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I am having trouble understanding this problem.

A proposed mechanism for the decomposition of $\ce{N2O5}$ is as follows

\begin{align} \ce{N2O5 &->[$k_1$]NO2 + NO3} &&\text{(slow step)} \tag1\\ \ce{NO2 + NO3 &->[$k_2$]NO2 + O2 + NO} &&\text{(fast step)} \tag2\\ \ce{NO + N2O5 &->[$k_3$]3NO2} &&\text{(fast step)} \tag3\\ \end{align} What is the rate law predicted by this mechanism?

A. $\quad \text{Rate} = k[\ce{N2O5}]$
B. $\quad \text{Rate} = k[\ce{NO2}][\ce{NO3}]$
C. $\quad \text{Rate} = k[\ce{NO}][\ce{N2O5}]$
D. $\quad \text{Rate} = k[\ce{N2O5}][\ce{NO2}][\ce{NO3}]$
E. $\quad \text{Rate} = k[\ce{N2O5}]^2$

I can determine the rate law for each individual reaction:

\begin{align} \text{Rate} &= k_1[\ce{N2O5}] \tag{1'}\\ \text{Rate} &= k_2[\ce{NO2}][\ce{NO3}]\tag{2'}\\ \text{Rate} &= k_3[\ce{NO}][\ce{N2O5}]\tag{3'} \end{align}

But I am having trouble understanding how to incorporate all of this reaction rates into one complete reaction rate.

Do I multiply all the rates?

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As the system is described, we can suppose $k_1 \ll k_2$ ans $k1 \ll k_3$, as Martin correctly noted and I have omitted to explicitly mention.

For such cases, we can consider for intermediate products to be in a steady state, i.e. $\frac {\mathrm{d}[A]}{\mathrm{d}t} \simeq 0$. So the rate of their creation is about equal to the rate of there destruction.

E.g. the rate of $\ce{NO3}$ production in reaction (1) is the same as the rate of its consumption in the reaction (2). Similarly, the rate of $\ce{NO}$ production in the reaction (2) is the same as the rate of its consumption in the reaction (3)

$$\frac {\mathrm{d}[\ce{NO3}]}{\mathrm{d}t} = k_1 [\ce{N2O5}] - k_2 [\ce{NO2}][\ce{NO3}]=0 \tag{1}$$

$$\frac {\mathrm{d}[\ce{NO}]}{\mathrm{d}t} = k_2 [\ce{NO2}][\ce{NO3}]- k_3 [\ce{NO}][\ce{N2O5}]=0 \tag{2}$$

Then try to express concenrations of intermediate products as function of concentration of reagents and final products.

$$[\ce{NO3}] = \frac{k_1 [\ce{N2O5}]}{k_2 [\ce{NO2}]} \tag{3}$$

$$ \ce{[NO}] = \frac{ k_2 [\ce{NO2}][\ce{NO3}] }{k_3 [\ce{N2O5}]} \tag{4}$$

$$ \ce{[NO}] = \frac{ k_2 [\ce{NO2}]\left( \frac{k_1 [\ce{N2O5}]}{k_2 [\ce{NO2}]} \right) }{k_3 [\ce{N2O5}]}=\frac{ k_1 [\ce{N2O5}] }{k_3 [\ce{N2O5}]}=\frac{k_1}{k_3} \tag{5}$$

From (5) and (2):

$$ k_2 [\ce{NO2}][\ce{NO3}]= k_1[\ce{N2O5}] \tag{6}$$

$$ [\ce{NO3}]= \frac{k_1[\ce{N2O5}]}{k_2 [\ce{NO2}]} \tag{7}$$

$$\frac {\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_1 [\ce{N2O5}] + 3 \cdot k_3 [\ce{NO}][\ce{N2O5}] = \\ k_1 [\ce{N2O5}] + 3 \cdot k_1 [\ce{N2O5}] = k [\ce{N2O5}] \tag{8}$$


So the answer is A.

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  • $\begingroup$ How does eq(4) come?, from eq (5), you say that concentration of NO is fixed!? $\endgroup$
    – user98209
    Aug 28 '20 at 18:17
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    $\begingroup$ Directly from (2) $\endgroup$
    – Poutnik
    Aug 28 '20 at 18:20

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