Question

The rate constant for the forward reaction $A\left(g\right)\rightleftharpoons 2B\left(g\right)$ is $1.5\times {10}^{-3}{s}^{-1}$ at 100 K. If ${10}^{-5}$ moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium, then rate constant for the backward reaction at this temperature is :

• A. $1.50\times {10}^{4}Lmo{l}^{-1}{s}^{-1}$
• B. $1.5\times {10}^{11}Lmo{l}^{-1}{s}^{-1}$
• C. $1.5\times {10}^{10}Lmo{l}^{-1}{s}^{-1}$
• D. $1.5\times {10}^{-11}Lmo{l}^{-1}{s}^{-1}$

Answer 1

The correct option is D $1.5\times {10}^{-11}Lmo{l}^{-1}{s}^{-1}$

As given,
$\underset{{10}^{-5}mole}{A\left(g\right)}\rightleftharpoons \underset{100mole}{2B\left(g\right)}$

$\left[A\right]=\frac{{10}^{-5}}{10}$

$\left[B\right]=\frac{100}{10}$

$K_f=1.5\times {10}^{-3}{s}^{-1}$

$K_c=\frac{[100{]}^2}{{10}^{-5}\times 10}=\frac{K_f}{K_b}$

$K_b=\frac{1.5\times {10}^{-3}\times {10}^{-4}}{{10}^4}$

$K_b=1.5\times {10}^{-11}Lmo{l}^1{s}^1$