ΔG° from K

Worked example

At 404 K, K = 6300. What is the standard Gibbs energy change?

1. 1. Standard Gibbs energy and K: solve for ΔG°

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

   = −29.39 kJ mol⁻¹

Answer: ΔG° = −29.4 kJ mol⁻¹

Method

ΔG° = −RT ln K. A K above 1 gives a negative ΔG°.

Formulas used

• Standard Gibbs energy and K

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

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