Q26 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Chemistry
For a reversible reaction $A \rightleftharpoons B$, pre-exponential factor is same for both the forward and backward reactions and has value of $20 \mathrm{~S}^{-1}$. If the enthalpy change along the forward reaction is $-41.5 \mathrm{~kJ} / \mathrm{mol}$, the value of equilibrium constant at $500 \mathrm{~K}$ is
- Correct answerA$e^{10}$
- B$e^9$
- C$e^8$
- D$e^7$
Solution
For the forward reaction of $A \rightleftharpoons B$,
$\begin{aligned}
\ln K & =-\frac{\Delta H}{R T} \quad \text { [van't Hoff isotherm equation] } \\
K & =e^{-\Delta H / R T} \\
& =e^{\frac{-(-41.5)}{8.314 \times 10^{-3} \times 500}}=e^{9.983} \simeq e^{10}
\end{aligned}$
$[\because \Delta H=$ Enthalpy change of the forward reaction $=-41.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$R=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, T=500 \mathrm{~K}$
$K=$ Equilibrium constant $]$
Hence, the correct answer is option: 'A'.