTULATE-III 6.6.4. POSTULATE-IV 6.6.4.1. STATIONARY STATES 6.7. AVERAGE OR EXPECTATION VALUES 6.7.1. UNCERTAINTIES IN POSITION AND MOMENTUM 6.8. LET US SUM UP 6.9. CHECK YOUR PROGRESS 6.10. LESSON - END ACTIVITIES 6.11. REFERENCES 6.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on operators to the students. On successful completion of this lesson the student should have: * Understand the operators. 6.1. INTRODUCTION An operator is a symbol for transforming a given mathematical function into another function. It has no physical meaning if written alone. For example, √ is an operator which in itself does not mean anything, but if a quantity or a number is put under it, it transforms that d quantity into its square root, another quantity. Similarly, is an operator which transforms a dx This watermark does not appear in the registered version - http://www.clicktoconvert.com 44 function into its first derivative with respect to x; for example d transform the function sin x dx into the function cos x. In general, if  denotes an operator which transforms the function f(x) into the function g(x), then we write Aˆ f ( x ) = g ( x ) . For example, d d , and f(x) = ax2 ; then Aˆ f ( x) = (ax 2 ) = 2ax , i.e., g(x) = 2ax. dx dx 1. Let  be 2. Let  be a. (i.e., multiplication by a) and f(x) = x2 + c, then Aˆ f ( x ) = a.( x 2 + c ) = ax 2 + ac , i.e., g(x) = ax2 +ac. 6.2. ALGEBRA OF OPERATORS Although operators do not have any physical meaning, they can be added, subtracted, multiplied and have some other properties. 6.2.1. ADDITION AND SUBTRACTION The addition or subtraction of operators yields new operators, the sum or the difference of operators being defined by, (Aˆ ± Bˆ ) f (x ) = Aˆ f (x ) ± Bˆ f (x ) For example, let  be loge and B̂ be d , and f ( x ) be ; then, dx (Aˆ ± Bˆ ) f (x ) = æç log ± dxd ö÷ x è e ø 2 = log e x 2 ± d 2 x dx ( ) = log e x ± 2 x = Aˆ f ( x ) ± Bˆ f ( x ) 6.2.2. MULTIPLICATION Multiplication of two operators means operations by the two operators one after the other, the order of operation being from right to left; for example, Aˆ Bˆ f ( x ) means that the function f ( x ) is first operated on by B̂ to yield a new function g ( x ) which is then operated on by  to yield the final function h( x ) , [ ] Aˆ Bˆ f ( x ) = Aˆ Bˆ f ( x ) = Aˆ g ( x ) = h( x ) For example, let,  be 4x 2 , B̂ be d ,and f ( x ) = ax 2 , then, dx This watermark does not appear in the registered version - http://www.clicktoconvert.com 45 d Aˆ Bˆ f ( x ) = 4 x 2 × ax 3 = 4 x 2 × 3ax 2 = 12ax 4 dx ( ) ( ) The square of an operator means that the same operator is applied successively twice, i.e., Aˆ 2 f ( x ) = Aˆ Aˆ f ( x ) . For example, 2 d æd ö Let Aˆ = and f ( x ) = sin x , then Aˆ 2 f ( x ) = ç ÷ sin x dx è dx ø 2 d éd æd ö (sin x )ùú = d (cos x ) = - sin x ç ÷ sin x = ê dx ë dx è dx ø û dx Or 6.3. COMMUTATOR OPERATOR For any two operators  and B̂ , the difference Aˆ Bˆ - Bˆ Aˆ , which is simply denoted by Aˆ Bˆ - Bˆ Aˆ or [A,B] is called “commutator operator” If  and B̂ commute then [A,B] = 0, where 0 is called the zero operator which means multiplying a function with zero. d ˆ In the earlier example, where, Aˆ = , B = 3 x 2 and f ( x ) = sin x ,the commutator is obtained as dx follows [A, B ] f (x ) = [Aˆ Bˆ - Bˆ Aˆ ] f (x ) = (6 x sin x + 3x 2 cos x ) - 3x 2 cos x = 6 x sin x = 6 xf (x ) or [A, B] f (x ) = 6 x 6.4. LINEAR OPERATOR An operator is said to be linear if its application on the sum of two functions gives the result which is equal to the sum of the operations on the two functions separately, i.e., if, Aˆ [ f ( x ) + g ( x )] = Aˆ f ( x ) + Aˆ g ( x ) or Aˆ cf ( x ) = C × Aˆ cf ( x ) , C = constant Example: d d d d (i) is a linear operator because ( ( ax m + bx n ) = ax m ) + (bx n ) dx dx dx dx (ii) , square root, is not a linear operator because f (x ) + g (x ) ¹ f (x ) + g (x ) This watermark does not appear in the registered version - http://www.clicktoconvert.com 46 6.5. EIGENVALUES AND EIGENFUNCTIONS If an operator  operates on a well-behaved (i.e., finite, continuous and single-valued) function f to give the same function but multiplied by a constant, then the constant is called the “eigenvalue” of the operator and the function f is called the “Eigenfunctions”. The equation is called “eigenvalue equation”. For example, if the function f = e - ax is acted upon by the d d - ax operator , the result is e = - a (e - ax ) . dx dx d Therefore, (- a ) is the eigenvalue and e - ax is the eigenfunction of the operator . dx ( ) 6.6. BASIC POSTULATES OF QUANTUM MECHANICS The fundamental postulates are four which are stated and explained below. 6.6.1. POSTULATE-I Every physical property of a system (i.e. a particle or a system of particles) has a corresponding mathematical (quantum mechanical) operator. Physically measurable quantities of a particle are a position ( x ) , momentum ( p ) , kinetic energy (T ) , potential energy (V ) , total energy (E ) , etc. The operators corresponding to these quantities are given in table I. Table-I. Quantum Mechanical OPERATORS Corresponding to Various Physical Quantities Physical quantity Position ( x ) Position (r ) x-Component of momentum ( p x ) momentum ( p ) kinetic energy (T ) x-Component of K.E. (Tx ) Quantum mechanical operator x r h ¶ 2pi ¶x h Ñ 2pi h2 - 2 Ñ2 8p m - h2 ¶2 2 8p 2 m ¶x This watermark does not appear in the registered version - http://www.clicktoconvert.com 47 Potential energy (V ) V Total energy (H ) - h2 2 Ñ2 + V . 8p m 6.6.1.1. RULES FOR SETTING UP A QUANTUM MECHANICAL OPERATORS (i) Write down the expression for the physical quantity in classical terms, i.e., in terms of Cartesian coordinates of position ( x, y, z ) and momenta ( p x , p y , p z ) . (ii) Replace these coordinates and momenta by their corresponding operators (vide Table I above). (iii) Operator for a coordinate of position (say x) is multiplication by that variable x itself (i.e.x.). (iv) Operators for a coordinate of momentum(say px ) is h ¶ . 2pi ¶x Take for example the kinetic energy (Tx ) of a single particle moving in one direction(x). 2 p 1 2 Tx = mv x = x 2 2m Therefore, the K.E. Operator 1 æ h ¶ ö Tˆx = ç ÷ 2m è 2pi ¶x ø =- 2 h2 ¶2 2 8p 2 m ¶x ----- (1) 2 p Similarly, potential energy operator Vˆ ( x ) = V ( x ) . The total energy (H) is x + V ( x ) ; 2m hence, the total energy operator h2 d2 ˆ H =- 2 + V (x ) 2 8p m dx ----- (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 48 6.6.1.2. SOME OPERATORS OF INTEREST 6.6.1.2.1. MOMENTUM OPERATOR The kinetic, potential and total energies are derived from coordinates of momentum and position. but how is momentum operator to be derived? This is done by using a more fundamental property of electron wave itself. For an electron wave the wave function may be represented by the function, y = A exp(± 2pix l ) ----- (3) Differentiating with respect to x, dy 2pi 2pi =± A exp(± 2pix l ) = ± y dx l l But by de Broglie relationship, h l= px Therefore, dy 2pi =± p xy dx h or, p xy = ± h dy 2P i dx ----- (4) Buty is a function which on being removed from the equation reduces it to a differential operator, h d ) px = ± 2P i dx The operator Px in the =x direction is ----- (5) h d h d and in the negative –x direction it is . 2pi dx 2pi dx 6.6.1.2.2. HAMILTONIAN OPERATOR The operator corresponding to the total energy of a system, written as a sum of kinetic Ù and potential energies, is called Hamiltonian operator æç H ö÷ . The total energy of a single particle è ø of mass m is H= p2 1 2 2 2 + V ( x, y , z ) = px + p y + pz + V 2m 2m ( ) ----- (6) This watermark does not appear in the registered version - http://www.clicktoconvert.com 49 where V is written for V(x,y,z). The corresponding operator will, therefore, be Ù 1 æÙ 2 Ù 2 Ù 2ö Ù ç px + py + pz ÷ +V . 2m è ø H= But Ù h2 ¶2 æ h ¶ ö 2 ; px = ç = ÷ 4p 2 ¶x 2 è 2pi ¶x ø similarly, Ù py Ù h2 ¶2 h2 ¶2 2 = - 2 2 and p z = - 2 2 4p ¶y 4p ¶z 2 Thus, Ù H =- =- h2 æ ¶2 ¶2 ¶2 ç + + 8p 2 m çè ¶x 2 ¶y 2 ¶z 2 ö Ù ÷÷ + V ø Ù h2 2 Ñ + V 8p 2 m ----- (7) For a system of n particles, h2 H =- 2 8p Ù 1 åm i Ù Ñi 2 + V ----- (7a) i where mi is the mass and Ñ i2 the Laplacian operator of the ith particle. 6.6.1.2.3. ANGULAR MOMENTUM OPERATOR The angular momentum (L) is a very important physical quantity for rotating systems. r r Classically, it is given by the vector product of position (r ) and linear momentum ( p ) , r r r ----- (8) L=r´p r r r If i , j and k are unit vectors along x, y and z coordinates respectively, we have r r r r r = i x + jy + kz , and r r r r p = i p x + j p y +kp z Therefore, r r r r r r r L = i x + jy + kz ´ i p x + jp y + kp z r r r = i ( yp z - zp y ) + j ( zp x - xp z ) + k (xp y - yp x ) Also, by definition, r r r r L = i Lx + j L y + k Lz ----- (9) ( ) ( ) where Lx , L y and Lz represent the three components of L. Replacing px , p y and pz by their corresponding operators we obtain the operators for the three components of angular momentum. This watermark does not appear in the registered version - http://www.clicktoconvert.com 50 Thus, h æ ¶ ¶ öü çç y - z ÷÷ ï Lˆ x = ( ypˆ z - zpˆ y ) = 2pi è ¶z ¶y ø ï h æ ¶ ¶ ö ïï Lˆ y = ( zpˆ x - xpˆ z ) = çz - x ÷ý 2pi è ¶x ¶z ø ï h æ ¶ ¶ öï çç x - y ÷÷ï Lˆ z = (xpˆ y - ypˆ x ) = 2pi è ¶y ¶x øïþ ----- (10) 6.6.2. POSTULATE-II The possible values of any physical quantity of a system (e.g. energy, angular momentum etc) are given by the eigenvalues a in the operator equation, ) Ay = a y ----- (11) ) where A i s the operator corresponding to that physical quantity and y is a well-behaved eigenfunction. The eigenvalues and eigenfunctions for a system can be obtained by solving the operator equation. While the Eigenfunctions and the operator may be real or complex, the eigenvalues must be real because they represent observable physical quantities. Moreover, quantum mechanical operators of interest, e.g. momentum, energy, etc., are Hermitian operators which have been shown earlier to have real eigenvalues. Consider, for example, the operator equation for momentum px (in one dimension), h dy ( x ) ) p xy ( x ) = = p xy ( x ) 2pi dx ----- (12) This is an ordinary differential equation whose solution may be easily obtained as y ( x) = A exp(± )2pip x x h ----- (13) where A is a constant. Now, any value of px is allowed that keeps the function y ( x ) well2pp x = k , then the eigenfunction of behaved, i.e., finite, single- valued and continuous. If we put h momentum operator becomes, y = A exp(± ikx ) ----- (14) with eigenvalues px = kh 2p ----- (15) This watermark does not appear in the registered version - http://www.clicktoconvert.com 51 6.6.2.1. SCHRODINGER EQUATION AS AN EIGENVALUE EQUATION A particularly important eigenvalue equation is the equation for energy of a system. The operator for energy E (K.E+P.E) is the Hamiltonian operator. For a single particle in three dimensions, the Hamiltonian is h2 Hˆ = - 2 Ñ 2 + Vˆ ( x, y, z ) 8p m ----- (16) The eigenvalue is, Hˆ y ( x, y , z ) = Ey ( x, y , z ) or simply Hˆ y = Ey i.e. - h2 Ñ 2y + Vˆy = Ey 2 8p m ----- (17) This is the familiar Scharodinger equation. It is a second order partial differential equation which has to be solved in order to obtain expressions for the eigenfunctions,y , and the observable quantity, energy (E), of the particle. Since energy is a very important observable property of a system, and can often be measured experimentally, a topic of major importance in the quantum chemistry is solving of Schrodinger equation. 6.6.3. POSTULATE-III The expected average (expectation) value of a physical quantity (M) of a system, whose state function is y , is given by ) òy *Mydt M = òy *ydt ----- (18) ) where M is the operator for M. ) ) It has to be noted that y is not necessarily an eigenfunction of M , though M has a set of orthonormal eigenfunction (j i ) and y can be expressed as y = å aij i i Hence, æ ö )æ ö M = ò ç å aij i ÷ * M ç å aij i ÷dt , if y is normalized è i ø è i ø ) = åå ai * a j ò j i * Mj j dt i j This watermark does not appear in the registered version - http://www.clicktoconvert.com 52 ) Mj j = l j j j But Therefore, M = åå (ai * a j )ò j i * l j j j dt i j = åå (ai * a j )l j ò j i * j j dt i j Since òj i * j j dt = 0 for i ¹ j , ( ji and j j are orthogonal) = 1 for i = j , ( ji and j j are normalized) M = å ai * ai li or i åa 2 i li i This expression means that each measurement of M must give one of the eigenvalues li (postulate-II) and the average of many such measurements is equal to M . Further, ai * ai or 2 ai represents the fraction of the total number of measurements that give the eigenvalue li , and å a i 2 = 1 .0 . Consider for example the expectation value of position, x, for a particle moving in one dimension, which is given by +¥ x= +¥ ò xy *ydx = òy * xydx -¥ -¥ because the operator for x is just x, for which the order of applying is immaterial. Similarly, the expected average value of momentum for the above system is, +¥ h dy px = y* dx ò 2pi -¥ dx ----- (19) +¥ and K.E., Tx = - h2 d 2y y * dx 8p 2 m -ò¥ dx 2 ----- (20) The limits of integration in each case will depend on the total length available for the particle’s motion. 6.6.4. POSTULATE-IV The time-dependent Schrodinger equation is given by, This watermark does not appear in the registered version - http://www.clicktoconvert.com 53 ) ih ¶y Hy = 2p ¶t ----- (21) ) where H is the Hamiltonian operator, y is a function of position (x,y,z) as well as time (t) and V is a function of position only. Since this a partial differential equation involving both position and time variables, it can be solved only by separating it into two differential equations, one containing position variable and the other containing time variable only. This can be done if we assume that, y = y (q, t ) = y (q ).j (t ) ----- (22) where q is the collective symbol for the three coordinates of position x, y and z; dropping q and t for simplicity, y = yf Then, ¶y ¶y =j ¶q ¶q or ¶ 2y ¶ 2y = j ¶q 2 ¶q 2 ----- (23) ¶y ¶y =y ¶t ¶t ----- (24) and Substituting equations (22),(23) and (24) into (21) and dividing the resulting equation by yj , we get é h 2 ¶ 2y ) ù 1 ih 1 ¶j . ----- (25) ê- 2 . 2 + Vy ú = ë 8p m ¶q û y 2p j ¶t and ih ¶j ----- (26) = Ej 2p ¶t Equation (25) is the familiar Schrodinger equation, also called the amplitude equation of the type ) Hy = Ey in which the function y ,the wave function, is found by solving the equation Equation (26) has a solution, æ 2piEt ö f = a expç ÷ h ø è So the time dependent wave function y (q, t ) is given by, ----- (27) This watermark does not appear in the registered version - http://www.clicktoconvert.com 54 æ 2piEt ö Y = ay expç ÷ h ø è ----- (28) 6.6.4.1. STATIONARY STATES States for which Y is given by (28) are called stationary states. A stationary state does not imply that the particle, or particles, of the system are at rest. It is stationary in the sense that the 2 probability density Y and the energy are independent of time as long as the state of the system is being described by the function Y . The probability density is given by, 2 æ 2piEt ö * * æ 2piEt ö YY * = Y = ay expç ÷.a y expç ÷ = aa*yy * ------ (29) h ø è è h ø æ 2piEt ö which is independent of time. Thus, we see that the factor expç ÷ is of no significance h ø è and the essential part of the state function for a stationary state is the time-independent wave function y . A knowledge of stationary states is of prime importance in understanding chemical problems. If the state of a system corresponds to one of these stationary states, every physical property of the system e.g. energy, charge density, dipole moment, etc., will be independent of time. Atomic and molecular spectra arise due to transitions between these stationary states brought about by the action of radiation. 6.7. AVERAGE OR EXPECTATION VALUES We can use the wave function to determine, in a state n, the average or expectation values of any physical quantity, e.g., position, momentum, energy, etc. Consider, for example, the position. Since the particle can be anywhere in the box, the result of a single measurement is never precise; but the average of a series of measurements has some meaning. This average can be computed by the formula, L L 2 npx x = òyx ydx = ò x sin 2 dx L0 L 0 ------ (30) Let npx Lu L , or x = , dx du; ' L np np when x = 0, u = 0 and when x = L, u = np u= Therefore, x= 2 L np Lu ò np sin 0 2 æ L ö uç ÷du è np ø ------ (31) This watermark does not appear in the registered version - http://www.clicktoconvert.com 55 = np 2L u sin (np ) ò 2 2 udu ------ (32) 0 Using the table of integrals np 2 L éæ u 2 u sin 2u cos 2u öù ÷ú x= êç 4 8 ÷øû 0 (np )2 ëçè 4 = 2 (np ) sin (2np ) - cos (2np )ü - ì0 - 0 - 1 üù 2 L éì (np ) ý í ýú 2 êí 4 8 8 þúû (np ) êëî 4 þ î =L 2L (np ) 2 2 ( np ) ´ = 4 L 2 ------ (33) Thus, quantum mechanical considerations show that the average value of the position of the particle in any state (n) lies in the middle of the box, which is the same as that obtained from classical point of view. This should not be surprising as the probability density curves (Fig. 1) indicate that the probability of finding the particle at a distance d to the left of the centre is exactly the same as that at the distance d to the right. We can likewise calculate other quantities. A quantity of interest is x 2 . 2 æ 2L ö x = ç ÷ L è np ø 2 3 np òu 2 sin 2 u du 0 np 3 2 æ L ö éu 3 æ u 2 1 ö u cos 2u ù = ç ÷ ê - çç - ÷÷ sin 2u ú L è np ø ë 6 è 4 8 ø 4 û0 This watermark does not appear in the registered version - http://www.clicktoconvert.com 56 3 = 2 æ L ö éæ n 3p 3 np -0ç ÷ êçç L è np ø ëè 6 4 ù ö ÷÷ - (0 - 0 - 0 )ú ø û 3 é n 2p 2 1 ù 2æ L ö = ç - ú ÷ (np ) ê L è np ø 4û ë 6 L2 L2 = 3 2n 2p 2 3 ù é = ê1 - 2 2 ú ë 2n p û ------ (34) 6.7.1. UNCERTAINTIES IN POSITION AND MOMENTUM Since the particle can be anywhere within the length L, the error (uncertainty) involved in any single measurement of position, Dx = L . the momentum lies between + so, the uncertainty, nh 2L and - nh ; 2L This watermark does not appear in the registered version - http://www.clicktoconvert.com 57 Dp x = nh æ nh ö nh - ç, ÷= 2L è 2L ø L for n = 1, px = h L, and the product Dx.Dp x = h ------ (35) which satisfy the Heisenberg uncertainty relation. 6.8. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Algebra of operators Addition and subtraction Multiplication Commutator operator Linear operator Eigenvalues and eigenfunctions Basic postulates of quantum mechanics Postulate-I Rules for setting up a quantum mechanical operators Some operators of interest Momentum operator Hamiltonian operator Angular momentum operator Postulate-II Schrodinger equation as an eigenvalue equation Postulate-III Postulate-IV Stationary states Average or expectation values Uncertainties in position and momentum 6.9. CHECK YOUR PROGRESS 1. 2. Prove that the Hamiltonian Operator for the total energy of a system is Hermitian provided that the wave function is well behaved. What are eigen functions and eigenvalues of an operator? 6.10. LESSON – END ACTIVITIES 1. 2. Outline the different postulates of quantum mechanics. Write notes on (a) Average or expectation values (b) Uncertainity in position and momentum This watermark does not appear in the registered version - http://www.clicktoconvert.com 58 6.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co. This watermark does not appear in the registered version - http://www.clicktoconvert.com 59 UNIT-IV LESSON 7: PARTICLE IN A ONE-DIMENSIONAL BOX CONTENTS 7.0. AIMS AND OBJECTIVES 7.1. INTRODUCTION 7.2. PARTICLE IN ONE DIMENSIONAL BOX 7.3. NORMALIZATION OF y 7.4. ORTHOGONALITY OF THE WAVE FUNCTIONS 7.5. PARTICLE IN A THREE-DIMENSIONAL CUBICAL BOX 7.4.1. SEPARATION OF VARIABLES 7.4.2. DEGENERACY 7.6. ONE-DIMENSIONAL SIMPLE HARMONIC OSCILLATOR (S.H.O) 7.5.1. ENERGY OF S.H.O 7.7. THE RIGID ROTOR (OR) ROTATOR 7.8. LET US SUM UP 7.9. CHECK YOUR PROGRESS 7.10. LESSON - END ACTIVITIES 7.11. REFERENCES 7.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on particle in a onedimensional box to the students. On successful completion of this lesson the student should have: * Understand the particle in a one-dimensional box. 7.1. INTRODUCTION Now consider the particle which is allowed to move in a limited space such a model, usually called “particle in a box” model, serves as the simplest case for the treatment of bound electrons in atoms and molecules. 7.2. PARTICLE IN A ONE-DIMENSIONAL BOX This is the simplest quantum mechanical problem. Here a particle of mass m is confined to move in a one-dimensional box of a length a, having infinitely high walls (Fig.1). It is This watermark does not appear in the registered version - http://www.clicktoconvert.com 60 assumed, for the sake of simplicity, that the potential energy of the particle is zero everywhere inside the box, that is V (x ) = 0 ----- (1) Thus, inside the box the Schrodinger equation, viz., é h2 2 ù d dx 2 + V ( x )úy ( x ) = Ey ( x ) êë 2m û ( takes the form - ) h2 2 d y dx 2 = Ey ( x ) 2m ( ) ----- (2) ----- (3) Our problem is solving this equation for energy E and the wave functiony ( x ) . Mathematically, Eq.3 may be rewritten as d 2y + 2mE h 2 y = 0 2 dx ----- (4) d 2y + k 2y = 0 2 dx ----- (5) ( or as ) where k2 (= 2mE h 2 ) is a constant, independent of x. Eq.5 is an ordinary second-order differential equation which has solution of the form y = A cos kx + B sin kx ----- (6) This watermark does not appear in the registered version - http://www.clicktoconvert.com 61 where A and b are constants. Before we proceed further, it should be pointed out that outside the box where V ( x ) = ¥ , the Schrodinger equation (2) is é h2 2 ù d dx 2 + ¥ úy ( x ) = Ey ( x ) êë 2m û ----- (7) d 2y 2m + 2 (E - ¥ )y = 0 dx 2 h ----- (8) ( or ) This equation is satisfied if y is zero everywhere outside the box. This is another way of saying that the particle cannot be found outside the box; it is confined within the box. This implies that y must be zero at the walls of the box, i.e., at x=0 and at x=a. Since the postulate of quantum mechanics requires that y must be a continuous function of y , we are forced to conclude that in Eq. 6, A must be zero. Thus, the solution of Eq. 5 is of the form ----- (9) y = B sin kx Since y = 0 at x=0 and at x=a, we have B sin ka = 0 or B sin ka = 0 or sin ka = 0 so that k = np a ----- (10) where n(=0,1,2,3,4.......,)is the quantum number. Hence, the allowed solutions of Eq. 4 are y º y n = B sin (npx a ) ; (n=1,2,3, ......) ----- (11) Notice that the n=0 value of the quantum number, though allowed, is not acceptable since it implies that the wave function y is zero everywhere inside the box. This is not corre4ct since the particle is taken to be inside the box, to begin with. From eqs.4 and 9, we have (2mE h ) = n p 2 or 2 2 a2 ----- (12) E = E n = n 2 h 2p 2 2ma 2 = n 2 h 2 8ma 2 ; n=1,2,3,........ (h = h 2p ) ----- (13) Eq.13 gives the expression for the energy of the particle in a one-dimensional box. It should be committed to memory. Since energy depends upon the quantum number n, which can have any integral value, the energy level of the particle in the box are quantize. Before we discuss eq.13 further, let us determine the coefficient b in the wave function given by Eq. 11. This can be done by normalizing the wave function. 