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thermodynamics B K Sharma Definition - Thermodynamics quantitative relationship between various forms of energies. Importance of Thermodynamics Predict the feasibility of a chemical reaction under a given set of conditions. Predict the extent to which the reaction is carried out before attainment of equilibrium. Basic Terms in Thermodynamics Universe system surroundings Types of systems Open system Closed system Isolated system Homogeneous system e.g. aqueous solution of NaCl, aqueous solution of sugar. Heterogeneous system e.g. ice in water, oil in water, etc. Properties of a system 1. Macroscopic property shown by large number of particles of a system. volume, pressure, temperature,density. 2. Microscopic property Property of individual atoms and molecules of a system at micro level atomic mass, molecular mass. Macroscopic Properties of System Macroscopic Properties extensive properties mass, volume intensive properties temperature, pressure, viscosity, density, State function Thermodynamic properties which depend only upon the initial and final states of a system For example, internal energy, enthalpy, etc. mass, volume, etc. State of a system The macroscopic properties which causes a change in the state of a system are called state variables. For example, pressure, temperature, etc. Types of Thermodynamic Process (i) Isothermal process :-A process which is carried out at constant temperature. (ii) Adiabatic process:-A process which is carried out in such a way that no heat flows from the system to surroundings and vice versa. (iii) Isochoric process :-A process which is carried out at constant volume. (iv) Isobaric process:-A process which is carried out at constant pressure. Types of Thermodynamic Process (v) Reversible process :- carried out infinitesimally slow so that all changes occurring can be reversed and the system remains almost in a state of equilibrium with the surroundings at every instant of time. (vi) Irreversible process:- Occurs rapidly but the system cannot be reversed to its initial state immediately and the system does not remain in equilibrium during transition. (vii)Cyclic process:- When a system undergoes different processes and finally returns to its initial state, is known as cyclic process. Internal energy Energy stored within a substance of a system internal energy or intrinsic energy. E = Ee + Ev + Er Internal energy Actual (absolute) value can not be measured change in internal energy (DE) can be measured It is an extensive property. Internal energy is a state function For a cyclic process, DE = 0 Work Work done = Force × Displacement W = P(V2 – V1) (At constant pressure) W = PDV Where, W = work done W = –ve, work done by the system (expansion) W = +ve, work done on the system (contraction) Work Work done in isothermal reversible expansion of an ideal gas Pext– dP Pext dv Pg Pg Wrev V2 V1 Wrev Work nRT dV V V2 nRT ln V1 Wrev = – 2.303 nRT log V2 V1 1 Since P V \ Wrev = – 2.303 nRT log P1 P2 First Law of Thermodynamics Energy can neither be created nor destroyed E2 = E 1 + q + w E2 – E1 = q + w DE = q + w If work is done by the system, (in case of expansion) w = - PDv \ DE = q - PDv \ q = DE + PDv First Law of Thermodynamics a. During isothermal expansion DE = 0 \ q = PDv b. In case of adiabatic process, q = 0 \ DE = - PDv = +w DE = +w c. In case, the process is carried out at constant volume First Law of Thermodynamics DV = 0 PDV = 0 or w = 0 \ qV = DE \ Quantity of heat supplied at constant volume, qv is equal to increase in the internal energy of the system. Session objectives 1. Enthalpy 2. Various types of enthalpy of reactions 3. Heat capacities of gases 4. Adiabatic process 5. Hess’s law 6. Bond energy 7. Lattice energy 8. Limitation of first law of thermodynamics Enthalpy Enthalpy is the total heat contents of the system at constant pressure. Enthalpy is shown by ‘H’. H = E + PV Enthalpy change at constant pressure DH = DE + PDV H2 – H1 = E2 – E1 + P(V2 – V1) Where H1, E1 and V1 are the enthalpy, internal energy and volume respectively in initial state while H2, E2 and V2 are the enthalpy, internal energy and volume respectively in final state. Enthalpy Enthalpy is a state function PV1 = n1RT At constant pressure (for initial state) PV2 = n2RT (for final state) P(V2 – V1) = RT(n2 – n1) PDV = Dn g RT Where Dng = np–nr (gaseous moles only) \ DH = DE + DngRT Enthalpy of formation It is the change in enthalpy when one mole of a compound is formed from its elements in their naturally occuring physical states. 