Gibbs free energy calculator

How do you calculate Gibbs free energy?

ΔG = ΔH − TΔS, with T in kelvins (OpenStax Chemistry 2e). Enter ΔH in kJ/mol and ΔS in J/(mol·K); the calculator divides ΔS by 1,000 so the units match.

What does the sign of ΔG mean?

Negative ΔG: the process is spontaneous at that temperature. Positive: it is not spontaneous (the reverse is). Zero: the system is at equilibrium.

Example

For ΔH = −92.2 kJ/mol and ΔS = −198.7 J/(mol·K) at 298 K: ΔG = −92.2 − 298 × (−0.1987) = −33.0 kJ/mol — spontaneous.

At what temperature does spontaneity change?

Where ΔG = 0: T = ΔH ÷ ΔS. In the example, 92,200 ÷ 198.7 = 464 K. Above it ΔG turns positive, because both ΔH and ΔS are negative.

Does spontaneous mean fast?

No — ΔG says whether a process can happen on its own, not how quickly. Some spontaneous reactions are extremely slow.

How do the signs of ΔH and ΔS decide the temperature behavior?

If ΔH is negative and ΔS positive, ΔG is negative at every temperature. If ΔH is positive and ΔS negative, it is positive at every temperature. If both are negative, the process is spontaneous at low temperature; if both are positive, at high temperature, and the switch happens at T = ΔH ÷ ΔS (OpenStax Chemistry 2e, section 16.4). The calculator’s table lists ΔG from 200 K to 1000 K so you can see the crossover.

What is the difference between ΔG and ΔG°?

ΔG° applies when every reactant and product is in its standard state; ΔG applies to the actual mixture and equals ΔG° + RT ln Q. This calculator computes ΔG° from ΔH° and ΔS°. OpenStax Example 16.12: for 2NH₃ → 3H₂ + N₂ with ΔG° = 33.0 kJ/mol at 25 °C and partial pressures of 0.870 atm N₂, 0.250 atm H₂ and 12.9 atm NH₃, Q = 8.17 × 10⁻⁵ and ΔG = 9.67 kJ/mol, so the reaction is still not spontaneous. Use the equilibrium constant calculator for Q.

The equation and its units

Gibbs free energy is G = H − TS, and at constant temperature and pressure the change is ΔG = ΔH − TΔS (OpenStax Chemistry 2e, section 16.4). A negative ΔG means the process is spontaneous, a positive one that it is not, and zero that the system is at equilibrium. Enthalpy changes are normally tabulated in kJ, but entropy changes in J/K, so the TΔS term has to be divided by 1,000 before it is subtracted; the calculator does this for you. The inputs ΔH° and ΔS° are values at 298 K and are treated as constant when you change the temperature, which is an approximation. ΔH° can come from the enthalpy of reaction calculator.

Worked example

OpenStax Example 16.7 evaluates H₂O(l) → H₂O(g) at 298 K. The enthalpy of formation is −285.83 kJ/mol for the liquid and −241.82 kJ/mol for the gas, so ΔH° = 44.01 kJ; the standard entropies are 70.0 and 188.8 J/(mol·K), so ΔS° = 118.8 J/K. At 298 K, ΔG° = 44.01 − 298 × 0.1188 = 8.6 kJ: positive, so evaporation at 298 K and standard pressure is not spontaneous. The crossover is T = ΔH ÷ ΔS = 44.01 × 1,000 ÷ 118.8 = 370.5 K (97.3 °C), the book’s estimate of the boiling point in Example 16.11; the accepted value is 373.2 K, and the small gap comes from treating ΔH and ΔS as independent of temperature. At 400 K, ΔG° is −3.5 kJ, so the vapor is favored.

Common mistakes

The classic one is skipping the 1,000: using ΔS in J/K against ΔH in kJ gives ΔG = −35,358 kJ in the example instead of 8.6 kJ. Next is entering the temperature in Celsius; the calculator lets you choose the unit, but in a hand calculation T must be in kelvin. Third, treating a negative ΔG as “fast”: it says nothing about rate. Fourth, using ΔG° for conditions that are not standard, where ΔG = ΔG° + RT ln Q can even have the opposite sign. Last, forgetting to multiply ΔH and ΔS by the number of moles when the equation is written for more than one mole.

Sources: OpenStax Chemistry 2e, section 16.4 “Free Energy” (ΔG = ΔH − TΔS, sign combinations, Examples 16.7, 16.11 and 16.12). Related: enthalpy of reaction calculator, equilibrium constant calculator, Nernst equation calculator and activation energy calculator.