7.3. NORMALIZATION OF y This watermark does not appear in the registered version - http://www.clicktoconvert.com 62 Since the total probability of finding the particle within the box is 1, therefore, according to Born’s interpretation of the wave function, the normalization of y requires that a òy 2 n dx = 1 ----- (14) 0 Substituting the value of y n from Eq.11, we have 2 a ò (B sin npx a ) dx = 1 ----- (15) 0 a or B 2 ò sin 2 (npx a ) dx = 1 ----- (16) 0 1 Since sin 2q = (1 - cos 2q ) , hence from Eqs. 14 and 16, we have 2 a a é a 1 æ 2npx ö ù 2 2 1 y dx = B dx cosç ÷dx ú ê ò ò0 n ò 20 è a ø û ë2 0 ----- (17) B 2 [a 2 - 0] = 1 ----- (18) 1 Hence, B = (2 a ) 2 ----- (19) Thus, the normalized wave functions for the particle in a one-dimensional box are 1 y n ( x ) = (2 a ) 2 sin (npx a ) ; n=1,2,3,....... ----- (20) A few normalized wave functions for a particle in a one-dimensional box are given in Fig.2. This watermark does not appear in the registered version - http://www.clicktoconvert.com 63 7.4. ORTHOGONALITY OF THE WAVE FUNCTIONS We have shown in lesson 6 that eigenfunctions of a Hermitian operator corresponding to different eigenvalues are necessarily orthogonal to each other, and that Hamiltonian is an Hermitian operator. For the problem of particle in a one-dimensional box, it can be demonstrated easily. Let y1 = 2 px sin , and y 2 = L L 2 2px sin L L be the two normalized eigenfunctions corresponding to the two eigenvalues E1 a n d E 2 (characterized by quantum numbers 1 and 2 ). Then, L L òy 1y 2 dx = 0 2 nx 2px sin sin ò L0 L L (2 - 1)px - cos (2 + 1)px ù é L cos ú 2 ê L L = òê údx L0ê 2 ú êë úû L (2 - 1)px - L sin (2 + 1)px ù 1é L = ê sin (2 + 1) L ë (2 - 1) L L úû 0 = 1 (0 - 0) = 0 L This watermark does not appear in the registered version - http://www.clicktoconvert.com 64 7.5. PARTICLE IN A THREE-DIMENSIONAL CUBICAL BOX Let us now consider the motion of a particle of mass m confined to a three-dimensional cubical box with edges of length a and volume equal to a 3 . The potential is zero within the box and is infinite outside the box and at its boundaries. The time- independent Schrodinger equation for the particle is é h2 æ ¶2 ¶2 ¶ 2 öù ç ----- (21) + + ê ç ¶x 2 ¶y 2 ¶z 2 ÷÷úy ( x, y, z ) = Ey ( x, y, z ) 2 m è ø ë û Eq. 21 may be rewritten as ¶ 2y ¶ 2y ¶ 2y æ 2mE ö + + +ç ÷y = 0 ¶x 2 ¶y 2 ¶z 2 è h ø ----- (22) 7.5.1. SEPARATION OF VARIABLES Assuming that the wave function y is a product of three parts which separately depend on x,y, z, we have Hence, y ( x, y, z ) = X ( x )Y ( y )Z ( z ) ----- (23) ¶ 2y ¶2 X = YZ ¶x 2 ¶x 2 ----- (24) ¶ 2y ¶ 2Y = XZ ¶y 2 ¶y 2 ----- (25) ¶ 2y ¶2Z = XY ¶z 2 ¶z 2 ----- (26) Substituting Eqs.24-26 in Eq.22 and dividing throughout by (2m h 2 )XYZ , we have - h 2 é 1 ¶ 2 X 1 ¶ 2Y 1 ¶ 2 Z ù + + ê ú+E =0 2m ë X ¶x 2 Y ¶y 2 Z ¶z 2 û ----- (27) We notice that the first term is a function of x only and is independent of y and z; the second term is a function of y only and is independent of x and z and the third term is a function of z only and is independent of x and y. The fourth term E is a constant. If energy E is written as the sum of three contributions associated with the three coordinates, then Eq. 27 can be separated into three equations. Thus, for instance, for motion along the X-axis (where X varies while Y and Z remain constant), the second and third terms on the left hand side of Eq.27 remain constant be Ex . Accordingly, we can write - h2 æ 1 ¶2 X ö ç ÷ = Ex 2m çè X ¶x 2 ÷ø ----- (28) This watermark does not appear in the registered version - http://www.clicktoconvert.com 65 - h 2 æ 1 ¶ 2Y ö ç ÷ = Ey 2m çè Y ¶y 2 ÷ø ----- (29) - h2 æ 1 ¶2Z ö ç ÷ = Ez 2m çè Z ¶z 2 ÷ø ----- (30) where E = E x + E y + E z . We see that each of the Eqs. 28-30 is of the form of Eq.3, the solution of which is given by Eq.20, i.e., we have 1 X (x ) = (2 / a )2 sin (nx px a ) ; nx = 1,2,3... 2 E x = n x h 2 8ma 2 and ----- (31) ----- (32) Similar solutions exists for Y(y) and Z(z). Hence ( y ( x, y, z ) = X ( x )Y ( y )Z ( z ) = 8 a 3 E = Ex + E y + Ez and ) 1 2 sin (n x px a )sin (n y py a )sin (n z pz a ) (n = 2 x 2 2 ) + n y + nz h 2 8ma 2 ----- (33) ----- (34) where nx , ny , nz =1,2,3,4,.......... 7.5.2. DEGENERACY It is seen from Eq.34 that the total energy depends upon the sum of the squares of three quantum numbers. It is evident that groups of different states, each specified by a unique set of quantum numbers, can have the same energy. In such a case, the energy level and the corresponding independent states are said to be degenerate. Consider, for instance, the energy level having energy = 14h2 /8ma2 . There are six combinations of nx ,ny and nz which can give this value of energy: nx 1 1 2 3 2 3 ny nz 2 3 3 2 1 3 1 2 3 1 2 1 This energy level is, therefore, 6- fold degenerate, i.e., its degeneracy is equal to 6. Proceeding in this manner we can calculate degeneracy of the energy levels of a particle in a three-dimensional cubical box. The results are shown in Fig. 3. This watermark does not appear in the registered version - http://www.clicktoconvert.com 66 7.6. ONE-DIMENSIONAL SIMPLE HARMONIC OSCILLATOR (S.H.O) A diatomic vibrating molecule can be represented by a simple model, the so-called simple harmonic oscillator (S.H.O). The force acting on the molecule is given by f = -kx, where x is the displacement from the equilibrium position and k is a constant called the force constant. The potential energy V(x) of this molecule is given by x x V ( x ) = - ò f dx = ò kxdx = 0 0 1 2 kx 2 ----- (35) Eq. 35 is the equation of a parabola. Thus, if we plot potential energy of a particle executing simple harmonic oscillations as a function of displacement from the equilibrium position, we get a curve as shown in Fig. 4. This watermark does not appear in the registered version - http://www.clicktoconvert.com 67 The vibrational frequency of the oscillator of mass m is given by 1 1 æ k ö2 u= ç ÷ 2p è m ø ----- (36) It is more accurate to define the vibrational frequency as 1 1 æ k ö2 ç ÷ u= 2p çè m ÷ø ----- (37) where m is the reduced mass of the diatomic molecule defined as 1 1 1 = + m m1 m 2 ----- (38) where m1 and m2 are the atomic masses of the two atoms. Using the potential energy given by Eq.35, for one dimensional S.H.O., the Schrödinger equation é h2 2 ù Ñ + V úy = Ey is represented as êë 2m û é h2 æ d 2 çç 2 êë 2m è dx ö 1 2ù ÷÷ + kx úy ( x ) = Ey ( x ) ø 2 û ----- (39) Mathematically,Eq. 39 may be written as d 2y 2m æ 1 ö + 2 ç E - kx 2 ÷y = 0 2 2 dx h è ø Defining a and b as a = 2mE h 2 Eq. 40 becomes and b = (mk ) ----- (40) 1 2 h, d 2y + a - b 2 x2 y = 0 dx 2 Defining a new variable x = b ( 1 2 ) ----- (42) x , the above equation becomes ( x = Xi ) ö d 2y æ a + çç - x 2 ÷÷y = 0 2 dx èb ø Eq. 42 has a solution of the form ----- (41) ----- (43) This watermark does not appear in the registered version - http://www.clicktoconvert.com 68 y (x ) = f (x )e -x 2 ----- (44) 2 Using this solution, Eq. 42 becomes ö d 2f df æ a - 2x + çç - 1÷÷f = 0 2 dx è b dx ø ----- (45) Eq. 45 is identical in form to a well known second –order differential equation, called the Hermite equation,. viz., d 2f df - 2x + 2nf = 0 2 dx dx ----- (46) The Hermite equation has solutions which depend upon the value of n. These solutions are called Hermite polynomials, Hn (x).i.e.,f (x) º Hn (x). The Hermite polynomial of degree n is defined as ( )üïý ìï ¶ n e -x 2 H n (x ) = (- 1) e í n ïî ¶x n x2 ----- (47) ïþ A few Hermite polynomials are given below: H0 (x) =1 H3 (x) =8x 3-12x H1 (x) =2x H4 (x) =16x 4-48x2 +12 H2 (x) =4x 2-2 The normalized wave functions of the one dimensional S.H.O. are then written as é b 12 y n (x ) = ê n êë 2 n! p 1 ù 2 22 ú e -x H n (x ) Where n= 0,1,2,3,........ úû ----- (48) The first three wave functions (with n= 0,1,2), the corresponding energy levels and the 2 probability functions y n for the S.H.O., are shown in Fig. 5. This watermark does not appear in the registered version - http://www.clicktoconvert.com 69 7.6.1. ENERGY OF S.H.O The energy of the S.H.O is obtained by comparing Eqs. 45 and 46, hence we find that a b = 2n + 1 ----- (49) Substituting for a and b from Eq. 41, we have (2mE h ) 2 1 æ E = çn + 2 è or From Eq. 36, (mk ) h = 2n + 1 (k m )1 2 12 ö 12 ÷h (k m ) ø = 2pn Combining Eqs. 51 and 52, we have 1ö æ E º E n = ç n + ÷h (2pn ) 2ø è 1ö æ = ç n + ÷hn ;n=0,1,2,3,.....( h = h 2p ) 2ø è ----- (50) ----- (51) ----- (52) ----- (53) The energy state with n=0 is the vibrational ground state with energy E0 = 1 hn 2 ----- (54) This watermark does not appear in the registered version - http://www.clicktoconvert.com 70 This energy is called the zero point energy of the oscillator. Classical mechanics predicts that the zero-point energy of the oscillator is zero whereas quantum mechanics predicts that the zero-point energy is non-zero. The occurrence of the zero-point energy is consistent with the Heisenberg uncertainty principle. 7.7. THE RIGID ROTOR (OR) ROTATOR A diatomic molecule rotating about an axis perpendicular to the intermolecular axis and passing through the center of gravity of the molecule constitutes an example of a rigid rotor, it being assumed that the internuclear distance does not change during rotation. The kinetic energy (K.E.) of the molecule is given by 1 2 ----- (55) Iw = L2 2 I 2 Where w is the angular velocity and I is the momentum of inertia of the rotating molecular. The angular momentum L =Iw. If no force acts on the rotor, we can set the potential energy V=0. Hence, the Hamiltonian is expressed as K.E. º T = H = T + V = L2 2 I ----- (56) The expression for L2 in spherical polar coordinates (r,q,f) is given by é 1 ¶ æ ¶ ö 1 ¶2 ù L2 = -h 2 ê sin q + ç ÷ ú ¶q ø sin 2 q ¶f 2 û ë sin q ¶q è ----- (57) The Schrödinger equation Hˆ y = Ey may thus be written as 1 2I é 2ì 1 ¶ æ ¶y ö 1 ¶ 2y üù ç sin q ÷+ ê- h í ýú = Ey ¶q ø sin 2 q ¶f 2 þúû êë î sin q ¶q è ----- (58) The above equation may be written as 1 ¶ æ ¶y ç sin q sin q ¶q è ¶q 1 ¶ 2y 8p 2 I ö + + 2 Ey = 0 ÷ 2 2 h ø sin q ¶f ----- (59) Eq. 59 contains two angular variables q and f. It can be solved by the method of separation of variables, i.e., we look for a solution of the form y (q , f ) = Q(q )F (f ) ----- (60) Substituting Eq. 60 into Eq. 59, we obtain sin q ¶ æ ¶Q ö 8p 2 IE 1 ¶ 2F 2 sin f = ç sin q ÷+ Q ¶q è ¶q ø F ¶f 2 h2 ----- (61) This watermark does not appear in the registered version - http://www.clicktoconvert.com 71 We can set both sides of Eq.61 equal to a constant, say m2 , thereby obtaining two differential equations each in one variable. These equations are: ¶ 2F + m 2 F = 0 and ¶f 2 1 ¶ æ ¶y ç sin q sin q ¶q è ¶q ----- (62) m2 ö ö æ ç ÷Q = 0 + b ÷ ç sin 2 q ÷ø ø è ----- (63) Where b = 8p 2 IE h 2 Eq.62 has the solution F (f ) = N exp(± imf ) , i = - 1 ----- (64) where N is the normalization constant. This wave function is acceptable provided m is an integer. This condition arises because F must be single-valued. Thus, F (f ) = F (f + 2p ) ----- (65) It follows, therefore, that exp (2pmi) =1 ----- (66) Since e x = cos x + i sin x (Euler’s relation) ----- (67) \ cos 2pm + i sin 2pm = 1 ----- (68) This can be true only if m=0, ± 1, ± 2, ± 3... etc. Let us now normalize the wave function F(f) to determine the normalization constant N. 2p * ò F Fdf = 1 (0 £ f £ 2p ) ----- (69) 0 2p N 2 ò e imf ´ e imf df = 1 Or ----- (70) 0 2p Or N 2 ò df = 1 , i.e., N2 (2p) =1 so that 0 N = (2p ) -1 2 ----- (71) Hence, the normalized wave functions become F ± m (f ) = (2p ) -1 2 exp(± imf ) ; m=0, 1, 2, 3... ----- (72) We shall not attempt to give a complete solution of Eq.63 but will merely state that if b=l(l+1) where l is the rotational quantum number, then this equation becomes a standard This watermark does not appear in the registered version - http://www.clicktoconvert.com 72 mathematical equation whose solutions are known to be associated Legendre polynomials m Pl (cos q ) where l is either zero or a positive integer and l> m . The normalized solutions are given by 12 é 2l + 1 (l - m )!ù m Q(q ) = Q l , ± m (q ) = ê ´ ú Pl cos q (l + m )!úû êë 2 ----- (73) The energy eigenvalues of the rigid rotor are obtained as follows: b = 8p 2 IE h 2 = l (l + 1) ----- (74) Thus, E= l (l + 1)h 2 ; l = 0,1,2,3 ... 8p 2 I ----- (75) In spectroscopy it is customary to use the symbol J rather than l for the rotational quantum number so that the rotational energy levels are given by the expression, J ( J + 1)h 2 EJ = ; J = 0, 1, 2, 3 ... 8p 2 I ----- (76) 7.8. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Particle in a one dimensional box Normalization of y Orthogonality of the wave functions Particle in a three-dimensional cubical box Separation of variables Degeneracy One-dimensional simple harmonic oscillator (S.H.O) Energy of S.H.O The rigid rotor (or) rotator 7.9. CHECK YOUR PROGRESS 1. 2. Derive an expression for the energy of a rigid rotor using the Schrodinger equation. Calculate the degeneracy of the energy level with energy equal to i. 11(h2 /8ma2 ) and ii 12(h2 /8ma2 ) for a particle in a cubical box. 7.10. LESSON – END ACTIVITIES This watermark does not appear in the registered version - http://www.clicktoconvert.com 73 1. 2. Set up and solve the Schrodinger wave equation for a particle in a one dimensional box, with potential energy zero inside the box. Normalize the wave function. Set up the Schrodinger wave equation for a simple harmonic oscillator, and solve it for the energy eigenvalues. 7.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co. This watermark does not appear in the registered version - http://www.clicktoconvert.com 74 UNIT-V LESSON 8: THE SCHRODINGER EQUATION FOR HYDROGEN ATOM CONTENTS 8.0. AIMS AND OBJECTIVES 8.1. INTRODUCTION 8.2. ANGULAR FUNCTIONS 8.3. PROBABILITY DENSITY AND RADIAL DISTRIBUTION FUNCTIONS 8.4. THE MOST PROBABLE DISTANCE OF THE H-ATOM (OR H-LIKE SPECIES) 1S ELECTRON 8.5. PHYSICAL INTERPRETATION OF THE HYDROGENIC ATOMIC ORBITALS 8.6. THE SCHRODINGER WAVE EQUATION FOR MULTI-ELECTRON ATOMS 8.6.1. TIME-INDEPENDENT PERTURBATION THEORY 8.6.2. APPLICATION OF FIRST-ORDER PERTURBATION THEORY TO HELIUM ATOM 8.7. VARIATION METHOD 8.8. LET US SUM UP 8.9. CHECK YOUR PROGRESS 8.10. LESSON - END ACTIVITIES 8.11. REFERENCES 8.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on the Schrödinger equation for hydrogen atom to the students. On successful completion of this lesson the student should have: * Understand the Schrödinger equation for hydrogen atom. 8.1. INTRODUCTION Hydrogen atom is the simplest of all atoms. It is a three-dimensional system and the Schrodinger equation for this system is where Hˆ y = Ey ----- (1) h2 2 Hˆ = Ñ + V ( x, y , z ) 2m ----- (2) and where the Laplacian operator in Cartesian coordinates is given by This watermark does not appear in the registered version - http://www.clicktoconvert.com 75 Ñ2 = ¶2 ¶2 ¶2 + + ¶x 2 ¶y 2 ¶z 2 ----- (3) The potential energy of interaction between the electron and the nucleus is given by V (r ) = - Ze 2 4pe 0 r ----- (4) Since this attractive potential has spherical symmetry depending only upon r, it is convenient to express the Schrodinger equation in terms of polar coordinates (r,q,f), rather than Cartesian coordinates (x,y,z). The Cartesian coordinates are related to polar coordinates (see Fig.1) as follows: x = r sin q cos f y = r sin q sin f z = r cos q The Schrodinger equation for hydrogen atom in terms of polar coordinates becomes 1 ¶ æ r 2 ¶y ç r 2 ¶r çè ¶r ö 1 ¶ æ ¶y ö 1 ¶ 2y 2 m æ Ze 2 ö ç ÷y = 0 ----- (5) ÷÷ + 2 + E + ç sin q ÷+ 2 ¶q ø r sin 2 q ¶f 2 h çè 4pe 0 r ÷ø ø r sin q ¶q è where m is the reduced mass of the electron and the nucleus, that is, m = me mn (mn + me ) . Assuming that V(r) is a function of r only, the above wave equation can be solved by separating the variables: y (r , q , f ) = R(r )Q(q )F (f ) ----- (6) where R(r) is the radial function which is a function of r only, and Q(q) and F(f) are angular functions. Substituting Eq. 6 into Eq. 5 then we get an equation which can be separated into three equations which are: This watermark does not appear in the registered version - http://www.clicktoconvert.com 76 1 d æ r 2 dR ö b 2m æ Ze 2 ö ç ÷R = 0 ç ÷ R + E + 4pe 0 r ÷ø r 2 dr çè dr ÷ø r 2 h 2 çè 1 d æ dQ ö m 2 Q + bQ = 0 ç sin q ÷sin q dq è dq ø sin 2 q or d 2F + m2F = 0 df 2 ----- (7) ----- (8) -----(9) where m and b are constants. The complete solution of Eq. 5 comprises the solutions of Eqs. 7,8 and 9. The energy values of hydrogen-like atoms are found to be given by En = - 2p 2 Z 2 me 4 ----- (10) (4pe 0 )2 (n 2 h 2 ) These energy levels are the same as those obtained from the Bohr theory. The normalized solutions of radial equation (9) are found to be where æ (2 Z na 0 )3 (n - l - 1)! ö ÷ exp(- r 2 )r l L2nl++l1 (r ) Rn ,l (r ) = -çç 3 ÷ 2n[n + l ] ! è ø ----- (11) r = (2 Z na 0 )r and a0 = (4pe 0 )h 2 me 2 ----- (12) L2nl++l1 (r ) are the associated Laguerre polynomials defined as Lsr (r ) = d s é e r d r r -r ù r e ú ê dr s ë dr r û ( ) ----- (13) where s=2l+1 and r = n+l and n=1,2,3,4,...... . The number n is called the principal quantum number which determines the energies of different atomic orbitals and the distances of the electron from the nucleus. The probability of finding the electron between r and r+dr is given by the quantity R*(r)R(r)r2 dr. 8.2. ANGULAR FUNCTIONS The solution of Eq. 8 gives 1 é (2l + 1)(l - m )!ù 2 m Q l ,m (q ) = ê ú Pl (cos q ) êë 2(l + m )! úû ----- (14) This watermark does not appear in the registered version - http://www.clicktoconvert.com 77 where Pl m (cos q ) are called associated Legendre polynomials defined as 1 Pl (cos q ) = l 1 - cos 2 q 2 l! ( m ) m 2 d l+ m (cos 2 d cos q ) q -1 l+ m l ----- (15) In the above equation, the quantum number l (=0,1,2....) is the azimuthal quantum number. It determines the shape of the atomic orbitals. The solution of Eq. 9 gives F m (f ) = 1 2p exp(imf ) ----- (16) where 1 2p is the normalization constant and m = 0,±1,±2,......... .... ± l is the magnetic quantum number. This name is given to this quantum number because in the presence of the external magnetic field, the states having different values of m have different energies. The phenomenon of removal of degeneracy of an energy state by application of external magnetic field is known as Zeeman effect, after P. Zeeman (1865-1943) who was the co-winner (with H.A Lorentz) (1853-1928) of the 1902 Physics Nobel Prize. These Dutch physicists were honored for their work on the influence of magnetism upon radiation phenomena. To sum up, the solution of the radial equation (7) gives quantum numbers n and l. The solution of the Q-equation (8) gives quantum numbers l and m, and the solution of the F-equation (9) gives quantum number m. In other words, the quantum numbers n, l, and m follow directly from the wave mechanical treatment. 8.3. PROBABILITY DENSITY AND RADIAL DISTRIBUTION FUNCTIONS As mentioned before, the value of y in the Schrodinger wave equation is important since y 2 gives the probability of finding an electron of a given energy (i.e., of known quantum numbers) at a given distance from the nucleus. Thus, by determining y 2 at different distances from the nucleus, it is possible to trace out a region of space around the nucleus where there is high probability of locating the electron. Each such region of space, as already pointed out, is called an orbital. The probability of locating the electron at different distances from the nucleus can be represented graphically by plotting probability (y 2 ) against the distance (r) from the nucleus of the atom. However, electron probability distribution is generally expressed in the form of radial probability distribution which means the probability of finding an electron within a small radial space around the nucleus. Let us suppose that the space around the nucleus is divided into a very large number of thin concentric shells of thickness dr at a distance r from the nucleus. The volume of a spherical shell between radii r and r + dr will be given by 4pr2 dr. The probability of finding electron within this spherical shell will be given by 4pr2y 2 dr. The radial probability at a distance r from the nucleus is thus given by the function, 4pr2y 2 dr. In this function, while the probability factor y 2 decreases, the volume factor 4pr2 dr increases with increase in the value of r. The radial This watermark does not appear in the registered version - http://www.clicktoconvert.com 78 probability distribution of the electron is thus obtained by plotting the function 4pr2y 2 against the distance r from the nucleus. The probability of finding an electron at zero distance from the nucleus is zero. The probability increases gradually as the distance increases, goes to a maximum and then begins to decrease. The peak of the curve gives the distance from the nucleus where the probability of finding the electron is maximum. This distance is called the radius of maximum probability. ° In the case of hydrogen atom, the radius of maximum probability is 0.529 A . Bohr’s model restricts the electron to a definite orbit at a fixed distance from the nucleus. The wave mechanical model, however, gives merely the maximum probability of locating the electron at a given distance from the nucleus. In the case of hydrogen atom, for instance, according to Bohr’s calculations, the electron under ordinary conditions always stays at a ° distance of 0.529 A from the nucleus. According to the wave mechanical model, however, the electron keeps on moving toward or away from the nucleus and the maximum probability of ° locating it lies at a radius of 0.529 A from the nucleus. 