Question

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

Answer 1

Answer:

1.125×10−3

Explanation:

Equilibrium constant (K) is given by

K=KfKb

Kf=K × Kb=1.5×7.5×10−4=1.125×10−3

Answer 2

Answer: The rate constant of forward reaction is $11.25\times 10^{-4}$

Explanation:

Equilibrium constant is defined as the ratio of rate of forward reaction to the rate of backward reaction. It is expressed as $K_[eq}$

Mathematically,

$K_{eq}=\frac{K_f}{K_b}$

where,

$K_{eq}$ = equilibrium constant of the reaction = 1.5

$K_f$ = rate constant of forward reaction = ?

$K_b$ = rate constant of backward reaction = $7.5\times 10^{-4}$

Putting values in above equation, we get:

$1.5=\frac{K_f}{7.5\times 10^{-4}}\\\\K_f=11.25\times 10^{-4}$

Hence, the rate constant of forward reaction is $11.25\times 10^{-4}$