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* 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=([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]`

`(K_f)/(K_b)=([NO_2])/([NO][O_2])`

`(K_f)/(K_b)=K_c`

Given:

`K_c=8.7* 10^6`

`K_b=0.54M^(-1)s^(-1)`

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

`(K_f)/(K_b)=K_c`

`(K_f)/(0.54)=8.7* 10^6`

`K_f=4.70* 10^6`

Therefore, the value of the rate constant for the forward reaction at 700 K is,
`4.70* 10^6`