8.4. THE MOST PROBABLE DISTANCE OF THE H-ATOM (OR H-LIKE SPECIES) 1S ELECTRON Consider the 1s orbital wave function of hydrogen atom given by ( y 1s = 1 pa 03 ) 1 -r 2 e a0 ----- (1) where r is the distance of the electron from the nucleus. The probability density for an electron in this orbital is given by ( ) y 12s = 1 pa 03 e -2 r a0 ----- (2) It may be noted that y 1s2 is independent of q and f and hence the electron distribution in 1s orbital is spherically symmetrical. Graphically the plot of y 1s2 against r is as shown by curve a in Fig. 2. This watermark does not appear in the registered version - http://www.clicktoconvert.com 79 There is also another procedure for describing the electron distribution in an atomic orbital. It consists in determining the probability of finding an electron in a spherical shell of thickness dr at a distance r from the nucleus. This is called radial probability. The radial probability distribution of electron is given by P (r )dr = 4p 2y 12s dr æ 1 ö = 4pr 2 çç 3 ÷÷e - 2 r a0 dr è pa 0 ø ----- (3) The plot of P(r), i.e., of 4p 2y 12s against r is given by curve b in Fig.1. The maximum of the curve is found by differentiating P(r) with respect to r and setting the derivative equal to zero, i.e., ö dP(r ) 4 æ 2r 2 = 3 çç + 2r ÷÷e - 2 r a0 = 0 ----- (4) dr a0 è a0 ø ( It readily follows from Eq. 4 that r = a 0 . Q e -2 r a0 ¹ 0 ) Thus, we find that the maximum radial probability density occurs at a distance a0 , the Bohr radius. We, therefore, conclude that the most probable distance of the 1s electron from the nucleus is precisely what has been predicted by the Bohr theory. As already mentioned, while the Bohr model restricts the electron to a definite orbit at a fixed distance from the nucleus, the wave mechanical model gives merely the maximum probability of locating the electron at a given distance from the nucleus. This watermark does not appear in the registered version - http://www.clicktoconvert.com 80 8.5. PHYSICAL INTERPRETATION OF THE HYDROGENIC ATOMIC ORBITALS According to the Bohr theory, the angular momentum L of an electron in an orbital is given by L = n(h 2p ) ; n=1, 2, 3, ..... ----- (5) However, according to wave mechanics, the value of L is given by L = [l (l + 1)] h = [l (l + 1)] 12 12 (h 2p ) ----- (6) Thus , while according to the Bohr theory the ground state angular momentum is equal to h/2p since n=1(i.e., it is finite), according to wave mechanics its value is zero(Since l=0). Consider an electron of mass me moving about the nucleus at a distance r in an orbit with velocity u. The centrifugal force which keeps the electron away from the nucleus is given by me u 2 me2 u 2 r 2 L2 F= = = r me r 3 me r 3 ----- (7) Where L= me ur is the angular momentum of the electron. In terms of atomic units frequently used in quantum mechanics, we said me as well as e and h =1. Hence, in atomic units we can write Eq. 6 as L2 = l (l + 1) ----- (8) Substituting for L2 in Eq.7, we have F = l (l + 1) r 3 ----- (9) The radial motion of an electron in an atom is controlled by the centrifugal force given by Eq. 9 as well as by the coulombic force of attraction, viz., - Ze 2 r 2 (or - Z r 2 in atomic units). For the atom to be stable, l (l + 1) Z ----- (10) - 2 = 0 or r = l (l + 1) Z r3 r Eq.10 shows that , other things being equal, an electron in an orbital of high angular momentum tends to stay farther from the nucleus then an electron in a state of lower angular momentum. Fig. 3 shows the radial parts of the Hydrogenic wave functions for the principal quantum number n = 3. We see that the higher value of l, the less likely it is to be found near the nucleus. Thus, for a given value of n, the size of the orbit increases with increasing the azimuthal quantum number, l. This watermark does not appear in the registered version - http://www.clicktoconvert.com 81 8.6. THE SCHRODINGER WAVE EQUATION FOR MULTI-ELECTRON ATOMS The Schrodinger wave equation cannot be solved exactly for atoms beyond hydrogen atom in the Periodic table. The troublesome term is the interelectron repulsion term in the Hamiltonian because of which the n-electron wave function cannot be split into n one-electron wave equations. Hence, methods have been developed for the approximate solution of the multielectron Schrodinger wave equation. One of these methods is based on the time-independent perturbation theory and the other, called the variation method. Involves the selection of a trial wave function. 8.6.1. TIME-INDEPENDENT PERTURBATION THEORY The Schrodinger wave equation to be solved is Hˆ y = Ey ----- (1) The Hamiltonian is decomposed into two parts as Hˆ = Hˆ ( 0 ) + lHˆ ' ----- (2) where Ĥ ( 0 ) is the unperturbed part and lĤ ' is the perturbation where l is the perturbation parameter which measures the deviation of the problem of interest from the unperturbed system. It is further assumed that lHˆ ' << Hˆ ( 0 ) ----- (3) In general, l is set equal to unity which means that perturbation is fully applied. Associated with Ĥ ( 0 ) are a set of eigenvalues E1( 0 ) , E 2( 0 ) , ........, E n( 0 ) and the corresponding eigenfunctions y 1( 0 ) ,y 2( 0 ) ,.......y n( 0 ) , i.e., Hˆ y n( 0 ) = E n( 0 )y n( 0 ) ----- (4) This watermark does not appear in the registered version - http://www.clicktoconvert.com 82 It is assumed that the eigenfunctions y n and the energy eigenvalues En of the total Hamiltonian Ĥ can be expressed in the form of a power series of l : y n = y n(0 ) + ly n(1) + l2y n(21) + l3y n(3 ) + .... ----- (5a) E n = E n(0 ) + lE n(1) + l2 E n(21) + l3 E n(3 ) + .... ----- (5b) The first term in Eq. 5a is the zeroth-order term, the second represents the first order correction; the third represents the second-order correction, etc., to the unperturbed zeroth-order term. The eigen functions y n(1) ,y n(2 ) ,.... and the eigenvalues E n(1) , E n(2 ) ,... are independent of l and y n(1) ,y n(2 ) ,.... are so chosen that they are orthogonal to y n(0 ) which is assumed to be normalized. Omitted details, When Eq. 5a and 5b are solved, the results to the first-order in l are: y n = y n(0 ) + y n(1) (0 ) =y n < y n(0 ) Hˆ 'y m(0 ) > (0 ) + lå (0 ) (0 ) y m En - Em m¹n ----- (6) E n = E n(0 ) + E n(1) where = E n(0 ) + l < y n(0 ) Hˆ 'y n(0 ) > ----- (7) (0 ) E n(0 ) =< y n(0 ) Hˆ (0 ) y n(0 ) >º Hˆ nn ----- (8) It can be further shown that and E n(1) =< y n(0 ) Hˆ 'y n(0 ) > ----- (9) E n(2 ) =< y n(0 ) Hˆ 'y n(1) > ----- (10) E n(3 ) =< y n(0 ) Hˆ 'y n(2 ) > ----- (11) E n(n ) =< y n(0 ) Hˆ 'y n(n -1) > ----- (12) These equations show that the calculation of nth order energy requires a knowledge of (n-1)th order eigenfunctions. Since the contribution of second order, third order, etc., terms goes on successively decreasing, we are primarily interested in the first order correction. 8.6.2. APPLICATION OF FIRST-ORDER PERTURBATION THEORY TO HELIUM ATOM This watermark does not appear in the registered version - http://www.clicktoconvert.com 83 We shall solve the Schrodinger wave equation for the ground state of helium atom using the first-order time- independent perturbation theory. In this case, Hˆ y = Ey ----- (13) Hˆ = Hˆ ( 0 ) + lHˆ ' ----- (14) where the unperturbed Hamiltonian, Ĥ (0 ) , is given by h2 2 Ze 2 æ 1 1 ö ç + ÷ Hˆ (0 ) = Ñ1 + Ñ 22 2m 4pe 0 çè r1 r2 ÷ø ( ) ----- (15) and the perturbation is the interelectron repulsion term : Hˆ ' = e 2 4pe 0 r12 ----- (16) Here r1 and r2 are the distances of the two electrons from the helium nucleus of charge Ze, and r12 is the interelectron distance. We shall use atomic units, a.u. (h = e = m » m = 1 4pe 0 = 1) , so that æ2 2ö 1 Hˆ (0 ) = - Ñ12 + Ñ 22 - çç + ÷÷ ----- (17) 2 è r1 r2 ø ----- (18) Hˆ ' = 1 r ( ) 12 Since Hˆ ' << Hˆ (0 ) , it is pertinent to use perturbation theory. Since Ĥ (0 ) is the sum of two one-electron Hamiltonians, the unperturbed wave function y (0 ) (r1 , r2 ) can be written as the product of two Hydrogenic wave functions: y (0 ) (r1 , r2 ) = y (0 ) (r1 )y (0 ) (r2 ) ----- (19) Where y (0 ) (ri ) is the wave function of the ith electron in a Hydrogenic atom with nuclear charge = Ze. Thus, y (0 ) æ Z 3 ö - Zr1 (r1 , r2 ) = çç 3 ÷÷e è pa 0 ø ( ) = Z3 p e ( - Z r1 + r2 12 a0 ) æ Z 3 ö - Zr1 ´ çç 3 ÷÷ e è pa 0 ø (in a.u.) a0 ----- (20) The unperturbed ground state energy, E 0(0 ) , is equal to the sum of the ground state energies of two Hydrogenic atoms : ( ) ( ) E o(0 ) = - Z 2 2 + - Z 2 2 = - Z 2 ----- (21) This watermark does not appear in the registered version - http://www.clicktoconvert.com 84 The first-order correction to the ground-state energy is E 0(1) =< y (0 ) Hˆ 'y (0 ) > = òòy *(0 ) (r1 , r2 )Hˆ 'y (0 ) (r1 , r2 )dt 1 dt 2 ----- (22) Substituting the value of y (0 ) (r1 , r2 ) from Eq. 20 into Eq. 22, we obtain E 0(1) = Z6 p2 òò e - 2 Z ( r1 + r2 )(1 r12 )dt 1dt 2 ----- (23) where the volume elements of the two electrons (in spherical polar coordinates are) : dt 1 = r12 sin q1 dr1 dq1 df1 dt 2 = r212 sin q 21 dr21 dq 21 df 21 The evaluation of the interelectron repulsion integral (Eq.23) is rather tedious; it can be shown that it is given by E 0(1) = (5 8)Z ----- (24) Notice that E 0(1) is positive, as was to be expected since the repulsion energy between two electrons is always positive. Adding Eqs. 21 and 24, we have æ5ö E 0 = E 0(0 ) + E 0(1) = - Z 2 + ç ÷ Z è8ø 5 ö æ = -ç Z 2 - Z ÷ 8 ø è ----- (25) Re- incorporating the original units we find that 5 ö me 2 æ E 0 = -ç Z 2 - Z ÷ 2 8 ø 2h è Recalling that the ground state energy of hydrogen atom is - ----- (25a) 1 a.u. or - me 4 2h 2 or -13.60eV, 2 we get for helium atom 5 ö æ E 0 = -ç Z 2 - Z ÷(27.2 )eV = -2.75 a.u. = -74.80 eV 8 ø è ----- (26) This watermark does not appear in the registered version - http://www.clicktoconvert.com 85 The experimental value is -2.904 a.u. or – 78.986 eV. The agreement between the theoretical and experimental values is not good. If the second-order and higher order contributions are included, the agreement improves but is still not very good. 8.7. VARIATION METHOD According to the variation method, also called the variation theorem, i f y is an approximate wave function of a quantum mechanical system, an atom or a molecule, described by the Hamiltonian Ĥ , then the energy eigenvalue of the system, approximately given by the integral < y Hˆ y > E= <y y > = òy * Hˆ ydt òy *ydt ----- (1) or, if y is normalized (so that <y /y >=1), by the integral E ³ E0 ----- (2) is an upper bound to the ground state energy, E0 , of the system, i.e., We shall not give the proof of the variation theorem here. If the approximate wave function, also called the trial function, happens to be exact (true) wave function, then E = E0 ----- (3) However, this is seldom the case since we have no idea of the exact wave function, to begin with. The application of the variation method involves the following steps: (i) Choose a trial wave function y dependent on variable parameters. (ii) Evaluate the integral < y Ĥ y > (iii) (iv) Minimise the above integral with respect to the variable parameters. The function y with the optimum value of the parameters is the best approximation to the true wave function and the lowest value of < y Ĥ y > is the nearest approximation to the true energy. 8.7.1. APPLICATION OF VARIATION METHOD TO HELIUM ATOM The Hamiltonian for He atom (in atomic units) is æ2 2ö 1 1 Hˆ = - Ñ12 + Ñ 22 - çç + ÷÷ + 2 è r1 r2 ø r12 ( ) ----- (4) This watermark does not appear in the registered version - http://www.clicktoconvert.com 86 where the symbols have their usual meanings. As a result of screening, each electron (1 and2) shields the other from the nucleus so that the two electrons do not ‘’see’’ the same nuclear charge. Here it is natural to use a wave function with nuclear charge Z less than 2. That is, Z is a variation parameter which we shall determine from the variational calculation. It is important to remember that the Hamiltonian in Eq. 4 contains the exact value of Z = 2.The vibrational parameter appears only in the trial wave function: 1 y (r1, r2 ) æ Z 3 ö 2 - Zr1 ÷÷ e = çç èp ø 1 æ Z 3 ö 2 - Zr2 ÷÷ e ´ çç èp ø ----- (5) = (Z 3 p )exp[- Z (r1 + r2 )] ----- (6) E =< y Hˆ y > ----- (7) Substituting the value of Ĥ from Eq. 4 in Eq. 7, E=- 1 1 1 1 1 < y Ñ12 y > - < y Ñ 22 y > -2 < y y > -2 < y y > + < y y > 2 2 r1 r2 r12 ----- (8) The Laplacian operator in spherical polar coordinates (r,q,f) is given by Ñ2 = 1 ¶ æ 2 ¶ ö 1 ¶ æ ¶ ö 1 ¶2 r + sin q + ç ÷ ç ÷ ¶q ø r 2 sin 2 q ¶f 2 r 2 ¶r è ¶r ø r 2 sin q ¶q è ----- (9) Since the trial function (Eq. 5) does not depend upon angles q and f, differentiations with respect to q1, f 1, q2 and f 2 vanish so that for the first kinetic energy operator in Eq. 8 , we get æZ3 ö 1 2 ÷÷ - < y Ñ1 y >= çç 2 p è ø 2 é 1 ¶ æ 2 ¶ öù - Zr1 - Zr1 - 2 Zr 2 2 e ò ê- 2r12 ¶r1 ççè r1 ¶r1 ÷÷øúe ´ r1 sin q1dr1dq1df1 ò e 2 r2 sin q 2 dr2 dq 2 df2 ë û ----- (10) The functions of r2 are not affected by differentiation. Carrying out the integrations overall angles we obtain 2 ¥ é 1 ¶ æ 2 ¶e - Zr1 æZ3 ö 1 çç r1 ÷÷ 4p 2 ò e - Zr1 ê- 2 - < y Ñ12 y >= çç 2 ¶r1 èp ø 0 ë 2r1 ¶r1 è ( ) ¥ öù 2 ÷÷ú r1 dr1 ´ ò e - 2 Zr2 r22 dr2 øû 0 ----- (11) The second integral, evaluated using the standard integral, is equal to 2!/(2Z)3 so that Eq. 11 becomes This watermark does not appear in the registered version - http://www.clicktoconvert.com 87 (4Z ) æç - 12 ö÷ 3 2 ¥ 2! òe ø (2 Z ) è - Zr1 3 0 é¶ - Zr12 e - Zr1 ê ¶ r ë 1 ( )ùú dr û 1 ¥ ( = - 2Z 3 ) ò e (- 2Zr + Z r )e - Zr1 2 1 2 1 - Zr1 dr1 0 ¥ ¥ é ù = - 2 Z 3 ê- 2 Z ò r1e - 2 Zr1 dr1 + ò r12 e - 2 Zr1 dr1 ú 0 0 ë û ( ) é 2Z Z 2 2! ù Z 2 + = (- 2 Z 3 )êú= 2 (2Z )3 úû 2 êë (2 Z ) ----- (12) Similarly, the second kinetic energy operator term in Eq. 8 gives -2Z. The fifth term, the interelectron repulsion energy integral, is some what more tedious. So, we will not go into details about its evaluation, merely stating that it evaluates to (5/8)Z. Combining all these contributions Eq. 8 becomes ( ) E = 2 Z 2 2 + 2(- 2 Z ) + (5 8)Z = Z 2 - (27 8)Z ----- (13) Using the variation method, we minimize the energy with respect to Z, obtaining ¶E ¶Z = 2 Z - 27 8 = 0, where Z = 27 16 = 1.69 Substituting this value of Z in Eq. 13, we obtain the ground state energy of helium atom: E0 = (27/16)2 - (27/8)(27/16) = -2.8476 a.u. = - (2.8476)(27.21 eV) = - 77.48 eV [ Q 1 a.u. = 27.21 eV ] ----- (14) This is much better than the first-order perturbation theory result, viz., - 74.80 eV, we obtained in Eq. 26, the experimental value being – 2.904 a. u., i.e., - 78.986 eV. in 1929, E.A. Hyleraas, using a more complicated wave function and the concept of the so-called electron correlation which stipulates that electrons try to avoid each other, obtained a result for the ground state energy of helium which is very close to the experimental value. 8.8. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Angular functions Probability density and radial distribution functions The most probable distance of the H-atom (or H-like species) 1s electron Physical interpretation of the hydrogenic atomic orbitals This watermark does not appear in the registered version - http://www.clicktoconvert.com 88 Ø Ø Ø Ø The Schrödinger wave equation for multi-electron atoms Time- independent perturbation theory Application of first-order perturbation theory to helium atom Variation method 8.9. CHECK YOUR PROGRESS 1. Show that the following radial wave function of hydrogen atom is normalized. 32 R1, 0 (r ) = (2 a 0 ) exp(- r a 0 ) . 2. Using the first order perturbation theory solve the Schrodinger wave equation for the ground state energy of Helium atom. 8.10. LESSON – END ACTIVITIES 1. 2. Using the variation method solve the Schrodinger wave equation for the ground state energy of Helium atom. Write the Schrödinger wave equation for Hydrogen atom in terms of polar coordinates. Separate the resultant equation in three equations using the technique of separation of variables. How do the quantum numbers n,l, and m emerge from the solution of the wave equation. 8.11. REFERENCES 1. R.K. Prasad, Quantum Chemistry, Wiley Eastern Limited, New Delhi. 2. IRA N. Levin, Quantum Chemistry, Prentice-Hall of India private limited, New Delhi. 3. G. Aruldhas, Quantum mechanics, Prentice-Hall of India private limited, New Delhi. 4. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, millennium edition, Vishal Publishing co. This watermark does not appear in the registered version - http://www.clicktoconvert.com 89 UNIT-VI LESSON 9: THERMODYNAMICS AND NON-IDEAL SYSTEMS: CHEMICAL POTENTIAL CONTENTS 9.0. AIMS AND OBJECTIVES 9.1. INTRODUCTION: CONCEPT OF CHEMICAL POTENTIAL 9.2. GIBBS- DUHEM EQUATION 9.3. SOME IMPORTANT RESULTS 9.4. VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE 9.5. VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE 9.6. CHEMICAL POTENTIAL IN CASE OF A SYSTEM OF IDEAL GASES 9.7. APPLICATION OF THE CONCEPT OF CHEMICAL POTENTIAL 9.7.1. CLAPEYRON–CLAUSIUS EQUATION 9.7.2. INTEGRATED FORM OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIUM 9 . 7 . 3 . A P P L I CATION OF CLAPEYRON-CLAUSIUS EQUATION LIQUID Û VAPOR EQUILIBRIA 9.7.4. CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û VAPOR EQUILIBRIUM 9.7.5. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û LIQUID EQUILIBRIA 9.8. LET US SUM UP 9.9. CHECK YOUR PROGRESS 9.10. LESSON - END ACTIVITIES 9.11. REFERENCES FOR 9.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on chemical potential to the students. On successful completion of this lesson the student should have: * Understand the concept of chemical potential. 9.1. INTRODUCTION: CONCEPT OF CHEMICAL POTENTIAL The thermodynamic properties, E, H, S, A and G are extensive properties because their values change with change in the mass (or) the numbers of moles of the system. In the derivation of various equations described earlier the change of was considered to be due to change in temperature and pressure only. This watermark does not appear in the registered version - http://www.clicktoconvert.com 90 A tactic assumption was made that the systems under consideration was a closed system. (i.e.) there could be no change in the mass of the system. However, if the case of an open system containing two or more components, there can be change in the number of moles of various components as well. In that case, an extensive property like G must be a function not only of temperature and pressure but of the number of moles o the various components present in the system as well. Let T and P be the temperature and pressure respectively, of a system and let n1 , n2 , n5 …….ni be the respective number of moles of the constituents; 1, 2, 3,….i, then in view of what has been said above, the Gibbs free energy, G, must be a function of temperature, pressure and the numbers of moles of the various constituents (i.e.) G = f (T, P, n1, n2, n3¼¼ ni) ------- (1) Where n1 + n2 + n3 + ¼¼ + ni = total number moles = N (say) Then, for a small change in temperature, pressure and the numbers of moles of the components, the change in free energy dG will be given by the expression é dG ù é dG ù é dG ù é dG ù é dG ù dG = ê ú dT + ê ú dP + ê dn i + ê dn 2 ..... + ê dn i ú ú ú ë dT û P , N ë dP û T , N ë dn 2 û T , P , n1, n 2...ni ë dn i û T , P , n 2...ni ë dn i û T , P , n1, n 2..... ni æ ¶G ö ÷ The quantity ç is called partial molal ç ¶n ÷ è 2 øT , P, n , n .......n 1 2 i more often, chemical potential (µi) of the concerned component i. Thus, æ ¶G ö ç ÷ = Gi = m i ç ¶n ÷ è 2 øT , P, n , n .......nj 1 2 free energy Gi or ------ (3) The term chemical potential was first introduced by Gibbs. The physical significance of chemical potential easily follows from eqn (3). The chemical potential of a given substance is, evidently, the change in free energy of the system that results on the addition of one mole of that particular substance at a constant temperature and pressure, to such a large quantity of the system that is no appreciable change in the overall composition of the system. Eqn (2) may be written as é dG ù é dG ù dG = ê ú dT + ê ú dP + m dn + m dn + m dn 1 1 2 2 i i ë dT û P, N ë dP û T , N ------ (4) Where µ1, µ2 ¼¼µj are chemical potential of the components 1, 2 and j, respectively. This watermark does not appear in the registered version - http://www.clicktoconvert.com 91 If temperature and pressure remains constant then, ------ (5) (dG) T, P = µ1dn 1 + µ 2 dn 2 + ¼¼.. µ j dn j If a system has a definite composition having n1 , n2 ….ni moles of the constituents respectively, then on integrating eqn (5) we have (G) T, P, N 1, 2 …j ------ (6) = n 1 µ1 + n 2µ 2 + ¼¼.. n jµ j From eqn (6) chemical potential may be taken as the contribution per mole of each particular constituent of the mixture to the total free energy of the system under conditions of constant temperature and pressure. It readily follows that for a total 1 mole of a pure substance, G = µ (i.e.) free energy is identical with chemical potential. 9.2. GIBBS- DUHEM EQUATION We know that the eqn (6) (G) T, P, N ------ (6) = n 1 µ1 + n 2µ 2 + ¼¼.. n jµ j Shows that the free energy of a system at constant temperature and pressure can be expressed as a sum of ‘nµ ‘terms for the individual components of the system. Diffentiating eqn (6), we obtain dG = µ1dn 1 + n 1dµ1 + µ 2 dn 2 + n 2 dµ 2 ¼¼ + µ j dn j + n j dµ j (or) dG = (µ1dn 1 + µ 2 dn 2 + .......µ j dn j ) + ( n 1dµ1 + n 2 dµ 2 + ........ n j dµ j ) ------ (7) But, according to eqn. (5) (dG) T, P = µ1dn 1 + µ 2 dn 2 + …….. µjdnj ------ (5) The first term on right hand side of (7) is equal to dG , at constant temperature and pressure. It follows therefore that at constant temperature and pressure, for a system of a definite composition n1dµ1 + n2dµ2 + ¼¼ njdµj = 0 (Or) å n i dµ i = 0 ------ (8) ------ (9) This simple relationship is known as Gibbs – Duhem equation. For a system having only two components ((e.g.) a binary solution) the above equation reduces to n 1dµ1 + n 2 dµ 2 = 0 ------- (10) This watermark does not appear in the registered version - http://www.clicktoconvert.com 92 (Or) dµ1 = (-) n2 dµ 2 n1 ------ (11) Eqn (11) shows that variation in chemical potential of one component affects the value for the other component as well. Thus, if dµ1 is positive, (i.e.) if µ1 increases then dµ2 must be negative, (i.e.) µ2 must decease and vice versa. 9.3. SOME IMPORTANT RESULTS In a special case when there is no change in the number of moles of the various constituents of a system, that is, when the system is a closed one, then, dn 1 , dn 2 , ¼¼ dn j are all zero. In such a case, eqn (4) dG = æ ¶G ö æ ¶G ö ç ÷ dT + ç ÷ + µ1dn 1 + µ 2 dn 2 + …….. µjdnj è ¶T ø P , N è ¶P ø T , N ------ (4) Reduces to dG = æ ¶G ö æ ¶G ö ç ÷ dT + ç ÷ è ¶T ø P , N è ¶P øT , N ------ (12) It also follows from the expression G = H- T∆S dG = dE + PdV + - TdS SdT ---- (A) TdS = dq = dE + PdV Combining eqns A & B That in a closed system ------ (B) we get, dG = VdP + SdT dG = VdP SdT Hence, by equating coefficients of dT and dP in the above two equations, we get æ ¶G ö ç ÷ = -S è ¶T ø P, N ------ (13) æ ¶G ö çç ÷÷ = V è ¶p ø T, N ------ (14) and These results are important as they help us in deriving expressions for the variation of chemical potential with temperature and pressure. 9.4. VARIATION OF CHEMICAL POTENTIAL WITH TEMPERATURE This watermark does not appear in the registered version - http://www.clicktoconvert.com 93 The variation of chemical potential of any constituent (i) of a system with temperature can be derived by differentiating eqn (3) with respect to temperature and eqn (13) with respect to ni æ ¶G ö ç ÷ = Gi = m i ç ¶n ÷ è i øT , P, n1 , n .......nj ------ (3) 2 æ ¶G ö ç ÷ = -S è ¶T ø P, N ------ (13) The results are and ¶ 2G æ ¶m ö =ç i÷ ¶ni ¶T è ¶T ø P , N ------ (15) æ ¶S ¶ 2G = -ç ç ¶n ¶n ¶T i è i ------ (16) ö ÷ = -Si ÷ øT , P, n ,....n 1 j Where Si , by definition, is the partial molal (or) molar entropy of the component i. It follows from equations (15) and (16) that æ ¶m i ö ç ÷ = -Si è ¶T ø P , N ------ (17) Chemical potential Eqn (17) gives the variation of chemical potential µi of any constituent (i) with temperature. Since the entropy of a substance is always positive, hence, according to eqn (17), the chemical potential would decrease with increase in temperature. This is illustrated in fig (1) for a substance in solid, liquid and gaseous states. It is evident from this figure that at the melting point (Tm), the chemical potentials of the solid and liquid phases are the same. Similarly, at the boiling point (Tb); the chemical potentials of liquid and gaseous phases are the same. These observations are extremely useful in the phase rule studies. Gas Solid Liquid Tm Tb Temperature This watermark does not appear in the registered version - http://www.clicktoconvert.com 94 Fig.1. Variation of chemical potential with temperature. 