Question

Methanol is formed by the following gas phase homogeneous reaction:

\[ \mathrm{CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)} \]

The standard Gibbs free energies of formation at \(298\ \mathrm{K}\) for CO and \(\mathrm{CH_3OH}\) are \(-137\ \mathrm{kJ\,mol^{-1}}\) and \(-162\ \mathrm{kJ\,mol^{-1}}\), respectively. The value of the universal gas constant is \(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\). The equilibrium constant for the given reaction at \(298\ \mathrm{K}\) is ______ \(\times 10^4\) (rounded off to one decimal place).

Hint:

Compute \(\Delta G_{rxn}^\circ\) from the formation energies (remembering \(\Delta G_f^\circ = 0\) for elemental \(\mathrm{H_2}\)), then use \(\Delta G_{rxn}^\circ=-RT\ln K\).

Answer 1

Correct Answer: 2.4

Step 1: Gibbs free energy change of reaction.

\[ \Delta G_{rxn}^\circ = (-162)-(-137)-0 = -25\ \mathrm{kJ/mol} \]

Step 2: Relate to K.

\[ \ln K = \frac{-\Delta G_{rxn}^\circ}{RT} = \frac{25000}{2477.57} = 10.0906 \]

Step 3: Exponentiate.

\[ K \approx 24119 = 2.4119\times 10^4 \]

\[ \boxed{K \approx 2.4\times 10^{4}} \]