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Gibbs Free Energy Equation Calculator

ΔG = ΔH − TΔS and ΔG° = −RT ln K, solved step by step
Find ΔG for ΔH = -92.2 kJ/mol, ΔS = -198.7 J/(mol·K) at 298 K
At what temperature does a reaction with ΔH = +178 kJ/mol and ΔS = +161 J/(mol·K) become spontaneous?
Find the equilibrium constant K when ΔG° = -33.0 kJ/mol at 298 K
Is a reaction with ΔH < 0 and ΔS > 0 ever non-spontaneous?

The Gibbs Free Energy Equation

Gibbs free energy combines a reaction's heat and its disorder into one number that decides whether it proceeds on its own:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Symbols and SI units:

  • ΔG\Delta G — change in Gibbs free energy, joules per mole (J/mol); usually quoted in kJ/mol
  • ΔH\Delta H — enthalpy change, J/mol (kJ/mol in tables)
  • TT — absolute temperature, kelvin (K)
  • ΔS\Delta S — entropy change, joules per mole per kelvin, J/(mol·K)

Reading the sign: ΔG<0\Delta G < 0 means spontaneous in the forward direction, ΔG>0\Delta G > 0 means non-spontaneous (the reverse is spontaneous), and ΔG=0\Delta G = 0 means the system is at equilibrium.

When it applies: a process at constant temperature and pressure. At constant volume the analogous quantity is the Helmholtz free energy ΔA=ΔU−TΔS\Delta A = \Delta U - T\Delta S.

The assumption people forget: ΔH\Delta H arrives in kJ/mol while ΔS\Delta S arrives in J/(mol·K). Convert one of them before subtracting — a factor of 10001000 is the single most common error on this equation.

Crossover Temperature and the Equilibrium Constant

Because TT multiplies only the entropy term, the sign of ΔG\Delta G can flip with temperature. Setting ΔG=0\Delta G = 0 gives the crossover temperature:

T=ΔHΔST = \frac{\Delta H}{\Delta S}

Four cases follow directly:

ΔH\Delta HΔS\Delta SBehaviour
−-++spontaneous at every temperature
++−-never spontaneous
−-−-spontaneous only below T=ΔH/ΔST = \Delta H/\Delta S
++++spontaneous only above T=ΔH/ΔST = \Delta H/\Delta S

Under standard conditions, ΔG∘\Delta G^\circ also fixes the equilibrium constant:

ΔG∘=−RTln⁡K⟺K=e−ΔG∘/(RT)\Delta G^\circ = -RT\ln K \quad\Longleftrightarrow\quad K = e^{-\Delta G^\circ/(RT)}

with R=8.314R = 8.314 J/(mol·K) and KK dimensionless. Away from standard conditions, ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT\ln Q.

The assumption people forget: ΔG∘\Delta G^\circ must be in joules per mole before dividing by RTRT, and TT in kelvin.

Common Mistakes to Avoid

  • Mixing kJ and J — ΔS\Delta S in J/(mol·K) times TT gives J/mol, which cannot be subtracted from a ΔH\Delta H left in kJ/mol.
  • Using Celsius for TT — the equation needs kelvin, so add 273.15273.15 first. At 2525 °C, T=298.15T = 298.15 K.
  • Reading spontaneous as fast — ΔG\Delta G says nothing about rate. Diamond turning to graphite has ΔG<0\Delta G < 0 and takes geological time.
  • Confusing ΔG\Delta G with ΔG∘\Delta G^\circ — the standard value applies at unit activities only; use ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT\ln Q elsewhere.
  • Forgetting the minus sign in ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K — dropping it inverts KK.
  • Assuming ΔH\Delta H and ΔS\Delta S are temperature-independent — a good approximation over modest ranges, not over hundreds of kelvin.

Examples

Step 1: Convert the entropy term to kJ: ΔS=−198.7 J/(mol\cdotpK)=−0.1987 kJ/(mol\cdotpK)\Delta S = -198.7\ \text{J/(mol·K)} = -0.1987\ \text{kJ/(mol·K)}
Step 2: TΔS=(298 K)(−0.1987 kJ/(mol\cdotpK))=−59.21 kJ/molT\Delta S = (298\ \text{K})(-0.1987\ \text{kJ/(mol·K)}) = -59.21\ \text{kJ/mol}
Step 3: ΔG=ΔH−TΔS=−92.2 kJ/mol−(−59.21 kJ/mol)\Delta G = \Delta H - T\Delta S = -92.2\ \text{kJ/mol} - (-59.21\ \text{kJ/mol})
Step 4: ΔG=−33.0 kJ/mol\Delta G = -33.0\ \text{kJ/mol}, which is negative, so the reaction is spontaneous at 298298 K
Answer: ΔG≈−33.0\Delta G \approx -33.0 kJ/mol (spontaneous)

Step 1: Spontaneity begins where ΔG=0\Delta G = 0, so ΔH=TΔS\Delta H = T\Delta S
Step 2: T=ΔH/ΔST = \Delta H / \Delta S, with both in the same energy unit
Step 3: T=(178 000 J/mol)÷(161 J/(mol\cdotpK))T = (178\,000\ \text{J/mol}) \div (161\ \text{J/(mol·K)})
Step 4: T=1106 KT = 1106\ \text{K}, about 833833 °C; above this the +ΔS+\Delta S term wins
Answer: T≈1.11×103T \approx 1.11 \times 10^3 K (≈833\approx 833 °C)

Step 1: ΔG∘=−RTln⁡K⇒ln⁡K=−ΔG∘/(RT)\Delta G^\circ = -RT\ln K \Rightarrow \ln K = -\Delta G^\circ/(RT)
Step 2: RT=(8.314 J/(mol\cdotpK))(298 K)=2477.6 J/molRT = (8.314\ \text{J/(mol·K)})(298\ \text{K}) = 2477.6\ \text{J/mol}
Step 3: ln⁡K=−(−33 000 J/mol)÷(2477.6 J/mol)=13.32\ln K = -(-33\,000\ \text{J/mol}) \div (2477.6\ \text{J/mol}) = 13.32
Step 4: K=e13.32=6.1×105K = e^{13.32} = 6.1 \times 10^{5} (dimensionless)
Answer: K≈6.1×105K \approx 6.1 \times 10^{5}, so products are strongly favoured

Frequently Asked Questions

ΔG = ΔH − TΔS, where ΔH is the enthalpy change, T the absolute temperature in kelvin and ΔS the entropy change. A negative ΔG means the process is spontaneous at that temperature and pressure.

Put ΔH and ΔS into the same energy unit first — table values are usually kJ/mol and J/(mol·K), a factor of 1000 apart. Convert the temperature to kelvin, multiply T by ΔS, then subtract that product from ΔH.

The forward reaction is not spontaneous under those conditions; the reverse direction is. It does not mean the reaction is impossible — coupling it to a strongly negative process, or changing the temperature, can still drive it.

ΔG° = −RT ln K, with R = 8.314 J/(mol·K) and T in kelvin, so K = e^(−ΔG°/RT). A ΔG° of −33.0 kJ/mol at 298 K gives K ≈ 6.1 × 10⁵. Express ΔG° in joules per mole before dividing.

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