9.5. VARIATION OF CHEMICAL POTENTIAL WITH PRESSURE The variation of chemical potential of any constituent I with pressure may be derived by differentiation eqn (3) with respect to pressure and eqn(14) with respect to n i , æ ¶G ö ç ÷ = Gi = m i ç ¶n ÷ è i øT , P, n1 , n .......nj ------ (3) 2 W.r.t. P and æ ¶G ö ç ÷ =V è ¶P øT , N ------ (14) W.r.t. n i The results are æ ¶ 2 G ö æ ¶m i ö çç ÷÷ = ç ÷ è ¶P¶ni ø è ¶P øT , N æ ¶ 2G ö æ ¶V ö ç ÷=ç ÷ =V i ç ¶n ¶P ÷ ç ¶n ÷ è i ø è i øT , P, n .....n 1 j ------ (18) And ------ (19) Here Vi by definition, is the partial molal (or) molar volume of the component i. It follows from the eqns. (18) (19) that æ ¶m i ö ------ (20) ç ÷ = Vi è ¶P øT , N Eqn. 20 gives the variation of chemical potential (µi) of any constituent i with pressure. 9.6. CHEMICAL POTENTIAL IN CASE OF A SYSTEM OF IDEAL GASES For a system of ideal gases, a faster development of equation æ ¶m i ö ç ÷ = -Vi ----- (A) is also possible. In an ideal gas, PV = nRT . Consider now a system è ¶P øT , N consisting of a number of ideal gases, let n1 , n2……, be the number of moles of each constituent present in the mixture. Then, in the ideal gas equation, n the total number of moles may be replaced by (n1 + n2 + ……). Hence, nRT RT ------ (1) V = = (n1 + n2 + ....) P P This watermark does not appear in the registered version - http://www.clicktoconvert.com 95 Differentiating equation (1) with respect to the ni , at constant temperature and pressure, we have æ ¶V çç è ¶ni ö RT ÷÷ = Vi = P øT , P ,n1 ,n2 .... ------ (2) æ RT ö Substituting the value of Vi ç = ÷ equation (A), we have è P ø RT æ ¶m i ö ç ÷ = P è ¶P øT , N ------ (3) For a constant composition of the gases and at a constant temperature, equation (3) ay also be expressed in the form dm i = RT dP = RTd ln P P ------ (4) Let pi be the partial pressure of the constituent ‘i’ present in the mixture, since, each constituent behaves as an ideal gas, therefore, piV = ni RT ------ (5) It follows from equation (5) and (1) that æn ö ------ (6) pi = ç i ÷ P ènø Since ni and n are constants, therefore, on taking logarithms and then differentiating, we get d ln pi = d ln P ------ (7) Substituting in equation (4) we have dm i = RTd ln pi ------ (8) On integrating equation (8), we get m i = m i0 ( P ) + RT ln pi ------ (9) Where m i0 ( P ) is the integration constant, the value of which depends upon the nature of the gas and also on the temperature. It is evident from equation (9) that the “chemical potential of any constituent of a mixture of ideal gases is determined by its partial pressure in the mixture”. If the partial pressure of the constituent ‘i’ is unity, (i.e.), pi = 1, then m i = m i0 ( P ) ------ (10) This watermark does not appear in the registered version - http://www.clicktoconvert.com 96 Thus, m i0 ( P ) gives the chemical potential of the gaseous constituent ‘i’ when the partial pressure of the constituent is unity, at a constant temperature. According to equation (5), æn ö ------ (11) pi = ç i ÷ RT èV ø æn ö Now ç i ÷ represents molar concentration (i.e.), the number of moles per unit volume of èV ø the constituent ‘i’ in the mixture. If this concentration is represented by Ci , then equation (11) gives pi = Ci RT ------ (12) Introducing this value of pi in eqn. (9), we have m i = m i0 ( P ) + RT ln (Ci RT ) = m i0 ( P ) + RT ln RT + RT ln Ci (Or) 1442443 cons tan t mi = m 0 i (C ) + RT ln Ci ------ (13) Where m i0 (C ) is a constant depending upon the nature of the gas and the temperature. If Ci = 1 , then, m i = m i0 (C ) ------ (14) Thus m i0 (C ) represents the chemical potential of the constituent (i) when the concentration of the æn ö constituent in the mixture is unity, at a constant temperature. Lastly, since ç i ÷ represents the ènø mole fraction ( xi ) of the constituent ‘i’ in the mixture, equation (6) may be represented as pi = xi P ------ (15) Substituting this value of pi in equation (9), we have m i = m i0 ( P ) + RT ln ( xi P ) ------ (16) = m i0 ( P ) + RT ln P + RT ln xi 1442443 cons tan t m i = m i0 ( x ) + RT ln xi ------ (17) Where, one quantity m i0 ( x ) is also a constant which depends both on the temperature the total pressure. If xi = 1 then, m i = m i0 ( x ) ------ (18) This watermark does not appear in the registered version - http://www.clicktoconvert.com 97 Thus, the quantity m i0 ( x ) represents the chemical potential of the constituent ‘i’ when its mole fraction, at a constant temperature and pressure is unity. 9.7. APPLICATION OF THE CONCEPT OF CHEMICAL POTENTIAL The concept of the chemical potential has been used in deriving a number of important generalizations such as the law of mass action the phase rule, the distribution law, the laws of osmotic pressure etc., 9.7.1. CLAPEYRON–CLAUSIUS EQUATION An equation of fundamental importance which finds extensive application in one component, two–phase systems, was derived by Clapeyron and independently by Clausius from the second law of thermodynamics and is generally knows as Clapeyron–Clausius equation. The two phases in equilibrium may be any of the following types: (i) Solid and liquid, S L at the melting point of the solid. (ii) Liquid and vapor, L V at the boiling point of the liquid. (iii)Solid and vapor, S V at the sublimation temperature of the solid (iv) One crystalline form and another crystalline form as, for example, Rhombic and monoclinic sulphur SR SM at the transition temperature of the allotropic forms Consider any two phase (say liquid and vapor) of one and the same substance in equilibrium with each other at a given temperature and pressure. It is possible to transfer any definite amount of the substance from one phase to another, in a thermodynamically reversible manner, (i.e.), infinitesimally slowly, the system remaining in a state of equilibrium all along. For example: By supplying heat infinitesimally slowly to the system, it is possible to change any desired amount of the substance from the liquid to vapor, phase at the same temperature and pressure. Similarly, by withdrawing heat infinitesimally slowly from the system, it is possible to change any desired amount of the substance from vapor to liquid phase with out change in temperature and pressure. Since, the system remains in a state of equilibrium, the free energy change of either process will be zero. We may conclude, therefore, that “equal amounts of a given substance must have exactly the same free energy in the two phases at equilibrium with each other”. Consider in general, the change of a pure substance from phase ‘A’ to another phase ‘B’ in equilibrium with in at a given temperature and pressure. A B at constant temperature and pressure. If GA is the free energy per mole of the substance in the initial phase ‘A’ and GB is the free energy per mole in the final phase ‘B’, then, since This watermark does not appear in the registered version - http://www.clicktoconvert.com 98 GA = GB Hence, there will be no free energy change. (i.e.,) G = GB – GA = 0 If the temperature of such a system is raised, say from T to T + dT, the pressure will also have to change. Say, from P to P + dP, in order to maintain the equilibrium. The relationship between dT and dP can be delivered from thermodynamics. Let the free energy per mole of the substance in phase ‘A’ at the new temperature and pressure be GA + dGA and that in phase ‘B’ be GB + dGB. Since, the two phases are still in equilibrium, hence, GA + dGA = GB + dGB ------ (1) According to thermodynamics, Since \ G = H – TS H = E + PV G = E + PV – TS ------ (2) dG = dE + PdV + VdP – TdS – SdT ------ (4) ------ (3) Upon differentiation, The first law equation for an infinitesimal change may be written as dq = dE + dw ------ (5) If the work done dw is only due to expansion, then dq = dE + PdV ------ (6) Now for a reversible process, dS = dq T TdS = dq = dE + PdV ------ (7) Combining equations 4 and 7, we have dG = VdP = SdT ------ (8) The equation (8) gives change of free energy when a system under goes reversible a change of temperature dT and a change of pressure dP equation (8) for phase ‘A’ may be written as ------ (9) dG A = V A dP - S A dT This watermark does not appear in the registered version - http://www.clicktoconvert.com 99 and for phase ‘B’ as dGB = VB dP - S B dT ------ (10) Since GA = GB, hence from equation (1), dG A = dGB ------ (11) V A dP - S A dT = VB dP - S B dT (or) dP S B - S A = dT VB - V A ------ (12) ------ (13) It may be noted that that since VA and VB are the molar volumes of the pure substance in the two phases ‘A’ and ‘B’ respectively, VB – V A represents the change in volume when one mole of the substance passes from the initial phase ‘A’ to the final phase ‘B’. It may be represented by DV . Similarly, SB – SA being the change in entropy for the same process, may be put as DS . Hence dP DS = dT DV ------ (14) If ‘ q ’ is the heat exchanged reversibly per mole of the substance during the phase transformation at temperature T , then the change of entropy DS in this process is given by q DS = T dP q Hence ------ (15) = dT TDV Thus dP q = dT T (VB - V A) ------ (16) This is Clapeyron-Clausius equation. This equation, evidently gives change in pressure dP which must accompany the change in temperature dT (or) vice verse, in the case of a system containing two phases of a pure substance in equilibrium with each other. Suppose the system consists of water in the two phases, viz., liquid and vapor in equilibrium with each other at the temperature T (i.e.) Water (liquid) The water (vapor) q = molar heat of vaporization, DH V VB = volume of one mole of water in the vapor state, say V g V A = Volume of one mole of water in the liquid state, say Vl Equation (16) therefore, taken the form This watermark does not appear in the registered version - http://www.clicktoconvert.com 100 DH V dP = dT T (Vg - Vl ) ------ (17) If the system consist of water its freezing point, then, the two phases in equilibrium will be water (solid) (ice) water (liquid) Equation (16) may then be written as DH f dP = dT T (Vl - VS ) ------ (18) Where DH f is the molar heat of fusion of ice. 9.7.2. INTEGRATED FORM OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID Û VAPOR EQUILIBRIUM The clapeyron- clausius equation dP DHv = dT T (Vg - Vl ) ------ (1) as applied to liquid vapor equilibrium can be easily integrated. The molar volume of a substance in the vapor state is considerably greater than that in the liquid state. In the case of water, for example the value of Vg at 100℃ is 18 × 1670 = 30060 ml while that of Vl is only a little more than 18 ml. thus Vg-Vl can be taken as vg without introducing any serious error. The Clapeyron equation (1) therefore, may be written as dP DHv = dT TV g ------ (2) Assuming that the gas law is applicable, (i.e.), PV = RT (Per mole) Vg = RT P ------ (3) Hence, dP DHv P DHv = ´ =P dT R RT RT 2 ------ (4) (or) dP DHv dT = ´ 2 P R T ------ (5) This watermark does not appear in the registered version - http://www.clicktoconvert.com 101 (i.e.) 1 dP DHv ´ = P dT RT 2 ------ (6) (or) d (ln P ) DHv = dT RT 2 ------ (7) Assuming that DHv remains constant over a small range of temperature, we have ò d ln P = DHv dT ò R T2 ------ (8) DHv æ 1 ö ç ÷+C R èT ø ------ (9) \ ln P = - (or) log P = - DHv æ 1 ö ç ÷ + C¢ 2.303R è T ø ------ (10) Where C and C ¢ are integration constants. Equation (10) is, evidently, the equation of a straight DHv æ1ö line. Hence, the plot ln P against ç ÷ should yield a straight line with slope and R èT ø intercept = C. This enables evaluation of DHv . Equation (7) can also be integrated between limits of pressure P1 and P2 corresponding to temperature T1 and T2 . Thus, P2 T2 ò d (ln P) = P1 DHv dT R Tò1 T 2 ------ (11) T P DHv é 1 ù 2 \ ln 2 = P1 R êë T úû T1 ------ (12) DHv é 1 1 ù ê - ú R ë T1 T2 û ------ (13) =+ = (or) DHv é T2 - T1 ù ê ú R ë T1T2 û 2.303 log P2 DHv é T2 - T1 ù = ê ú P1 R ë T1T2 û ------ (14) ------ (15) The equation (15) is known as Clapeyron-Clausius equation. 9.7.3. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR LIQUID VAPOR EQUILIBRIA Û This watermark does not appear in the registered version - http://www.clicktoconvert.com 102 Equation (13) can be used for calculating the molar heat of vaporization, DHv liquid if we know the vapor pressures at two temperatures. Further, if DHv is known, vapor pressure at a desired temperature can be calculated from the knowledge of a single value of vapor pressure at a given temperature. It can also be used for calculating the effect of pressure on the boiling point of a liquid. A few examples are given below: 1. Calculation of molar heat of vaporization, DHv : The molar heat of vaporization of a liquid can be calculated if its vapor pressure at two different temperatures is known. Eg. 1: Vapor pressure of water at 95℃ and 100℃ are 634 and 760mm, respectively. Calculate the molar heat of vaporization, DHv , of water between 95℃ and 100℃ . Solution: Substituting the given data in eqn (13) we have ln é 373k - 368k ù 760mm DHv = = DHv = 41363 Jmol-1 ú -1 -1 ê 634mm 8.314 JK mol ë (368k )(373k ) û ------ (12) 2. Effect of temperature on vapor pressure of a liquid: If vapor pressure of at one temperature is known, that another temperature can be calculated. a liquid Eg 2: The vapor pressure of water at 100℃ is 760mm. what will be the vapor p r e s s u r e a t 95℃ ? The heat of vaporization of water in this temperature range is 41.27 KJ per mole Solution: Substituting the given data in eqn (13) we have ln p2 41.27 ´ 10 3 Jmol -1 = 760mm 8.314 JK -1mol -1 é 373k - 368k ù ê (368k )(373k ) ú ë û ∴ p2 = 634.3mm 3. Effect of pressure on boiling point: If boiling point of a liquid at one known, that at another pressure can be calculated. pressure is Eg 3: Ether boils at 33.5℃ at one atmosphere pressure. At what temperature will it boil at a pressure of 750mm, given that the heat of vaporization of ether is 369.86 Joules per gram? Solution: Substituting the given data in eqn (13), we have ln ( )( ) 750mm 369.86 Jg -1 74 gmol -1 é T2 - 306 Jk ù = ê ú 760mm 8.314 JK -1mol -1 ë (306.5k )(T2 ) û T2 = 305.9K = 32.9℃ This watermark does not appear in the registered version - http://www.clicktoconvert.com 103 9.7.4. CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û VAPOR EQUILIBRIUM The clapeyron- clausius equation for solid Û vapor equilibrium may be put as DH S dP = dT T (Vg - VS ) ------ (1) Where DH S stands for the molar heat of sublimation of the substance. Since the molar volume of a substance is the gaseous state as very much greater than that in the solid, Vg - VS can be safely taken as V g . Eqn (1) can thus be easily integrated, as before to give the following expression: log DH S é T2 - T1 ù P2 = ê ú P1 2.303R ë T2T1 û 9.7.5. APPLICATION OF CLAPEYRON-CLAUSIUS EQUATION FOR SOLID Û LIQUID EQUILIBRIA The Clapeyron-Clausius equation (1) for solid Û liquid equilibrium cannot be integrated easily since VS cannot be ignored in comparison with Vl . Also the laws of liquid state are not as simple as those for gaseous state. However, this equation can be used for calculating the effect of pressure on the freezing point of a liquid. This aspect will be discussed in chapter phase rule equilibria and phase rule. Eqn (1) can also be used for calculating heat of fusion from vapor pressure data obtained at different temperatures. Eg: 1 The vapor pressure of ice Û water system at 0.0075℃ is 4.58mm and at 0℃ is 759.80mm of mercury. Calculate the molar heat of fusion of ice, given that the specific volumes of ice and water at 0℃ are 1.0907cc and 1.0001cc, respectively Density of mercury at 0℃ = 13.6gm per cc. Solution: dT Vl = 759.8-4.58 = 755.22mm=75.52cm Hg = 75.52 ´ 13.6 ´ 981 dynes cm-2 = 0.0075-0=0.00750 = 18 ´ 1.001 VS = 18 ´ 1.0907 cm3 dP This watermark does not appear in the registered version - http://www.clicktoconvert.com 104 Since vapor pressure increases on decreasing the temperature will have a negative sign. Substituting various values in eqn (1) we have DH f 75.52 ´ 13.60 ´ 981 dynes cm -2 = 0.0075 K 273 ´ 18(1.0001 - 1.0907 )cm -3 Hence, DH f = 5.98 ´ 1010 ergs mol-1 (1 dyn cm = 1 erg) = 5.98 ´ 1010 ergs mol -1 = 1429 cal mol-1 4.184 ´ 10 7 ergs cal -1 9.8. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Concept of chemical potential Gibbs- Duhem equation Some important results Variation of chemical potential with temperature Variation of chemical potential with pressure Chemical potential in case of a system of ideal gases Application of the concept of chemical potential Clapeyron–Clausius equation Integrated form of clapeyron-clausius equation for liquid Û vapor equilibrium Application of Clapeyron-Clausius equation for liquid Û vapor equilibria Clapeyron-Clausius equation for solid Û vapor equilibrium Application of Clapeyron-Clausius equation for solid Û liquid equilibria 9.9. CHECK YOUR PROGRESS 1. What is chemical potential? 2. How does chemical potential vary with temperature and pressure? Derive the Duhem equation. Gibbs- 9.10. LESSON – END ACTIVITIES 1. 2. Derive Claperyon-Clausius equation in the form dP/dT= DHv /(TVg). Under what conditions can this equation be integrated and how? What are the applications of Claperyon-Clausius equation? Explain with examples. 9.11. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand Delhi. 2. & Co., B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. This watermark does not appear in the registered version - http://www.clicktoconvert.com 105 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. This watermark does not appear in the registered version - http://www.clicktoconvert.com 106 LESSON 10: THERMODYNAMICS AND NON-IDEAL SYSTEMS: FUGACITY CONTENTS 10.0. AIMS AND OBJECTIVES 10.1. INTRODUCTION: CONCEPT OF FUGACITY 10.2. FUGACITY AT LOW PRESSURES 10.3. DETERMINATION OF FUGACITY OF A GAS 10.4. CALCULATION OF FUGACITY AT LOW PRESSURE 10.5. FUGACITY OF A GAS IN A GASEOUS MIXTURE 10.6. FUGACITY OF A LIQUID COMPONENT IN A LIQUID MIXTURE 10.7. PHYSICAL SIGNIFICANCE OF FUGACITY 10.9. LET US SUM UP 10.10. CHECK YOUR PROGRESS 10.11. LESSON - END ACTIVITIES 10.12. REFERENCES 10.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on fugacity t o the students. On successful completion of this lesson the student should have: * Understand the concept of fugacity. 10.1. INTRODUCTION: CONCEPT OF FUGACITY Making use of the free energy function G Lewis introduced the concept of fugacity for representing the actual behavior of real gases which is distinctly different from the behavior of ideal gases. We know that variation of free energy with pressure at constant temperature is given by equation Viz. æ dG ö ç ÷ =V è dP øT ------ (1) This equation is applicable to all gases whether ideal or non- ideal. If one mole of a gas is under consideration, then V refers to molar volume. For an ideal gas, the above equation may be written as (dG )T and for n moles as, dP = nRT P éQ PV = RT ù ê ú êV = RT ú P ë û ------ (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 107 (dG )T = nRT dP = nRT (ln P ) P ------ (3) Integration of equation (3) yields, G = G * + nRT ln P ------ (4) Where G * , the integration constant, is the free energy of n moles of ideal gas at temperature, when the pressure P is unity. Equation (4), evidently, gives the free energy of an ideal gas at temperature T and pressure P . Integration of equation (3) between pressures P1 a n d P2 at constant temperature T , yields, P2 P dP = nRT ln 2 P P1 DG = ò nRT P1 ------ (5) The corresponding equation for 1 mole of the gas would be DG = RT ln P2 P1 ------ (6) Equations (4) and (6) are not valid for real gases since V is not exactly equal to RT . P In order to make these simple equations applicable to real gas, Lewis introduced a new function f is called fugacity function. It takes the place of P in equation (3), which, for real gases, may be expressed as (dG )T = nRT ln df f ------ (7) And equation (7) may be represented as G = G * + nRT ln f ------ (8) Where G * is the free energy of n moles of real gas at temperature T , when its fugacity happens to be 1. Thus fugacity is a sort of fictitious pressure which is used in order to retain for real gases simple forms of equation which are applicable to ideal gases. Equation (8), evidently, gives the free energy of a real gas at temperature T and pressure P at which its fugacity can be taken as f . Equation (7) on integration between fugacities f1 and f 2 , at constant temperature T , yields, DG = nRT ln f2 f1 ------ (9) This watermark does not appear in the registered version - http://www.clicktoconvert.com 108 The corresponding equation for one mole of the gas would be DG = RT ln f2 f1 ------ (10) As discussed above, equations (9) and (10) are applicable to real gases. Eg: 1 Calculate the free energy change accompanying the compression of 1 mole of a gas at 57 ℃ from 25 to 200 atm. The fugacities of the gas at 57 ℃ may be taken as 23 and 91 atm, respectively, at pressure 25 and 200 atm. Solution: DG = RT ln P2 P1 = 1mole (8.314 JK -1mol -1 ) 330 K ln 200 atm = 5702.8 J 25atm for more accurate value, we should use the equation involving fugacities. Thus, DG = RT ln f2 f1 = 1mole (8.314 JK -1mol -1 ) 330 K ln 91atm = 3730.0 J 23atm 10.2. FUGACITY AT LOW PRESSURES f , where P is the actual pressure, approaches unity when P approaches zero, P since in that case a real gas approximates to ideal behavior. The fugacity function, therefore, may be defined as The ratio lim it P®0 f =1 P Evidently, at low pressures, fugacity is equal to pressure. The two terms differ materially only at high pressures. 