Perspectives
S = k_{B} lnΩ (1) 
Δ S = 
 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 = S_{final} − S_{initial} = k_{B} lnΩ_{final} − k_{B} lnΩ_{initial} = 
 (3) 
The entropy of a perfect crystal at a temperature of absolute zero (T = 0 K) is zero (S^{0 K} = 0 J/K).
0 J/K = k_{B} 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.
We don’t carry out chemistry at 0 K, more like 298 K. We need the entropy of chemicals at 298 K, S^{298 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, C_{s}, and enthalpies of phase transitions, Δ H_{fus/vap}, but that’s harder than it looks and involves calculus:
Δ S_{0 K}^{298 K} = S^{298 K} − S^{0 K} = S^{298 K} − 0 = S^{298 K} =  ∫ 

 dT 
We will understand the tabulated standard molar entropies, S^{∘} = Δ S_{0 K}^{298 K}, given in the back of the book.
Δ S_{rxn}^{∘} = S_{products}^{∘} − S_{reactants}^{∘} (4) 
How do different molecules store/use heat in their standard states?
S_{solid}^{∘} < S_{liquid}^{∘} < S_{gas}^{∘} (5) 
S_{H2O (l)}^{∘} = 70.0 J/mol · K and S_{H2O (g)}^{∘} = 188.8 J/mol · K 
S_{lighter}^{∘} < S_{heavier}^{∘} (6) 
gas  MM (g/mol)  S^{∘} (J/mol·K) 
He  4.003  126.2 
Ne  20.18  146.1 
Ar  39.95  154.8 
Kr  83.80  163.8 
Xe  131.29  169.4 
gas  MM (g/mol)  Vibrational Modes  S^{∘} (J/mol·K) 
Ar  39.948  3(1)−3=0  154.8 
NO  30.006  3(2)−5=1  210.8 
N_{2}O  44.0128  3(3)−6=3  220.0 
NO_{2}  46.0055  3(3)−6=3  240.1 
N_{2}O_{4}  92.011  3(6)−6=12  304.4 
Allotrope  S^{∘} (J/mol·K) 
C (s, graphite)  5.7 
C (s, diamond)  2.4 
C_{60} (s)  426.0 
P (s, white)  41.1 
P (s, red)  22.8 
P (g)  163.2 
P_{2} (g)  218.1 
P_{4} (g)  280.0 
S (g)  167.8 
S_{2} (g)  228.2 
S_{8} (g)  430.9 
S^{∘} (J/mol·K)  
KClO_{3} (s)  143.1 
KClO_{3} (aq)  265.7 
Reaction examples
For any spontaneous process the entropy of the universe increases. Δ S_{universe} > 0 for spontaneous processes. Δ S_{universe} < 0 for nonspontaneous processes.
Δ S_{universe} = Δ S_{system/rxn} + Δ S_{surroundings} 
Δ S_{surroundings} = 
 = 
 =^{Δ P = 0}= 
 (7) 
Δ S_{system/rxn} > 
 only true for a irreversible, isothermal, spontaneous change (8) 
Δ S_{rxn}^{∘} = S_{products}^{∘} − S_{reactants}^{∘} (9) 
Δ S_{universe} can be calculated using only system/rxn variables while still accounting for both system/rxn and surroundings.
Δ S_{universe}^{∘} = Δ S_{rxn}^{∘} − 
 (10) 
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 S_{rxn} = Δ S_{universe} 
Δ G_{rxn} = −T Δ S_{universe} (13) 
spontaneous  Δ S_{universe} > 0  Δ G_{rxn} < 0 
nonspontaneous  Δ S_{universe} < 0  Δ G_{rxn} > 0 
equilibrium  Δ S_{universe} = 0  Δ G_{rxn} = 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 S_{universe} = 0 J/K.
Δ G_{rxn} = Δ H_{rxn} − T Δ S_{rxn} (14) 
H_{2}O (l) ⇋ H_{2}O (g), using tabulated data at 25^{∘}C.
Δ S_{rxn}^{∘} = 188.8 − 70.0 = 118.8 J/K = 0.1188 kJ/K 
Δ H_{rxn}^{∘} = −241.8 − −285.8 = +44 kJ 
Δ G_{rxn}^{∘} = Δ H_{rxn}^{∘} − T Δ S_{rxn}^{∘} = 44 kJ − T(0.1188 kJ/K) 
For T = 0 K, Δ G_{rxn}^{∘} = Δ H_{rxn} = +44 kJ, meaning the reaction is nonspontaneous at that temperature.
For T = 298 K, Δ G_{rxn}^{∘} = 44 − (298.15)(0.1188) = 8.58 kJ, vaporization of water is nonspontaneous at 25^{∘}C.
At what temperature does this reaction become spontaneous?
Δ G_{rxn}^{∘} must go from being positive to negative, and so therefore must pass through zero, Δ G_{rxn}^{∘} = 0 kJ.
