Chemical EquilibriumhardMCQ SINGLE⭐ Must-Do

For a given reaction at , and . Based on this information (using ), what is the value of ?Chemical Equilibrium Chemistry Question

Question

For a given reaction at $298\text{ K}$, $\Delta H^\circ = -54.07\text{ kJ mol}^{-1}$ and $\Delta S^\circ = 10\text{ J K}^{-1}\text{mol}^{-1}$. Based on this information (using $2.303 \times R \times 298 \approx 5705\text{ J mol}^{-1}$), what is the value of $\log_{10} K$?

Answer: B

💡 Solution & Explanation

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -54070 - (298)(10) = -57050\text{ J mol}^{-1}$. Using $\Delta G^\circ = -2.303 RT \log_{10} K$, we get $-57050 = -5705 \log_{10} K$, which yields $\log_{10} K = 10$.

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