Question

These elementary steps are proposed for a mechanism:

1) 2NO(g) → N2O2(g)

2) 2H2(g) → 4H(g)

3) N2O2(g) + H(g) → N2O(g) + HO(g)

4) 2HO(g) + 2H(g) → 2H2O(g)

5) H(g) + N2O(g) → HO(g) + N2(g)

a) Write the rate law for the first step.

b) Determine the molecularity of the third step.

c) Write the balanced equation for the overall reaction.

Answer 1

a) The rate law depends on the type of molecularity. In these case, you have one reactant and generates one product, so the rate law would be like this:

rate = k[A]²

the first step is:

2NO_(g) ----------> N₂O_2(g)

This reaction could be seen like this: NO + NO ---------> N₂O₂

This is for a bimolecular reaction of this form: A + A --------> products so the rate law for this bimolecular reaction is:

rate = k[NO]²

b) The third reaction is: N₂O_2(g) + H_(g) -----------> N₂O_(g) + HO_(g)

This would be a reaction of the type: A + B --------> Products

Writting the rate law: k[A][B] and for this reaction

rate = k[N₂O₂][H]

So the molecularity of this reaction is bimolecular.

c) The overall reaction, with all the steps would be:

2NO(g) → N₂O₂(g)

2H₂(g) → 4H(g)

N₂O₂(g) + H(g) → N₂O(g) + HO(g)

2HO(g) + 2H(g) → 2H₂O(g)

H(g) + N₂O(g) → HO(g) + N₂(g)

2NO_(g) + 2H_2(g) --------------------> 2H₂O_(g) + N_2(g)