Phase changes are also isothermal, reversible processes meaning at the phase change, Δ G_{rxn}^{∘} = 0 kJ.
Δ G_{rxn}^{∘} = 0 kJ = 44 kJ − T(0.1188 kJ/K) 
T = 
 = 
 = 370.4 K = 97.2^{∘}C 
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 = nonspontaneous.
Below the boiling point, the entropy lost by the surroundings is not offset by the entropy gain of the vaporization.
How do Δ H_{rxn} and Δ S_{rxn} affect the spontaneity?
Δ H  Δ S  Δ H − T Δ S  
Δ H_{rxn} > 0 (Δ S_{surr} < 0)  Δ S_{rxn} > 0  (+) − T(+)  Δ G < 0 at high T, spontaneous at high T 
Δ H_{rxn} > 0 (Δ S_{surr} < 0)  Δ S_{rxn} < 0  (+) − T(−)  Δ G > 0, spontaneous at no T 
Δ H_{rxn} < 0 (Δ S_{surr} > 0)  Δ S_{rxn} > 0  (−) − T(+)  Δ G < 0, spontaneous at all T 
Δ H_{rxn} < 0 (Δ S_{surr} > 0)  Δ S_{rxn} < 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 G_{rxn}^{∘} = 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 nonspontaneous when surroundings and system are decreasing in entropy (Δ H=+, Δ S=−).
How spontaneous (or nonspontaneous) 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.
———————————————————————–
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 25^{∘}C.
Δ G_{rxn}^{∘} = Δ G_{f,products}^{∘} − Δ G_{f,reactants}^{∘} = Δ H_{rxn}^{∘} − T Δ S_{rxn}^{∘} for T data was tabulated at (298.15 K = 25^{∘}C) (15) 
Δ G_{f} = 0.0 kJ/mol for the most stable elemental forms, e.g. magnesium as a solid, Mg (s).
Mg (s) ⇋ Mg (s) Δ H_{rxn}^{∘} = 0.0 kJ/mol and Δ S_{rxn}^{∘} = 0.0 kJ/mol, therefore Δ G_{f}^{∘} = 0.0 kJ/mol
Mg (s) ⇋ Mg (g) Δ H_{rxn}^{∘} = 147.1 kJ/mol and Δ S_{rxn}^{∘} = 0.1486 kJ/K·mol, therefore Δ G_{f}^{∘} = 112.5 kJ/mol
1/2 O_{2} (g) + H_{2} (g) ⇋ H_{2}O (l)
Δ G_{rxn}^{∘} = Δ G_{f,H2O (l)}^{∘} = Δ H_{f,H2O (l)}^{∘} − Δ H_{f,H2}^{∘} − 
 Δ H_{f,O2}^{∘} − (25^{∘} + 273.15)(S_{H2O}^{∘} − S_{H2}^{∘} − 
 S_{O2}^{∘}) 
Δ G_{f,H2O (l)}^{∘} = −285.8 − 0 − 
 0 − (298.15)(0.070 − 0.1307 − 
 0.2052) = −237.1 kJ/mol 
At 25^{∘}C, you could calculate Δ G_{rxn}^{∘} without Δ H_{rxn}^{∘} or Δ S_{rxn}^{∘}, explicitly.
CH_{4} (g) + 2 O_{2} (g) ⇋ 2 H_{2}O (l) + CO_{2} (g)
Δ G_{rxn}^{∘} = (2)Δ G_{f,H2O (l)}^{∘} + Δ G_{f,CO2}^{∘} − Δ G_{f,CH4}^{∘} − (2)Δ G_{f,O2}^{∘} 
Δ G_{rxn}^{∘} = 2(−237.1) + (−394.4) − (−50.5) − 2(0.0) = −818.1 kJ 
Manipulating Reactions and Δ G_{rxn}^{∘}
Gibbs Free Energy (Gibbs Energy/Free Energy) is the maximum, theoretical amount of energy available to do nonexpansion work (inefficiencies will cause w_{actual} < w_{max}.
w_{max} = Δ G_{rxn}^{∘} (16) 
If we reorganize the equation to solve for Δ H_{rxn}^{∘} 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 H_{rxn}^{∘} = Δ G_{rxn}^{∘} + T Δ S_{rxn}^{∘} 
Processes with positive Δ G_{rxn}^{∘} require work to be supplied for the process to occur.
spontaneous  Δ S_{universe}^{∘} > 0  Δ G_{rxn}^{∘} < 0  K > 1  product favored 
nonspontaneous  Δ S_{universe}^{∘} < 0  Δ G_{rxn}^{∘} > 0  K < 1  reactant favored 
Δ G_{rxn}^{∘} = −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 K_{p}.
Don’t worry about converting between K_{c} and K_{p}, just report K.
