Question

Nitric oxide and nitrogen dioxide are found in photochemical smog. Nitrogen dioxide is formed from nitrogen monoxide in the exhaust of automobile engines. A possible mechanism for this reaction is given below. What is the rate law predicted by the mechanism? Reaction: 2NO(g) + O2(g) ® 2NO2(g) Step 1 (fast): NO + NO ® N2O2 Step 2 (slow): N2O2 + O2 ® 2NO2 a. Rate = k[NO]2 b. Rate = k[NO2]2 c. Rate = k[NO][O2] d. Rate = k[NO]2[O2] e. Rate = k[No2]^2/[No]^2[o2]

Answer 1

The answer is d:

Rate = k[NO]²[O₂]

When identifying the rate of reaction you always have to look at the slowest step in the reaction in this case the slowest step in the reaction is step2:

Step 2: N₂O₂ + O₂ --> 2NO₂ (slow)

Then the rate of reaction depends "theoretically" on the concentration of the reagents

rate of reaction = k'[N₂O₂][O₂]

from step 1 we know that

Step 1: NO + NO --> N2O2 (fast), this step is fast so it will reach an equilibrium

Kc = [N₂O₂] / [NO] [NO]

this can be expressed like:

Kc = [N₂O₂] / [NO]²

[N₂O₂] = Kc *  [NO]²

The rate of reaction will be:

rate of reaction = k' * Kc *  [NO]²[O₂]

k = k' * Kc so:

rate of reaction = K  [NO]²[O₂]

*hope it helps =)