Reaction Mechanisms


Practice Problems


Problem 1

Given the following information, determine if the reaction mechanism is plausible as written. If it isn’t, derive a rate law that agrees with experiment.

Overall reaction: 2H\(_2\)(\(g\)) + 2NO(\(g\)) \(\longrightarrow\) 2H\(_2\)O(\(g\)) + N\(_2\)(\(g\))

Step 1: 2NO(\(g\)) \(\underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}}\) = N\(_2\)O\(_2\)(\(g\))

Step 2: H\(_2\)(\(g\)) + N\(_2\)O\(_2\)(\(g\)) \(\overset{k_2}{\longrightarrow}\) H\(_2\)O(\(g\)) + N\(_2\)O(\(g\))

Step 3: N\(_2\)O(\(g\)) + H\(_2\)(\(g\)) \(\overset{k_3}{\longrightarrow}\) N\(_2\)(\(g\)) + H\(_2\)O(\(g\))

Experimental rate law: rate \(= k\)[H\(_2\)][NO]\(^2\)

Predicted rate law: rate \(= k_2\)[H\(_2\)][N\(_2\)O\(_2\)]

Solution

Step 1 reaches equilibrium: rate\(_{\mathrm{f}}\) = rate\(_{\mathrm{r}}\)

Set forward and reverse rates equal: \[\begin{align*} \mathrm{rate} = k_1[\mathrm{NO}]^2 \qquad \mathrm{rate} = k_{-1}[\mathrm{N_2O_2}] ~~~~~~ \longrightarrow ~~~~~~ k_1[\mathrm{NO}]^2 = k_{-1}[\mathrm{N_2O_2}] \end{align*}\] Isolate the intermediate, N\(_2\)O\(_2\): \[\begin{align*} \mathrm{N_2O_2} = \dfrac{k_1}{k_{-1}}[\mathrm{NO}]^2 \end{align*}\] Substitute the previous expression into the predicted rate law: \[\begin{align*} \mathrm{rate} &= k_2[\mathrm{H_2}][\mathrm{N_2O_2}] \\ &= k_2[\mathrm{H_2}]\dfrac{k_1}{k_{-1}}[\mathrm{NO}]^2 \\ &= \dfrac{k_2k_1}{k_{-1}} [\mathrm{H_2}][\mathrm{NO}]^2 \\ &= k[\mathrm{H_2}][\mathrm{NO}]^2 \end{align*}\]

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