In a chemical equilibrium the rate constant of the backward reaction is 7.5 × 10-4 and the equilibrium constant is 1.5. So the rate constant of the forward reaction is

Question

In a chemical equilibrium the rate constant of the backward reaction is 7.5 × 10-4 and the equilibrium constant is 1.5. So the rate constant of the forward reaction is (A) 2 × 10-3 (B) 15 × 10-4 (C) 1.125 × 10-3 (D) 9.0 × 10-4

Tardigrade Admin · Accepted Answer

We know, K =( K f / K b ) or, K f = K × K b =1.5 × 7.5 × 10-4 =1.125 × 10-3

Q. In a chemical equilibrium the rate constant of the backward reaction is $7.5 \times 1 0^{- 4}$ and the equilibrium constant is $1.5$ . So the rate constant of the forward reaction is

$2 \times 1 0^{- 3}$

$15 \times 1 0^{- 4}$

$1.125 \times 1 0^{- 3}$

$9.0 \times 1 0^{- 4}$

We know,
$K = \frac{K _{f}}{K _{b}}$
or, $K_{f} = K \times K_{b}$
$= 1.5 \times 7.5 \times 1 0^{- 4}$
$= 1.125 \times 1 0^{- 3}$

Q. In a chemical equilibrium the rate constant of the backward reaction is 7.5×10−4 and the equilibrium constant is 1.5. So the rate constant of the forward reaction is

2×10−3

15×10−4

1.125×10−3

9.0×10−4