Calculate Logarithm of Equilibrium Constant from Reaction Enthalpy and Entropy

To find the value of $\log_{10}\text{K}$ for the reaction $\text{A} \rightleftharpoons \text{B}$, we use the fundamental thermodynamic relationships between standard Gibbs free energy change ($\Delta_r\text{G}^\circ$), standard enthalpy change ($\Delta_r\text{H}^\circ$), standard entropy change ($\Delta_r\text{S}^\circ$), and the equilibrium constant ($\text{K}$).

Step 1: Calculate the standard Gibbs free energy change ($\Delta_r\text{G}^\circ$)

The Gibbs free energy equation is given by:
$\Delta_r\text{G}^\circ = \Delta_r\text{H}^\circ - \text{T}\Delta_r\text{S}^\circ$

Given:

• $\text{T} = 298\text{ K}$
• $\Delta_r\text{H}^\circ_{298\text{K}} = -54.07\text{ kJ mol}^{-1} = -54070\text{ J mol}^{-1}$
• $\Delta_r\text{S}^\circ_{298\text{K}} = 10\text{ J K}^{-1}\text{mol}^{-1}$

Substitute these values into the equation:
$\Delta_r\text{G}^\circ = -54070\text{ J mol}^{-1} - (298\text{ K} \times 10\text{ J K}^{-1}\text{mol}^{-1})$
$\Delta_r\text{G}^\circ = -54070\text{ J mol}^{-1} - 2980\text{ J mol}^{-1}$
$\Delta_r\text{G}^\circ = -57050\text{ J mol}^{-1}$

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Step 2: Relate $\Delta_r\text{G}^\circ$ to the equilibrium constant ($\text{K}$)

The relationship between standard Gibbs free energy change and the equilibrium constant is:
$\Delta_r\text{G}^\circ = -2.303 \text{RT} \log_{10}\text{K}$

Rearranging for $\log_{10}\text{K}$:
$\log_{10}\text{K} = \frac{-\Delta_r\text{G}^\circ}{2.303 \text{RT}}$

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Step 3: Calculate $\log_{10}\text{K}$

Given that $2.303 \times \text{R} \times \text{T} = 2.303 \times 8.314 \times 298 = 5705\text{ J mol}^{-1}$:

$\log_{10}\text{K} = \frac{-(-57050\text{ J mol}^{-1})}{5705\text{ J mol}^{-1}}$
$\log_{10}\text{K} = \frac{57050}{5705} = 10$

Thus, the value of $\log_{10}\text{K}$ is 10.

Correct Option: (B)