What is the Gibbs free energy change?
Enter the enthalpy change ΔH, the entropy change ΔS and the temperature. The calculator gives the Gibbs free energy change ΔG and whether the process is spontaneous, or works out ΔH, ΔS or T from the other three.
- Answer
- ΔG = 8.6076 kJ/mol
ΔG = 8.6076 kJ/mol: not spontaneous (ΔG > 0).
- Spontaneous?
- not spontaneous (ΔG > 0)
- ΔG (kJ/mol)
- 8.6076
- ΔH (kJ/mol)
- 44.01
- ΔS (J/(mol·K))
- 118.8
- T (K)
- 298
- TΔS (kJ/mol)
- 35.4024
- Spontaneous when
- Above 370.455 K (ΔH > 0 and ΔS > 0)
- T where ΔG = 0 (K)
- 370.455
- ln K
- −3.47401
- K
- 0.0309924
Answer: ΔG = 8.6076 kJ/mol. ΔG = 8.6076 kJ/mol: not spontaneous (ΔG > 0).
How to calculate
Solves ΔG = ΔH − TΔS for the Gibbs free energy change, the enthalpy change, the entropy change or the temperature, says whether the process is spontaneous, and gives the temperature where ΔG = 0 and the equilibrium constant.
Example with the default inputs (Find ΔG, ΔH, kJ/mol 44.01, ΔS, J/(mol·K) 118.8, Temperature 298 K): ΔG = 8.6076 kJ/mol: not spontaneous (ΔG > 0).
Method: ΔG = ΔH − T × ΔS ÷ 1000 (ΔH and ΔG in kJ/mol, ΔS in J/(mol·K), T in K); the temperature where ΔG = 0 is T = 1000 × ΔH ÷ ΔS; K = e^(−1000 × ΔG ÷ (R × T)) with R = 8.31446261815324 J/(mol·K).
- ΔH and ΔS are taken as constant over the temperature range.
- Spontaneous means ΔG < 0 at constant temperature and pressure; it says nothing about how fast the process is.
- K is shown only when ΔG is the standard change ΔG°; the page cannot tell which you typed.
Worked examples
Each example is checked against the calculator on every build.
- Find ΔG, ΔH, kJ/mol 44.01, ΔS, J/(mol·K) 118.8, Temperature 298 K gives ΔG (kJ/mol) 8.6076, TΔS (kJ/mol) 35.4024, Spontaneous? not spontaneous (ΔG > 0).Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.7 Evaluation of ΔG° Change from ΔH° and ΔS°: TΔS° = 35.4 kJ, ΔG° = 8.6 kJ
- Find ΔG, ΔH, kJ/mol 90.46, ΔS, J/(mol·K) 107.4, Temperature 298 K gives ΔG (kJ/mol) 58.4548.Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.8 (b): ΔG° = 90.46 − 298 × 0.1074 = 58.45 kJ/mol
- Find T, ΔH, kJ/mol 44.01, ΔS, J/(mol·K) 118.8, ΔG, kJ/mol 0 gives T (K) 370.454545, Answer T = 370.455 K (97.3045 °C).Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.11 Equilibrium Temperature for a Phase Change: 370.5 K (97.3 °C)
- Find ΔG, ΔH, kJ/mol 0, ΔS, J/(mol·K) 0, Temperature 298.2 K gives ΔG (kJ/mol) 0, K 1, Spontaneous? at equilibrium (ΔG = 0).Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Table 16.4: K = 1 when ΔG° = 0
- Find ΔH, ΔG, kJ/mol 55.7, ΔS, J/(mol·K) 0, Temperature 298.2 K gives ΔH (kJ/mol) 55.7, K 0.Source: OpenStax, Chemistry 2e, §16.4 Free Energy (ΔG = ΔH − TΔS; ΔG° = −RT ln K), https://openstax.org/books/chemistry-2e/pages/16-4-free-energy, Example 16.13: Ksp = e^(−55,700 ÷ (8.314 × 298.15)) = 1.74 × 10⁻¹⁰ (the page uses R = 8.31446261815324)
How it works
Formula (OpenStax §16.4).
ΔG = ΔH − T × ΔS ÷ 1000
with ΔG and ΔH in kJ/mol, ΔS in J/(mol·K) and T in kelvins (°C + 273.15; °F converts exactly). Pick the value to find; the page rearranges the same formula:
- ΔH = ΔG + T × ΔS ÷ 1000
- ΔS = 1000 × (ΔH − ΔG) ÷ T
- T = 1000 × (ΔH − ΔG) ÷ ΔS
Typed values are exact decimals; each result is rounded once, to 6 significant figures with trailing zeros dropped. A value below 0.000001 or of 10²¹ or more is written with a power of ten, in the sentences as in the number rows (ΔG = 9.99 × 10⁻⁷ kJ/mol). A temperature worked out is shown in kelvins and in °C (T − 273.15).