10.3. DETERMINATION OF FUGACITY OF A GAS Equation G = G * + nRT ln f for one mole of a gas, may be put as G = G * + nRT ln f ------ (1) This watermark does not appear in the registered version - http://www.clicktoconvert.com 109 Differentiation of eqn (1) with respect to pressure at constant temperature and constant number of moles of the various constituents, (i.e.) in a closed systems, gives æ dG ö æ d ln f ö ç ÷ = RT = ç ÷ è dP øT è dP ø ------ (2) since æ dG ö ç ÷ =V è dP øT it follows that æ d ln f ö V ç ÷= è dP ø RT ------ (3) Thus, at definite temperature, eqn (3) may be written as ------ (4) RTd (ln f ) = VdP Since one mole of the gas is under consideration, ‘V’ is the molar volume of the gas. Knowing that for an ideal gas, V = RT P ------ (5) The quantity a , defined as departure from ideal behavior at a given temperature, is given by a= RT -V P ------ (6) Eqn (6) is multiplied by dP throughout, we get adP = RT dP - VdP P ------ (7) Combining equations (4) and (6), we have RTd (ln f ) = RT dP - adP P Both sides dividing by RT (or) d (ln f ) = d ln P - a dP RT ------ (8) This watermark does not appear in the registered version - http://www.clicktoconvert.com 110 Integrating eqn (8) between pressures ‘ O ’ and ‘ P ’, we have P ln f 1 =adP P RT ò0 ------ (9) Now ‘ a ’ is given by eqn (6), can be determined experimentally, at different pressures. These values of ‘ a ’ are then plotted against corresponding pressures, as shown in fig. The area under the curve between pressure e=0 and any given pressure ‘ P ’, yields the value of P the integral ò adP , as illustrated by the shaded portion in fig. negative positive 0 Plot of a = +ve area pressure 'P' -ve area RT - V vs. P for the determination of fugacity of the gas P Incorporating this value in eqn (9) the fugacity ‘ f ’ can be evaluated at any given pressure ‘ p ’of the gas. Since ‘ a ’ of the departure from ideal behavior, can be both positive as well as negative, the area under the curve (with respect to the pressure axis) can be both positive as well as negative. Thus, the fugacity of the gas can be both less than or more than the pressure. As is P evident from fig the area and hence the value of ò a (dP) is positive at low pressure and negative 0 at high pressures. Hence, in accordance with eqn (9) fugacity ‘ f ’ of the gas would be less than the pressure ‘ p ’ at low pressures and more than the pressure at very high pressures. This is borne out by the data given in the table (1) for nitrogen gas at various pressures at 0℃ . Fugacity of nitrogen gas at various pressures at 0℃ Pressure atmosphere Fugacity atmosphere This watermark does not appear in the registered version - http://www.clicktoconvert.com 111 50 100 200 400 800 1000 48.9 96.7 194.2 424.4 1191 1834 Fugacity of hydrogen gas at various pressures at 0℃ Pressure atmosphere 25 50 100 200 500 1000 Fugacity atmosphere 25.4 51.5 106.1 225.8 685 1899 It may be recalled that for hydrogen and helium PV is greater than RT for all pressures at ordinary temperatures. Hence fugacity in these gases always remains greater than the pressure. This is borne out by the data in table (2) for fugacity of hydrogen gas at 0℃ . This watermark does not appear in the registered version - http://www.clicktoconvert.com 112 10.4. CALCULATION OF FUGACITY AT LOW PRESSURE It has been found that the experimental value of a at low pressure assumes almost constant value. Under such conditions, therefore, eqn P f 1 =a (dP ) P RT Oò f P ln = a P RT ln Becomes ------ (1) Now at low pressure, since gases tend to be ideal, or f =p f =1 P ------ (2) Making use of the fact that ln x is approximately equal to x - 1 when x approaches unity we have, ln Hence, f f = -1 P P f f = 1 + ln P P P =1-a RT = ------ (3) ------ (4) PV RT ------ (5) Hence, f = P 2V RT ------ (6) This equation is useful in calculating fugacity at moderately low pressures. Eg: For a VanderWaals gas, express the fugacity as a function of V, T, R and the VanderWaals constants. This watermark does not appear in the registered version - http://www.clicktoconvert.com 113 Solution: From eqn RTd ln ( f ) = Vdp --- (i) for one mole of the gas Integrating eqn (1) RT ò d ln ( f ) = ò Vdp ------ (ii) It is advisable to change the variable on the right hand side of the eqn (ii) by resorting to integration by parts. Accordingly, ò VdP = PV - ò ------ (iii) Vdp Now, for one mole of the VanderWaals gas, P= RT a - 2 V -b V ------ (iv) (Where V is the molar volume) hence, from eqns (ii), (iii) and (iv), a ù é RT RT ln f = PV - ò ê - 2 údV = RT - RT ln(V - b) - dV + C ëV - b V û ------ (v) Where C is the integration constant. To evaluate C, we recall that f ® p as p ® o . Also, as p®o at constant temperature T ,V ® a so that (V - b ) ® V a n d 1 ® 0 .Hence, V RT ln P = RT - RT ln V + C = RT - RT ln (RT / P ) + C = RT - RT ln RT + RT ln P + C \ C = RT ln RT - RT QP = 0 ------ (vi) Substituting for C in eqn (v), we get æaö RT ln f = PV - RT ln (V - b ) - ç ÷ + RT ln RT - RT èV ø æ RT æ a öö = PV - RT + RT lnçç - ç ÷ ÷÷ è (V - b ) è V ø ø ------ (vii) ------ (viii) Rearranging the Vander-Waals equation, after expansion, neglecting the term ab /V 2 and later the term a / V 2 as well, we obtain This watermark does not appear in the registered version - http://www.clicktoconvert.com 114 PV - RT = RTb a V -b V ------ (ix) Hence, ln f = b /(V - b) + ln RT /(V - b) - 2a / RTV this is the required expression for fugacity. 10.5. FUGACITY OF A GAS IN A GASEOUS MIXTURE Remembering that for one mole of a pure substance, the free energy (G) is identical with the chemical potential m eqn (dG )T = nRTd (ln f ) c for one mole of dm i = RTd (ln f i ) ------ (1) and eqn G = G * + nRT ln f may be written as * m i = m i + RT ln f i ------ (2) Where m i is the chemical potential of the gaseous component i at its unit fugacity. 10.6. FUGACITY OF A LIQUID COMPONENT IN A LIQUID MIXTURE The eqn * m i = m i + RT ln f i is valid not only for the fugacity of a gas in gaseous mixture but also for the fugacity of a pure liquid in a liquid mixture. This easily follows from the following discussion. As is well known, in the case of phase equilibria, the chemical potential of any given component is the same in all phases. Thus if there are three phases a, b, and c containing a component i then at equilibrium, m i (a ) = m i (b) = m i (c) Consider, for example, a liquid i in equilibrium with its vapor. The chemical potential of the liquid, ( m i ) l = ( m i ) g Now according to the eqn G = G * + nRT ln f chemical potential of a vapor and a gas may be written as ( m i ) g = ( m * i ) g + RT ln f i ------ (1) (m ) Where as before i g is the chemical potential of the vapor (when its fugacity is equal to 1. therefore, chemical potential of a liquid may, evidently, be written as ( m i ) l = ( m * i ) l + RT ln f i ------ (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 115 Where ( m *i ) l stands for the chemical potential of the liquid when its fugacity is equal to 1. It follows from the above discussion that the fugacity of a pure liquid would be the same as that of its vapor in equilibrium with it, at a given temperature. 10.7. PHYSICAL SIGNIFICANCE OF FUGACITY In order to understand the physical significance of the term fugacity, consider a system consisting of liquid water in contact with its vapor water molecule in the liquid phase will have a tendency to escape into the vapor phase by evaporation, while those in the vapor phase will have a tendency to escape into the liquid phase by condensation. At equilibrium, the two escaping tendencies will be equal. It is now accepted that each substance in a given state has a tendency to escape from that state. This escaping tendency was termed as fugacity. 10.9. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Concept of fugacity Fugacity at low pressures Determination of fugacity of a gas Calculation of fugacity at low pressure Fugacity of a gas in a gaseous mixture Fugacity of a liquid component in a liquid mixture Physical significance of fugacity 10.10. CHECK YOUR PROGRESS 1. 2. Explain the term fugacity. How is it related to chemical potential? Explain why the fugacity of Helium or Hydrogen is always more than the pressure. This watermark does not appear in the registered version - http://www.clicktoconvert.com 116 10.11. LESSON – END ACTIVITIES 1. How can you determine the fugacity of gas using (i) Graphical method and (ii) From equations of state? 2. What is the physical significance of fugacity? How does fugacity vary with temperature? 10.12. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand Delhi. & Co., 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. This watermark does not appear in the registered version - http://www.clicktoconvert.com 117 LESSON 11: THERMODYNAMICS AND NON-IDEAL SYSTEMS: ACTIVITY CONTENTS 11.0. 11.1. 11.2. 11.3. 11.4. 11.5. 11.6. AIMS AND OBJECTIVES INTRODUCTION: CONCEPT OF ACTIVITY ACTIVITY COEFFICIENT TEMPERATURE COEFFICIENT REFERENCE STATES OR STANDARD STATES RATIONAL AND PRACTICAL APPROACHES DETERMINATION OF ACTIVITY OF SOLVENT COLLIGATIVE PROPERTIES 11.7. LET US SUM UP 11.8. CHECK YOUR PROGRESS 11.9. LESSON - END ACTIVITIES 11.10. REFERENCES AND SOLUTE FROM 11.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on activity to the students. On successful completion of this lesson the student should have: * Understand the concept of activity. 11.1. INTRODUCTION: CONCEPT OF ACTIVITY It may be pointed out that since the absolute value of free energy (or) chemical potential is not known, it is impossible to evaluate of a substance. This difficulty has been overcome by referring all free energy and chemical potential measurements for any given substance to a standard reference point. Let m i be the chemical potential of a substance ‘i’ in pure state and let f i be its fugacity \ Equation * m i = m i + RT ln f i May then be put as * 0 m i = m i + RT ln f i 0 ------ (1) Let m i be the chemical potential of the same substance in some other state. Then * m i = m i + RT ln f i ------ (2) The difference between chemical potential of a substance in any state and that in the pure state is given by This watermark does not appear in the registered version - http://www.clicktoconvert.com 118 æ f ö 0 m i - m i = RT lnçç i0 ÷÷ Or è fi ø æ f ö 0 m i = m i + RT lnçç i0 ÷÷ è fi ø ------ (3) We may introduce here a new term, activity a and define it as a= f f0 ------ (4) fi 0 fi ------ (5) or for a substance i , as a= Activity of a substance in any given state is thus defined as the ratio of the fugacity of the substance in that state to the fugacity of the same substance in the pure state. The equation 3, therefore, reduces to 0 m i = m i + RT ln a ------ (6) Let a system consisting of one mole of a substance change from a state in which its chemical potential (or) free energy, both being identical since we are dealing with one mole of the substance is m i , to another state in which its chemical potential is m 2 , the change in chemical potential Dm , is then given by ( Dm = m 2 - m1 = m 0 + RT ln a2 - m 0 + RT ln a1 or æa ö Dm = RT lnçç 2 ÷÷ è a1 ø ) ------ (7) Comparing the above equation with that for an ideal gas, viz. æP ö DG = Dm = RT lnçç 2 ÷÷ è P1 ø ------ (8) It is evident that in the case o real gases, activity replaces pressure. Thus activity of a gas like fugacity, serves as a thermodynamic counter part of a gas pressure. In case of solutions it serves as a counter part of concentration of the solute in the given solution. 11.2. ACTIVITY COEFFICIENT For an ideal gas, activity is numerically equal to its pressure (i.e.) a = P For real gases, however, activity is only proportional to its pressure (i.e.) a µ P or a = gP where g is known as the activity coefficient Eg: This watermark does not appear in the registered version - http://www.clicktoconvert.com 119 The activity of 2.5 moles of a substance changes from 0.05 to 0.35. What would be the change in its free energy at 27 ℃ ? Solution: The change of free energy for one mole of a substance is given by the relation æa ö DG = RT lnçç 2 ÷÷ è a1 ø æa ö \ For n moles, DG = nRT lnç 2 ÷ = (2.5 mol) (8.314 JK -1mol -1 ) (300K) lnæç 0.25 ö÷ ça ÷ è 0.05 ø è 1ø = 12133.65 J 11.3. TEMPERATURE COEFFICIENT Temperature coefficient of a chemical substance is defined as the ratio of rate constants of a reaction at two different temperatures separated by 100 C. The two temperatures are generally taken as 350 and 250 C. Thus, the temperature coefficient is expressed as Temperature coefficient = k 350 0 k 25 In general the temperature coefficient is expressed as Temperature coefficient = k t + 10 0 kt Where kt is the specific rate of the reaction at t0 C and kt + 10 is the specific rate of the same reaction at t+100 . 11.4. REFERENCE STATES OR STANDARD STATES It may be emphasized again that there is no means of finding absolute values of free energy (G) or Chemical potential (m) of any substance. It is necessary, therefore, to make such measurements with reference to the value obtained for some convenient though arbitrary reference state called standard state. The standard state for a gas at any given temperature is defined as that state in which the fugacity of the gas is unity. Since a = f/f0 , it follows that if f0 = 1, a = f. Evidently, the standard state for a gaseous component is such that fi = ai =1 The standard state for a liquid is the pure state of the liquid at one atmosphere pressure at any given temperature. In this state of the liquid, f = f0 and a = 1.The standard state for a liquid component I in a liquid solution is the pure state of the component such that ai = 1. This watermark does not appear in the registered version - http://www.clicktoconvert.com 120 The standard state for a solid is the pure state of the solid at one atmosphere pressure at any given temperature. In this state, its activity a = 1. 11.5. RATIONAL AND PRACTICAL APPROACHES The concentration of a solution is usually expressed as mole fraction, molality or molarity. The standard state chosen depends on the concentration unit used. If the choice is mole fraction, it is referred to as the rational system, whereas if molality or molarity (practical units) are used it is called the practical system. (A) RATIONAL SYSTEM If the mole fraction of the solute, x2 , is taken as the measure of the concentration of the solute, Henry’s law being applicable to solutes, f2 = K x2 , where K is the Henry’s law constant. It is therefore desirable to choose the standard state for the solute in such a way that in a dilute solution, the activity becomes equal to the mole fraction of the solute. Thus it is desirable that a2 ® 1 as x2 ® 0 ----- (1) x2 The fugacity- mole fraction relationship over the entire range of mole fraction, 0 to 1, can be represented as the solid line in Fig1. For the very dilute solution as x2 ® 0 , it is seen that the actual curve merges with the Henry’s law line. Since a 2 = f 2 f 20 , Eq. 1 can be written as This watermark does not appear in the registered version - http://www.clicktoconvert.com 121 æa ö æ f ö Limçç 2 ÷÷ = Limçç 2 ÷÷ = 1 ----- (2) x x ®0 x è 2 ø x x ®0 è f 2 x 2 ø Since Henry’s law is also applicable to the solute, in the very dilute solution, Lim [f2 /x2 ] for solid line = limiting slope = K. For Henry’s law line (dotted) x 2 ® 0 . æ f2' ö ÷=K. Limçç x x ®0 x ÷ è 2 ø Therefore [ ] Lim[ f 2 x 2 ] = Lim f 2 x 2 = K x x ®0 x x ®0 ' ----- (3) Since f2 = Kx2 , eqn. (3) can be written as é f ù Lim ê 2 ú = 1 x2 ®0 Kx ë 2û ----- (4) If equations 2 and 4 should hold good simultaneously f2 0 =K, the Henry’s law constant for this solute. From Fig. 1., it is seen that this state can be found by extrapolating the dotted line to a concentration x2 = 1. From Henry’s law f2 ’ =Kx2 , it is seen that when x2 = 1, f2 ’ = k. This fugacity is taken as the standard fugacity for the solute. It should be noted that the standard fugacity (f2 0 ) is a hypothetical quantity and is not equal to the fugacity (f2 ’) of the pure solute. The standard state for the solute is chosen as the hypothetical liquid solution at the given temperature and 1 atm total pressure, in which the mole fraction of the solute is unity and yet it behaves ideally, obeying Henry’s law. If this law is obeyed over the entire range of composition (x2 = 0 to 1). Kx 2 a 2 = f 2 f 20 = f 2 K = = x2 ----- (5) K Thus as x 2 ® 1 , a2 becomes unity. The activity at any other concentration will be equal to x2 , the mole fraction of the solute. If the activity of the solute, a2 = f2 /f2 ’ is plotted against x2 , the curve obtained (Fig.2) is similar to the one in Fig.1, since each value of the fugacity is divided by the same constant, f2 0 . For an ideal solution, the slope of the curve should be unity (Fig.2.). The activity of the pure solute, a2 is seen to be different from a2 0 . For any mole fraction, xj the activity coefficient gx is aj /xj. In Fig.2, Y=a2 and since the slope of the dotted line is unity, xj =X. Thus gx =Y/X. For a solution behaving ideally over the whole range of concentration, the activity will always be equal to its mole fraction. For non-ideal solutions, the standard state has no reality and it is preferable to define the standard state in terms of a reference state. It is seen from Fig.2 that the activity coefficient gx becomes unity as x 2 ® 0 . It is thus possible to choose the infinitely dilute solution as the reference state, such that as x2 ® 0 , n x ® 1 or a 2 ® x 2 . This watermark does not appear in the registered version - http://www.clicktoconvert.com 122 (B) PRACTICAL SYSTEM Molality is more widely used to express concentrations than mole fraction. In very dilute solutions, molality is proportional to mole fraction. Henry’s law is valid under these conditions (i.e.) f2 =Km2 . If f2 is plotted against m2 , the Henry’s law constant, K, can be obtained from the limiting slope of the curve (I) (Fig.3.). The choice of the standard fugacity should be such that as éa ù m2 ® 0, ê 2 ú ® 1 or ë m2 û éa ù é f ù Lim ê 2 ú = Lim ê 0 2 ú = 1 m2 ® 0 m ë 2 û m2 ®0 ë f 2 m2 û ----- (1) Under such limiting conditions, Henry’s law is valid (i.e.) f2 =Km2 or é f ù Lim ê 2 ú = 1 m2 ®0 Km 2 û ë ----- (2) For the same region (m2 ® 0 ) of the solid line curve (Fig.3) equations 1 and 2 should hold good simultaneously. This is true when f2 0 =K. In Fig.3., this condition is realized by finding the fugacity corresponding to m2 =1. The interpolation is done on the Henry’s law plot (dotted line) since this law is obeyed by ideal systems under all conditions. The standard state of the solute is the state, which has the fugacity that the solute in a solution of unit molality would This watermark does not appear in the registered version - http://www.clicktoconvert.com 123 have, if Henry’s law is obeyed at this concentration. With increasing dilution, a solute approaches ideal behavior specified by the Henry’s law. However it is misleading to say that the standard state of the solute is the infinitely dilute solution. In such a case f2 =0. If f2 0 =0, the activity at any finite concentration a2 = f2 /f2 0 , would be infinite. A curve similar to that in Fig.3 can be obtained by plotting activity, a2 of the solute against molality, m2 (Fig. 4). Since the mole fraction scale has limits of 0 and 1, the choice of x2 =1 as a standard state, is quite natural. Theoretically molality has no upper limit, but in practice the upper limit is the solubility of the substance. This watermark does not appear in the registered version - http://www.clicktoconvert.com 124 The choice of standard state, m2 0 = 1 mole/kg is arbitrary. The standard state is the hypothetical 1 molal solution obtained by extrapolating the Henry’s law line to m2 =1. If the concentration of the solute is expressed in molarity (c) the standard state is chosen as the hypothetical state obtained when Henry’s law plot is extrapolated to c2 = 1 mol/l. 11.6. DETERMINATION OF ACTIVITY OF SOLVENT AND SOLUTE FROM COLLIGATIVE PROPERTIES VAPOUR PRESSURE MEASUREMENTS (A) SOLVENT The activity of a constituent, j, in solution is given by fj /fj0 , where fj is its fugacity in the solution and fj0 is the value at the standard state (pure liquid at the same temperature as the solution and at 1 atm total pressure). If the vapour pressures are sufficiently low, ideal behaviour may be assumed; The fugacities may therefore be replaced by the respective partial pressures (i.e.) a j = p j p 0j ----- (1) Since the effect of external pressure on activity is negligible, Eq. – can also be written as a j = Pj p j ----- (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 125 Here pj is the partial vapour pressure of the solvent in equilibrium with the solution in which its activity is aj, and p is the vapour pressure of the pure solvent at the same temperature and 1atm pressure i.e. at the standard state. This method is useful in determining the activity of the solvent in aqueous solutions, mixtures of organic liquids etc. If the mole fraction or the concentration of the solvent is known, the activity coefficient of the solvent can be calculated. (B) SOLUTE If the solute is completely miscible with the solvent, the standard state is chosen as the pure liquid. The activity may be determined by using Eq. a j = Pj p j If the solute is sufficiently volatile to permit the determination of its vapour pressure over the solution. For dilute solutions it is preferable to use the infinitely dilute solution as the reference state. The activity is proportional to its fugacity, the value of the proportionality constant depending on the standard state used or reference state used. If 1/k is the proportionality constant for the chosen reference state, a2 =f2 /k. In dilute solutions, as x 2 ® 0, [a 2 x 2 ] ® 1 or f 2 = a2 k = x2 k as x2 ® 0 For a dilute solution f2 = p2 , the partial pressure of the solute. Hence a2 » p2 k ----- (1) The value of k can be obtained by utilizing the fact that at high dilutions, the activity of the solute, a2 * is equal to the mole fraction x2 *. If p2 * is the partial pressure under this condition, a 2 * = x 2 * = p 2 * k or k = p2 * x2 * ----- (2) The activity of the solute a2 , in the given solution is given by Eq.1. Substituting for k, a2 = p2 = p2 k æ p2 * ö çç ÷÷ x * è 2 ø ----- (3) The activity coefficient (g x )2 being a2 /x2 , from Eq. 3. a2 = (g x )2 x2 æ p2 ö çç ÷÷ x =è 2 ø æ p2 * ö çç ÷÷ è x2 * ø ----- (4) A graphical evaluation of k is possible, by plotting p2 /x2 against x2 and extrapolating it to x2 = 0. Once k is known, by using Eq. 3 the activity can be calculated. From Eq. 4 the activity coefficient can be calculated. This watermark does not appear in the registered version - http://www.clicktoconvert.com 126 11.7. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Concept of activity Activity coefficient Temperature coefficient Reference states or standard states Ø Rational and practical approaches Ø Determination of activity of solvent and solute from colligative properties 11.8. CHECK YOUR PROGRESS 1. 2. Define activity and activity coefficient. What are the applications of activity concept to solutions? 11.9. LESSON – END ACTIVITIES 1. Explain briefly the rational and practical approaches of activity. 2. How can you determine (i) the activity of solvent from colligative properties (ii) the activity of solute. 11.10. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand Delhi. & Co., 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. LESSON 12: THERMODYNAMICS AND NON-IDEAL SYSTEMS: THIRD LAW OF THERMODYNAMICS CONTENTS 12.0. AIMS AND OBJECTIVES 12.1. INTRODUCTION: THIRD LAW OF THERMODYNAMICS 12.3. NERNST HEAT THEOREM 12.4. CONSEQUENCES OF THE NERNST HEAT THEOREM 12.5. THIRD LAW OF THERMODYNAMICS: PLANK’S FORMULATION 12.6. STATEMENT OF LEWIS AND RANDALL 12.7. UNATTAINABILITY OF ABSOLUTE ZERO 12.8. THERMODYNAMIC QUANTITIES AT ABSOLUTE ZERO 12.9. APPLICATIONS OF THIRD LAW OF THERMODYNAMICS 12.10. THERMODYNAMIC PROBABILITY AND STATISTICAL THERMODYNAMICS 12.10.1. DERIVATION OF THE BOLTZMANN ENTROPY EQUATION This watermark does not appear in the registered version - http://www.clicktoconvert.com 127 12.11. APPARENT EXCEPTIONS 12.12. LET US SUM UP 12.13. CHECK YOUR PROGRESS 12.14. LESSON - END ACTIVITIES 12.15. REFERENCES 12.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on third law of thermodynamics to the students. On successful completion of this lesson the student should have: * Understand the third law of thermodynamics. 