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Chapter 8  Thermodynamics

8.1  Entropy, S

Perspectives

  1. Information Theory
      S = kB lnΩ     (1)
  2. Thermodynamics
      Δ S = 
    q
    T
      only true for a reversible, isothermal change     (2)

Connecting the two definitions is not a part of this course as we will not generally ever calculate Ω:

Δ S = Sfinal − Sinitial = kB lnΩfinal − kB lnΩinitial = 
q
T
    (3)

8.2  The 3rd Law of Thermodynamics

The entropy of a perfect crystal at a temperature of absolute zero (T = 0 K) is zero (S0 K = 0 J/K).

0  J/K = kB lnΩ 
Ω = 1 

There is only 1 microstate for a perfect crystal at T = 0 K. There is no ambiguity about where the energy is because there is no energy to distribute (randomize); there is only 1 microstate = 1 way to arrange the energy in the perfect crystal.

8.3  Standard Molar Entropy, S

We don’t carry out chemistry at 0 K, more like 298 K. We need the entropy of chemicals at 298 K, S298 K. We are not going to calculate Ω at 298 K because that involves Quantum Statistical Mechanics. We could calculate how much heat, q, is absorbed to increase our chemicals from 0 K to 298 K using the heat capacities, Cs, and enthalpies of phase transitions, Δ Hfus/vap, but that’s harder than it looks and involves calculus:

Δ SK298 K = S298 K − SK = S298 K − 0 = S298 K = 
298 K


K
q
T
dT 

We will understand the tabulated standard molar entropies, S = Δ S0 K298 K, given in the back of the book.

Δ Srxn = Sproducts − Sreactants     (4)

How do different molecules store/use heat in their standard states?

  1. Phase
  2. Molar Mass
  3. Molecular complexity
  4. Allotropes = various elemental forms of an element (not a difference of phase)
    AllotropeS (J/mol·K)
    C (s, graphite)5.7
    C (s, diamond)2.4
    C60 (s)426.0
    P (s, white)41.1
    P (s, red)22.8
    P (g)163.2
    P2 (g)218.1
    P4 (g)280.0
    S (g)167.8
    S2 (g)228.2
    S8 (g)430.9
  5. Solutions
     S (J/mol·K)
    KClO3 (s)143.1
    KClO3 (aq)265.7

Reaction examples

8.4  The 2nd Law of Thermodynamics

For any spontaneous process the entropy of the universe increases. Δ Suniverse > 0 for spontaneous processes. Δ Suniverse < 0 for non-spontaneous processes.

Δ Suniverse can be calculated using only system/rxn variables while still accounting for both system/rxn and surroundings.

Δ Suniverse = Δ Srxn − 
Δ Hrxn
T
    (10)

8.5  Gibbs Free Energy, G

G = H − TS     (11)

The change in Gibbs Free Energy at constant T

Δ G = Δ H − T Δ S − S Δ T = Δ H − T Δ S     (12)

If we divide Δ G by −T and apply the formula to our reaction:

−Δ Grxn
T
 = 
−Δ Hrxn
T
 + Δ Srxn = Δ Suniverse 
Δ Grxn = −T Δ Suniverse     (13)
spontaneousΔ Suniverse > 0Δ Grxn < 0
non-spontaneousΔ Suniverse < 0Δ Grxn > 0
equilibriumΔ Suniverse = 0Δ Grxn = 0

The last line is required since reactions never cease even at equilibrium, and their equilibrium state must not generate entropy.

Reversible processes (which are always at equilibrium) always have Δ Suniverse = 0 J/K.

Δ Grxn = Δ Hrxn − T Δ Srxn     (14)

H2O (l) ⇋ H2O (g), using tabulated data at 25C.

Δ Srxn = 188.8 − 70.0 = 118.8  J/K = 0.1188  kJ/K 
Δ Hrxn = −241.8 − −285.8 = +44  kJ 
Δ Grxn = Δ Hrxn − T Δ Srxn = 44  kJ − T(0.1188  kJ/K

For T = 0 K, Δ Grxn = Δ Hrxn = +44 kJ, meaning the reaction is non-spontaneous at that temperature.

For T = 298 K, Δ Grxn = 44 − (298.15)(0.1188) = 8.58 kJ, vaporization of water is non-spontaneous at 25C.

At what temperature does this reaction become spontaneous?

Δ Grxn must go from being positive to negative, and so therefore must pass through zero, Δ Grxn = 0 kJ.

Phase changes are also isothermal, reversible processes meaning at the phase change, Δ Grxn = 0 kJ.

Δ Grxn = 0  kJ = 44  kJ − T(0.1188  kJ/K
T = 
Δ Hrxn
Δ Srxn
 = 
44  kJ
0.1188  kJ/K
 = 370.4  K = 97.2

Complete vaporization of a liquid will occur above the boiling point = spontaneous.

Above the boiling point, the entropy lost by the surroundings is offset by the entropy gain of the vaporization.

Incomplete vaporization of a liquid will occur below the boiling point = non-spontaneous.

