Question

At 700 K the equilibrium constant KC for the reaction between NO(g) and O2(g) forming NO2(g) is 8.7 × 106. The rate constant for the reverse reaction at this temperature is 0.54 M–1s–1. What is the value of the rate constant for the forward reaction at 700 K?

Answer 1

Answer : The value of the rate constant for the forward reaction at 700 K is, $4.70\times 10^6$

Explanation :

The given chemical equilibrium reaction is:

$NO(g)+O_2(g)\rightleftharpoons NO_2(g)$

The expression for equilibrium constant is:

$K_c=\frac{[NO_2]}{[NO][O_2]}$

The expression for rate of forward and backward reaction is:

$R_f=K_f[NO][O_2]$

and,

$R_b=K_b[NO_2]$

As we know that at equilibrium rate of forward reaction is equal to rate of backward reaction.

$R_f=R_b$

$K_f[NO][O_2]=K_b[NO_2]$

$\frac{K_f}{K_b}=\frac{[NO_2]}{[NO][O_2]}$

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

Given:

$K_c=8.7\times 10^6$

$K_b=0.54M^{-1}s^{-1}$

Now put all the given values in the above expression we get:

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

$\frac{K_f}{0.54}=8.7\times 10^6$

$K_f=4.70\times 10^6$

Therefore, the value of the rate constant for the forward reaction at 700 K is, $4.70\times 10^6$