Question

What do you mean by Rate determining step?

Answer 1

Hint: To answer this question, dive right into the sea of chemical kinetics. The reaction mechanisms determine the reaction rate law. From the name itself you can understand that the step which gives the rate of the reaction, will be the rate determining step. There might not be a rate determining step for all reactions, and if it has there is only a rate determining step.
Formula for rate law: Rate law= $k\text{ }\!\![\!\!\text{ A}{{\text{ }\!\!]\!\!\text{ }}^{x}}\text{ }\!\![\!\!\text{ B}{{\text{ }\!\!]\!\!\text{ }}^{y}}$

Complete answer:
We got an idea about the meaning of the rate determining step. But the actual step in the reaction which determines the rate of the reaction is the step which occurs slowly.
So, the rate determining step is the slowest elementary step in the set of chemical reactions and decides the speed at which the general reaction occurs.
To decide which step of the reaction is slow, it must have more than one reaction step. So the reaction mechanism is set up to determine the slowest step of all. Some reactions do not take place in a single step but need multiple elementary steps.
Let’s understand it with the help of an example-
In the gas phase reaction between $N{{O}_{2}}$and $CO$, $C{{O}_{2}}$ and $NO$are the final products.
$N{{O}_{2}}\text{ + CO }\to \text{ C}{{\text{O}}_{2}}\text{ + NO}$
If this reaction would have taken place in a single step the rate law would be,
Rate law= $k\text{ }\!\![\!\!\text{ }N{{O}_{2}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ CO }\!\!]\!\!\text{ }$
But the actual rate law is given by,
Rate law = $k\text{ }\!\![\!\!\text{ N}{{\text{O}}_{2}}{{]}^{2}}$
This states that the reaction takes place in two elementary steps which are as follows,
Step I- $N{{O}_{2}}$forms $N{{O}_{3}}\text{ and NO}$.
$2N{{O}_{2}}\text{ }\to \text{ }N{{O}_{3}}\text{ + NO}$ (Slow step)
The product of step I is the reactant in step II. They are known as the Intermediates in the reaction.
Step II- $\text{ }N{{O}_{3}}\text{ }$reacts with $CO$to form $N{{O}_{2}}\text{ and C}{{\text{O}}_{2}}\text{ }$as the final product.
$N{{O}_{3\text{ }}}+\text{ CO }\to \text{ N}{{\text{O}}_{2}}\text{ + C}{{\text{O}}_{2}}$ (Fast step)
According to the reaction mechanism, here Step I is the slow step and is responsible for determining the rate of the reaction.
If the first step in the mechanism is the slow step it becomes easy to determine the rate, whereas with the second step being the slowest, determining the rate law becomes more complicated.
In a multistep or complex reaction, the slowest step among them is the Rate determining step.

Note:
A reaction cannot take place faster than the slowest step, so the overall reaction rate will be limited by the rate determining step. Do remember that, an addition of a catalyst will speed up the reaction. So the catalyst will change the reaction that will determine the rate.