Below the boiling point, the entropy lost by the surroundings is not offset by the entropy gain of the vaporization.
How do Δ Hrxn and Δ Srxn affect the spontaneity?

Δ HΔ SΔ HT Δ S 
Δ Hrxn > 0 (Δ Ssurr < 0)Δ Srxn > 0(+) − T(+)Δ G < 0 at high T, spontaneous at high T
Δ Hrxn > 0 (Δ Ssurr < 0)Δ Srxn < 0(+) − T(−)Δ G > 0, spontaneous at no T
Δ Hrxn < 0 (Δ Ssurr > 0)Δ Srxn > 0(−) − T(+)Δ G < 0, spontaneous at all T
Δ Hrxn < 0 (Δ Ssurr > 0)Δ Srxn < 0(−) − T(−)Δ G < 0 at low T, spontaneous at low T

When Δ H and Δ S have the same sign, indicating competing changes in entropy between system and surroundings, spontaneity will have a dependence on T.

You can approximately calculate the transition temperature by setting Δ Grxn = 0 kJ.

If Δ H and Δ S have different signs, regardless of the temperature, the reaction is always spontaneous when both surroundings and system are increasing in entropy (Δ H=−, Δ S=+) or always non-spontaneous when surroundings and system are decreasing in entropy (Δ H=+, Δ S=−).

How spontaneous (or non-spontaneous) would still be affected by T, but it wouldn’t switch from being one to the other.

If you try to calculate a transition temperature, you end up with a negative Kelvin which is impossible.

———————————————————————–

8.6  Gibbs Free Energy of Formation, Δ Gf

The change in free energy for the formation of 1 mole of compound from its elements in their most stable elemental forms at the given T (assumed 25C.

Δ Grxn = Δ Gf,products − Δ Gf,reactants = Δ Hrxn − T Δ Srxn  for T data was tabulated at (298.15 K = 25C)     (15)

Δ Gf = 0.0 kJ/mol for the most stable elemental forms, e.g. magnesium as a solid, Mg (s).

Mg (s) ⇋ Mg (s) Δ Hrxn = 0.0 kJ/mol and Δ Srxn = 0.0 kJ/mol, therefore Δ Gf = 0.0 kJ/mol

Mg (s) ⇋ Mg (g) Δ Hrxn = 147.1 kJ/mol and Δ Srxn = 0.1486 kJ/K·mol, therefore Δ Gf = 112.5 kJ/mol
1/2 O2 (g) + H2 (g) ⇋ H2O (l)

Δ Grxn = Δ Gf,H2O (l) = Δ Hf,H2O (l) − Δ Hf,H2 − 
1
2
 Δ Hf,O2 − (25 + 273.15)(SH2O − SH2 − 
1
2
 SO2
Δ Gf,H2O (l) = −285.8 − 0 − 
1
2
0 − (298.15)(0.070 − 0.1307 − 
1
2
 0.2052) = −237.1  kJ/mol 

At 25C, you could calculate Δ Grxn without Δ Hrxn or Δ Srxn, explicitly.

CH4 (g) + 2 O2 (g) ⇋ 2 H2O (l) + CO2 (g)

Δ Grxn = (2)Δ Gf,H2O (l) + Δ Gf,CO2 − Δ Gf,CH4 − (2)Δ Gf,O2 
Δ Grxn = 2(−237.1) + (−394.4) − (−50.5) − 2(0.0) = −818.1  kJ

Manipulating Reactions and Δ Grxn

Gibbs Free Energy (Gibbs Energy/Free Energy) is the maximum, theoretical amount of energy available to do non-expansion work (inefficiencies will cause wactual < wmax.

wmax = Δ Grxn     (16)

If we reorganize the equation to solve for Δ Hrxn we see that the enthalpy (the total heat) has two components: 1) what you can do work with 2) what is lost to entropy:

Δ Hrxn = Δ Grxn + T Δ Srxn 

Processes with positive Δ Grxn require work to be supplied for the process to occur.

8.7  Equilibrium

spontaneousΔ Suniverse > 0Δ Grxn < 0K > 1product favored
non-spontaneousΔ Suniverse < 0Δ Grxn > 0K < 1reactant favored
Δ Grxn = −RT lnK          K = e−Δ Grxn/RT     (17)

K is not specific to any particular equilibrium constant, though one may consider this K to be representative of Kp.

Don’t worry about converting between Kc and Kp, just report K.

If we solve this equation for when Δ Grxn = 0 (as we did before to solve for the transition temperature between spontaneous and non-spontaneous) you get K = 1.

The transition from non-spontaneous, K < 1, to non-spontaneous, K > 1 occurs at the temperature where Δ Grxn = 0 which is the temperature at which K = 1.

While Δ Grxn is switching from positive to negative, K is switching from less than 1 to greater than 1.

If we move everything over to the left hand side we find the sum is equal to zero:

Δ Grxn + RT lnK = 0 

What does that zero represent? It represents the slope of the plot of G versus Q at Q = K.

At equilibrium the slope is zero.

What do we get when QK? Slope ≠ 0.

