The reaction of nitric oxide (NO(g)) with molecular hydrogen (H_2(g)) results in

Question

The reaction of nitric oxide (NO(g)) with molecular hydrogen (H_2(g)) results in the formation of nitrogen and water as follows: 2NO (g) + 2H_2(g) rightarrow N_2(g) + 2H_2O(g) The experimentally determined rate-law expression for this reaction is first order in H_2(g) and second order with NO(g). Is the reaction above, as written an elementary reaction? One potential mechanism for this reaction is as follows: H_2(g) + 2NO(g) rightarrow N_2O (g) + H_2O (g) k_1 H_2(g) + N_2O (g) rightarrow N_2(g) + H_2O(g) k_2 Is this mechanism consistent with the experimental rate law? If not, why? An alternative mechanism for the reaction is: 2NO(g) leftarrow rightarrow N_2O_2(g) (fast) k_1(fwd) k_-1(rev) H_2(g) + N_2O_2 (g) rightarrow N_2O(g) + H_2O(g) k_2 H_2(g) + N_2O (g) rightarrow N_2(g) + H_2O(g) k_3 Show that this mechanism is consistent with the experimental law.

Explanation / Answer

(a) Given that, the rate law is first order with H2 and second order in NO.

So,

The rate law = k(H2)1(NO)2

(b)

Yes, this mechanism is consistent with the given rate law.

The rate determining step is equation (1) with rate constant k1

(c)

Add reactions (1) ,(2) and (3).

Net rate constant = (k1/k-1) x(k2)x(k3) =K

Rate law will be = K (NO)2(H2)

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