AP Chemistry 5.9: Mechanism-Based Rate Laws

Practice AP Chemistry questions on deriving an overall rate law when the first elementary step is not rate limiting, including pre-equilibrium treatment of intermediates.

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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.

Figure for Question 5.9.A—Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting question 1 — AP Chemistry

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

Table for Question 5.9.A—Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting question 1 — AP Chemistry

6 \mathrm{H}^{+}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightarrow 10 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Mn}^{2+}(a q)

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.

6 \mathrm{H}^{+}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightarrow 10 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Mn}^{2+}(a q)

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5.9.A—Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting question 1