If we replace K with Q we get something other than zero, we get non-standard Gibbs free energy, Δ Grxn.

Δ Grxn = Δ Grxn + RT lnQ     (18)

Non-standard Gibbs free energy is the slope of the function of G versus Q at point Q.

K<1, reactant favored reaction showing how Δ Grxn > 0 at Q=1>K, indicating that the reaction needs to go downhill in the backward direction because of the positive slope, i.e. backward making more reactants, to get to equilibrium from when Q=1

The equilibrium state is a maximum entropy state because of the 2nd Law of Thermodynamics.

In order for a reaction to "cease" macroscopically upon reaching and sustaining an equilibrium state, the continued forward and reverse reactions must cancel each other out entropically for a net zero change in entropy.

Δ Grxn tells you the same information as Q.

reaction goes forward to reach equilibriumQ < KΔ Grxn < 0
reaction goes backward to reach equilibriumQ > KΔ Grxn > 0
reaction is at equilibriumQ = KΔ Grxn = 0

If everything is at standard conditions (1 atm or 1 M), what is Q? Q = 1.

Δ Grxn = Δ Grxn + RT ln(1) 
Δ Grxn = Δ Grxn + 0 

From the middle ground, Q=1 (where [products]≈[reactants]), Δ Grxn = Δ Grxn, tells you whether the reaction is reactant or product favored based on this slope.

A negative slope indicates a product favored reaction as equilibrium is "downhill" in the forward direction.

A positive slope indicates a reactant favored reaction as equilibrium is "downhill" in the backward direction.

K=1, neither reactant or product favored reaction showing how Δ Grxn = 0 at Q=1=K, indicating that the reaction is at equilibrium when Q=1 (this scenario is VERY rare, and weird; basically you’re imagining a reaction at the temperature at which it switches between being reactant and product favored, because K=1 is what K has to pass through in order to go from K<1 to K>1, AND we’re also imagining Q=1 where there’s "equal" amount of reactants and products)

K>1, product favored reaction showing how Δ Grxn < 0 at Q=1<K, indicating that the reaction needs to go downhill in the forward direction because of the negative slope, i.e. forward making more products, to get to equilibrium from when Q=1

We earlier calculated the boiling point of water using Δ Grxn = Δ HrxnT Δ Srxn = 0 which we now understand to be the temperature at which K = 1, T=370.4 K = 97.2C.

Calculate K for T = 25C = 298.15 K, remembering that Δ Hrxn = 44 kJ/mol and Δ Srxn = 0.1188 kJ/mol·K.

Δ Grxn = 44 − (298.15)(0.1188) = 8.58  kJ/mol 
K = e−8.58/RT = 0.0314 

Vaporization at this temperature is non-spontaneous, or rather reactant favored.

Realize that K=Kp=PH2O (the vapor pressure) for the vaporization of water, H2O (l) ⇋ H2O (g).

Kp = PH2O = 0.0314  atm = 23.85  mmHg (23.76 actual value at 25C) 

Lubbock has a dry climate. We’ll assume a relative humidity of 50% at 25C in order to calculate the partial pressure of H2O in our atmosphere.

PH2O
PH2O
 = 
PH2O
0.0314
 = 0.5 
PH2O = 0.0157  atm 

What we have calculated is Lubbock’s Qp = PH2O = 0.0157.

What is non-standard Gibbs free energy for Lubbock?

Δ Grxn = Grxn + RT lnQ = 8.58 + (8.314 × 10−3)(298.15  K) ln(0.0157) = −1.72  kJ/mol 

This means that the vaporization of H2O in Lubbock will happen because Qp < KP as confirmed by Δ Grxn < 0.

This vaporization will go in the forward direction up until Qp = Kp = 0.0314 and Δ Grxn = 0.

However, the vaporization reaction as a whole, at 25C, is considered non-spontaneous because the equilibrium is reactant favored, Kp = 0.0314 < 1 and Δ Grxn = 8.58 kJ > 0.

T determines Δ Grxn. Δ Grxn determines K. Q at T determines Δ Grxn.

8.8  van’t Hoff equation

Δ Grxn = −RT lnK 
Δ Hrxn − T Δ Srxn = −RT lnK 
Δ Hrxn
RT
 − 
T Δ Srxn
RT
 = lnK 

This gives the van’t Hoff equation.

Δ Hrxn
RT
 + 
Δ Srxn
R
 = lnK     (19)
Δ Hrxn
R



1
T



Δ Srxn
R
 = lnK 
mx + b = y 

Setting the slope of the plot of lnK versus 1/T gives us the 2 point form of the van’t Hoff equation.

m = −
Δ Hrxn
R
 = 
rise
run
 = 
y2 − y1
x2 − x1
 = 
lnK2 − lnK1
1
T2
 − 
1
T1
 
ln


K2
K1



= −
Δ Hrxn
R
 


1
T2
 − 
1
T1
 


    (20)

Which, for the vaporization reaction where K = P, leads to the Clausius-Clapeyron Equation:

ln


P2
P1



= −
Δ Hvap
R
 


1
T2
 − 
1
T1
 


    (21)

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