For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = -92.2 kJ and ΔS° = -198.7 J/K. (a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous? (c) At what temperature does ΔG = 0?
Master Gibbs free energy, predict reaction spontaneity, and understand the relationship between ΔG, ΔH, ΔS, and temperature.
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Spontaneity Criterion
ΔG predicts spontaneity:
ΔG
Process
ΔG < 0
Spontaneous (favorable)
ΔG = 0
At equilibrium
ΔG > 0
Non-spontaneous (unfavorable)
Important: Spontaneous ≠ fast
Thermodynamics (ΔG) vs kinetics (rate)
Diamonds → graphite: spontaneous but infinitely slow
Temperature Dependence
Four cases from ΔG = ΔH - TΔS:
Case 1: ΔH < 0, ΔS > 0
ΔG < 0 at all temperatures
Spontaneous always
Example: Combustion reactions
Case 2: ΔH > 0, ΔS < 0
ΔG > 0 at all temperatures
Never spontaneous
Example: Reverse of combustion
Case 3: ΔH < 0, ΔS < 0
ΔG < 0 at low T
ΔG > 0 at high T
Spontaneous only when cold
Case 4: ΔH > 0, ΔS > 0
ΔG > 0 at low T
ΔG < 0 at high T
Spontaneous only when hot
Example: Melting, vaporization
Summary Table
ΔH
ΔS
ΔG
Spontaneous?
-
+
Always -
Always
+
-
Always +
Never
-
-
- at low T, + at high T
Low T only
+
+
+ at low T, - at high T
High T only
Calculating ΔG°_{rxn}
Method 1: From ΔH° and ΔS°
ΔG°=ΔH°−TΔS°
Watch units! Convert ΔS° from J/K to kJ/K
Method 2: From ΔG°_f values
ΔG°rxn=∑nΔG°f(products)−∑nΔG°f(reactants)
Like ΔH°_f:
Elements in standard state: ΔG°_f = 0
Tabulated for compounds
Standard Free Energy of Formation
ΔG°_f: Free energy change to form 1 mole from elements (25°C, 1 atm)
Examples:
H₂(g): ΔG°_f = 0
H₂O(l): ΔG°_f = -237.1 kJ/mol
CO₂(g): ΔG°_f = -394.4 kJ/mol
Temperature and Equilibrium
At equilibrium: ΔG = 0
0=ΔH−TeqΔS
Teq=ΔSΔH
This gives transition temperature:
Below: one direction spontaneous
Above: reverse direction spontaneous
At T_eq: equilibrium (both directions equal)
Free Energy and Equilibrium Constant
Relationship:
ΔG°=−RTlnK
Or:
ΔG=ΔG°+RTlnQ
Where:
R = 8.314 J/(mol·K)
K = equilibrium constant
Q = reaction quotient
Interpretation:
ΔG° < 0: K > 1 (products favored)
ΔG° = 0: K = 1 (equal amounts)
ΔG° > 0: K < 1 (reactants favored)
Coupling Reactions
Non-spontaneous reaction can be driven by spontaneous one:
For the vaporization of water at 100°C: H₂O(l) → H₂O(g), ΔH°{vap} = 40.7 kJ/mol. (a) Calculate ΔS°{vap}. (b) Calculate ΔG° at 90°C, 100°C, and 110°C. (c) Explain results.
💡 Show Solution
Given:
Process: H₂O(l) → H₂O(g)
ΔH°_{vap} = 40.7 kJ/mol
Normal boiling point: 100°C = 373 K
(a) Calculate ΔS°_{vap}
At boiling point: ΔG = 0 (equilibrium)
ΔG=ΔH−TΔS=0
ΔS=TΔH
At T = 373 K:
ΔS°vap=373 K40.7 kJ/mol
ΔS°vap=0.109 kJ/(mol\cdotpK)=109 J/(mol\cdotpK)
Answer (a): ΔS°_{vap} = 109 J/(mol·K)
(b) Calculate ΔG° at different temperatures
Use: ΔG° = ΔH° - TΔS°
ΔH° = 40.7 kJ/mol
ΔS° = 0.109 kJ/(mol·K)
At 90°C (363 K):ΔG°=40.7−(363)(0.109)ΔG°=40.7−39.6
At 100°C (373 K):ΔG°=40.7−(373)(0.109)ΔG°=40.7−40.7
At 110°C (383 K):ΔG°=40.7−(383)(0.109)ΔG°=40.7−41.7
Answer (b):
90°C: ΔG° = +1.1 kJ/mol
100°C: ΔG° = 0 kJ/mol
110°C: ΔG° = -1.0 kJ/mol
(c) Explain results
At 90°C (below boiling point):
ΔG° = +1.1 kJ (positive)
Non-spontaneous
Liquid is stable phase
Water won't boil at 1 atm
At 100°C (boiling point):
ΔG° = 0
Equilibrium
Liquid ⇌ gas
Both phases coexist at 1 atm
Normal boiling point!
At 110°C (above boiling point):
ΔG° = -1.0 kJ (negative)
Spontaneous
Gas is stable phase
Water vaporizes spontaneously
Can't maintain liquid at 1 atm
General principle:
Phase transitions:
ΔH > 0, ΔS > 0 (liquid → gas)
Low T: ΔG > 0 (liquid favored)
T = T_{boiling}: ΔG = 0 (equilibrium)
High T: ΔG < 0 (gas favored)
Boiling point is where ΔG = 0 for vaporization at 1 atm
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