K from ΔH° and ΔS°

Worked example

For a reaction, ΔH° = 38.0 kJ mol⁻¹ and ΔS° = 96.7 J K⁻¹ mol⁻¹. Calculate K at 199 °C, assuming both are independent of temperature.

1. 1. Gibbs energy from ΔH and ΔS: solve for ΔG

   $\Delta G = \Delta H - T\Delta S$

   = −7.657 kJ mol⁻¹

2. 2. Standard Gibbs energy and K: solve for K

   $K = \exp\!\left(-\dfrac{\Delta G^\circ}{RT}\right)$

   = 7.032

Answer: K = 7.03

Method

Two steps: ΔG° = ΔH° − TΔS°, then K = exp(−ΔG°/RT).

Formulas used

• Gibbs energy from ΔH and ΔS

  $\Delta G = \Delta H - T\Delta S$

• Standard Gibbs energy and K

  $\Delta G^\circ = -RT\ln K$

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