# What is the order of the reaction with respect to NO?

The following reaction mechanism has been proposed for the reaction of $\ce{NO}$ with $\ce{Br2}$ to form $\ce{NOBr}$
\begin{align} \ce{NO(g) + Br2(g)&<=>NOBr2(g)}\\ \ce{NOBr2(g) + NO(g) &-> 2NOBr(g)} \end{align} If the second step is the rate determining step, what is the order of the reaction with respect to $\ce{NO}$?

Since the 2nd step is the rate determining, shouldn't the order with respect to $\ce{NO}$ be 1? According to my book though, the correct answer is 2. What am I missing here?

The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:

$$R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].$$

The problem is, the overall order of the reaction is determined from the rate expression involving only the reactants. The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.

To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,

$$K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},$$

to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:

$$[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].$$

The expression for $R = R_2$ can thus be rewritten as

$$R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],$$

which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.

Even though the 2nd elementary step is the rate determining, you cannot just consider the stoichiometry of a species in one single elementary step to find out the overall order, which is what you are asked to find out in this question. This a complex reaction and thus the order of each reactant will depend on all the elementary steps of the reaction mechanism.

We will use the equilibrium constant method(I actually forgot what this method is called) to find out the rate law for this reaction taking the help of the mechanism that you have mentioned.

The rate law will be $${r}=k\ce{[NO][NOBr2]}$$ Since $\ce{NOBr2}$ is the intermediate we are going to exclude it from the final rate law. (Remember, intermediates should never appear in the final rate law.) $$K= \frac{\ce{[NOBr2]}}{\ce{[Br2]}{[NO]}}$$

$$\ce{[NOBr2]}=K\ce{[Br2][NO]}$$

Therefore, $$r=k\ce{[NO]}K\ce{[Br2][NO]}=kK\ce{[NO]^2[Br2]}=k'\ce{[NO]^2[Br2]}$$

Thus you can see that the order with respect to $\ce{NO}$ is 2 and not 1 as you initially though.