Question

2 NO(g)  N2O2(g) (fast)
N2O2(g) + H2(g)  N2O(g) + H2O(g) (slow)
N2O(g) + H2(g)  N2(g) + H2O(g) (fast)

Draw a circle around any reaction intermediates. Draw a box around any species that can be considered a catalyst. Write the rate law for this process in the space below:

Answer 1

Intermediates: N2O2 and N2O (circle these — each is formed in one step and consumed in a subsequent step).

Catalyst: none (no species is consumed then regenerated).

Rate law:
- The slow step is step 2: rate = k2[N2O2][H2].
- Step 1 is fast (pre‑equilibrium): 2 NO ⇌ N2O2, so [N2O2] = Keq[NO]^2.
- Substitute to get the observable rate law:

rate = k2 Keq [NO]^2 [H2] = k[NO]^2[H2].