# How to determine rate law for a reaction?

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?

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):

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

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

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