Question

The following mechanism has been proposed for the reaction of $NO$, with Br to form $NOBr$:
\[NO(g)+B{{r}_{2}}(g)\rightleftharpoons NOB{{r}_{2}}(g)\,\,;fast \\
B{{r}_{2}}(g)+NO(g)\rightleftarrows 2NOBr(g)\,;\,slow \\ \]
If the second step is the rate determining step, the order of the reaction with respect to $NO(g)$ is:
A. 1
B. 0
C. 3
D. 2

Answer 1

Hint:The slowest step will be the rate determining step and this reaction will be used to calculate the order of the reaction.

Complete step-by-step answer:In order to answer our question, we need to know about what order of a reaction means. Order of a reaction is an important parameter for every chemical reaction. It is always determined experimentally and cannot be written from the balanced chemical equation. It may be defined as "the sum of powers or exponents to which the concentration terms are raised in the rate law expression."
For a hypothetical reaction:
\[aA+bB\to \Pr oducts\]
If the rate law expression for this reaction is $Rate=k{{[A]}^{m}}{{[B]}^{n}}$, then we can conclude the following:
1. The order of the above reaction is equal to (m + n).
2. The powers or exponents, i.e., m and n have no relation to the stoichiometric coefficients a and b of the balanced chemical equation.
3. Order of the reaction with respect to A is m and that with respect to B is n.
4. If the sum of the power is equal to one, the reaction is called first order reaction. If the sum of the powers is two or three, the reaction is second order or third order reaction respectively.
Now, let us come to the question. Order can be determined from the rate determining step and the reaction can be written as:
\[
2NO+B{{r}_{2}}\rightleftarrows 2NOBr\,\,\,;\,slow \\
Rate\,\alpha \,{{[NO]}^{2}}{{[B{{r}_{2}}]}^{1}} \\
\]
So, the order of the reaction would be 2+1 which is 3. So, we get option C as the correct answer for this question.

Note: It is to be noted that order of a reaction can be fractional, as well as 0 too, however, molecularity of a reaction is always a whole number, hypothetically.