12.1. INTRODUCTION: THIRD LAW OF THERMODYNAMICS In the earlier development of chemical thermodynamics, the difficult task was to ascertain a quantitative relation between DG and DH in a chemical reaction and to find out DG from thermal data (i.e.), DH . A preliminary discussion will now be given which will deal with this problem thereby ultimately leading to the Nernst heat theorem and third law of thermodynamics. 12.3. NERNST HEAT THEOREM Nernst postulate “for a process in a condensed system that the value o f d (DG ) dT approaches zero asymptotically as the absolute zero is approached”. In other words it means that DG and DH curves meet each other at a short region above the absolute zero and run coinciding with each other upto absolute zero. This behavior is being shown by full lines and not by dotted lines (Fig. 1). Fig. 1. Variation of DG and DH with change of temperature. The dotted lines curve reveals that the two values, DG and DH not only become equal to each other at absolute zero but their approach to each other becomes rapid and not gradual. This watermark does not appear in the registered version - http://www.clicktoconvert.com 128 The mathematical expression of Nernst’s postulate may be expressed analytically as Lt d (DG )=T ®Lt0 (DH ) = 0 T ®0 dT ----- (1) This mathematical expression expresses the Nernst heat theorem. 12.4. CONSEQUENCES OF THE NERNST HEAT THEOREM We know second law of thermodynamics that ìd (DG ) ü í ý = - DS î dT þ P and ì d (DH ) ü í ý = DC P (Kirchoff’s equation) î dT þ P When we set up the conditions demanded by the Nernst heat theorem, we get Lt T ®0 DS = T ®Lt0 DC P = 0 ------ (2) The significance of the equation (2) is that the entropy change tends to approach zero and the difference between the heat capacities of products and reactants also tends to approach zero as the temperature is lowered towards the absolute zero. This is the statement of Nernst heat theorem. Nernst heat theorem holds good only in the case of solids. Since no gas exists at Nernst heat theorem cannot apply to gases. It can also not be applied to liquids. 0K. 12.5. THIRD LAW OF THERMODYNAMICS: PLANK’S FORMULATION The absolute value of entropy of a pure solid (or) a pure liquid approaches zero at 0K. Lt T ®0 ------ (1) DS = 0 If equation (1) is assumed correct, Nernst equation (2) follows immediately for Lt Lt T ®0 (- DS )= æ dDG ö ç ÷=0 T ®0 è DT ø ------ (2) pure solids and liquids. Plank’s formulation is also consistent with the treatment of entropy which is introduced in statistical mechanics. Plank statement asserts that S 0 K is zero for only pure solids and liquids whereas Nernst assumed that his theorem was applicable to condensed This watermark does not appear in the registered version - http://www.clicktoconvert.com 129 phases, including solutions. According to plank “solutions at 0 K have positive entropy equal to entropy of mixing”. 12.6. STATEMENT OF LEWIS AND RANDALL Lewis and Randall modified the statement of the third law of thermodynamics as follow. “Every substance has positive entropy but at the absolute zero the entropy may become zero, and it dose so become in the case of perfect crystalline substance”. From the point of view of a chemist it is the most practical tool for calculating free energy. Consider for example the transition S (rhombic, 0 K) S (monoclinic 0 K) DS 0 K from heat capacity measurement is zero. Hence, both rhombic and monoclinic sulphur are assigned zero entropy at 0 K. Thus no difficulty is involved for calculating absolute entropy of sulphur at a temperature T. 12.7. UNATTAINABILITY OF ABSOLUTE ZERO It is impossible to attain absolute zero in a finite series of operations. For example, if we consider entropy change at constant pressure (or) dS = DQP C P = dT T T Lt Lt ------ (1) æ dS ö æPö ç ÷= ç ÷ = ¥ T ®0 è dT ø T ®0 è T ø ------ (2) Equation (2) shows the entropy of any substance should tend towards infinity as T approaches absolute zero. Since this cannot happen clear no infinite series of processes lead to absolute zero. 12.8. THERMODYNAMIC QUANTITIES AT ABSOLUTE ZERO (i) Equivalence of G and H Since S 0 for any substance is zero, according to Lewis and Randall statement, is follows that for any substance G0 K = H 0 K - TS 0 K = H 0 K (or) G0 K = H 0 K ------ (1) This watermark does not appear in the registered version - http://www.clicktoconvert.com 130 (ii) DC P in a chemical transformation From Gibbs-Helmholtz equation æ dDG ö DG = DH + T ç ÷ è dT ø P ------ (1) æ æ dDG ö ö Q çç - DS = ç ÷ ÷÷ è dT ø P ø è DG - DH æ dDG ö ç ÷ = T è dT ø P Lt Lt DG0 K - DH 0 K æ dDG ö =0 ç ÷= T T ®0 è dT ø T ®0 ------ (2) Q DG0 K - DH 0 K = 0 A close examination of (eqn. 2) shows that the result comes out to be an intermediate form 0 at 0 T=0. To resolve an indeterminate expression we may apply the mathematical rule of differentiating numerator and denominator with respect to independent variable ‘T’. Thus we have Lt æ dDG ö æ dDH ö ç ÷-ç ÷=0 è dT ø T ®0 è dT ø ------ (3) (or) Lt Lt æ dDG ö æ dDH ö Lt ç ÷= ç ÷=T ®0 DC P d T d T è ø è ø T ®0 T ®0 ------ (4) It follows then Lt T ®0 DC P = 0 many investigations D have shown that DC P does approaches zero at T approaches zero. (iii). Limiting value of Cp and Cv For a reversible temperature change in a substance at constant pressure, æ DQ p ö C p dT ÷÷ = dS p = çç T è T ø On integrating at constant pressure This watermark does not appear in the registered version - http://www.clicktoconvert.com 131 T ò dS = S = ò 0 C p dT T + S0 Where S 0 is integration constant. An examination of equation (2) Lt Lt DG0 K - DH 0 K æ dDG ö =0 ç ÷= T T ®0 è dT ø T ®0 T Shows that if C p has a finite value at T = 0. Integral ò C p dT will not give a finite value since T the denominator T approaches zero. Hence ‘S’ would not be finite. But according to Lewis and Randall statements, ‘S’ must be finite at all temperature. Hence T ®Lt0 C P = 0 following is an 0 analogous procedure, it can be shown that T ®Lt0 C P = 0 . 12.9. APPLICATIONS OF THIRD LAW OF THERMODYNAMICS The third law of thermodynamics is used to calculate the absolute entropy of the solids, liquids and gases at different temperatures and the entropy changes of chemical reactions and other processes. Let us describe these one by one. 1. DETERMINATION OF ABSOLUTE ENTROPY OF SOLIDS It is useful in calculating the absolute entropies of pure substance at different temperature by using thermal data. dT We know dS = C p , this on integrating between temperature limits 0 and ‘T’ K T becomes as follows T ST - S 0 = ò C p d ln T 0 On setting up the condition demanded by the third law (i.e.)., S 0 = 0 at T = 0. We get T ST = ò C p d ln T ------ (1) 0 Where ST = absolute entropy of the solid at temperature T. One can calculate the integrant in eqn. (1) graphically by plotting C p against T (or) C p against ln T and the area of curve so obtained gives the value of integrant (i.e.), ST between T = 0 to any desired temperature (Fig. 2). This watermark does not appear in the registered version - http://www.clicktoconvert.com 132 Fig. 2. Determination of absolute entropy 2. EVALUATION OF ABSOLUTE ENTROPIES OF LIQUIDS AND GASES In all these cases we generally start from the crystalline (solid) state of a given substance at absolute zero when its absolute entropy is taken as zero and then supplying heat to this solid; it can be converted into the required state of the substance at a given temperature. The sum of the entropy changes involving these conversions will give the value of absolute entropy of the specific substance at the given temperature. In all these calculations, the absolute entropy at 0 o K has been taken to be zero. When we have to calculate the absolute entropy of a liquid, the following steps are involved. (1). First of all measurements are made on the solid form at the melting point. The entropy of the solid at this temperature (DS1 ) is given as follows. Tm DS1 = ò (C ) p 0 (C ) p s s dT T ------ (1) is the heat capacity of solid. (2). The integrant in eqn. (1) can be obtained graphically. Changing one solid into the liquid state at the melting point Tm. The entropy of this process (entropy of fusion) is given by DS 2 = Where DH f DH f Tm ------ (2) is the molar heat of fusion of the substance. (3). Heating the liquid from its melting point (Tm) to this boiling point (Tb). Te entropy involved in this case is given by This watermark does not appear in the registered version - http://www.clicktoconvert.com 133 Tb DS 3 = ò (C ) p l 0 dT T ------ (3) Where (C p )l is the heat capacity of the substance in the liquid state. This can be evaluated by plotting (C p )l vs ln T between temperature Tm and Tb and noting the area below the graph as described before. (4). Changing the liquid into the gaseous state at the temperature Tb. The entropy involved here DS 4 is the entropy of vaporization and is given by DS 4 = DH v Tb ------ (4) Where DH v is the heat of vaporization per mole of the substance. (5). Heating the gaseous from Tb to the required temperature (i.e.), 25 o C entropy involved in this process, DS 5 is given by 298.15 DS 5 = ò (C ) p Tb g dT T (298.15 K). The ------ (5) where (C p )g is the heat capacity of the substance in the gaseous state at constant pressure. Thus the absolute entropy of the gas at 25 o C ST is equal to the sum of the all the entropies listed above that is Tm Tb 298.15 dT DH f dT DH v dT ST = DS1 + DS 2 + DS 3 + DS 4 + DS 5 = ò (C p ) + + ò (C p )l + + ò (C p ) T Tm T Tb T 0 0 Tb s g (3). CALCULATION OF FREE ENERGY CHANGES THE REACTIONS We know the simplified form of Gibbs-Helmholtz eqn. is DG = DH - TDS ------ (1) If the reactants and products are in their standard states equation (1) may be written as follows DG 0 = DH 0 - T DS 0 The standard entropy changes may be determined experimentally while the values of standard entropy changes may be obtained from the literature incorporating these values in eqn. (1) the standard free energy of a reaction may be easily calculated. Ex: Calculate the standard free energy change DG 0 of the following reaction CO ( g ) + 1 O2 ( g ) ® CO2 ( g ); DH o = -282.84kJ 2 This watermark does not appear in the registered version - http://www.clicktoconvert.com 134 Solution: In this case DH o = -282.84kJ = - 282840 J. The value of DS 0 is given by the expression. DS 0 = S 0 products - S 0 reac tan t The standard entropy of CO2 ( g ) is 213.80 JK -1mol -1 while the values for CO (g) and O (g) are 197.90 and 205.01 JK -1mol -1 respectively. Hence, { } DS 0 = S 0 [CO2 ] - S 0 [CO ] + S 0 [O2 ] = 213.80 – (197.90 + 102.50) = - 86.60 JK -1mol -1 DG 0 = DH 0 - T DS 0 = - 282840 – (298 K) (- 86.60 JK -1 ) = - 257033.2 J. 12.10. THERMODYNAMIC PROBABILITY AND STATISTICAL THERMODYNAMICS 12.10.1. DERIVATION OF THE BOLTZMANN ENTROPY EQUATION The Boltzmann entropy equation, viz., S = k ln W is probably the most famous equation in statistical thermodynamics. Its derivation is naturally simple and appealing. Before we derive it, we must distinguish between two kinds of probabilities, viz., mathematical probability and thermodynamic probability. Mathematical probability is a ratio of the number of case favorable to the occurrence of an event to the total number of equally probable cases. This probability always lies between 0 and 1. Thermodynamic probability is the number of microstates corresponding to a given macro state when we are dealing with the distribution of molecules amongst an extremely large number of energy levels. This probability W is a very large number, tending to be infinite. Boltzmann suggested that entropy can be related to the thermodynamic probability W as S = f (W ) ------ (1) Consider two systems A and B whose entropies and microstates (also called complexions) are S A and S B , W A andWB , respectively, when the two systems are combined, ------ (2) S = S A + SB By definition, the probability of a thermodynamic state is proportional to the number of complexions (macro states) required to achieve it. Mathematically, we know that the total probability is the product of the probabilities of the independent events. This is also true of the complexions. Hence, W = W A × WB ------ (3) This watermark does not appear in the registered version - http://www.clicktoconvert.com 135 S A + S B = f (W AWB ) and ------ (4) This relationship suggests that entropies are additive and probabilities are multiplicative. This can be true only if S is a logarithmic function of W , (i.e.), S = k ln W ------ (5) This can be Boltzmann entropy equation. Eqn. (5) gives the quantitative definition of entropy of disorder. In equation (5) W is defined as the total number of different ways in which a given system in the specified thermodynamic state may be realized. 12.11. APPARENT EXCEPTIONS Entropies calculated using the third law is called thermal entropies. However, the statistical entropies, calculated by the method of statistical mechanics are more rigorous. It is found that the thermal entropies are somewhat smaller than the statistical entropies, the deviation ranging from .1 to 4.8 JK -1mol -1 . We thus conclude that entropies of substance (such as H2 , D2 , CO, NO, N2 O, H2O, etc.) are not zero at 0 K, as the third law formulates, but are finite. These entropies are called residual entropies. The existence of residual entropy in a crystal at 0 K is presumably due to the alterative arrangements of molecules in the solid. Such arrangements are Fig. for CO and N2 O: CO CO CO CO CO CO CO CO CO CO CO OC OC OC OC CO NNO NNO NNO NNO NNO NNO NNO NNO NNO NNO NNO ONN ONN ONN ONN NNO (a) (b) Fig. Alternative molecular arrangements (a) perfect crystal (b) actual crystal Since both the arrangements are equally likely, from the Boltzmann entropy equation (5), S = k ln W , with W = 2 N A (where N A is Avogadro’s number), we find that S = k ln 2 N A = N A k ln 2 = R ln 2 = (8.324 JK -1mol -1 ) (2.303) (0.3010) = 5.76 JK -1mol -1 Since the residual entropies are found experimentally to be less than this value, it is evident that the two alternative orientations of the CO and N2 O molecules in the solid state at 0 K are not completely random. For H2 and D2 , too, thermal entropies at 0 K are less than the corresponding statistical entropies. The calculation of the statistical entropy assumes that there exists an equilibrium between ortho and para H2 at all temperatures. The DS mix of ortho H2 and para H2 is found to be This watermark does not appear in the registered version - http://www.clicktoconvert.com 136 18.37 JK -1mol -1 in the vicinity of 0 K. When this value is added to the thermal entropy (calculated from heat capacity measurements), the arrangement with the statistical entropy is very good. 12.12. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Third law of thermodynamics Nernst heat theorem Consequences of the Nernst heat theorem Third law of thermodynamics: Plank’s formulation Statement of Lewis and Randall Unattainability of absolute zero Thermodynamic quantities at absolute zero Applications of third law of thermodynamics Thermodynamic probability and statistical thermodynamics Derivation of the Boltzmann entropy equation Residual entropy 12.13. CHECK YOUR PROGRESS 1. 2. State third law of thermodynamics. Explain the relation between probability and third law. 12.14. LESSON – END ACTIVITIES 1. Explain the Nernst heat theorem. How does it lead to the enunciation of the third l a w o f thermodynamics? 2. Write short notes on (a) Thermodynamic quantities at absolute zero (b) Statistical meaning of third law and its apparent exceptions. 12.15. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand Delhi. & Co., 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. This watermark does not appear in the registered version - http://www.clicktoconvert.com 137 UNIT-VII LESSON 13: QUANTUM STATISTICS 13.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on quantum statistics to the students. On successful completion of this lesson the student should have: * Understand the quantum statistics. 13.1. INTRODUCTION Statistical thermodynamics is concerned with the calculation of quantities such as heat capacity, entropy, etc., in terms of atomic and molecular parameters. The discipline which deals with the computation of the microscopic properties of matters from the data on the microscopic properties of individual atoms or molecules is called statistical mechanicals or statistical thermodynamics. 13.2. PROBABILITY The word is commonly used to indicate the like hood of an event taking place. For example: - suppose a coin is flipped or tossed only two results are possible, i.e.., either the head or the tail will show up. The probability of the head showing up is ½ this does not mean that if a person tosses a coin 10 times or tosses10 coins. The chance of getting head is 5. The probability of any event occurring is given by Number of cases favoring a given occurrence C Probability = ------------------------------------------------------- = -----Total number of equality possible cases r 13.3. TYPES OF STATISTICS Different physical situations encounter in nature are described by three types of statistics, Viz., the Maxwell–Boltzmann or (M-B) statistics, the Bose–Einstein or (B-E) statistics and the Fermi–Dirac or (F-D) statistics. The M–B statistics, developed long before the advent of quantum mechanics, is called classical statistics whereas; the B- E and F–D statistics are collectively called quantum statistics. The characteristics of the three types of statistics are summarized as follows. 1) In M-B statistics: - This watermark does not appear in the registered version - http://www.clicktoconvert.com 138 The particles are assumed to be distinguishable and any number of particles may occupy the same energy level. Particles obeying M-B statistics are called boltzmannons or maxwellons. 2) In B-E statistics:The particles are indistinguishable and any number of particles may occupy a given energy level. This statistics is obeyed by particles having integrated spin, such as hydrogen H2 deuterium D2 , nitrogen N2 , Helium – 4 (He4 ) and photons. Particles obeying B-E statistics are called bosons. 3) In F-D statistics:The particles are in distinguishable but only one particle may occupy a given energy level. This statistics is obeyed by particles having half–integrated spin. Example: The protons, electrons, helium 3(He3 ) and nitric oxide (NO) particles obeying F-D statistics are called fermions. 13.4. THERMODYNAMIC PROBABILITY (W ) Consider a system of ‘ N ’distinguishable particles occupying energy levels e 0 , e 1 , e 2 ...... etc., the total number of arrangements for playing ‘ n0 ’ particles in the ground state energy level e 0 , n1 particles in the first excited energy level e 1 , n2 particles in the second excited energy level e 2 and so on, is known as the thermodynamic probability, W of the given macro state. It is in general a very large number. Our problem is to determine ‘ W ’ (i.e.) to determine how many microstates correspond to a given macro state. It can be shown that ‘ W ’ is given by W= N! N! = n1!n2 !n3 !.......n j ! Õ ni ! ------ (1) Where N = å ni In eqn (1) ‘ N ’ is the total number of particles and the summation is over all the energy levels. It is possible to realize a given energy level in more than one way (i.e.) more than one quantum state has the same energy. When this happens, the energy levels is said to be degenerate. Let g i be the degeneracy (or) multiplicity of the energy level e i . This means that if there is one particle in the i th energy level, there are g i ways of distributing it. For two particles in the i th level, there are g i2 possible distributions. Thus, for ni particles in the i th level there n are g i possible distributions. Hence the thermodynamic probability for the system of N i particles is given by This watermark does not appear in the registered version - http://www.clicktoconvert.com 139 n g i W = N !Õ i ´ Constant i ni ! ------ (2) 13.5. BOLTZMANN EXPRESSION FOR ENTROPY Boltzmann suggested that entropy can be related to the thermodynamic probability W as S = f (W ) ------ (1) Consider two systems A and B whose entropies and microstates (also called complexions) are S A and S B ,W A andWB , respectively, when the two systems are combined, S = S A + SB ------ (2) By definition, the probability of a thermodynamic state is proportional to the number of complexions (macro states) required to achieve it. Mathematically, we know that the total probability is the product of the probabilities of the independent events. This is also true of the complexions. Hence, W = W A × WB S A + S B = f (W AWB ) And ------ (3) ------ (4) This relationship suggests that entropies are additive and probabilities are multiplicative. This can be true only if S is a logarithmic function of W , (i.e.), S = k ln W ------ (5) This can be Boltzmann entropy equation. Eqn. (5) gives the quantitative definition of entropy of disorder. In equation (5) W is defined as the total number of different ways in which a given system in the specified thermodynamic state may be realized. 13.6. STIRLING’S APPROXIMATION The derivation Consider the figure where ln x is plotted against x . In addition the figure also contains steps both above and below the smooth curve with the values equal to ln 1 , ln 2 , ln 3 etc. The area under the smooth curve from x = 1 to x = N is given by eqn (1) N ò ln xdx = N ln N - N + 1 ------ (1) l Now, the same area can be approximated by adding up the areas of the rectangles formed by these steps. If we choose the rectangles whose tops lie above the curve, we find for the area ln 2 + ln 3 + ..... + ln N = ln(2 ´ 3 ´ ........ ´ N ) This watermark does not appear in the registered version - http://www.clicktoconvert.com 140 = ln N ! lnx ------ (2) 1 2 3 x 4 5 6 On the other hand if we choose the rectangles below the curve, we get ln 1 + ln 2 + ln 3 + ..... + ln( N - 1) = ln[1 ´ 2 ´ 3 ´ ........ ´ ( N - 1)] = ln( N - 1)! = ln N !- ln N ------ (3) 1 ´ 2 ´ .......( N - 1) × N N !ù é = ú êë( N - 1)!= 1 ´ 2 ´ .......( N - 1) = N Nû The correct value of the area lies between those given by eqn (2) and (3). Thus, N ln N - N + 1 á ln N ! (Actual area) (Top rectangle) ln N !- ln N á N ln N - N + 1 (Bottom rectangle) (Actual area) ------ (4) (Or) ln N !á N ln N - N + 1 + ln N = ln N !( N + 1) ln N - N + 1 ------ (5) Combining (4) and (5) we have N ln N - N + 1á ln N !á ( N + 1) ln N - N + 1 ------ (6) Now, if ‘ N ’ is very large (» 10 23 ) , we can neglect unity as compared with ‘ N ’ without appreciably affecting the calculated value for the upper and the lower limits of ln N ! with this approximation, the upper and the lower limit become identical and therefore, ln N != N ln N - N This watermark does not appear in the registered version - http://www.clicktoconvert.com 141 The check: N ln N ! N ln N - N Error 102 103 104 105 363.7 5912.1 82,108.9 10,51,999 360.5 5907.8 82,103.4 10,51,293 -0.8% -0.07% -0.007% -0.0006% Note: Each additional power of ‘ W ’ in ‘ N ’ decreases the % error by roughly a factor of 10. Therefore % error is entirely negligible for N » 10 23 . 