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?


1 Answer 1


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.

  • $\begingroup$ How does eq(4) come?, from eq (5), you say that concentration of NO is fixed!? $\endgroup$
    – user98209
    Aug 28, 2020 at 18:17
  • 4
    $\begingroup$ Directly from (2) $\endgroup$
    – Poutnik
    Aug 28, 2020 at 18:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.