Download Section 1 and 2

Chemistry 223H1 Fall Term 2010
Instructor:
Dr. D.F. McIntosh
Email: [email protected]
Office hour: Thursdays 3:00 – 4:00 pm Rm 223 in Lash Miller
Lectures:
Textbook:
T : LM 159
3:00 – 4:30 pm
F : LM 159
2:00 – 3:30 pm
Physical Chemistry for the Biological Sciences
by Raymond Chang. (in the Bookstore)
Midterm:
Wednesday, November 03, 2010
Other Suggested Reading……
….these additional texts also provide very useful
discussions of topics covered in this course …..
1. Physical Chemistry 7th Edition
Atkins P., de Paula J.
2. The Elements of Physical Chemistry with
Applications in Biology
Atkins P.
*** Please note that if not specified the figures within the
slides are taken from the Chang Text.
*** The majority of the problems are also taken from the
Chang Text.
Problem Sets & Tutorials
Problems
Recommended problems will be assigned for
each chapter. They will not be graded.
Full solutions will be covered in detail at the TUTORIALs.
TUTORIALS: Mondays 3:00-4:00 pm
(A–L)
GB 244
Megan Oh
(M–Z)
GB 248
Brian Lam
A teaching assistant will go over solutions to problems and
answer any specific questions.
Course Evaluation
45 %
Midterm
Wednesday, November 03, 2010
55 %
Final Exam
To Be Scheduled
Outline of the Course
1)
Review and Definitions
2) Molecules and their Energies
3) 1st Law of Thermodynamics
4) 2nd Law of Thermodynamics
5) Gibbs Free Energy
6) Phase Diagrams and REAL Phenomena
7) Non-Electrolyte Solutions & Simple Mixtures
8) Chemical Equilibrium
9) Kinetics
SECTION 1.0. Review and Definitions
Relevant Sections in Text
…….Chapter 2 (sections 2.1 – 2.8)
SECTION 1.0. Review and Definitions
GOALS:
1) to review some basic definitions
2) to review the gas laws (i.e. Boyle’s law, Charles’ law,
Avogadro’s law) and the “ideal” gas equation
3) to understand difference between an ideal gas and
a real gas (i.e. Z, van der Waals equation)
Why is this important: we need these basic concepts to
understand molecules and their energies…….
1.1. Review and Definitions
A) Define System:
Su
rr
o
un
din
gs
OPEN: mass and energy exchange
CLOSED: only energy exchange
ISOLATED: no mass nor energy exchange
SYSTEM
S
ro
ur
un
s
ng
i
d
Surroundings
CLOSED
OPEN
ISOLATED
B) Define Properties of Matter
Extensive Properties……properties whose values are directly
proportional to amount of material in system
e.g. mass, area, volume, energy
Intensive Properties…….properties whose values are not
dependent on amount of material in system
e.g. temperature, pressure, density
….Intensive properties are usually defined as ratio of two
Extensive properties……
Pressure = Force/Area
Density = Mass/Volume
C) Temperature
How do we define temperature ?
….. hot or cold
Need more precise “scientific” definition……..
- temperature is a variable
- consider closed systems A & B
(no mass exchange)
PAVA
A
PBVB
B
PA’VA’
A
PB’VB’
B
heat exchange
“ An ISOTHERM”
1.2. Review of GAS Laws
A) Boyle’s Law
…relationship between Pressure (P) and Volume (V)
V1/P
…….PV = constant
so P1V1 = P2V2 (constant n and T)
B) Charles’ Law
…relationship between V and T
V T
so V1/T1 = V2/T2
ABSOLUTE ZERO
• plot of V versus T yields straight line (at a given P)
• extend line to zero V …..intercept is – 273.15°C
• - 273.15 °C is known as absolute zero
Kelvin Temperature Scale:
T(K) = t (°C) + 273.15
 Absolute zero = - 273.15 °C or 0 K
** Absolute temperatures are always used in gas law and
thermodynamic calculations !
C)
Avogadro’s Law
………..equal volumes of gas at the same T and P contain
the same number of molecules………
Vn
OK ….we have three laws (Boyle’s, Charles’ and
Avogadro’s……let’s put them together………
IDEAL GAS EQUATION
D)
Ideal Gas Equation
PV = nRT
R* = 0.082057 L atm mol-1 K-1
R = 8.3144 J mol-1 K-1
