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Chemistry · EQUILIBRIUM
हिन्दीThe standard Gibbs energy change for a reaction at \( 298\text{ K} \) is \( \Delta G^\circ = 13.8\text{ kJ mol}^{-1} \). What is the value of \( K_c \) for this reaction? (Given: \( R = 8.314\text{ J mol}^{-1}\text{K}^{-1} \), \( e^{-5.569} \approx 3.81 \times 10^{-3} \))
Correct answer · Option 1
\( 3.81 \times 10^{-3} \)
Explanation
\( \ln K_c = -\frac{\Delta G^\circ}{RT} = -\frac{13800}{8.314 \times 298} = -5.569 \implies K_c = e^{-5.569} \approx 3.81 \times 10^{-3} \).
