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)

We know,
$K =\frac{ K _{ f }}{ K _{ b }}$
or, $K _{f} = K \times K _{ b } $
$=1.5 \times 7.5 \times 10^{-4} $
$=1.125 \times 10^{-3}$