* How did we get this value for R ?
• Experimentally it is found that 1 mole of an ideal gas
occupies 22.414 L at 1 atm and 273.15 K (STP)
R = the gas constant
R = (PV)/(nT)
=
(1 atm)(22.414L)
(1 mol)(273.15K)
=
0.082057 L atm mol-1 K-1
Sample Problem
Nitrogen is heated to 500 K in a vessel of constant volume.
If it enters the vessel at P = 100 atm and T = 300 K, what
pressure would it exert at the working temperature if it
behaves as an ideal gas ?
3.0. REAL gases versus IDEAL gases
Ideal Gas:
• Gas molecules possess negligible volume
• No interaction between molecules
(attractive or repulsive)
No Such Gases EXIST !
* Ideal Gas Equation is only useful for gases at high T, low P
Why ???
Real Gases
• deviations from the ideal gas law are most important at high
pressures and low temperatures
…..especially when a gas is on the verge of becoming
a liquid…..
• Real gases deviate from ideal gas law
due to intermolecular interactions
Attractive Intermolecular Interactions
….exist when molecules are close
together but not in actual contact.
Repulsive Intermolecular Interactions
….important when molecules are in
actual contact.
….these are short-range interactions.
Figure 1.12 - Physical Chemistry 7th Edn. Atkins P., de Paula J.
INTERMOLECULAR INTERACTIONS
ATTRACTIVE
• these interactions are present over distance equal to a few
molecular diameters
• attractive interactions are responsible for condensation of
gases into liquids at low T
….at low T, molecules don’t have enough KE to escape
from each other so they stick together.
……when molecules are separated by a few “molecular
diameters” they attract each other….when they are in direct
contact they repel each other……
REPULSIVE
• these interactions are reason that solids and liquids have
volume and don’t completely collapse.
Z = Compressibility Factor
…..….plot Z versus P to measure deviation from ideality
Z = PV/nRT = PV/RT
V is molar volume (V/n)
• For an ideal gas Z = 1
for any value of P at a given T
• in limit of P approaching zero
Z = 1 for all gases
Why do all gases have Z > 1 at high pressures ????
….at high P real gases have a larger molar volume than ideal
gases due to repulsive intermolecular interactions.
Van der Waals Equation……….
• attempts to account for
(1) finite volume of individual molecules
(2) the attractive forces between molecules
[ P + (an2)/(V2) ] [V – nb] = nRT
1st term accounts for
attractive intermolecular
forces
2nd term accounts for
the finite volume of
molecules
• a and b are constants characteristic of the gas under study
• rough correlation between a and boiling point
* VdW equation becomes unreliable at very high T and low P
Values for CO2 determined using ideal gas and VdW equation
P
observed/atm
P
50
75
100
56.9
92.2
136
ideal/atm
P
vdWaals/atm
49.4
73.4
96
Sample Problem
To synthesize ammonia using the Haber process, 2000 moles
of N2 are heated in an 800 L vessel to 625°C. Calculate the
pressure of the gas if N2 behaves (a) as a van der Waals gas
and (b) as an ideal gas.
Note: a = 1.35 atm L2 mol-2
b = 0.0386 L mol-1
Problem continued………………..
*** Pressure calculated by van der Waals equation is approx.
6 % higher……..for industrial processes it’s important to have
the accurate pressure so to ensure the reaction vessel can
handle it. (page 22 in Chang Text)
Sample Problem……
Find the density of F2 at 20°C at 0.247 atm.
Mol. Wt. = 38.00 g/mol
Solution
1- the unknown is the density () of F2
2- since the gas is at relatively low pressure we assume it behaves
ideally (so use equation PV = nRT)
Our Progress to DATE………..
• Definitions…..system, properties and Temperature
• Gas Laws…….and we put them together to get the
IDEAL GAS equation
• Difference between IDEAL and REAL gases
• Van der Waals equation
