Question

The standard free energy change of a reaction is \(\Delta G^\circ = -115\) kJ at 298 K. Calculate equilibrium constant \(K_p\) in \(\log K_p\). (R = 8.314 J k\(^{-1}\) mol\(^{-1}\))

• A. 20.16
• B. 2.303
• C. 2.016
• D. 13.83

Hint:

\(\log K = \frac{-\Delta G^\circ}{2.303RT}\). Negative \(\Delta G^\circ\) means large \(K\).

Answer 1

The Correct Option is A

Step 1: Use relation between \(\Delta G^\circ\) and equilibrium constant.
\[ \Delta G^\circ = -RT\ln K_p \]
Step 2: Convert \(\Delta G^\circ\) into J.
\[ \Delta G^\circ = -115\,kJ = -115000\,J \]
Step 3: Substitute values.
\[ -115000 = -(8.314)(298)\ln K_p \]
\[ \ln K_p = \frac{115000}{8.314 \times 298} \approx \frac{115000}{2477.6} \approx 46.41 \]
Step 4: Convert \(\ln K_p\) to \(\log K_p\).
\[ \log K_p = \frac{\ln K_p}{2.303} = \frac{46.41}{2.303} \approx 20.16 \]
Final Answer:
\[ \boxed{20.16} \]