Gibbs–Helmholtz Equation: How ΔG Changes with Temperature
The standard Gibbs energy change ΔG° is usually given at 298 K. But reactions run at other temperatures: a furnace, a reactor, a cold room. How does ΔG° change when T changes? The Gibbs–Helmholtz equation answers this using only the enthalpy change. It is a standard result in the thermodynamics sections of IIT-JAM, GATE, CSIR-NET and CUET-PG chemistry, and it connects ΔG, ΔH, ΔS and the equilibrium constant in one line.
The equation
If ΔH is nearly constant over the temperature range, integrate between T1 and T2:
- ΔG1, ΔG2 = Gibbs energy change at T1 and T2 (J mol−1).
- ΔH = enthalpy change, assumed constant (J mol−1).
- T = temperature in kelvin.
Keep ΔG and ΔH in the same unit (both J or both kJ) for the whole calculation. The result is the Gibbs energy of the reaction as written, so a reaction balanced for 2 mol of product has twice the value of the one balanced for 1 mol.
Reading the equation
- If ΔH is negative (exothermic), ΔG/T rises as T rises. The reaction becomes less favourable at higher temperature.
- If ΔH is positive (endothermic), ΔG/T falls as T rises. The reaction becomes more favourable at higher temperature.
This is the same conclusion as Le Chatelier's principle, and the same as the van't Hoff equation for K, because ΔG° = −RT ln K. Putting ΔG°/T = −R ln K into the Gibbs–Helmholtz equation gives d ln K/dT = ΔH°/(RT²).
Step 1: ΔG1/T1 = −32,900 / 298 = −110.40 J K−1.
Step 2: 1/T2 − 1/T1 = 1/500 − 1/298 = 0.002000 − 0.003356 = −0.001356 K−1.
Step 3: ΔH × (−0.001356) = (−92,200) × (−0.001356) = +125.0 J K−1.
Step 4: ΔG2/T2 = −110.40 + 125.0 = +14.59 J K−1.
Step 5: ΔG2 = 14.59 × 500 = +7297 J ≈ +7.30 kJ.
Second route: ΔS = (ΔH − ΔG)/T = (−92,200 + 32,900)/298 = −199.0 J K−1. Then ΔG(500 K) = ΔH − TΔS = −92,200 − 500 × (−199.0) = −92,200 + 99,497 = +7297 J ✓. (Both routes assume constant ΔH and ΔS, so they must agree.)
Answer: ΔG(500 K) ≈ +7.3 kJ. The reaction, which is spontaneous at 298 K, is no longer spontaneous under standard conditions at 500 K.
Step 1: At 298 K: ln K1 = −ΔG1/(RT1) = 32,900 / (8.314 × 298) = 32,900 / 2477.6 = 13.28, so K1 = 5.85 × 105.
Step 2: At 500 K: ln K2 = −7297 / (8.314 × 500) = −7297 / 4157 = −1.755, so K2 = 0.173.
Step 3: Check with the van't Hoff equation: ln K2 = ln K1 − (ΔH/R)(1/T2 − 1/T1) = 13.28 − (−92,200/8.314)(−0.001356) = 13.28 − 15.04 = −1.76 ✓.
Answer: K falls from about 5.9 × 105 to 0.17 as T rises from 298 K to 500 K. The two thermodynamic routes agree.
Step 1: ΔS = (ΔH − ΔG)/T = (178,300 − 130,100)/298 = 48,200/298 = 161.7 J K−1.
Step 2: ΔG = 0 when T = ΔH/ΔS = 178,300 / 161.7 = 1102 K.
Step 3: Check with Gibbs–Helmholtz at 1100 K: ΔG/T = 130,100/298 + 178,300 × (1/1100 − 1/298) = 436.58 + 178,300 × (−0.0024465) = 436.58 − 436.2 = +0.35 J K−1. Then ΔG = 0.35 × 1100 ≈ +0.4 kJ, which is almost zero, as expected just below 1102 K ✓.
Answer: the decomposition becomes spontaneous (under standard conditions) above about 1100 K. An endothermic reaction with ΔS > 0 is favoured by high temperature.
Where it applies, and where it does not
The simple integrated form assumes that ΔH does not change with temperature. Strictly, ΔH changes with T according to ΔCp. Over a range of a few tens of kelvin the error is small. Over several hundred kelvin, a numerical problem will either give ΔCp or tell you to treat ΔH as constant. Read the question before choosing.
In industrial design, this relation shows why a process chooses a compromise temperature. For an exothermic reaction like Example 1, higher T lowers the equilibrium yield (Example 2), but lower T makes the reaction too slow. The thermodynamics gives the limit on yield, and the kinetics decides how fast you get there.
- Mixing J and kJ. ΔG in kJ and ΔH in J gives an answer wrong by a factor of 1000.
- Using °C. T must be in kelvin everywhere, including 1/T.
- Interpolating the wrong quantity. The equation is linear in ΔG/T against 1/T when ΔH is constant. Do not interpolate ΔG itself against 1/T.
- Forgetting the sign convention for 1/T2 − 1/T1. For T2 > T1 this term is negative.
- Ignoring the equation as written. Halving the reaction halves ΔG and ΔH, but K changes to its square root.
Exam relevance
| Question type | Key step |
|---|---|
| Find ΔG at a new temperature from ΔG and ΔH | ΔG2/T2 = ΔG1/T1 + ΔH(1/T2 − 1/T1) |
| Predict the effect of T on spontaneity | Sign of ΔH decides the direction of the change in ΔG/T |
| Find K at a new temperature | Use ΔG = −RT ln K, or the van't Hoff equation |
| Find the changeover temperature | T = ΔH/ΔS when ΔG = 0 |
Check the current official syllabus of your exam for the exact form in which the Gibbs–Helmholtz equation is asked. Do not assume a fixed weightage.
Use the calculator to check ΔG = ΔH − TΔS at each temperature after you have worked the Gibbs–Helmholtz route by hand.
Open the Gibbs Free Energy Calculator →