# 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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