# Chapter 8  Thermodynamics

## 8.1  Entropy, S

Perspectives

1. Information Theory
 S = kB lnΩ     (1)
• kB = R/NA = 8.314 J/mol · K/6.022 × 10−23 mol−1 = 1.38 × 10−23 J/K = Boltzmann’s constant = energy per 1 K
• Ω = number of ways a system can arrange itself on the microscopic level but still result in the same macroscopic state; number of microstates that lead to the same observed macrostate as defined by T, P, V, and n; microstate = specification of where the energy is among the components of the system
• Entropy is how much energy, or rather how many ways energy can be "randomly" distributed microscopically to achieve the macroscopic state at some temperature, T
• Entropy is the amount of microscopic information we do not know about a system because we are macroscopic observers
2. Thermodynamics
Δ S =
 q T
only true for a reversible, isothermal change     (2)
• a Change in Entropy is equal to the heat exchanged divided by the temperature; units are still J/K.
• reversible = the path taken has infinitesimally small step sizes such that the process is always in equilibrium with itself and surroundings; a process that proceeds along the same path in the forward and reverse directions; theoretical ideality; the path obeys the state function
• isothermal = the temperature is constant throughout the process
• a Change in Entropy is a change in the amount of information you don’t know about a changing macroscopic system which is proportional to the amount of heat exchanged at some constant temperature, T

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 0 K
 q T
dT

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

• S has units of J/mol·K so it is an extensive property, just like Δ H; it depends on how much of the chemical you have.
• S is a state function; i.e. Δ S = SfinalSinitial ≠ Δ Spath
• means standard state = unspecified temperature (determined by the table you’re looking at, our book uses 25C), pure gases are at 1 atm, pure condensed phases have Psurroundings = 1 atm, and solutions have concentration of 1 M.
 Δ Srxn∘ = Sproducts∘ − Sreactants∘     (4)

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

1. Phase
• solids = vibrate in place, no translation or rotation = not many places to put energy = fewer microstates
• liquids = vibrate, translate and rotate slowly
• gases = vibrate, translate and rotate slowly or quickly = lots of places to put energy = more microstates  Ssolid∘ < Sliquid∘ < Sgas∘     (5)
 SH2O (l)∘ = 70.0 J/mol · K  and  SH2O (g)∘ = 188.8 J/mol · K
2. Molar Mass
• Heavier molecules have closer spaced translational energy levels = more microstates  Slighter∘ < Sheavier∘     (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
3. Molecular complexity
• N atoms in a molecule have 3N degrees of freedom = ways to move/use energy; x, y, and z for each atom
• 3 of those 3N are for translation of the molecule
• 2 of the 3N for diatomic molecules and 3 of the 3N for polyatomic molecules are for rotation of the molecule
• The rest, 3N−5 for diatomics or 3N−6 for polyatomics, are for vibration; e.g. H2O has 3(3)−6 = 3 vibrational modes
• More vibrational modes means more microstates = higher entropy
 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 N2O 44.0128 3(3)−6=3 220.0 NO2 46.0055 3(3)−6=3 240.1 N2O4 92.011 3(6)−6=12 304.4
4. Allotropes = various elemental forms of an element (not a difference of phase)
 Allotrope S∘ (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

• H2O (l) ⇋ H2O (g); Δ Srxn > 0; Δ Srxn = 188.8 − 70.0 = 118.8 J/K
• NH4I (s) ⇋ NH3 (g) + HI (g); Δ Srxn > 0; Δ Srxn = 192.8 + 206.6 − 117.0 = 282.4 J/K
• N2 (g) + 2 O2 (g) ⇋ 2 NO2 (g); Δ Srxn < 0; Δ Srxn = 2(240.1) − 191.6 − 2(205.2) = −121.8 J/K

## 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.

• Entropy is not conserved
• Spontaneous = occurs without external influence, product favored ≠ immediate
• Non-spontaneous = reactant favored
• Have to worry about the universe, not just system/reaction.  Δ Suniverse = Δ Ssystem/rxn + Δ Ssurroundings
• Surroundings ≈ Cosmic Heat Bath with infinite heat capacity and constant temperature = reversible heat exchanges
Δ Ssurroundings =  qsurroundings Tsurroundings
=  −qsystem/rxn Tsurroundings
=Δ P = 0 −Δ Hsystem/rxn Tsurroundings
(7)
• Reactions are not reversible. That we start with only reactants in non-equilibrium with their products indicates it is not a reversible reaction since reversible processes are always at equilibrium.
Δ Ssystem/rxn >  qsystem/rxn T
only true for a irreversible, isothermal, spontaneous change     (8)
• Must use the standard molar entropies to calculate Δ Srxn.  Δ Srxn∘ = Sproducts∘ − Sreactants∘     (9)

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)
 spontaneous Δ Suniverse > 0 Δ Grxn < 0 non-spontaneous Δ Suniverse < 0 Δ Grxn > 0 equilibrium Δ Suniverse = 0 Δ 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 H Δ S Δ H − T 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 = 25∘C)     (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

• Reverse a reaction, reverse the sign of Δ Grxn
• Multiply a reaction by a factor, multiply Δ Grxn by the same factor

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

 spontaneous Δ Suniverse∘ > 0 Δ Grxn∘ < 0 K > 1 product favored non-spontaneous Δ Suniverse∘ < 0 Δ Grxn∘ > 0 K < 1 reactant 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 equilibrium Q < K Δ Grxn < 0 reaction goes backward to reach equilibrium Q > K Δ Grxn > 0 reaction is at equilibrium Q = K Δ 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, 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 25∘C)

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)