# Effect of concentration on rate of reaction

I came across this problem:

The $$\ce{2NO + Br2 -> 2NOBr}$$ follows the mechanism below:

\begin{align} \ce{NO + Br2 &-> NOBr2} & &\text{fast} \tag{1} \\ \ce{NO + NOBr2 &-> 2NOBr} & &\text{slow} \tag{2} \end{align}

If the concentration of both $$\ce{NO}$$ and $$\ce{Br2}$$ are increased two times the rate of the reaction would become

A) 4 times
B) 6 times
C) 16 times
D) 8 times

Now I know that overall order of a reaction is equal to the molecularity of the slowest step, hence order is 2 and since the concentrations of both the reactants are being increased by the same factor, i.e. 2 , the rate should be $$2^2$$ times irrespective of what is the order with respect to any reactant.

But the answer given is 8. How is this possible?

• The concentration of $\ce{NOBr2}$ is not the thing that is increased explicitly in the statement of the problem. You should express that in terms of reactant concentrations. – Zhe Apr 3 '19 at 18:14
• Actually, reading your reasoning again, you're just very confused about how to compute the rate of the reaction. For this kind of a simple system, the rate of the reaction is the product of concentrations of reactants raised to the power of their corresponding coefficients for the slow step. The overall molecularity for the step has nothing to do with the answer, except maybe in limited cases where the concentrations are all scaled by the same amount. – Zhe Apr 3 '19 at 18:17

The fast reaction should be written as reversible reaction if $$\ce{NOBr2}$$ is a high energy intermediate: $$\ce{NO + Br2 <=> NOBr2}$$

Then, the concentration of $$\ce{NOBr2}$$ can be estimated as:

$$[\ce{NOBr2}] = K_{eq} [\ce{Br2}] [\ce{NO}]$$

If the second reaction is an elementary step, the rate law follows the stoichiometry:

$$\text{rate} = k [\ce{NOBr2}] [\ce{NO}]$$

Substituting the concentration of the intermediate $$\ce{NOBr2}$$ into that rate law gives the overall rate law:

$$\text{rate} = k K_{eq} [\ce{Br2}] [\ce{NO}] [\ce{NO}]$$

and the dependency of the rate on doubling all reactants.

Now I know that overall order of a reaction is equal to the molecularity of the slowest step

This is only true if the slowest step is the first step. Otherwise, you have to consider all the reactants that make the intermediates necessary for the slow step.

What if the first reaction goes to completion? In this case, the concentration of the intermediate would depend on the limiting reactant. Changing all reactant concentration by a factor of two, the concentration of intermediate would increase by a factor of two as well (not like the factor of 4 if the first step is a pre-equilibrium).

What else do we need to know to solve this problem? The problem description does not tell us the physical state of the reactants, intermediates and products. If $$\ce{Br2}$$ is a pure liquid, we would not be able to change its concentration. I would guess the intention is to have all species in the gas phase, though.

For the reaction:

$$\ce{2NO + Br2 <=> 2NOBr}$$

the initial forward rate is given by

$$r_1 = k\ce{[NO]_1^2[Br]_1}$$

now if you double $$\ce{[NO]_1}$$ and $$\ce{[Br]_1}$$ in a second experiment, then

$$r_2 = k\ce{[NO]_2^2[Br]_2} = k\ce{(2[NO]_1)^2(2[Br]_1)} = 8k\ce{[NO]_1^2[Br]_1} = 8r_1$$

In (i), the concentration of both reactants is doubled so the rate of formation of $$\ce{NOBr2}$$ quadruples.

Since net rate is dependant on RDS which is (ii), the rate become 8 times as 4(As for $$\ce{NOBr2}$$) * 2 (due to the NO involved directly) = 8.

And I have done these calculations while assuming that all reactions are first order in each of the reactant.