13.7. STATE OF MAXIMUM THERMODYNAMIC PROBABILITY (Equilibrium State) The state of maximum thermodynamic probability is also the state of maximum absolute probability and it is also the equilibrium state. W = N! Õ ni ! ------ (1) Taking logarithms on both sides ln W = ln N !- å ln ni ------ (2) ln W = N ln N - N - å(ni ln ni - ni ) ------ (3) (Or) We would like to adjust the ni to make ln W (and hence W ) a maximum while keeping constant the total number of particles N = å ni ------ (4) Diffentiating the eqn (4) dN = å dni = 0 ------ (5) (Since N is constant) Similarly, d ln W - d = dni dni =å å (n i ln ni - ni ) = 0 d (ni ln ni - ni ) dni This watermark does not appear in the registered version - http://www.clicktoconvert.com 142 = -å (ln ni + ni 1 - 1) ni dni Þ Cross multiply (Or) d ln W = -å ln ni .dni = 0 ------ (6) Now the restriction given in eqn (5) has to be incorporated into eqn (6). This is done by using Lagrange’s multiplies ‘ a ’. Multiplying eqn (5) by a we get, adN = å adni = 0 ------ (7) d ln W - adN = -å (ln ni + a )dni = 0 ------ (8) (6)- (7) [In getting the right hand side of eqn (8) we have subtracted term by term to right hand side of eqn (6)] (Or) å (ln n i + a )dni = 0 Since dni represent the variations in ni and hence are not necessarily zero, eqn (8) requires the coefficient to be equal to zero. (i.e.) (ln ni + a ) = 0 (Or) ln ni = -a (or ) ni = e -a But N = å ni ------ (9) \ N = å e -a = se -a \ ni = e -a = ni = N s N s ------ (10) Eqn (10) says the state and maximum thermodynamic probability is one with equal number of particles in each one of the ‘ s ’ aspects 13.8. MAXWELL-BOLTZMANN DISTRIBUTION LAW If we consider our investigation to a closed system of independent particles, it would meet the following two requirements This watermark does not appear in the registered version - http://www.clicktoconvert.com 143 (i)The total number of particles is constant (i.e.) N = å ni = Constant ------ (1) (i i)The total energy ‘ U ’ of the system is constant U = å ni e i= Constant ------ (2) The constancy of the total number of particles implies that dN = å dni = 0 ------ (3) i The constancy of the total energy implies that dU = å dni e i = 0 ------ (4) It is well known that a system of ‘ N ’ distinguishable particles occupying energy levels of e 1 , e 2 etc. The total number of arrangements for planning no particles in the ground state energy level e 0 , n1 particles in the first excited level e 1 and so on is known as thermodynamic probability ‘ W ’ is given by W = N! n0 ! n1!n 2 !.........ni ! ------ (5) On taking logarithms on both sides of eqn (5), we get ln W = ln N !-(ln n0 !+ ln n1!+ ln n2 !+......) = ln N !-å ln ni ! ------ (6) Stirling’s formula can be used for the factorial of large numbers (i.e.) ln N != N ln N - N ln ni = ni ln ni - ni ------ (7) å ln n != å n ln n - å n = å n ln n i i i i i i -N ------ (8) ------ (9) Combining eqns (7), (8), & (9) we get ln W = N ln N - N - å ni ln ni + N ln W = N ln N - å ni ln ni Differentiating and bearing in mind that ‘ N ’ is constant, we get ------ (10) This watermark does not appear in the registered version - http://www.clicktoconvert.com 144 dn ln W = -å ln ni dni - å ni d ln ni ------ (11) Now, å n d ln n = å n i i i dni = å dni = 0 ni ------ (11a) Hence at equilibrium, d ln W = -å ln ni dni = 0 ------ (12) Eqn (12) gives the change in ln W which results when the number of particles in each energy level is varied. If our system were open, then ni would vary without restriction and the variation would be independent of one another. It would then be possible to solve eqn (12) by setting each of the coefficients of the dni in eqn (12) equal to zero. However our system is not open but closed and since ‘ N ’ is constant, the value of dni are not independent of one another, as is seen from eqn (11a). Again, the energy of the system is constant too. How, then can we solve eqn (12) subject to the constraints of eqn (1) and (2)?. The desired solution is obtained by applying the method of Lagrange’s undetermined multipliers. Multiplying eqns (3) & (4) by the arbitrary constants a and b respectively and subtracting the eqn (12) we get, d ln W - adN - bdU = -å (ln ni + a + be i )dni = 0 ------ (13) As the variables ¶ni , dni ..... are independent of each other, so that the eqn (13) is to hold good, then the summation must be zero ln ni + a + be i = 0 ------ (14) Q dni ¹ 0 (Or) (Or) ln ni = -(a + be i ) - (a + be ) i n =e i - be i n = e- a e i ------ (15) This equation is known as Maxwell- Boltzmann distribution law. For complete generality, it is now necessary to make extension of this distribution law. In deriving eqn (15), assumption is made that each energy level is said to be non-degenerate. It is possible that there may be a number of quantum levels of almost identical energies, and for this, a statistical weight factor ‘ g i ’ is introduced for level e i . Hence the eqn (15) becomes as n = g e- a e i i - be i ------ (16) This watermark does not appear in the registered version - http://www.clicktoconvert.com 145 (or) Where (or) e - i n = g e - a e KT i i 1 b = KT ------ (17) e - i å n = å g e - a e KT i i Q å ni = N e - i \ N = å g e - a e KT i ------ (18) ------ (19) Dividing eqn (17) by (19) we obtain e - i n g e - a e KT i = i e N - i - a e KT å gi e ------ (20) This is the general form of Maxwell–Boltzmann distribution law. 13.9. EVALUATION OF LAGRANGE’S UNDETERMINED MULTIPLIERS Evaluation of a : We now proceed to determine a and b , we know that N = åg e i Since - a - be i ------ (1) N = å ni e- a = N - be i å gi e ------ (2) Defining a quantity ‘ q ’ called partition function, as q = åg e i - be i ------ (3) This watermark does not appear in the registered version - http://www.clicktoconvert.com 146 We obtain e- a = N q ------ (4) Accordingly, the Boltzmann distribution law becomes - be i Ng e i n = i q Partition function ‘ q ’ is a quantity of immense importance in the statistical thermodynamics. We shall see presently that by evaluating the partition function for a system we can calculate the value of any thermodynamic function for that system. Evaluation of b : The constant b , can be evaluated as follows g ni We know that [ W = N !Õ i ´ constant] n! i ln W = ln N !+ å (ni ln g i - ln ni !) ------ (1) Applying Stirling’s approximation to ln N and ln N !, we have = N ln N - N + å (ni ln g i - ni ln ni + ni ) i = N ln N + å ni ln g i -å ni ln ni i ------ (2) i Taking log of equation - be i Ng e i n = i q ------ (20) We have ln ni = ln N - ln q + ln g i - be i Substituting this value into eqn (2), we get ln W = N ln N + å ni ln g i - å ni (ln N - ln q + ln g i - be i ) i i = N ln N + å ni ln g i - N ln N + N ln q - å ni ln g i + be i å ni i = + N ln q + b U Substituting this result into the Boltzmann equation ------ (3) This watermark does not appear in the registered version - http://www.clicktoconvert.com 147 S = K ln W = NK ln q + Kb U ------ (4) From the combined statement of the first and the second laws of thermodynamics, we know that for a simple system, dU = TdS - PdV ------ (5) At constant volume ( V = constant; dV = 0 ) dU = TdS 1 æ ¶S ö \ç ÷ = è ¶V øV T ------ (6) ------ (7) Differentiating eqn (4) with respect to U at constant V , we get NK æ ¶S ö æ ¶S ö æ ¶b ö ç ÷ = ç ÷ + Kb + KU ç ÷ q è ¶U øV è ¶U øV è ¶U øV = NK dq æ ¶b ö æ ¶b ö ç ÷ + Kb + KU ç ÷ q db è ¶U øV è ¶U øV ------ (8) Also, from eqn - be i q = åg e i i dq Uq =db N ------ (8a) ------ (9) Substitution of eqn (9) in eqn (8) results in cancellation of the first and the last terms, giving æ ¶S ö ç ÷ = Kb è ¶U øV ------ (10) Comparing eqns (7) and (10), we find that b = 1 KT Hence, from eqn (8a) the molecular partition function q becomes ei q = å g e KT i i - ------ (11) This watermark does not appear in the registered version - http://www.clicktoconvert.com 148 And the Maxwell- Boltzmann distribution equation (eqn 2a) becomes e - i Ng e KT i n = i q ------ (12) From eqn (12) we can easily obtain the ratio of the populations, (i.e.) the number of particles in any two energy levels e i and e j . Thus, n i n j (e - e ) i j g i KT = e g j - ------ (13) 13.10. BOSE-EINSTEIN STATISTICS In Maxwell- Boltzmann distribution, particles are distinguished from one another. It means that if two particles interchange their positions or energy states, a new microstate or complex ion would arise. But in Bose Einstein statistics one considers the particles to be indistinguishable (i.e.) on interchanging. The two particles between two energy states, no new microstate or complex ion will arise. Let us consider four particles distributed between two cells x and y . (i.e.) three in ‘ x ’ and one in ‘ y ’. If these four particles are distinguishable as assumed in Boltzmann statistics. Four complex ion or microstates would arise. On the other hand, if four particles are distinguishable as assumed in Bose Einstein statistics. Only one complex ion or microstate would arise. Cell X abc abd Cell Y d c W=4 acd bcd aaa b a a W=1 Suppose that cell is further divided into four compartments by portions, the sections are representing the energy states. The possible distributions for these are shown in fig. (2). There will be 20 possible distributions for three particles in cell‘ x ’ and four ways of distributions of one particle in cell‘ y ’. Therefore, the total possible distributions for three particles in a cell‘ x ’ and one particle in cell ‘ y ’will be 4 ´ 20 = 80 . Let us now attempt a general expression of the microstates under such conditions. This watermark does not appear in the registered version - http://www.clicktoconvert.com 149 Suppose there are ni particles with energy e 1 in which there are g states of energy. We shall need ( g - 1) positions to place the ni particles in ‘ g ’sections, each sections corresponding to an energy state. It means that ‘ g ’ is the degeneracy of the level. Now the question arises how to distribute ni particles into ‘ g ’ sections without any restriction. Then the permutations of ni particles and ( g - 1) partitions simultaneously will be given by (ni + g - 1)! . However this also includes permutations of ni particles amongst themselves and also ( g - 1) particles amongst themselves because both these groups are internally indistinguishable. Hence the actual number in which ni particles may be allocated in ‘ g ’ states is given by a a a aa a Cell: Y a a a W = 20 aa a a a a a a a a a a aa aa a aa a a a a aa aa a aa aaa aa aa aa aaa aa a aaa aaa Cell: X a a a a W=4 (ni + g - 1)! ni !( g i - 1)! ------ (1) As in the case of Maxwell- Boltzmann statistics, we assume that in the present case also the total number of particles is constant and the total energy of the system is also constant. (i.e.) N = å ni = Constant ------ (2) U = å ni e i= Constant ------ (3) Thus, the thermodynamic probability ‘ W ’ for the system of ‘ N ’ particles (i.e.) the number of ways of distributing ‘ N ’ particles among various energy levels is given by This watermark does not appear in the registered version - http://www.clicktoconvert.com 150 W= Õ(ni + g i - 1)! ´ Constant ni !( g i - 1)! ------ (4) Taking logarithm on both sides of eqn (4) we get ln W = å (ln ni + g i - 1)! - ln ni !- ln( g i - 1)!+ Constant ------ (5) Here, too, since ni a n d g i are very large numbers, we can invoke Stirling’s approximation viz. ln x!= x ln x - x , neglecting unity Compared to ni and g i , we get, ln W = å (ni + g i ) ln ni + g i -ni + g i - ni ln ni + ni - g i ln ni + g i ln W = å (ni + g i ) ln(ni + g i ) - ni ln ni i - g i ln ni ------ (6) Where we have set ni + g i - 1 = ni + g i and g i -1 = g i . Since ni is very large, it can be treated as a continuous variable. Differentiation of eqn (6) with respect to ni and setting the differential equal to zero gives for the most probable thermodynamic state of the system. ¶ ln W = å[ln ni - ln(ni + g i )] × ¶ni = 0 (or) é ni ù å êln ú ¶ni = 0 ë ni + g i û ------ (7) From eqns (2) and (3) ¶N = å ¶ni = 0 ------ (8) ¶U = å e i ¶ni = 0 ------ (9) Applying the method of Lagrange’s undetermined multipliers (a & b ) to eqns (8) and (9) respectively we get é ù ni å êln + a + be i ú dni = 0 ë ni + g i û Each term of the summation is equal to zero é ù ni + a + be i ú dni = 0 êln ë ni + g i û This watermark does not appear in the registered version - http://www.clicktoconvert.com 151 Since the variations ¶ni are independent of one another, (¶ni ¹ 0) it means that the expression within the square bracket would be zero. Hence, ln ni + a + be i = 0 ni + g i ég ù ln ê i + 1ú = a + be i ë ni û ég ù a + be i ê i + 1ú = e n êë i úû g (or) ni = i (a + be ) i -1 e This equation is known as Bose-Einstein law of distribution. 13.11. ENTROPY OF BOSE-EINSTEIN GAS Unlike in the case of Maxwell-Boltzmann statistics, in both Bose- Einstein and FermiDirac statistics the indistinguishability of particles is assumed right in the beginning and the distribution laws described. Hence, no correction needs to be made at a latter stage. This is evident from the following derivation of the expression for the entropy of a Bose-Einstein gas. The thermodynamic probability of a Bose-Einstein system is W= Õ(ni + g i - 1)! ´ Constant ni !( g i - 1)! ------ (1) Maximization of W leads to the Bose-Einstein distribution law g ni = Now, i (a + be ) i -1 e S = K lnWmax ------ (2) ------ (3) Substituting eqn (1), after neglecting unity in comparison to g i , into eqn (3) gives S = K å (ln ( g i + ni )!i - ln g i !- i ln ni !) i Application of Stirling’s approximation results in S = K å {( g i + ni ) ln ( g i + ni ) - ( g i + ni ) - g i ln g i + g i - ni ln ni + ni } i This watermark does not appear in the registered version - http://www.clicktoconvert.com 152 or ì æ g + ni S = K å íni lnçç i i î è ni ö æ g + ni ÷÷ + g i lnçç i ø è gi öü ÷÷ý øþ ì æg ö æ n öü S = K å íni lnçç i + 1÷÷ + g i lnçç1 + i ÷÷ý g i øþ è ni ø è î æg ö from equation (2) lnçç i + 1÷÷ = a + be i è ni ø ì æ n öü \ S = K å íni (a + be i ) + g i lnçç1 + i ÷÷ý g i øþ i î è ì ï æ ï S = K å n í a + be + g lnç1 + i i i ç i ï è ï î ( Let us now denote ) g n i g i ö ÷ ÷ ø i ü n ï iï ý ï ï þ ------ (4) ni as Z . Thus the argument of logarithm in equation (4) becomes gi (1+ Z ) 1 z . The function (1+ Z ) 1 z approaches, the base of natural logarithm, as its limit when Z approaches zero. If g i ññ N i , it follows that Z will be small. Accordingly we can say that g i æ n ö ni ç1 + i ÷ » e , since for higher translational energy levels Z will be small. Thus, ç g ÷ iø è ( S = Kni å a + be + ln e i or S = nKa + Kbe + nk ) ------ (5) Equation (5) is identical to that obtained from Boltzmann statistics after correcting the latter for the indistinguishability. 13.12. FERMI-DIRAC STATISTICS In the Boltzmann or in the Bose-Einstein statistics, no restriction was made to the number of particles present in any energy state. But in applying Fermi-Dirac statistics to particles like electrons, the Pauli’s exclusion principle is taken into consideration; (i.e) two electrons (particles) in an atom cannot possess the same energy level. In simple words, it implies that not more than one particle can be assigned to a particular energy state. This watermark does not appear in the registered version - http://www.clicktoconvert.com 153 In the example consider earlier (in Bose Einstein statistics), in which four particles were distributed in two cells, three in the cell ‘ x ’ and one in ‘ y ’, one will have a distribution as in figure. a Cell: X a a a a a a a a a a a W=4 a Cell: Y a a a W=4 When one applies the Fermi- Dirac statistics the total number of possible arrangements, with the restriction that one particle can only occupy a compartment is ( 4 ´ 4) (i.e.) 16. ( ni á g i ) g Consider that the ni particles are distributed among the g i states where i , as before, is the degeneracy of the i th energy level. Imagine that the particles are distinguishable. This implies that the first particle may be placed in any one of the g i states and for each one of these choice, the second particle may be placed in any one of the remaining ( g i - 1) states, and so on. Thus the number of arrangements is given by the expression Number of arrangements = gi! ( g i - ni )! Since, however the particles are indistinguishable, the above expression has to be divided by the possible number of permutations of ni particles, viz, ni ! . Hence, the number of arrangements of ‘ ni ’ particles in the i th energy level is given by the expression gi! [ni !( g i - ni )!] Thus, the thermodynamic probability ‘ W ’for the system of ‘ N ’ particles (i.e.) the number of ways of distributing ‘ N ’ particles among the various energy level is given by W= Õ gi! ´ Constant ni !( g i - 1)! ------ (1) Taking logarithms of both sides of eqn (1) we have ln W = å [ln g i !- ln ni !- ln( g i - ni )!+ ] + Constant i Applying stiriling’s formula to evaluate the factorial ------ (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 154 ln N != N ln N - N ln W = å g i ln g i - g i - (ni ln ni - ni ) - [( g i - ni ) ln( g i - ni ) - g i - ni )] = å g i ln g i - g i - ni ln ni + ni - ( g i - ni ) ln( g i - ni ) + g i - ni = å g i ln g i - ni ln ni - ( g i - ni ) ln( g i - ni ) ------ (3) Differentiating eqn (3) with respect to ni é 1 ù 1 d ln W = å - êni + ln ni - ( g i - ni ) - ln( g i - ni )ú dni ------ (4) ( g i - ni ) ë ni û i å [ln n i - ln( g i - ni )]dni ------ (5) i The maximum value of ‘ W ’ is obtained by equating d ln W to zero å [ln n i - ln( g i - ni )]dni = 0 ------ (6) But dN = å dni = 0 ------ (7) adN = å adni = 0 dU = å e i dni = 0 ------ (8) bdU = å e i bdni = 0 Multiplying eqn (7) by a and eqn (8) by b and adding to eqn (6) we get, å [ln n i - ln( g i - ni ) + a + be i ]dni = 0 ------ (9) Each term of the summation is equal to zero (Q dni ¹ 0 ) Since the variation ¶ni are independent of one another, hence å [ln n i - ln( g i - ni ) + a + be i ]dni = 0 (Q dni ¹ 0 ) [ln ni - ln( g i - ni ) + a + be i ] = 0 ln (or) ni = -a - be i ; g i - ni (or) ég - n ù i ú = e (a + be i ) ; (or) ê i êë ni úû n i = e - a - be i g -n i i ég ù (a + be ) i +1 ê iú=e n êë i úû This watermark does not appear in the registered version - http://www.clicktoconvert.com 155 (or) gi n = i (a + be ) i +1 e This is known as Fermi-Dirac distribution. 13.13. ENTROPY OF FERMI-DIRAC GAS The thermodynamic probability of a Fermi-Dirac system is W =Õ gi! ´ Constant (g i - ni )!ni ------ (1) Maximization of W leads to the Fermi-Dirac distribution law g ni = Now, i (a + be ) i +1 e ------ (2) S = K lnWmax ------ (3) Substituting eqn (1), after neglecting unity in comparison to g i , into eqn (3) gives S = K å ln g i !-(ln ( g i + ni )!i - i ln ni !) i Application of Stirling’s approximation results in S = K å {( g i ln g i - g i - ( g i - ni )) ln ( g i - ni ) + ( g i + ni ) - ni ln ni + ni } i or ì æ g - ni S = K å íni lnçç i i î è ni ö æ g - ni ÷÷ - g i lnçç i ø è gi öü ÷÷ý øþ ì æg ö æ n öü S = K å íni lnçç i - 1÷÷ - g i lnçç1 - i ÷÷ý g i øþ è ni ø è î æg ö from equation (2) lnçç i -1÷÷ = a + be i è ni ø ì æ n öü g \ S = K å íni (a + be i ) - i lnçç1 - i ÷÷ý ni g i øþ i î è This watermark does not appear in the registered version - http://www.clicktoconvert.com 156 ì ï æ ï S = K å n í a + be - lnç1 i i ç i ï è ï î ( Let us now denote (1 - Z ) 1 ) g n i g i ö ÷ ÷ ø i ü n ï iï ý ï ï þ ------ (4) ni as Z . Thus the argument of logarithm in equation (4) becomes gi 1 . The function (1 - Z ) z approaches, the base of natural logarithm, as its limit when Z approaches zero. If g i ññ N i , it follows that Z will be small. Accordingly we can say that g i æ n ö ni 1 ç1 - i ÷ » , since for higher translational energy levels Z will be small. Thus, ç g ÷ e iø è z æ æ 1 öö S = Kni å çç a + be - lnç ÷ ÷÷ i è e øø è S = nKa + Kbe ' - nk or ------ (5) Equation (5) is identical to that obtained from Boltzmann statistics after correcting the latter for the indistinguishability of particle. 13.14. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Ø Probability Types of statistics Thermodynamic probability (W ) Boltzmann expression for entropy Stirling’s approximation State of maximum thermodynamic probability (equilibrium state) Maxwell- Boltzmann distribution law Evaluation of Lagrange’s undetermined multipliers Bose-Einstein statistics Entropy of Bose-Einstein gas Fermi-Dirac statistics Entropy of Fermi- Dirac gas 13.15. CHECK YOUR PROGRESS 1. 2. Derive the expression of Bose-Einstein statistics. Derive Stirling’s approximation. This watermark does not appear in the registered version - http://www.clicktoconvert.com 157 13.16. POINT FOR DISCUSSION 1. 2. Maximizing the thermodynamic probability of a macro state and invoking Lagrange’s undetermined multipliers derive the expression for Maxwell- Boltzmann statistics. Derive the expression for Fermi-Dirac statistics. 13.17. SOURCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. UNIT-VIII LESSON 14: PARTITION FUNCTION 14.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on partition function to the students. On successful completion of this lesson the student should have: * Understand the partition function. 14.1. INTRODUCTION: DEFINITION AND JUSTIFICATION OF NOMENCLATURE The partition function may be defined as the sum of the probability factors for different energy states or more conveniently it can be stated as the way in which the energy of a system partitioned among the molecules constituting the system. It is expressed as Q ( or ) q = S g i e - ei kT where gi is the statistical weight factor and is equal to the degree of degeneracy, (i.e.), the number of super- imposed energy levels k is the Boltzmann constant and equals to the ratio of the gas This watermark does not appear in the registered version - http://www.clicktoconvert.com 158 constant R to the Avogadro’s number, e I is the energy of the quantum state in excess of the lower possible value and T is the temperature on Calvin scale. 14.2. MICROCANONICAL AND CANONICAL ENSEMBLES MICROCANONICAL ENSEMBLE It is a collection of a large number of essentially independent assemblies each of which possesses the same energy E, volume V and the number of systems N. For sake of simplicity, it is assumed that all the systems are of the same type. Such an ensemble in which the individual assemblies are separated by rigid and well insulated walls so that E,V and N for a particular assembly are not affected by the presence of other assemblies. CANONICAL ENSEMBLE It is defined as a collection of a large number of independent assemblies, having the same temperature T, volume V, and number of identical system, N. As all the assemblies possess the same temperature, T, it means that one could bring them in thermal contact with each other and also a large heat reservoir at the same temperature T. Thus, in canonical ensemble, systems can exchange energy but not particles. The canonical ensemble in which the individual assemblies are separated by rigid, impermeable but conducting walls. 