Question

For a reversible reaction, the rate constants for the forward and backward reactions are $2.38\times {10}^{-4}$ and $8.15\times {10}^{-5}$ respectively. The equilibrium constant for the reaction is:

• A. $0.342$
• B. $2.92$
• C. $0.292$
• D. $3.42$

Answer 1

Answer: (B)

$2.92$

We have, $K_{eq}=\frac{K_f}{K_b}$

Now, $K_f=2.38\times {10}^{-4};K_b=8.15\times {10}^{-5}$

$\Rightarrow K_{eq}=\frac{2.38\times {10}^{-4}}{8.15\times {10}^{-5}}$

$=2.92$