Question

Consider the following elementary reaction:

N2O—> N2 + O
suppose we let K1 stand for the rate constant of this reaction, and k-1 stand for the rate constant of the reverse reaction. write an expression that gives the equilibrium concentration of N2O in terms of k1, k-1 and the equilibrium concentrations of N2 and O

Answer 1

Answer : The expression that gives the equilibrium concentration of N₂O in terms of K₁, K⁻¹ and the equilibrium concentrations of N₂ and O₂ is:

`[N_2O]=(K^(-1))/(K_1)* [N_2][O_2]`

Explanation :

As we are given that:

The rate of reaction in forward direction =
`R_f=K_1* [N_2O]`

The rate of reaction in backward direction =
`R_b=K^(-1)* [N_2][O_2]`

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

So,

`R_f=R_b`

`K_1* [N_2O]=K^(-1)* [N_2][O_2]`

The expression that gives the equilibrium concentration of N₂O in terms of K₁, K⁻¹ and the equilibrium concentrations of N₂ and O₂ is:

`[N_2O]=(K^(-1))/(K_1)* [N_2][O_2]`