Question

The activation energies for the forward and reverse elementary reactions in the system $A\rightleftharpoons B$ are 10.303 and 8.000 k cal respectively at 500K. Assuming the pre-exponential factor to be the same for both the forward and reverse steps and R = 2 cal $K^{-1}mo{l}^{-1}$, calculate equilibrium constant of the reaction :

• A. 1.00
• B. 10.0
• C. 100
• D. 0.1

Answer 1

The correct option is D

0.1

$E_a$ - activation energy for forward reaction
$E_a^{\prime }$ for backward reaction
$E_a-E_a^{\prime }=(8-10.303)kcal=-2.303\times {10}^{3}cal$
$K_c=\frac{-2.303\times {10}^3}{2\times 500}=e^{-2.303}$
$2.303logK=-2.303,$ $logK=-1,K=0.1$