Gibbs Free Energy Calculator

Results

ΔH and ΔS are both positive: the process is spontaneous above the crossover temperature and not spontaneous below it.

If ΔH and ΔS are standard values (ΔH°, ΔS°), then ΔG° = ΔG and the equilibrium constant is K = 0.03127 (log₁₀ K = -1.505) at this temperature.

How to Use the Gibbs Free Energy Calculator

1. 1. Pick the mode. Use "ΔG from ΔH and ΔS" for spontaneity and temperature questions, or "ΔG° and the equilibrium constant" to link free energy to K and Q.
2. 2. Choose the unknown and enter the rest. Enthalpy and free energy can be in kJ/mol, J/mol, kcal/mol or cal/mol; entropy in J/(mol·K), kJ/(mol·K) or cal/(mol·K); temperature in K, °C or °F. Mixed kJ and J entries are converted for you.
3. 3. Read the verdict. The result says whether ΔG is negative (spontaneous), positive (not spontaneous) or zero (at equilibrium), gives the crossover temperature where it changes, and plots the line.
4. 4. Follow the working. Every conversion and substitution is written out, with the exact gas constant R = 8.314 462 618 J/(mol·K) in the equilibrium mode.

The Equations

ΔG = ΔH − TΔS

T = ΔH / ΔS  (where ΔG = 0)

ΔG° = −RT ln K

ΔG = ΔG° + RT ln Q

ΔG is the Gibbs free energy change at constant temperature and pressure. A negative ΔG means the process can proceed spontaneously, a positive ΔG means it cannot without outside work, and ΔG = 0 means equilibrium. Spontaneous in this sense says nothing about speed: diamond turning to graphite is spontaneous and takes geological time. T must be the absolute temperature in kelvin, and the calculator converts Celsius or Fahrenheit and rejects anything at or below 0 K.

When Is a Reaction Spontaneous?

This is the same classification OpenStax Chemistry 2e gives in its free-energy chapter. The crossover temperature separates the "low" and "high" regimes; it is only meaningful when ΔH and ΔS have the same sign. It assumes ΔH and ΔS stay constant with temperature, which is a good approximation over a modest range but not across a large one — a real reaction's ΔH and ΔS drift, so the answer is an estimate.

Worked Examples

Water vaporizing at 298.15 K (the default inputs). With ΔH° = 44.01 kJ/mol and ΔS° = 118.8 J/(mol·K), ΔG = 44 010 − 298.15 × 118.8 = 8 589.8 J/mol, which is 8.590 kJ/mol. It is positive, so liquid water does not boil spontaneously at 25 °C and 1 bar. The crossover is 44 010 ÷ 118.8 = 370.5 K (97.3 °C), close to water's real boiling point of 373.15 K; the gap exists because ΔH and ΔS are not exactly constant.

Ice melting at 263.15 K (−10 °C). ΔH = 6.01 kJ/mol and ΔS = 22.0 J/(mol·K) give ΔG = 6 010 − 263.15 × 22.0 = +0.221 kJ/mol. It is positive, so ice does not melt at −10 °C; it would at any temperature above 273.2 K.

K from ΔG°. For ΔG° = 55.7 kJ/mol at 298.15 K, ln K = −55 700 ÷ (8.314 462 618 × 298.15) = −22.47, so K = 1.74 × 10⁻¹⁰ — the solubility product of silver chloride, as in OpenStax's worked example. The reverse also works: K = 1.0 × 10⁵ at 298.15 K means ΔG° = −RT ln K = −28.5 kJ/mol.

ΔG away from equilibrium. With ΔG° = −20.0 kJ/mol and a reaction quotient Q = 50.0 at 298.15 K, ΔG = −20 000 + 2 478.96 × ln 50.0 = −10.3 kJ/mol. It is negative, so the reaction runs forward until Q rises to K.

Common Mistakes

• Subtracting kJ from J. ΔH is usually quoted in kJ/mol and ΔS in J/(mol·K). Convert one before computing TΔS; the calculator does it from the unit menus. Forgetting is the textbook error that makes TΔS look a thousand times too small.
• Using Celsius for T. T in ΔG = ΔH − TΔS is in kelvin. 25 °C is 298.15 K.
• Mixing up ΔG and ΔG°. ΔG° applies to standard conditions (1 bar for gases, 1 mol/L for solutes) and gives K. ΔG applies to the actual mixture and depends on Q.
• Reading "spontaneous" as "fast". ΔG says whether a process can go, not how quickly. Rates belong to kinetics; see the Arrhenius Equation Calculator.
• Trusting the crossover temperature too far. It assumes constant ΔH and ΔS. Use it as an estimate, especially far from 298 K.

Frequently Asked Questions

What is the Gibbs free energy formula?

ΔG = ΔH − TΔS, where ΔH is the enthalpy change, T the absolute temperature in kelvin, and ΔS the entropy change. At standard conditions it is written ΔG° = ΔH° − TΔS°. Both energy terms must be in the same unit before you subtract.

How do I know if a reaction is spontaneous from ΔG?

If ΔG is negative the reaction is spontaneous in the forward direction; if it is positive the reverse direction is spontaneous; if it is zero the system is at equilibrium. Spontaneous means thermodynamically favored, not fast.

How do I find the temperature at which a reaction becomes spontaneous?

Set ΔG = 0 and solve: T = ΔH ÷ ΔS. Use the sign table: when both are positive the reaction is spontaneous above that temperature, when both are negative it is spontaneous below it. If the signs differ there is no such temperature.

How is ΔG° related to the equilibrium constant?

ΔG° = −RT ln K, so K = e^(−ΔG°/RT). A negative ΔG° gives K greater than 1 and favors products; a positive ΔG° gives K less than 1 and favors reactants. Every 5.7 kJ/mol at 298 K changes K by about a factor of 10.

What does ΔG = ΔG° + RT ln Q mean?

It gives the free energy change for the mixture you actually have. When Q is smaller than K, RT ln Q is negative enough to make ΔG negative and the reaction runs forward; when Q is larger than K it runs in reverse; at Q = K, ΔG = 0.

Why is my ΔG off by a factor of 1000?

That nearly always means kJ and J were mixed. ΔH of 44.01 kJ/mol is 44 010 J/mol; if ΔS is in J/(mol·K), convert ΔH to joules (or ΔS to kJ/(mol·K)) first. Choose the units in the menus and the calculator converts for you.