14.3. RELATION BETWEEN THE TOTAL PARTITION FUNCTION MOLECULE AND THE SEPARATE PARTITION FUNCTION OF A Consider a set of energy levels, represented by e i with degeneracies. The partition function is defined by ei ------ (1) q = å g e KT i i If we substitute translational energies into this equation, we obtain the translational partition function qtrans . If we substitute rotational or vibrational energies, we get q rot or q vib . Actually there is only one partition function for a molecule, namely, the one obtained by putting into the equation (1) all combinations of allowed energies and their corresponding degeneracies.Thus - qtot = åg tot ×e allenergy levels - e tot KT ------ (2) Now, e tot = e tr + e rot + e vib + e elec Most of the time zero (if all energies measured with respect to ground electronic level) g tot = g tr × g rot × g vib × g elec And This watermark does not appear in the registered version - http://www.clicktoconvert.com 159 åg \q = × g rot × g vib × g elec e tr - ( e tr +e rot +e vib +e elec ) KT allenergy value qtot = å g tr × g rot × g vib × g elec.e - e tr KT .e - e rot KT .e - e vib KT .e - e elec KT æ ö e öæ e öæ e öæ e - tr ÷ç - rot ÷ç - vib ÷ç - elec ÷ ç qtot = ç å g tr × e KT ÷ç å g rot × e KT ÷ç å g vib × e KT ÷ç å g elec × e KT ÷ allrot allvib allelec ç .alltra ÷ç levels ÷ç levels ÷ç levels ÷ è levels øè øè øè ø Define q tr = åg tr ×e - e tr KT tra levels q rot = åg rot ×e - e rot KT - e vib KT allrot levels q vib = å g vib × e allvib levels q elec = å g elec × e - e elec KT allelec levels 14.4. RELATION BETWEEN MOLECULAR PARTITION FUNCTION (q) AND CANONICAL PARTITION FUNCTION (Q) Consider a system of N non- interacting molecules. The Hamiltonian for such a system is the sum of separate term for the individual molecule and there will be no interaction terms. Hˆ = Hˆ 1 + Hˆ 2 + ............ + Hˆ N 1 ------ (1) The energy of such a system is the sum of the energies of all the N individual molecules E i = e 1( i ) + e 2( i ) .......... + e N( i ) ------ (2) This watermark does not appear in the registered version - http://www.clicktoconvert.com 160 Where, e 1(i ) is the energy of molecules 1 in a system with total energy ET . The canonical partition function is defined as, åg Q= i ×e - ei KT ------ (3) i Where the sum is over all the energy levels i of the system. Now (i ) (i ) g = g1 .g 2 ........g N i (i ) ------ (4) (2) and (4) in (3) (i ) æ e ( i ) + e 2( i ) ..... e N( i ) -ç 1 ç KT (i ) è (i ) Q º å g1 × g 2 ..... g N e ö ÷ ÷ ø i Q º å i g1 (i) æ e 1( i ) .e çç è KT - (i) ö (i) - æ e 2 ÷÷ × å g 2 e çç ø i è KT (i) ö (i) - æ e N ÷÷ .... å g N e çç i ø è KT ö ÷÷ ø If all the molecules in the system are of the same kind, it is no larger necessary to distinguish their energy levels and \ equation (5) reduces to (i) é (i) - æ e1 Q = ê å g 1 .e çç è KT ë i (i) öù é (i) - æ e 1 ÷÷ ú × ê å g 2 e çç øû ë i è KT Or é æ (i ) (i ) - e Q = êå g1 .e ç ç KT êë i è öù ÷ú ÷ú øû N Now, since the molecular partition function is defined as q = åg i (i ) æ e (i ) ö ÷ .e ç ç KT ÷ è ø - Our last equation becomes Q = qN 14.5. TRANSLATIONAL PARTITION FUNCTION ( qtrans ) (i) öù é (i) - æ e 1 ÷÷ ú .... ê å g 2 e çç øû ë i è KT öù ÷÷ ú øû This watermark does not appear in the registered version - http://www.clicktoconvert.com 161 For a particle of mass m, moving in an infinite three-dimensional box of sides, a, b and c, assuming that the potential is zero within the box, the energy levels obtained by the solution of the Schrödinger equation are given by the expression: q tr = åg tr ×e - e tr KT ------ (1) all × tr levels But e tr = h2 (nx 2 + ny 2 + nz 2 ) 2 8ma ------ (2) Assuming the vessel to be cubical e tr = nx 2 h 2 ny 2 h 2 nz 2 h 2 + + 8ma 2 8ma 2 8ma 2 ------ (3) (3) in (1) Þ q tr = åg tr , x g tr , y g tr , z × e nx 2 h 2 8 ma 2 KT - ×e ny 2 h 2 8 ma 2 KT - ×e nz 2 h 2 8 ma 2 KT all × tr levels Or æ nx 2 h 2 ç 2 q tr = ç å g tr , x e 8 ma KT all × tr , x ç levels è öæ ny 2 h 2 ÷ç 8 ma 2 KT ÷ ç å g tr , y e all × tr , y ÷ ç levels øè q tr = q tr , x × q tr , y × q tr , z öæ nz 2 h 2 ÷ç 8 ma 2 KT ÷ ç å g tr , z e all × tr , z ÷ ç levels øè ö ÷ ÷× ÷ ø ------ (4) qtr , x = qtr , y = qtr , z Let us evaluate qtr , x q tr , x æ nx 2 h 2 ç 2 = ç å g tr , x e 8 ma KT all ×tr , x ç levels è ö ÷ ÷ ÷ ø Where g tr , x =1 Since the translational energy levels for all practical purposes, are continuous, the summation in the above definition can be replaced by integration. This watermark does not appear in the registered version - http://www.clicktoconvert.com 162 ¥ - qtr , x = ò e nx 2 h 2 8 ma 2 KT .dn x 0 ¥ Õ 2 Standard integral= -ax e ò = 0 1 2 2a 1 2 Thus qtr , x = Õ 1 2 1 2 æ - h2 2ç e 8 ma KT ç è ö2 ÷ ÷ ø 1 qtr , x 1 1 æ 8 Õ ma 2 KT ö 2 æ 8 Õ mKT ö 2 ÷÷ = ç = çç ÷ ×a 2 2è h2 4 h è ø ø 1 2 æ 2 Õ mKT ö q tr , x = ç ÷ ×a 2 è 4h ø ------ (5) Substituting (5) in (4) and remembering that all the individual comments are equal, we have 3 q tr æ 2 Õ mKT ö 2 3 = ç ÷ ×a 2 4h è ø (Or) 3 2 æ 2 Õ mKT ö q tr = ç ÷ ×V 2 4 h è ø PV = RT ; V = RT P 14.6. ROTATIONAL PARTITION FUNCTION ( q rot ) - DIATOMIC MOLECULESRIGID ROTORS The simplest system that undergoes rotational motion is a diatomic molecule. The rotational energy levels of a rigid diatomic rotor (i.e., a rotor whose inter- nuclear distance remains fixed during rotation), obtained by solving the Schrödinger wave equation are, This watermark does not appear in the registered version - http://www.clicktoconvert.com 163 ¥ q å = rot g J ×e - e J KT ------ (1) J = 0 Where eJ = J ( J + 1)h 2 2I J = 0,1,2….. and g J = (2 J + 1) [degeneracy or statistical weight of the Jth level] ¥ å \ q rot = ( 2 J + 1 ). e - J ( J +1) h 2 IKT 2 ------ (2) J =0 For quantum mechanical rigid rotor equation (2) has to be evaluated partially to get q rot . If, however, we assume the rotation energy levels to be almost continuous (which will be true for large diatomic molecule at fairly high temperatures) the summation in equation (2) can be replaced by integration. ¥ q rot = ò ( 2 J + 1) e - J ( J +1) h 2 2 IKT dJ 0 Now, J (J+1) = J2 + J d [J ( J + 1)] = 2 J + 1 or dJ d [J ( J + 1)] = (2 J + 1)dJ \ Equation (3) becomes, ¥ q rot òe - h2 2 IKT J ( J + 1 ). d [ J ( J + 1 )] 0 ¥ Recall: ¥ òe - ax 0 q rot 1 - ax dx = e ò -a 0 é - 2 IKT = ê h 2 êë e - - J ( J +1)h 2 IKT 2 ù ú úû ¥ o ------ (3) This watermark does not appear in the registered version - http://www.clicktoconvert.com 164 2 IKT h2 q rot = For homonuclear diatomic molecule expression for the partition function becomes, 2 IKT sh2 q rot = where s is called the symmetry number has a value of 2. 14.7. VIBRATIONAL PARTITION HORMONIC OSCILLATOR FUNCTION: DIATOMIC MOLECULES – For diatomic molecule vibrating as a simple harmonic oscillator (S.H.O), the vibrational energy levels, obtained by the solution of the Schrödinger wave equation are given by E vib = V + 1 hn 2 ------ (A) where n is the vibrational frequency and V is the vibrational quantum number which has the values V = 0,1,2,3,....... The energy levels are non-degenerate, (i.e.) the degeneracy, g is unity. q = S i g i e - be i ------- (B) using equations A and B the vibrational partition function of the diatomic molecule is given by ¥ å q vib = gV ×e - eV KT V =0 where, e V = (V + 1 )hn 2 n = 1 2p V = 0,1,2….. k m gV = 1 : ¥ q vib = å 1 .e 0 Let, hn =x KT (V + 1 ) h n 2 KT ------ (1) This watermark does not appear in the registered version - http://www.clicktoconvert.com 165 ¥ + 1 )x 2 å e - (V q vib = 0 Term by term, =e =e -x -x 2 2 +e -3 2x +e -5 2x + ..... (1 + e - x + e -2 x + .....) -x Since, (1 + e - x + e -2 x e 2 + .....) = 1 - e-x -x We have, Or qvib e 2 = 1 - e-x q vib = e - hn 1- e 2 KT - hn KT 14.8. ELECTRONIC PARTITION FUNCTION Though it is possible in principle to solve the Schrödinger equation for the electronic states and energies of a molecule, it is more convenient to obtain this information from the spectroscopic data. For most of the molecules, the excited electronic energy levels lie so for above the ground state compared with kT (a typical excited state value being greater than 2eV » 3 ´ 10 -19 J ) that all the molecules may be considered to be in the ground state at ordinary temperatures. Thus, contributions to the electronic partition function arising from excited electronic states may be neglected. The electronic partition function is given by qele º q ele = g ele , 0 e - åg e - e elec KT ele allelectro nicenergylevels e ele , 0 KT + g ele ,1 e - e ele ,1 Taking the lowest level as our reference level that is, KT ------ (1) This watermark does not appear in the registered version - http://www.clicktoconvert.com 166 e ele , 0 = 0 we have, q ele = g ele , 0 1 + g ele ,1 e - e ele , 1 KT ------ (2) There is no general formula for the e ele . The series is added term by term using spectroscopically observed electronic energies. For nearly all diatomic molecules, e ele 1 is very much greater than KT at room temperature and therefore all terms in equation (2) after the first term contribute negligibly (unless the temperature is ñ10,000 K ). Therefore, q ele = g ele ,0 ------ (3) For most diatomic molecules the ground electronic level is non-degenerate (ie) q ele = g = 3 due to spin degeneracy. Another exception is NO, which has an odd number of electron g ele , 0 = 2 , ele , 0 = 1 an important exception is O2 . For O2 , g ele , 0 due to the two possible orientations of the spin of the unpaired electron. 14.9. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Ø Ø Ø Ø Ø Introduction Justification of nomenclature Microcanonical and canonical ensembles Relation between the total partition function of a molecule and the separate partition function Relation between molecular partition function (q) and canonical partition function (q) Translational partition function ( qtrans ) Rotational partition function ( q rot ) - diatomic molecules-rigid rotors Vibrational partition function: diatomic molecules – hormonic oscillator Electronic partition function 14.10. CHECK YOUR PROGRESS 1. 2. Derive an expression for the molecular translational partition function of an ideal gas. The fundamental vibrational frequency of F2 is 2.76´1013 HZ. Calculate the vibrational partition function of F2 at 250 c? 14.11. POINT FOR DISCUSSION 1. 2. Derive an expression for the molecular rotational partition function of an ideal gas. Derive an expression for the molecular vibrational and electronic partition function of an ideal gas. This watermark does not appear in the registered version - http://www.clicktoconvert.com 167 14.12. SOURCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand & Co., Delhi. 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut. This watermark does not appear in the registered version - http://www.clicktoconvert.com 168 LESSON 15: HEAT CAPACITY OF SOLIDS CONTENTS 13.0. 13.1. 13.2. 13.3. 13.4. 13.5. 13.6. 13.7. AIMS AND OBJECTIVES INTRODUCTION PROBABILITY TYPES OF STATISTICS THERMODYNAMIC PROBABILITY (W ) BOLTZMANN EXPRESSION FOR ENTROPY STIRLING’S APPROXIMATION STATE OF MAXIMUM THERMODYNAMIC PROBABILITY (Equilibrium State) 13.8. MAXWELL-BOLTZMANN DISTRIBUTION LAW 13.9. EVALUATION OF LAGRANGE’S UNDETERMINED MULTIPLIERS 13.10. BOSE-EINSTEIN STATISTICS 13.11. ENTROPY OF BOSE-EINSTEIN GAS 13.12. FERMI-DIRAC STATISTICS 13.13. ENTROPY OF FERMI-DIRAC GAS 13.14. LET US SUM UP 13.15. CHECK YOUR PROGRESS 13.16. LESSON - END ACTIVITIES 13.17. REFERENCES 15.0. AIMS AND OBJECTIVES The aim is to motivate and enable a comprehensive knowledge on heat capacity of solids to the students. On successful completion of this lesson the student should have: * Understand the heat capacity of solids. 15.1. INTRODUCTION In 1819, Dulong and Petit found that at constant pressure, the molar heat capacity at constant volume of most of the solid elements at room temperature was given by CV » 6calK -1 mol -1 In order to rectify this problem, two theories of heat capacities were developed. The first by Albert Einstein in 1907 and the second (which is a modification of the Einstein theory) by peter Debye in 1912. 15.2. THE EINSTEIN THEORY OF HEAT CAPACITY This watermark does not appear in the registered version - http://www.clicktoconvert.com 169 Einstein made the following assumptions in consulting his theory of heat capacities of mono atomic crystals. 1. The atoms in a crystal lattice undergo small oscillations (vibrations) about their equilibrium configurations. In fact, an ideal crystal can be considered as a system of ‘ N ’ non- interacting particles (i.e.) atoms. 2. Each atom vibrates independently of the others and has three independent vibrational degrees of freedom. Thus the crystal may be treated as a system of 3 N independent and distinguishable harmonic oscillations. 3. There are no electronic, translational or rotational modes of motions in a mono crystal. atomic Using assumptions 2 & 3, the molar vibrational partition function of the crystal can be written as QVib = ( q vib ) 3 N ------ (1) e ùN é - i ú ê From eqn Q = êå exp KT ú = q N ê ú ê ú ë û Q=Molar partition function; q = Molecular partition function ln QVib = 3 N ln qVib (or) Where (- ) QE æ ç 3 NQE = - 3 N lnç1 - e T 2 T çç è ö ÷ ÷ ÷÷ ø ------ (2) E is the characteristics Einstein temperature for the vibration. QE = hn K Where n is the vibrational frequency of the oscillator. Thus the internal energy of an ideal Einstein crystal is given by æ d ln QVib U = KT 2 ç ç dT è ö 3 3 NK QE ÷ = Nhn + ÷ QE øV , N 2 e T -1 QE (Or) U - U 0 = 3RT T QE e T -1 ------ (4) This watermark does not appear in the registered version - http://www.clicktoconvert.com 170 Where U 0 is the zero point-energy 3 Q ö æ 3 ç = Nhn = R E ÷ . Hence the molar heat capacity 2 è 2 ø QE 2 e T æ QE ö C = 3Rç ÷ 2 V è T ø æ QE ö ç ÷ ç e T - 1÷ çç ÷÷ è ø in ------ (5) Experimentally it is found that at temperature approaching zero, CV approaches zero and the limit of high temperatures, CV approaches the Dulong-Petit value of 3R (i.e.) (6CalK -1 Mol -1 ) .Einstein theory predicts these limiting values CV of quite successfully. Thus as, T¾ ¾® 0 QE QE e T -1 = e T ------ (6) So that QE 2 æ QE ö T¾ ¾® 0 C = 3Rç ÷ e T =0 V T è ø Again as T ¾ ¾® ¥ , ------ (7) QE »1 æ QE ö e T +ç ÷ So that è T ø 2 T ¾lim ¾ ¾it ® 0 QE æ QE ö e T ÷ C » 3Rçç » 3R 2 V T ÷ æ Q E ö÷ è ø ç 1+ -1 ç ÷ T è ø ------ (8) The above results are illustrated in Fig.1 for a number of metallic and non- metallic crystals. This watermark does not appear in the registered version - http://www.clicktoconvert.com 171 Fig.1. Temperature-dependence of CV for monotomic solids Einstein theory however is not successful in predicting the CV values in the lower and intermediate temperature ranges. The values predicted by it are lower than those actually observed. 15.3. DEBYE THEORY OF HEAT CAPACITY According to Debye’s theory the crystal is treated as a macromolecule containing ‘ L ’ atoms, for which there are (3L - 6) vibrational modes. Since 6áá 3L ,this assembly can be taken to have 3L vibrational modes. The frequencies of such vibrational range from n 0 (ground state) to a maximum valuen max . This limit arises when the wavelength of the oscillations is of the same order as the inter atomic distances. Debye considered the solid to be a continuous elastic medium. The vibrations of the atoms could be regarded as the elastic waves propagated through the medium. The vibrational motion of the atoms can be considered to be similar to sound waves propagated through this elastic medium. The vibrational energy is quantized and in order to calculate the total energy of vibration it is necessary to know the number of vibrations at each value of the allowed frequency (energy). This number, a function of the frequency (i.e.) f (n ) is known to be proportional to n 2 (from the elastic theory of solids). f (n ) = a ¢n 2 ------ (1) This watermark does not appear in the registered version - http://www.clicktoconvert.com 172 In eqn (1) a ’ is the constant of proportionality. Considering a narrow range of frequency dn , the number of vibrations is f (n ) dn . The frequency ranges from n 0 (equal to zero) to n max (Debye cut off frequency) and the energy levels of the vibrating solid are closely spaced. Hence the total modes of vibration in the crystal can be evaluated by integration and equated to 3L . The total vibrational energy of the crystal is equal to the average energy of an oscillation of frequency ‘ n ’ multiplied by the number of oscillations at that frequency. This must be summed up or integrated over all the ranges of allowed frequenciesn 0 andn max . n n max ò f (n )dn = 3L n 0 max 2 ò a ¢n dn = 3L n 0 ------ (2) Evaluating this integral, a¢ = 9L 3 n max f (n ) = ------ (3) 9L 3 n max ------ (4) Q dn = n , a ¢n 2 n = 3L , a ¢ = 9 L 3 3L 3L n max áU ñ has already been calculated (Einstein heat theory). Since n 0 is equal to zero. max U= ò áU ñ f (n )dn ------ (5) 0 max = ò 0 Since CV = hn e hn KT -1 9 L n 2 dn 3 n max ------ (6) dU , on differentiating eqn (6) with respect to temperature, dT CV = dU d = dT dT max ò 0 æ hn ö 2 ç ÷9 Ln dnKT è KT ø (e hn KT - 1)n 3 max ------ (7) This watermark does not appear in the registered version - http://www.clicktoconvert.com 173 hn e KT n 9 R max æ hn ö 2 n 2 dn ÷ ò ç 3 2 è KT ø æ hn n ö max 0 ç ÷ ç e KT - 1÷ çç ÷÷ è ø ------ (8) hn has the dimensions o temperature and whenn = n max , the temperature K hn max hn corresponding to is called the Debye characteristic temperature q D .Since is KT KT dimensionless, it can be replaced by u , The quantity U max = Since hn max q D = KT T qD = hn hn KT ;u= =n = ×u K KT h dn = KT × dU h And so 2 3 æ KT ö 2 KT æ KT ö 2 n 2 dn = ç dU = ç ÷ u × ÷ u du h è h ø è h ø Equation (8) can thus be written as CV = 3 9R 3 æ KT ö 3 ç ÷ u è h ø e u u 2 æ KT ö 2 ç ÷ u du (e u - 1) 2 è h ø (or) 3 æ h ö æç T = 9 Rç ÷ è KT ø çè q D 3 ö U max u 2 e u æ KT ö 3 2 ÷ ç ÷ u du ÷ ò u 2 h ø ø 0 (e - 1) è (or) æ T = 9 Rç çq è D (or) ö ÷ ÷ ø 3 qD 3 3 4 u æ h ö æ KT ö T u e du ç ÷ ç ÷ ò è KT ø è h ø 0 (e u - 1) 2 This watermark does not appear in the registered version - http://www.clicktoconvert.com 174 æ T = 9 Rç çq è D ö ÷ ÷ ø 3 qD T u 4 e u du ò u 2 0 (e - 1) ------ (9) The high and low temperature limits of CV (V ) can be obtained ö u2 4æ ç u 1 + u + + ....... ÷÷ ç 4 u 2! u e du ø = è 2 2 æ ö u (e u - 1) 2 çç u + + ....... ÷÷ 2! è ø Since u is small this reduces to æ T If T >> q D , CV (V ) » 9 Rç çq è D u4 or u 2 2 u q 3 D ö T ÷ u 2 du ÷ ò ø 0 2 q Q u 2 = æç hn ö÷ = æç D ö÷ è KT ø èT ø dU = 3 hn q D = KT T æ T \ CV (V ) = 9 Rç çq è D 3 ö 1 æ q ö3 ÷ ç D÷ ÷ 3ç T ÷ ø ø è CV (V ) » 3R The Debye formula has the same high temperature limit as the Einstein equation. At very 4p 4 q ¾® ¥ and the value of the integral become low temperature D ¾ and is independent of T . T 15 3 æ T ö ÷ . The Debye equation for very low The heat capacity varies under these conditions with ç çq ÷ è Dø 3 temperatures is thus CV = a ¢T (Debye-T-cubed law). This law is used to calculate CV at very low temperatures and to extrapolate the experimental results from the lowest accessible This watermark does not appear in the registered version - http://www.clicktoconvert.com 175 temperature to the absolute zero. Since CV (V ) is a function of T for all solids should fall on a curve q D T only, a plot of CV against q D (Fig. 2). Fig. 2. The Debye curve for atomic heat capacity T is q D is approximately 300. At temperatures near 300 K Dulong From Fig. 2 it is seen that CV (V ) attains the classical value of 3R when approximately one, (i.e.) when q D and Petit’s law holds good for carbon, q D is 1860 K and for other light elements also q being D T is small and CV (V ) value falls on the ascending portion of the curve (i.e.) the value is q D less than 3R . A higher value for CV is reduced only at high temperatures. large, 15.4. PLANCK DISTRIBUTION LAW FOR BLACK BODY RADIATION In the derivation of Planck’s radiation formula, we will consider a small cavity in an opaque material as a theoretical model of a black body. Suppose the cavity has volume ‘ V ’and it is full of electromagnetic radiation at temperature T .An electromagnetic wave of frequency ‘n ’ hn and velocity ‘ C ’ consists of particles called photons of energy hn , momentum . c This watermark does not appear in the registered version - http://www.clicktoconvert.com 176 The cavity in an oppose material is assumed to be having a large number of photons, each one of them is indistinguishable from the other. The assembly of photons in the cavity may be compared to the assembly of gas molecules in vessels. Therefore, the assembly of photons is supposed to constitute a gas known as photon gas. The photon gas possess the following properties 1. Photons are particles of rest mass zero 2. Unlike the gas molecules the photons may be created and destroyed. In this case we have å dni ¹ 0 . This is strictly in accordance with the Bose Einstein statistics because the number of photons is not fixed in the cavity due to their absorption and re-emission by the walls of the cavity. 3. There are two possible spin states which are associated with the photons of spin 1. In other words the photon possess two allowed modes, the number of allowed states or modes per unit volume for the photons with momentum lying in the range P to dP will lie 2 within a cell of volume 4pP dP .Therefore, the total number of eigen states will be given by g ( p )dP = 2.4pP 2 dP h3 V ------ (1) For a photon, P= h hn = l c ------ (2) Differentiating equation (2), we get dP = hdn c ------ (3) On substituting equations (3) in (1), we get g (n )d (n ) = = = 8pVP 2 hdn × c h3 8pVh 2n 2 hdn × c h 3c 2 8pVn 2 dn c3 QP = h 2n 2 hn , \ P2 = 2 c c ------ (4) Equation (4) represents the total number of Eigen states which are lying in the frequency range n andn + dn . Substituting eqn (4) in the Bose Einstein distribution law, we get This watermark does not appear in the registered version - http://www.clicktoconvert.com 177 g i (a + be ) i -1 e g (n )d (n ) n2 ¶n dn = = 8pV 3 e e c i i KT KT e -1 e -1 ni = (or) dn n2 = 8pV 3 V c ¶n e i e KT - 1 ------ (5) The left hand side of the equation (5) represents the number of photons per unit volume. When it is multiplied by hn , the energy of photon gives density Endn in the specified range. On substituting E = hn , we get En d n = 8phn 3 c3 ¶n hn e KT - 1 ------ (6) Equation (6) is the Planck distribution law for black body radiation. Endn is also known as the spectral energy density at the frequency n . 15.5. LET US SUM UP In this lesson, we: Pointed out Ø Ø Ø Ø Introduction The Einstein theory of heat capacity Debye theory of heat capacity Planck distribution law for black body radiation 15.6. CHECK YOUR PROGRESS 1. 2. For Al, QE = 240K, Calculate Cv for Al on the Einstein model at (a) 50K and (b) 240K For I2 at 10K, Cv = 4.02JK-1 mol-1 . Calculate (a) QD and (b) Cv of I2 at 12K. 15.7. LESSON – END ACTIVITIES This watermark does not appear in the registered version - http://www.clicktoconvert.com 178 1. Discuss the salient features of the Einstein theory of the heat capacity of monoatomic crystals. How did Debye modified it? Show the results of the Einstein and the Debye theories on a plot and comment briefly. 2. Explain briefly the heat capacities of solids. 15.8. REFERENCES 1. J.C. Kuriacose and J. Rajaram, Thermodynamics, Shoban lal Nagin Chand Delhi. & Co., 2. B.R. Puri, L.R. Sharma and M.S. Pathania, Principles of Physical Chemistry, Millennium edition, 2006-2007. 3. Gurdeep Raj, Advanced Physical Chemistry, Goel Publishing House, Meerut.