2.0. Molecules and their Energies
Relevant Sections in Text….Chapter 3 (section 3.8)
Outline of the Course
1) Review and Definitions
2) Molecules and their Energies
3) 1st Law of Thermodynamics
4) 2nd Law of Thermodynamics
5) Gibbs Free Energy
6) Phase Diagrams and REAL Phenomena
7) Non-Electrolyte Solutions
8) Chemical Equilibrium
9) Chemical Kinetics and Rates of Processes
2.0. Molecules and their Energies
Goals
(1) to understand how molecules store and use energy
(2) to define specific heat capacity and heat capacity
(3) to test equipartition of energy theorem
using heat capacity measurements
Why is this important……..need these basic concepts
to understand thermodynamics….
Molecules and their Energies (Section 3.8)
The average kinetic energy for one molecule is (3/2) kB T
………..where kB is the Boltzmann Constant = 1.381 x 10-23 JK-1
………..since kBNA = R ………
then average kinetic energy for one mole of a gas is
KE = ( 3/2 ) RT
How is this Energy distributed ?
to understand lets consider a theory
based on Classical Mechanics.
The Equipartition of Energy Theorem……..
• energy of a molecule is equally divided among all types of
motion or degrees of freedom
• energy for each molecule can be divided:
ETOT = ETRANSLATION + EROTATION + EVIBRATION + EELECTRONIC
Let’s consider a monatomic gas first….(simplest case…)
• for a monatomic gas (e.g. He, Ne or Ar) each atom needs three
coordinates to completely define its position
 an atom has three translational degrees of freedom, each of
which has an energy of ½ kBT : 3 (½kBT ) = (3/2)(R/NA)T
so ETOT = (3/2) RT (no rotation or vibration for monatomic)
U = (3/2) RT
where U is the molar internal energy
(no rotation or vibration for monatomic)
Consider molecules…..a little more complicated…..
• for molecules must consider rotational, vibrational and
translational motion………
• for a molecule with N atoms we need a total of 3N coordinates to
describe its motion
3
2 (linear) or 3 (non-linear)
(3N – 5) linear or (3N – 6) non-linear
translational components
rotational components
vibrational components***
*** it is difficult to count the number of vibrational modes so just
subtract translational and rotational modes…..
….now we can calculate the total energy for atoms and molecules
Mechanical Degrees of Freedom
and the Equipartition of Energy
Translational:
Non–Linear and Linear Molecules:
3 Degrees of Freedom
ie. movement in the x , y and z directions
Rotational:
Non–Linear Molecules:
3 Degrees of Freedom
ie. rotation about the x , y and z axes
Rx , Ry , Rz
Linear Molecules:
2 Degrees of Freedom
ie. rotation about the x and y axes
Rx , Ry
Mechanical Degrees of Freedom
and the Equipartition of Energy
Vibrational:
Non–Linear Molecules with N atoms:
(3N – 6) Degrees of Freedom
ie. (3N – 6) vibrations
Q1 , Q2 , Q3 , … , Q3N–6
Linear Molecules with N atoms:
(3N – 5) Degrees of Freedom
ie. (3N – 5) vibrations
Q1 , Q2 , Q3 , … , Q3N–5
Kinetic Theory of Gases
(Translational Motions)
All Kinetic Energy terms
3
Utrans =   RT
2
___
1
=   M v2 =
2
___
___
___
1
2
2
2
  M ( vx + v y + vz )
2
Therefore:
1
  RT per Degree of Translational Freedom
 2
Same is True for Rotational Motions
All Kinetic Energy terms
1
1
1
Urot =   I1 ω12 +   I2 ω22 +   I3 ω32
2
2
2
where:
N
Ij = Moment of Inertia =  mi ri2
i 1
mi = mass of ith atom
ri = distance of ith atom from Centre of Mass
ωi = angular velocity of rotation (radians per sec)
Therefore:
1
  RT per Degree of Rotational Freedom
2
However Vibrations Have Both Kinetic And Potential Energies
2
Uvib
 1   dx 
1
=   μ   +   k x2
 2   dt 
2
where:
μ = Reduced Mass =
m1m 2
(m1  m 2 )
(diatomic molecule)
k = Force Constant
Therefore:
(1) RT per Degree of Vibrational Freedom
Heat Capacity At Constant Volume
 U 
CV = 