Spontaneous? ΔG below 0: spontaneous. Above 0: not spontaneous. Exactly 0: at equilibrium.
When it is spontaneous (signs of ΔH and ΔS).
| ΔH | ΔS | Spontaneous |
|---|---|---|
| below 0 (or 0) | above 0 (or 0), not both 0 | at every temperature |
| above 0 (or 0) | below 0 (or 0), not both 0 | at no temperature |
| above 0 | above 0 | above T = 1000 × ΔH ÷ ΔS |
| below 0 | below 0 | below T = 1000 × ΔH ÷ ΔS |
ΔH and ΔS are taken as constant with temperature.
Equilibrium constant. If ΔG is the standard change ΔG°, ln K = −1000 × ΔG ÷ (R × T) and K = e^(ln K), with R = N_A × k = 8.31446261815324 J/(mol·K) exactly. OpenStax rounds R to 8.314, which moves K in the third significant figure at most. ln K is worked out exactly and rounded once. K is left out when it is too large to hold as a number (above about 1.8 × 10³⁰⁸, ln K above about 709.78) or too small to hold with all its digits (below about 2.2 × 10⁻³⁰⁸, ln K below about −708.40), so a K that is shown has its 6 significant figures right. ln K is shown with every answer.
Rules the page checks.
- ΔH and ΔG from −1,000,000 to 1,000,000 kJ/mol; ΔS from −1,000,000 to 1,000,000 J/(mol·K); T above 0 K and at most 100,000 K.
- Finding T: ΔS must not be 0, and the result must be above 0 K and at most 100,000 K.
- A ΔH or ΔS worked out beyond ±1,000,000 gives no answer, and so does a result or a ln K that is not 0 but too close to 0 to hold as a number (its nearest number is 0: below about 2.5 × 10⁻³²⁴).
Worked examples by hand
Water boiling at 298 K (OpenStax Example 16.7). TΔS = 298 × 118.8 ÷ 1000 = 35.4024 kJ/mol. ΔG = 44.01 − 35.4024 = 8.6076 kJ/mol: not spontaneous.
Yellow HgO decomposing at 298 K (Example 16.8 b). ΔG = 90.46 − 298 × 107.4 ÷ 1000 = 90.46 − 32.0052 = 58.4548 kJ/mol.
Boiling point of water (Example 16.11). With ΔG = 0: T = 1000 × 44.01 ÷ 118.8 = 370.455 K (97.3045 °C). OpenStax gives the accepted value as 373.2 K (100.0 °C); the estimate takes ΔH° and ΔS° as constant with temperature.
ΔG = 0 at 298.15 K. ln K = 0, so K = 1 (Table 16.4).
AgCl at 298.15 K (Example 16.13). ΔG° = 55.7 kJ/mol: ln K = −55,700 ÷ (8.31446261815324 × 298.15) = −22.4691, K = 1.745 × 10⁻¹⁰ (OpenStax, with R = 8.314: 1.74 × 10⁻¹⁰).
Other questions people ask
How do I calculate ΔG?
Use ΔG = ΔH − TΔS with T in kelvins. ΔS is usually given in J/(mol·K) and ΔH in kJ/mol, so divide TΔS by 1000 first. For water boiling at 298 K: 44.01 − 298 × 118.8 ÷ 1000 = 44.01 − 35.40 = 8.61 kJ/mol.
How do I know if a reaction is spontaneous?
At constant temperature and pressure a process is spontaneous when ΔG is negative, not spontaneous when it is positive, and at equilibrium when it is 0. Water boiling at 25 °C has ΔG = +8.6 kJ/mol, so it does not happen on its own at that temperature.
At what temperature does a reaction become spontaneous?
When ΔH and ΔS have the same sign, ΔG changes sign at T = ΔH ÷ ΔS (with ΔS in kJ/(mol·K)). If both are positive the process is spontaneous above that temperature; if both are negative, below it. If ΔH is negative and ΔS positive it is spontaneous at every temperature, and with the opposite signs at none.
How is ΔG related to the equilibrium constant?
For the standard change ΔG°, ΔG° = −RT ln K, so K = e^(−ΔG° ÷ RT) with ΔG° in J/mol and R = 8.314 J/(mol·K). ΔG° below 0 gives K above 1 (products favored); ΔG° = 0 gives K = 1.
Does spontaneous mean fast?
No. ΔG says whether a process can happen on its own, not how quickly. Diamond turning into graphite is spontaneous at room conditions, but its rate is extremely slow.
Why must the temperature be in kelvins?
ΔG = ΔH − TΔS needs the absolute temperature: at 0 °C the TΔS term is not zero. The page reads °C and °F too and converts them (K = °C + 273.15).