Question

At 298 K, \(\Delta_r G^\circ\) for the reaction \(\frac{3}{2} O_2(g) \to O_3(g)\) is 165.469 kJ mol\(^{-1}\). What is the equilibrium constant for this reaction?

• A. \(10^{29}\)
• B. \(10^{-29}\)
• C. \(5 \times 10^{-27}\)
• D. \(5 \times 10^{27}\)

Hint:

Relate Gibbs free energy change to equilibrium constant using \(\Delta G^\circ = -RT \ln K\).

Answer 1

The Correct Option is B

Step 1: Use Gibbs free energy relation
\[ \Delta G^\circ = -RT \ln K \] where \(R = 8.3 J/mol\cdot K\), \(T=298 K\).
Step 2: Calculate \(\ln K\)
\[ \ln K = -\frac{\Delta G^\circ}{RT} = -\frac{165469}{8.3 \times 298} = -66.8 \] Step 3: Calculate equilibrium constant \(K\)
\[ K = e^{-66.8} \approx 10^{-29} \] Step 4: Conclusion
Equilibrium constant is \(10^{-29}\).