Chemical EquilibriumHard

Question

The activation energies for the forward and reverse elementary reactions in the system A$\rightleftharpoons$B are 10.303 and 8.000 kcal, respectively at 500 K. Assuming the pre-exponential factor to be the same for both the forward and reverse steps, the equilibrium constant of the reaction at 500 K is

Options

A.1.00
B.10.0
C.100
D.0.1

Solution

$\Delta H = \varepsilon a_{f} - \varepsilon a_{b} = 10.303 - 8.000 = 2.303\text{ kcal}$

$\text{Now, }K_{eq} = \frac{A_{f}}{A_{b}}.e^{- \Delta H/RT} = 1 \times e^{- \frac{2.303 \times 10^{3}}{2 \times 500}} = 0.1$

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