Question

The rate constant for the forward reaction $A\left(g\right)\rightleftharpoons B\left(g\right)$ is $2\times {10}^{-3}{s}^{-1}$ at $300K$. If $10\text{moles}$ of $A$ and $1000$ mole of $B$ are present in a $100L$ container at equilibrium then rate constant for the backward reaction will be:

• A. $2\times {10}^{-5}{s}^{-1}$
• B. $2\times {10}^{-3}{s}^{-1}$
• C. $2\times {10}^{-6}{s}^{-1}$
• D. $2\times {10}^{-4}{s}^{-1}$

Answer 1

The correct option is A $2\times {10}^{-5}{s}^{-1}$

$A\left(g\right)\underset{k_b}{\stackrel{k_f}{\rightleftharpoons }}B\left(g\right)$
$k=\frac{\left[B\right]}{\left[A\right]}=\frac{k_f}{k_b}$
$\therefore k_b=k_f.\frac{\left[A\right]}{\left[B\right]}$
$=2\times {10}^{-3}\times \left(\frac{10/100}{1000/100}\right)$
$=2\times {10}^{-5}{s}^{-1}$