2C(s) 2H2(g) C2H4 DHf 52kJ Enthalpy of combustion It is the change in enthalpy when one mole of the substance undergoes complete combustion. CH4 g 2O2 g CO2 g 2H2O(g) ; DH 890.3 kJ Application of heat of combustion Calorific value Amount of heat produced per gram of a substance (food or fuel) is completely burnt. CH4 g 2O2 CO2 g 2H2O l DH = -890.3 kJ/mol Calorific value of CH4 (g) = – 890 – 55.6 kJ / g 16 Hydrogen has the highest calorific value (150 kJ/g) Enthalpy of solution Amount of heat evolved or absorbed per mole of the substance in excess of water, KCl s H2O KCl aq DH – 4.4 Kcal KOH s H2O KOH aq DH – 13.3 Kcal Enthalpy of fusion One mole of solid substance changes to its liquid state at its melting point. Melting H2O s H2O l Freezing DH 1.44 Kcal H2O l H2O s DH – 1.44 Kcal Enthalpy of vaporization One mole of the substance changes from liquid state to gaseous state at its boiling point. H2 O l H2 O g Boiling DH 10.5 Kcal H2 O g H2 O l DH – 10.5 Kcal Cooling Enthalpy of sublimation Enthalpy change per mole of a solid converts directly to its vapours sublimation NH4Cl (s) NH4Cl g DH 14.9 Kcal Heat capacity Specific heat capacity is the heat required to raise the temperature of unit mass by one degree. q = c × m × DT m = Mass of the substance q = Heat required DT = Temperature difference c = Specific heat capacity. Specific heat capacity of water is 4.18 J/g K. Hess’s Law According to Hess’s law q = q1 + q 2 q 1 C q 2 q A B Determination of lattice energy D H MX D Hsub M 1 (D H 2 ) diss X2 (IP)M ( EA)X ( U)MX Figure Determination of bond energies or bond enthalpies Energy required to break the bond or energy released during the bond formation is called bond energy. H2S H g SH g DH 100 kJ / mole SH g S g H g DH 200 kJ / mole The average of these two bond dissociation energies gives the value of bond energy of S — H. Bond energy of S — H bond 100 200 150 kJ / mole 2 Limitations of first law 1. The first law of thermodynamics states that one form of energy disappears, an equivalent amount of another form of energy is produced. But it is salient about the extent to which such conversion can take place. 2. It does not tell about the direction of flow of heat. 3. It does not tell about spontaneity of reaction. Class exercise 2 When 1 gram of methane (CH4) burns in O2 the heat evolved (measured under standard conditions) is 13.3 kcal. What is the heat of combustion? (a) –13.3 k cals (b) +213 k cals (c) – 213 k cals (d) – 416 k cals Soluti on 13.3 Kcal/gm evolved DHo 13.3 16 Kcal / mol comb = – 213 Kcal/mol Hence, the answer is (c). Class exercise 3 is transferred When 4.184 J of heat to 1 g of water at 20° C, its temperature rises to 21° C. The molar heat capacity at this temperature is (a) 18 JK–1 (c) 75.4 JK–1 (b) 18 4 .184 JK -1 (d) 4.184 JK–1 Solution: n CDT = 4.184 C = (4.184) × 18 = 75.4 J/K Hence, the answer is (c). Class exercise 4 (B.P 80° C) When 0.532 g of benzene is burnt in a constant volume system with an excess of oxygen, 22.3 kJ of heat is given out. forDthe H combustion process is given by (a) – 21 kJ (b) – 1234.98 kJ (c) – 221 kJ (d) – 3273.26 kJ Soluti on DH 22.3 Mol. wt C6H6 22.3 78 0.532 0.532 = – 3269.5 kJ Hence, the answer is (d). Class exercise 5 Consider the reaction 1 SO2(g) O2(g) SO3(g) DH 2 = – 98.3 kJ. If the enthalpy of formation of SO3(g) is – 395.4 kJ, then the enthalpy of formation of SO2(g) is (a) 297.1 kJ (b) 493.7 kJ (c) – 493.7 kJ (d) – 297.1 kJ Solution: DHo DHo 98.3 kJ f SO3 f SO2 DHo 98.3 395.4 kJ 297.1 kJ f SO2 Hence, the answer is (d). Class exercise 6 Calculate the heat change for the following reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH0for CH4 , H2O and CO2 are –17.89, –68.3 f and–94.05 kcal/mole. Soluti on CH g 2O 4 CO2 g 2H2O l 2 g o o o o DHreaction 2DHo D