|S = kB 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 = Sfinal − Sinitial = kB lnΩfinal − kB lnΩinitial =|
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.
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 S0 K298 K = S298 K − S0 K = S298 K − 0 = S298 K =||∫|
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?
|Ssolid∘ < Sliquid∘ < Sgas∘ (5)|
|SH2O (l)∘ = 70.0 J/mol · K and SH2O (g)∘ = 188.8 J/mol · K|
|Slighter∘ < Sheavier∘ (6)|
|gas||MM (g/mol)||S∘ (J/mol·K)|
|gas||MM (g/mol)||Vibrational Modes||S∘ (J/mol·K)|
|C (s, graphite)||5.7|
|C (s, diamond)||2.4|
|P (s, white)||41.1|
|P (s, red)||22.8|
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 = Δ Ssystem/rxn + Δ Ssurroundings|
|Δ Ssurroundings =|
|=Δ P = 0=|
|Δ Ssystem/rxn >|
|only true for a irreversible, isothermal, spontaneous change (8)|
|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∘ −|
|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 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 25∘C.
|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 25∘C.
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)|
|= 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 = 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.
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 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∘ −|
|Δ Hf,O2∘ − (25∘ + 273.15)(SH2O∘ − SH2∘ −|
|Δ Gf,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 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.
|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 Q ≠ K? 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 Hrxn∘ − T Δ Srxn∘ = 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 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 25∘C in order to calculate the partial pressure of H2O in our atmosphere.
|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 25∘C, 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.
|Δ Grxn∘ = −RT lnK|
|Δ Hrxn∘ − T Δ Srxn∘ = −RT lnK|
This gives the van’t Hoff equation.
|= lnK (19)|
|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 = −|
Which, for the vaporization reaction where K = P∘, leads to the Clausius-Clapeyron Equation: