Consider the following reaction at 298 K.
\(\frac{3}{2}O_{2(g)}⇌O_{3(g)},K_p=2.47×10^{−29}\).
\(Δ_rG^{\theta}\) for the reaction is ______ kJ. (Given R=\(8.314\;J K^{−1}mol^{−1}\))
\(\frac{3}{2}O_{2(g)}⇌O_{3(g)},K_p=2.47×10^{−29}\).
\(Δ_rG^{\theta}\) for the reaction is ______ kJ. (Given R=\(8.314\;J K^{−1}mol^{−1}\))
Correct Answer: 163
Approach Solution - 1
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 \( \Delta_r G^\circ = 163.19 \, \text{kJ} \).
Approach Solution -2
\[ \Delta G^\circ = -RT \ln K_p \]
\[ \Delta G^\circ = -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29}) \]
\[ = -8.314 \times 10^{-3} \times 298 \times (-65.87) \]
\[ = 163.19 \, \text{kJ} \]
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