AP Chemistry Unit 5.9 Mechanism-Based Rate Laws

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Nitrogen dioxide, $\mathrm{NO}_{2}(g)$ , is produced as a by-product of the combustion of fossil fuels in internal combustion engines. At elevated temperatures $\mathrm{NO}_{2}(g)$ decomposes according to the equation below.

2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)

The concentration of a sample of $\mathrm{NO}_{2}(g)$ is monitored as it decomposes and is recorded on the graph directly below. The two graphs that follow it are derived from the original data.

Consider two possible mechanisms for the decomposition reaction.

Is the rate law described by mechanism II shown below consistent with the rate law you wrote in part (b) ? Justify your answer.
Mechanism II

Yes. Step 2 is slow;
therefore, it is the rate-determining step of this
mechanism. The rate law of this elementary reaction is
rate $=k\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]$ . Because $\mathrm{N}_{2} \mathrm{O}_{4}$ is an intermediate, it cannot appear in
the rate law of the overall reaction. Because $K_{e q}=\frac{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\left[\mathrm{NO}_{2}\right]^{2}}$ in step 1,
$\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=K_{e q}\left[\mathrm{NO}_{2}\right]^{2}$ . Then, substituting $K_{e q}\left[\mathrm{NO}_{2}\right]^{2}$ for $\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]$ in the
rate law of step 2 gives rate $=\left(k K_{e q}\right)\left[\mathrm{NO}_{2}\right]^{2}$ , which is consistent with
the rate law in part (b).

1 point is earned for the correct answer with justification.

AP Chemistry 5.9 Pre Equilibrium Approximation Questions

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