Question

Consider the following reaction at 298 K.
\(\frac{3}{2}​O_{2(g)}​⇌O_{3(g)}​,K_p​=2.47×10^{−29}\). 
\(Δ_r​G^{\theta}\) for the reaction is ______ kJ. (Given R=\(8.314\;J K^{−1}mol^{−1}\))

Answer 1

Correct Answer: 163

\[ \Delta G^\circ = -RT \ln K_p \]
\[ \Delta G^\circ = -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29}) \]
\[ = -8.314 \times 10^{-3} \times 298 \times (-65.87) \]
\[ = 163.19 \, \text{kJ} \]