Question

How does the rate determining step affect the observed rate law?

Answer 1

Hint:A chemical reaction always proceeds according to the reaction mechanism. The reaction mechanism includes step by step description of how a reaction occurs. In every reaction, molecules break and form new bonds. Each step of mechanism is termed as elementary reaction.

Complete step-by-step answer:The addition of each elementary step results into the overall reaction equation. Rate determining step is the slowest step in the reaction mechanism which determines the rate of overall reaction. The rate law is defined as an expression that gives a relationship between rate of reaction and concentration of the reactants.
Let us see the general way of writing rate law expression.
$aA+bB\to cC+dD$
Where, $a,b,c,d$ are stoichiometric coefficients of products and reactants.
$Rate=k{{[A]}^{x}}{{[B]}^{y}}$
Where, $x$ and $y$ are the orders of the reaction, $[A]$ and $[B]$ are the concentrations of the reactants, and $k$ is the rate constant.
Let us see the relation between rate law and rate determining step.
If the reaction occurs in a single step, then we can easily determine the rate law as the first step is the slowest step and therefore, it is a rate determining step. But if a reaction contains multiple steps in its mechanism then it becomes difficult for us to calculate rate law using rate determining steps.
If only one step is there in a mechanism then we can easily calculate or determine the rate law using stoichiometry of the reactants. For example,
${{H}_{2(g)}}+2IC{{l}_{(g)}}\to {{I}_{2(g)}}+2HC{{l}_{(g)}}$
The reaction mechanism is as follows:
Slow reaction: ${{H}_{2}}+ICl\to HI+HCl$
${{R}_{1}}={{k}_{1}}[{{H}_{2}}O][ICl]$
Fast reaction: $HI+ICl\to {{I}_{2}}+HCl$
${{R}_{2}}={{k}_{2}}[HI][ICl]$
Here, the first step is the rate determining step. Hence, the overall reaction rate for this reaction will be:
$R=k[{{H}_{2}}][ICl]$
If the rate determining step is not the first step in the reaction mechanism then the derivation of rate law becomes more complex. Let us see an example,
$2N{{O}_{(g)}}+{{O}_{2(g)}}\to 2N{{O}_{2(g)}}$
The reaction mechanism is as follows:
Fast reaction: $2N{{O}_{(g)}}\rightleftharpoons {{N}_{2}}{{O}_{2(g)}}$
${{R}_{1}}={{k}_{1}}{{[NO]}^{2}}$
Slow reaction: ${{N}_{2}}{{O}_{2(g)}}+{{O}_{2(g)}}\to 2N{{O}_{2(g)}}$
${{R}_{2}}={{k}_{2}}[{{N}_{2}}{{O}_{2}}][{{O}_{2}}]$
Here, the second step is slow and therefore it is the rate determining step, but the rate law will be completely different, that is, it will be the overall rate law for the reaction. Here, the rate law contains ${{N}_{2}}{{O}_{2}}$ , which is a reaction intermediate and not a final product. The overall rate law does not contain any intermediates therefore, we need to go back and consider the first step. It can be written as:
At equilibrium, ${{k}_{1}}{{[NO]}^{2}}={{k}_{-1}}[{{N}_{2}}{{O}_{2}}]$
Now, if we rearrange for $[{{N}_{2}}{{O}_{2}}]$ , we get
$[{{N}_{2}}{{O}_{2}}]=\dfrac{{{k}_{1}}}{{{k}_{-1}}}{{[NO]}^{2}}$
Now, if we substitute this expression into the second step, that is the rate determining step, we get
\[R={{k}_{2}}\left( \dfrac{{{k}_{1}}}{{{k}_{-1}}}{{[NO]}^{2}} \right)[{{O}_{2}}]\]
$R=k{{[NO]}^{2}}[O{{ }_{2}}]$

Note:In this question, we have concluded that rate determining step is the slowest step in a reaction mechanism. If only a single step is there in the reaction mechanism, then that step is said to be the rate determining step. Therefore, the rate law is expressed on the basis of the first step. If a reaction contains multiple steps in mechanism, then it becomes complex and it is not necessary that the rate law is similar to the first step.