Gibbs free energy calculator
Enter the enthalpy and entropy change of a reaction and a temperature to get the Gibbs free energy change, whether the reaction runs on its own, the temperature where that changes and the equilibrium constant. Or start from a known equilibrium constant to get the standard Gibbs energy.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Choose to calculate from ΔH and ΔS, or from an equilibrium constant.
- Enter ΔH in kJ/mol and ΔS in J/(mol·K) — the units most tables use — or the value of K.
- Enter the temperature and pick °C, K or °F.
- Read ΔG, whether the reaction is spontaneous now, at which temperatures it is, and K.
- ΔG = ΔH − T × ΔS (T in kelvin, ΔS in kJ/(mol·K))
- K = e^(−ΔG° ÷ RT) and ΔG° = −RT × ln K, R = 8.314462618 J/(mol·K)
- Crossover temperature T = ΔH ÷ ΔS, meaningful when ΔH and ΔS have the same sign
| ΔH |
ΔS |
Spontaneous |
| − |
+ |
at every temperature |
| − |
− |
below ΔH ÷ ΔS |
| + |
+ |
above ΔH ÷ ΔS |
| + |
− |
at no temperature |
Worked example
Ammonia synthesis N₂ + 3 H₂ → 2 NH₃: ΔH = −92.2 kJ/mol, ΔS = −198.7 J/(mol·K), 25 °C = 298.15 K.
TΔS = 298.15 × (−0.1987) = −59.242 kJ/mol, so ΔG = −92.2 + 59.242 = −32.958 kJ/mol — spontaneous. K = e^(32,958 ÷ (8.314462618 × 298.15)) = 594,184. Both ΔH and ΔS are negative, so the reaction stays spontaneous only below 92,200 ÷ 198.7 = 464.02 K (190.87 °C).
From the other side: K = 1000 at 25 °C gives ΔG° = −8.314462618 × 298.15 × ln 1000 = −17.124 kJ/mol.
Common mistakes
- Mixing kJ and J. ΔS is usually listed in J/(mol·K); forgetting to divide by 1000 makes the entropy term a thousand times too big. The calculator takes ΔS in J and converts it.
- Using °C in the formula. T must be absolute; 25 °C is 298.15 K.
- Confusing ΔG with ΔG°. The K relation uses the standard value ΔG°; the actual ΔG under non-standard concentrations is ΔG° + RT ln Q.
Frequently asked questions
How do you calculate Gibbs free energy?
ΔG = ΔH − T × ΔS, with T in kelvin. Tables give ΔH in kJ/mol and ΔS in J/(mol·K), so ΔS is divided by 1000 first. For ammonia synthesis at 25 °C: −92.2 − 298.15 × (−0.1987) = −32.958 kJ/mol.
What does a negative ΔG mean?
The reaction is spontaneous at that temperature: it can proceed without outside work. It says nothing about speed — a spontaneous reaction can still be extremely slow if its activation energy is high. A positive ΔG means the reverse reaction is the spontaneous one.
At what temperature does a reaction become spontaneous?
When ΔH and ΔS have the same sign, ΔG changes sign at T = ΔH ÷ ΔS. If both are negative the reaction is spontaneous below that temperature; if both are positive, above it. Decomposing limestone (ΔH = 178.3 kJ/mol, ΔS = 160.6 J/(mol·K)) becomes spontaneous above about 1110 K.
How are ΔG° and the equilibrium constant related?
ΔG° = −R × T × ln K, so K = e^(−ΔG°/RT) with R = 8.314462618 J/(mol·K). K = 1 means ΔG° = 0; every 5.7 kJ/mol more negative multiplies K by about ten at 25 °C.
Are ΔH and ΔS assumed constant?
Yes. The calculator uses the values you enter at any temperature, which is the usual approximation. Over a wide temperature range both change a little, so the crossover temperature is an estimate.