Question

The rate constant of forward and backward reactions for certain hypothetical reaction are $1.1\times {10}^{-2}$ and $1.5\times {10}^{-3}$ respectively. The equilibrium constant of the reaction is:

• A. 7.33
• B. 0.733
• C. 73.3
• D. 733

Answer 1

The correct option is A 7.33

As we know,

$k_{eq}=\frac{k_f}{k_b}=\frac{1.1\times {10}^{-2}}{1.5\times {10}^{-3}}=\mathrm{7.33.}$

Hence, the correct option is $\text{A}$