T

V
Monatomic Gases:
No rotations or vibrations – only translations
 U

CV =  trans 
 T  V
=
  3  
 RT 

T  2   V
3
=  R
2
Linear Molecules with N Atoms
3 translational degrees of freedom
2 rotational degrees of freedom
(3N–5) vibrational degrees of freedom
 1  
 1  
Utotal = 3  RT  + 2  RT  + (3N–5)[(1)RT]
 2  
 2  

 5 
= 3N    RT
 2 

 
 U 
 5  
CV =  total  =
3
N


  RT 

T 
 T  V
 2   V

 5 
=  3N     R
 2 

Non-Linear Molecules with N Atoms
3 translational degrees of freedom
3 rotational degrees of freedom
(3N–6) vibrational degrees of freedom
 1  
 1  
Utotal = 3  RT  + 3  RT  + (3N–6)[(1)RT]
 2  
 2  
= [3N–3] RT

 U 
3N  3RT V
CV =  total  =
T
 T  V
=
3N  3 R
Equipartition of Energy for ONE MOLE of GAS
Kind of motion
Translational
No. of modes
Energy per mole
Monatomic Molecule
3
3/2 RT
Polyatomic Molecule
Linear
Non-linear
3
3/2 RT
Rotational
No. of modes
Energy per mole
0
0
2
RT
Vibrational
No. of modes
Energy per mole
0
0
3N – 5
(3N-5)RT*
TOTAL ENERGY
3/2 RT
(3N – (5/2))RT
3
3/2 RT
3
3/2 RT
3N – 6
(3N-6)RT*
(3N-3)RT
* For vibrational energy there are two terms one for PE (½ RT)and one for KE (½ RT)
What does this mean ???
(1) Molecules can store energy………………..
If you heat up a gas the energy provided by the increase
in temperature should be taken up by each of the 3N components.
………………..Molecules can TRANSLATE, ROTATE, VIBRATE, or
EXCITE to store energy
(2) We can now calculate the Total Energy for a monatomic or
polyatomic gas……………..
Example:
Total Energy (U) for 1 mole of a linear diatomic gas is then:
U = (3/2) RT + RT + RT = (7/2) RT
remember you are calculating U using values from previous table
which are for ONE MOLE of gas……..thus this is molar internal
energy
But what we really want to know is……..
how much energy can molecules hide as HEAT at any temperature ?
What is the molecule’s HEAT CAPACITY ?
Heat Capacity is defined as the amount of energy ( U) required to
raise the temperature of a substance by 1 K
Heat Capacity (C) = ΔU / ΔT
At constant volume…. CV = [U / T ] v
(see p. 58 and 66 in text)
Is C = U/ T or is Cv = U/ T ?
ANSWER:
C = U/ T page 90
or expressed in partial derivatives:
CV = [U / T ] v
page 58 & 90
……PLEASE NOTE…..when you use the table of values (p.58) given
to calculate U……these values are per mole of gas……therefore
you are calculating molar internal energy (U)
……..which in turn will give you Cv
Cv = {[(3/2)RT] / T} v = (3/2)R
Please refer p. 58-59 in text for more on this.
EXAMPLES:
For a monatomic gas ….
molar Heat Capacity is Cv = {[(3/2)RT] / T} v = (3/2)R = 12.47 JK-1mol-1
For a diatomic gas…….
molar Heat Capacity is Cv = [ [(7/2)RT] / T ] v = (7/2)R = 29.10 JK-1mol-1
can also calculate for polyatomic……..
To this point………
1) we reviewed the Equipartition of Energy Theorem
2) used this theorem to calculate the total Energy (U)
for atoms or molecules
3) used calculated value for U to calculate Cv
4) compared measured values with “our” calculated
values for Cv
- only see agreement for monatomics not for molecules
Why ?
• to this point we have used classical mechanics to predict
values for Cv
• discrepancies for molecules can be explained using quantum
mechanics…..
According to Quantum Mechanics……….
• Electronic, Vibrational and Rotational energies of a
molecule are quantized
meaning: different molecular energy levels are associated
with each type of motion
Spacing between
energy levels varies
depending on type of
energy
• spacings between electronic energy levels  vibrational  rot.  trans.
• spacings between translational energy levels are very small
….can be treated as a continuum……as in classical mechanics
Why is this important ?
How does this relate to HEAT CAPACITY ?
• when a system absorbs heat…..the energy is used to promote
various kinds of motion
remember: the heat capacity of a system tells us it’s ability to store
energy
• energy may be stored partly in vibrational, rotational motion etc.
…..i.e. the molecules may be promoted to a higher vibrational energy
level or a higher rotational energy level…etc.
• since spacings between rotational energy levels are less than
those between vibrational energy levels (Fig. 3.13)
…………..it’s easier to promote to a higher rotational energy
level than to a higher vibrational energy level
The ratio of the populations of molecules in any two energy
levels (E1 and E2) is given by the Boltzmann Distribution Law:
N2 = e –(E)/(kBT)
N1
………where  E = E2 – E1 (where we assume E2  E1)
kB = Boltzmann Constant = 1.381 x 10-23 JK-1
……N2/N1 is always  1……….meaning fewer molecules in upper level
(at thermal equilibrium at a finite temperature)
N2 = e –(E)/(kBT)
N1
Using the Boltzmann Equation we
can make estimates of populations in
lower and higher energy levels………
If E is large
compared with kBT
molecules will crowd
into lower energy level

