The value of $\log \mathrm{K}$ for the reaction $\mathrm{A} \rightleftharpoons \mathrm{B}$ at $298 \mathrm{~K}$ is ___________. (Nearest integer)
Given: $\Delta \mathrm{H}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta \mathrm{S}^{\circ}=10 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
(Take $2.303 \times 8.314 \times 298=5705$ )
Answer (integer)
10
Solution
<p>Given:<br/><br/>
$$
\begin{align}
\Delta H^0 & = -54.07 \, \text{kJ/mol} = -54070 \, \text{J/mol} \\\\
\Delta S^0 & = 10 \, \text{J/K}\cdot \text{mol} \\\\
T & = 298 \, \text{K}
\end{align}
$$</p>
<p>We find the change in Gibbs free energy $\Delta G^0$:<br/><br/>
$$
\begin{align}
\Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\
& = -54070 - (10 \times 298) \\\\
& = -54070 - 2980 \\\\
& = -57050 \, \text{J/mol}
\end{align}
$$</p>
<p>Now, we'll use the given expression with the correct constant for $ R $ as 8.314 J/(mol·K):<br/><br/>
$$
\begin{align}
\Delta G^0 & = -2.303 \times 8.314 \times 298 \times \log K \\\\
\log K & = \frac{-57050}{2.303 \times 8.314 \times 298} \\\\
& \approx 10
\end{align}
$$</p>
<p>So, the answer is $\log K = 10$.</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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