Question

What is the rate law for the following mechanism? What is the intermediate in this reaction?

1.) 2NO(g) ---> N₂O₂(g) fast)

2.) N₂O₂(g)+H₂(g) ----> N₂O(g)+H₂O(g) (slow)

	a.	rate = k[NO]2, N2O2
	b.	rate = k[NO]2[H2], NO
	c.	rate = k[N2O2], NO
	d.	rate = k[NO]2[H2], N2O2

Answer 1

Sol :-

Given steps are :

Step.1 :- 2NO (g) ----------> N₂O₂ (g) (fast)

Step.2 :- N₂O₂(g) + H₂ (g) ----> N₂O(g) + H₂O(g) (slow)

Intermediates are the reactive species which appears in the products side of first step and then in reactants side of the next step.

Because, N₂O₂ (g) appears in this manner, therefore N₂O₂ (g) is reaction intermediate.

Also, slowest step is the rate determining step, so rate law expression is :

Rate = K₁.[N₂O₂].[H₂] ...................(1)

Reaction intermediates are not written in the rate law expression, therefore expression of equilibrium constant (K_eq) for the first step is :

K_eq = [N₂O₂] / [NO]²

[N₂O₂] = K_eq. [NO]²

Substitute the value of [N₂O₂] in equation (1) :

Rate = K₁K_eq. [NO]².[H₂]

Let, K (rate constant) = K₁K_eq, therefore

Rate law expression will be :

Rate = K [NO]².[H₂]

Hence, option (d) " K[NO]2[H2], N2O2 " is the correct answer.