6th edition, 15.89

$K = \frac{k_{forward}}{k_{reverse}}$

Hai-Lin Yeh 1J
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6th edition, 15.89

The rate law of the reaction 2 NO(g) + 2 H2(g) -> N2(g) + 2 H2O(g) is Rate = k[NO]2[H2], and the mechanism that has been proposed is
Step 1 NO + NO -> N2O2
Step 2 N2O2 + H2 -> N2O + H2O
Step 3 N2O + H2 -> N2 + H2O
(a) Which step in the mechanism is likely to be rate determining? Explain your answer.

Can someone explain why we had to find the equilibrium of step 1 and then substitute into our proposed rate law. How did they get step 2 as the rate determining step?

Chem_Mod
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Re: 6th edition, 15.89

Begin by writing rate law for each step. You will see none of them match the real rate law, which indicates to you that we must do some "rearranging" of one of our suggested rate laws to match the one in the problem.

For step 1: if we assume Step 1 is in equilibrium, then forward and reverse reaction are equal:
k1 [NO2] ^2 = k'1 [N2O2]
k1/k1' [NO2]^2 = [N2O2]

This is promising because we want our rate law to have [NO2]^2 in it. However, we still need H2. If Step 1 is in equilibrium, then we can say Step 2 is the slowest step.
rate (2) = k2[N2O2][H2}

substitute what we found earlier to get: rate = k2k1/k'1 [N2O2][H2] = k[N2O2][H2]

This matches, so step 2 is indeed your rate determining step. It's trial and error in the sense that you have to assume step 1 is in equilibrium. If you said step 2 was in equilibrium (and step 3 was rate determining), you wouldn't get the correct rate law, so you would have to start over and make a new assumption.