E
OK….let’s estimate how many molecules are in lower
versus higher energy level for translational motion………
N2 = e –(E)/(kBT)
N1
…. (E)/(kBT) at 298 K is
10 –37 J
(1.381 x 10-23 JK-1)(298K)
thus
=
N2/N1 
2.4 x 10-16
1
Therefore, for translational motion the number of molecules
in the higher energy level is same as the number in the lower
energy level.
Meaning…….the Energy is not Quantized ………….
• for rotational motion E is also small compared to kBT
so ratio of N2/N1 is close to but smaller than 1
Energies are quantized for rotational (but not translational) motion
• for vibrational motion, spacings between energy levels is large
so E is  kBT…….and ratio of N2/N1 is much smaller than 1
at 298 K most molecules are in lower energy level and only a small
fraction in the higher levels
• for electronic motion…the spacings between levels are so large
that almost all molecules are in lowest energy level at 298 K
What should we understand from this……………..
1) Remember using classical mechanics the equation for Total
Energy (U) was:
U = ETRANSLATION + EROTATION + EVIBRATION + EELECTRONIC
heat capacity = Cv = [U / T ] v
….but when we used this equation the values calculated for Cv did
not match experimental values for Cv (for diatomic or polyatomic
molecules)
2) So we tried to use quantum mechanics to explain this discrepancy
and to calculate new values for Cv
-
from quantum mechanics approach we see that at room
temperature only the translational and rotational motions would
contribute to the heat capacity
• So the equation for Total Energy (U) becomes:
U = Etrans + Erot
= (5/2) RT
Cv = (5/2)R = 20.79 JK-1mol-1
……..this value is close to the experimental value of 21.05 JK-1mol-1
(see Table 3.3 in Chang text)
**this discrepancy means that vibrational motion does make a
small contribution
If we increase the temperature we see that vibrational motions
makes even more of a contribution………
T(K)
298
600
1000
Cv
21.05
23.78
26.56
Sample Problem
What is the total energy of CO2 and NH3 ?
(Calculate using classical mechanics.)
Solution
CO2 is a linear molecule
bond angles = 180 °
Total Energy = (3N – 5/2)RT = (9 – 5/2)RT = 6.5 RT
NH3 is a non-linear molecule H
N
H
bond angles = 109.5 °
H
Total Energy = (3N – 3)RT = (12 – 3)RT = 9 RT
e.g. For H2
• at room temperature only translational motion and rotational
motion contribute to the heat capacity
• at elevated temperatures vibrational energy must be taken
into account and at even higher temperatures electronic
motion must be accounted for
Our Progress to DATE………..
GOALS
(1) to understand how molecules store and use energy
(2) to define heat capacity
(3) to test equipartition of energy theorem using heat capacity
measurements
PROGRESS
(1) molecules store heat or energy by translational, rotational
vibrational and electronic motion
(2) heat capacity is a measure of the ability of a substance to
store energy
(3) must use quantum mechanics to estimate Cv ………
only translational and rotational motion make contribution
at room temperature