60) 2NO(g) + H2(g) N2O(g) + H2O(g)
NO(g) + NO(g) N2O2(g)
N2O2(g) + H2(g) N2O(g) + H2O(g)
(a) NO(g) + NO(g) N2O2(g) (Step 1)
N2O2(g) + H2(g) N2O(g) + H2O(g) (Step 2)
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2NO(g) + H2(g) N2O(g) + H2O(g)
(b) Rate1 = k[NO][NO] = k[NO]2
Rate2 = k[N2O2][H2]
(c) N2O2(g) is the intermediate because it is formed in one elementary step and consumed in the next. An intermediate is neither a reactant nor a product in the overall reaction.
(d) Rate = k[NO]2[H2]
Because [H2] exists in the rate law, the second step must be the slow (rate-determining step). The first step must be the fast step.