Question

The forward rate constant for the elementary reversible gaseous reaction
$C_2{H}_6\rightleftharpoons 2{CH}_3$ is $1.57\times {10}^{-3}{s}^{-1}$ at $100K$
What is the rate constant for the backward reaction at this temperature if ${10}^{-4}$ moles of ${CH}_3$ and $10$ moles of $C_2{H}_6$ are present in a $10$ litre vessel at equilibrium?

• A. $1.57\times {10}^9{L{mol}^{-1}s}^{-1}$
• B. $1.57\times {10}^{10}{L{mol}^{-1}s}^{-1}$
• C. $1.57\times {10}^{11}{L{mol}^{-1}s}^{-1}$
• D. $1.57\times {10}^7{L{mol}^{-1}s}^{-1}$

Answer 1

The correct option is D $1.57\times {10}^7{L{mol}^{-1}s}^{-1}$

As we know,
$K_{eq}=k_f/k_b;\left[{10}^{-4}/10{]}^2/\right(10/10)=1.57\times {10}^{-3}/k_b,so{k}_b=1.57\times {10}^7$