Temperature Dependence of Spontaneity and Equilibrium
Because ΔG = ΔH − TΔS, the spontaneity of a reaction can change with temperature, and the switching temperature is T = ΔH/ΔS; the equilibrium constant is related to the standard free energy by ΔG° = −RT ln K. In brief: the decomposition of CaCO3 (ΔH = +178 kJ, ΔS = +161 J/K) becomes spontaneous above about 1100 K, and a reaction with ΔG° = −10 kJ/mol at 298 K has an equilibrium constant of about 57.
The temperature dependence of ΔG explains why limestone has to be heated in a kiln, why ice melts when the weather warms up and why the Haber process uses a compromise temperature. The link between ΔG° and K connects thermodynamics with the equilibrium topics. This article shows how to find switching temperatures, how to calculate K from ΔG°, and how K changes with temperature.
How do you find the temperature at which spontaneity changes?
When ΔH and ΔS have the same sign, ΔG changes sign at the temperature where ΔG = 0:
T = ΔH / ΔS
This assumes that ΔH and ΔS are independent of temperature, which is a reasonable approximation over a moderate range because both change in a similar way. For a reaction with positive ΔH and ΔS, the process is spontaneous above T (the entropy term TΔS exceeds ΔH). For negative ΔH and ΔS, the process is spontaneous below T. For a phase change, the crossover temperature is the equilibrium temperature: the melting or boiling point.
Crossover temperatures for some processes
| Process | ΔH (kJ) | ΔS (J/K) | T = ΔH/ΔS | Spontaneous |
|---|---|---|---|---|
| H2O(s) → H2O(l) | +6.01 | +22.0 | 273 K (0 °C) | Above T |
| H2O(l) → H2O(g) | +40.7 | +109 | 373 K (100 °C) | Above T |
| CaCO3(s) → CaO(s) + CO2(g) | +178 | +161 | about 1100 K | Above T |
| N2 + 3H2 → 2NH3 | −92.2 | −198.7 | 464 K (191 °C) | Below T |
The boiling point of water is calculated as 40,700 / 109 = 373 K, in accordance with the experimental value. For the decomposition of limestone, 178,000 / 161 = 1106 K, which is about 830 °C. This is why lime kilns operate at temperatures of 900–1000 °C, so that the decomposition goes quickly, and the CO2 escapes.
How is ΔG° related to the equilibrium constant?
The standard free energy change is the difference in free energy between pure reactants and pure products in their standard states. A reaction mixture that is not at equilibrium has a free energy change ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. At equilibrium, ΔG = 0 and Q = K, so:
ΔG° = −RT ln K, or K = e−ΔG°/RT
A negative ΔG° gives K > 1, with products favored at equilibrium. A positive ΔG° gives K < 1. If ΔG° = 0, then K = 1. At 298.15 K, RT = 8.314 × 298.15 = 2479 J/mol, or 2.479 kJ/mol. Note that ΔG° is about the standard state and is not the same as ΔG at equilibrium, which is always zero.
Relationship between ΔG° and K
| ΔG° (kJ/mol) | ln K | K | At equilibrium |
|---|---|---|---|
| −40 | +16.1 | 1.0 × 107 | Almost only products |
| −20 | +8.07 | 3.2 × 103 | Mostly products |
| −10 | +4.03 | 57 | Products favored |
| 0 | 0 | 1 | Equal amounts |
| +10 | −4.03 | 0.018 | Reactants favored |
| +40 | −16.1 | 1.0 × 10−7 | Almost only reactants |
A change of just 5.7 kJ/mol in ΔG° at 298 K changes K by a factor of 10, so the equilibrium constant is very sensitive to ΔG°.
How does K change with temperature?
Combining ΔG° = −RT ln K with ΔG° = ΔH° − TΔS° gives:
ln K = −ΔH°/(RT) + ΔS°/R
A plot of ln K against 1/T is a straight line with a slope of −ΔH°/R (the van ‘t Hoff equation). For an exothermic reaction (ΔH° < 0), K decreases when the temperature rises, and for an endothermic reaction K increases. This is Le Châtelier’s principle seen from the standpoint of thermodynamics. Between two temperatures: ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1).
Worked examples
Example 1: switching temperature of limestone decomposition
CaCO3(s) → CaO(s) + CO2(g), ΔH° = +178 kJ and ΔS° = +161 J/K. T = 178,000 J / 161 J/K = 1106 K, or 833 °C. Below this temperature ΔG° > 0 and the carbonate is stable at 1 atm of CO2.
Example 2: K for ammonia synthesis at 298 K
ΔG° = −33.0 kJ. ln K = 33,000 / (8.314 × 298.15) = 13.31, so K = e13.31 = 6.0 × 105. At room temperature the equilibrium lies far to the products, but the reaction is extremely slow without a catalyst.
Example 3: K at a higher temperature
Assume that ΔH° and ΔS° are constant. At 500 K, ΔG° = −92.2 − (500)(−0.1987) = −92.2 + 99.35 = +7.1 kJ. ln K = −7100 / (8.314 × 500) = −1.71, so K = 0.18. The equilibrium constant has fallen from 6.0 × 105 to 0.18, as expected for an exothermic reaction.
Example 4: the compromise in the Haber process
Low temperature favors the yield of ammonia, but the rate is then too small. Industrial plants therefore use about 400–500 °C, an iron catalyst and pressures of 150–250 atm, since high pressure shifts the equilibrium toward the side with fewer moles of gas. The ammonia is removed continuously, and unreacted gases are recycled.
How were these relationships established?
Jacobus van ‘t Hoff derived the equation for the temperature dependence of the equilibrium constant in 1884, as part of his work on chemical dynamics, for which he received the first Nobel Prize in Chemistry in 1901. The link between free energy and the equilibrium constant came from Gibbs and Helmholtz. The industrial synthesis of ammonia was developed by Fritz Haber and Carl Bosch in the early 20th century, with the first plant starting production in 1913. It made large-scale fertilizer production possible, and thermodynamic analysis was central in the choice of conditions.
Which mistakes are common?
- Using Celsius in T = ΔH/ΔS: T must be in kelvins.
- Unit mismatch: ΔH in kJ and ΔS in J/K must be converted before dividing.
- Assuming that K does not change with temperature: K depends on temperature, and only a catalyst leaves it unchanged.
- Thinking that ΔG° = 0 at equilibrium: it is ΔG that is zero; ΔG° is related to K.
- Applying the crossover formula when the signs of ΔH and ΔS differ: the reaction is then spontaneous at all or at no temperatures.