If we solve this equation for when Δ G_{rxn}^{∘} = 0 (as we did before to solve for the transition temperature between spontaneous and nonspontaneous) you get K = 1.
The transition from nonspontaneous, K < 1, to nonspontaneous, K > 1 occurs at the temperature where Δ G_{rxn}^{∘} = 0 which is the temperature at which K = 1.
While Δ G_{rxn}^{∘} 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 G_{rxn}^{∘} + 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 Q ≠ K? Slope ≠ 0.
If we replace K with Q we get something other than zero, we get nonstandard Gibbs free energy, Δ G_{rxn}.
Δ G_{rxn} = Δ G_{rxn}^{∘} + RT lnQ (18) 
Nonstandard Gibbs free energy is the slope of the function of G versus Q at point Q.
K<1, reactant favored reaction showing how Δ G_{rxn}^{∘} > 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 G_{rxn} tells you the same information as Q.
reaction goes forward to reach equilibrium  Q < K  Δ G_{rxn} < 0 
reaction goes backward to reach equilibrium  Q > K  Δ G_{rxn} > 0 
reaction is at equilibrium  Q = K  Δ G_{rxn} = 0 
If everything is at standard conditions (1 atm or 1 M), what is Q? Q = 1.
Δ G_{rxn} = Δ G_{rxn}^{∘} + RT ln(1) 
Δ G_{rxn} = Δ G_{rxn}^{∘} + 0 
From the middle ground, Q=1 (where [products]≈[reactants]), Δ G_{rxn} = Δ G_{rxn}^{∘}, 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, product favored reaction showing how Δ G_{rxn}^{∘} < 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 G_{rxn}^{∘} = Δ H_{rxn}^{∘} − T Δ S_{rxn}^{∘} = 0 which we now understand to be the temperature at which K = 1, T=370.4 K = 97.2^{∘}C.
Calculate K for T = 25^{∘}C = 298.15 K, remembering that Δ H_{rxn}^{∘} = 44 kJ/mol and Δ S_{rxn}^{∘} = 0.1188 kJ/mol·K.
Δ G_{rxn}^{∘} = 44 − (298.15)(0.1188) = 8.58 kJ/mol 
K = e^{−8.58/RT} = 0.0314 
Vaporization at this temperature is nonspontaneous, or rather reactant favored.
Realize that K=K_{p}=P_{H2O}^{∘} (the vapor pressure) for the vaporization of water, H_{2}O (l) ⇋ H_{2}O (g).
K_{p} = P_{H2O}^{∘} = 0.0314 atm = 23.85 mmHg (23.76 actual value at 25^{∘}C) 
Lubbock has a dry climate. We’ll assume a relative humidity of 50% at 25^{∘}C in order to calculate the partial pressure of H_{2}O in our atmosphere.
 = 
 = 0.5 
P_{H2O} = 0.0157 atm 
What we have calculated is Lubbock’s Q_{p} = P_{H2O} = 0.0157.
What is nonstandard Gibbs free energy for Lubbock?
Δ G_{rxn} = G_{rxn}^{∘} + RT lnQ = 8.58 + (8.314 × 10^{−3})(298.15 K) ln(0.0157) = −1.72 kJ/mol 
This means that the vaporization of H_{2}O in Lubbock will happen because Q_{p} < K_{P} as confirmed by Δ G_{rxn} < 0.
This vaporization will go in the forward direction up until Q_{p} = K_{p} = 0.0314 and Δ G_{rxn} = 0.
However, the vaporization reaction as a whole, at 25^{∘}C, is considered nonspontaneous because the equilibrium is reactant favored, K_{p} = 0.0314 < 1 and Δ G_{rxn}^{∘} = 8.58 kJ > 0.
T determines Δ G_{rxn}^{∘}. Δ G_{rxn}^{∘} determines K. Q at T determines Δ G_{rxn}.
Δ G_{rxn}^{∘} = −RT lnK 
Δ H_{rxn}^{∘} − T Δ S_{rxn}^{∘} = −RT lnK 
 − 
 = lnK 
This gives the van’t Hoff equation.
− 
 + 
 = lnK (19) 
− 
 ⎛ ⎜ ⎜ ⎝ 
 ⎞ ⎟ ⎟ ⎠  + 
 = 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 = − 
 = 
 = 
 = 

ln  ⎛ ⎜ ⎜ ⎝ 
 ⎞ ⎟ ⎟ ⎠  = − 
 ⎛ ⎜ ⎜ ⎝ 
 − 
 ⎞ ⎟ ⎟ ⎠  (20) 
Which, for the vaporization reaction where K = P^{∘}, leads to the ClausiusClapeyron Equation:
ln  ⎛ ⎜ ⎜ ⎝ 
 ⎞ ⎟ ⎟ ⎠  = − 
 ⎛ ⎜ ⎜ ⎝ 
 − 
 ⎞ ⎟ ⎟ ⎠  (21) 