H D H 2 D H f H2O f CO2 f CH4 f O2 = – 2 × 68.3 – 94.05 + 17.89 – 0 = – 212.76 kcal/mol Class exercise Calculate the heat7of combustion of benzene from the following data: 6C (s) + 3H2(g) C6H6(l) DH = 11720 cal 1 H2(g) + O2(g) H2O(l) DH = -68320 cal 2 C(s) + O2(g) CO2(g) DH =-93050 cal Soluti onrequired reaction is The C6H6 15 O2 3H2O 6CO2 (1) 2 6C + 3H2 C6H6 (2) C + O2 CO2 (3) 1 H2 O2 H2O 2 3 × (4) + 6 × (3) – (2) o o o DH1 3DHo 6 D H D H 4 3 2 = 3(– 68320) + 6(– 93050) – (1720 Cal) = – 774.980 kcal/mol Class exercise 8 H the following Calculate Dfor reaction at 27o C. C2H4 (g) + 3O2(g) 2CO2 (g) + 2H2O(l) Given D=H– 337 kcal R= 1.987 cal deg–1 mole–1 Soluti on DH = DE + PDV = DE + DnRT {Dn = -2} DE =DH - DnRT =– 337 – (–2)(1.987)(300) × 103 = – 335.8078 kcal/mol Class exercise Calculate the heat9of combustion of acetic acid at 25o C if the heat of formation of CH3COOH(l),CO2(g) and H2O(l) are –116.4, –94.0 and –68.3 kcal mole–1 respectively. Solution: CH3COOH l 3O2 g 2CO2 g 2H2O l DHreactions = 2 . DHf (CO2) + 2 . DHf (H2O) -DHf (CH3COOH) - 3DHf (O2 ) = 2(– 94.0) + 2(– 68.3) – (–116.4) = – 208.2 Kcal/mol Spontaneous process A process which takes place by itself is called a spontaneous process. For example, 1. Dissolution of common salt in water 2. Evaporation of water in open vessel 3. Flow of heat from hot end to cold end of a metal rod Non-spontaneous process A process which cannot take place by itself is called a non-spontaneous process. For example, 1. Flow of water up a hill 2. Flow of heat from a cold body to a hot body Entrop y Entropy is measurement of randomness.It is denoted by S Unit of entropy is JK–1 q Entropy Change DS T Entropy Disorderness Crystalline solid – Lowest entropy Gaseous state – Highest entropy Entropy Entropy is state function DS SFinal SInitial standard entropy change DSo So (Products) So (Reactants) Entropy changes during phase transformations Entropy of fusion One mole of a solid changes into liquid at its melting point. DSfusion Sm (liquid) Sm (solid) DHfusion T ΔSfusion is entropy change of fusion per mole Entropy of vaporization One mole of a liquid changes into vapours at its boiling point. DSvap Sm (vapour) Sm (liquid) DHvap T DSvap is entropy of vaporization per mole Entropy of sublimation One mole of solid changes into vapours DSsub Sm (vapours) Sm (solid) DHsub T DSsub is entropy of sublimation per mole Second law of thermodynamics Various statements of second law of thermodynamics Total entropy change of system and surroundings is positive for all spontaneous processes. The entropy of the universe is continuously increasing. Total heat absorbed by a system cannot be converted completely into work Gibbs free energy Amount of energy available for useful work. It is denoted by G. G = H – TS … (i) Gibbs free energy for isothermal process, For initial state G1 H1 TS1 For final state G2 H2 TS2 \ G2 G1 H2 H1 T S2 S1 DG DH TDS This is Gibbs-Helmholtz equation. Gibbs-Helmholtz equation and spontaneity DG DH TDS DH (a) If is positive is non-spontaneous. DH TDS, then process DG (b) If negative DH isTD S, DGand thus, the process is spontaneous. (c) If is D zero DH T S, and DGthus, the process is in equilibrium. Such a process would be spontaneous at high temperatures DS Physical significance of Gibb’s free energy Calculation of electrical work done If work involved is the electrical work (like in Galvanic cells), DGo nFEo n is number of electrons involved in the cell reaction Eo is electromotive force (EMF) of the cell F is Faraday’s constant Calculation of equilibrium constant (Kc) DGo 2.303RTlogK (where R is the gas constant) Third Law of Thermodynamics Nernst in 1906 gave third law of thermodynamics. It states that the entropy of all pure and perfectly crystalline solids may be taken as zero at absolute zero temperature. Class exercise 1 Which of the following has the highest entropy? (a) Normal egg (b) Half boiled egg (c) Hard boiled egg (d) (b) and (c) have same entropy Solution: There is maximum randomness in normal boiled egg due to denaturation of proteins in the egg which involves conversion of helical form into more random coil form. Hence, the answer is (a). Thank you