Spontaneity And Gibbs Energy

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For a reaction, $\Delta G^\circ = -30$ kJ/mol at 298 K. Which of the following is the correct equilibrium constant $K$ for this reaction? (R = 8.314 J/mol·K)

RankGuru · Accepted Answer

The relationship between standard Gibbs free energy and the equilibrium constant is: $\Delta G^\circ = -RT\ln K$, or equivalently $K = e^{-\Delta G^\circ / RT}$. This arises from the condition of equilibrium in chemical thermodynamics. Given $\Delta G^\circ = -30{,}000$ J/mol, $R = 8.314$ J/mol·K, $T = 298$ K: $\frac{\Delta G^\circ}{RT} = \frac{-30000}{8.314 \times 298} = \frac{-30000}{2477.6} = -12.11$. Therefore $K = e^{12.11}$. Converting: $e^{12.11} = e^{12} \times e^{0.11} \approx 162755 \times 1.116 \approx 1.7 \times 10^5$. Option $3.3 \times 10^{-5}$ is $1/K$ — it arises from using the wrong sign in the exponent. Option $6.0 \times 10^4$ results from using $\log_{10}$ without converting to natural logarithm. Option $2.1 \times 10^3$ uses an incorrect value of $RT$. This is a JEE standard application of the $\Delta G^\circ = -RT\ln K$ relationship from NCERT Chapter 6. Plausibility check: a large negative $\Delta G^\circ$ implies a strongly product-favored equilibrium, consistent with $K \gg 1$.

Spontaneity And Gibbs Energy

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