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

Given Reaction:

The reaction is as follows: 

\[ \frac{3}{2} O_2 (g) \rightleftharpoons O_3 (g), \quad K_p = 2.47 \times 10^{-29} \]

Calculation of \(\Delta_r G^\circ\):

The formula to calculate the standard Gibbs free energy change \( \Delta_r G^\circ \) is:

\[ \Delta_r G^\circ = -RT \ln K_p \]

Substitute the known values:

\[ \Delta_r G^\circ = - (8.314 \times 10^{-3} \, \text{kJ/mol/K}) \times 298 \, \text{K} \times \ln(2.47 \times 10^{29}) \]

Now calculate the value of \( \ln(2.47 \times 10^{29}) \):

\[ \ln(2.47 \times 10^{29}) = -65.87 \]

Substitute this back into the equation:

\[ \Delta_r G^\circ = - (8.314 \times 10^{-3} \times 298 \times -65.87) = 163.19 \, \text{kJ} \]

Conclusion:

The standard Gibbs free energy change is \( \Delta_r G^\circ = 163.19 \, \text{kJ} \).

Answer 2

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