Question

In a chemical equilibrium the rate constant of the backward reaction is $7.5\times {10}^{-4}$ and the equilibrium constant is 1.5 So the rate constant of the forward reaction is?

• A. $2\times {10}^{-3}$
• B. $15\times {10}^{-4}$
• C. $1.125\times {10}^{-3}$
• D. $9.0\times {10}^{-4}$

Answer 1

Answer: (C)

$1.125\times {10}^{-3}$

Equilibrium constant $K$is given by:

$K=\frac{K_f}{K_b}$

So, $K_f=K\times K_b$

$=1.5\times 7.5\times {10}^{-4}$

$=1.125\times {10}^{-3}$

Option C is correct.