Question

The reaction between chloroform $\text{CHC}{{\text{l}}_{\text{3}}}(g)$ and chlorine\[\text{C}{{\text{l}}_{\text{2}}}(g)\] to form $\text{CC}{{\text{l}}_{\text{4}}}(g)$ and $\text{HCl(g)}$ is believed to occur by this series of steps:
Step1:$\text{C}{{\text{l}}_{\text{2}}}(g)\,\,\to \,\,\text{Cl}(g)\,\,+\,\,\text{Cl}(g)$ $\text{C}{{\text{l}}_{\text{2}}}(g)\,\,\to \,\,\text{Cl}(g)\,\,+\,\,\text{Cl}(g)$
Step 2:$\text{CHC}{{\text{l}}_{\text{3}}}(g)\,+\,\,\text{Cl}(g)\,\,\,\,\,\to \,\,\,\text{CC}{{\text{l}}_{\text{3}}}(g)\,\,+\,\,\text{HCl}(g)$ $\text{CHC}{{\text{l}}_{\text{3}}}(g)\,+\,\,\text{Cl}(g)\,\,\,\,\,\to \,\,\,\text{CC}{{\text{l}}_{\text{3}}}(g)\,\,+\,\,\text{HCl}(g)$
Step 3:$\text{CC}{{\text{l}}_{\text{3}}}\text{(}g)\,\,+\,\,\text{Cl}(g)\,\,\to \,\,\text{CC}{{\text{l}}_{\text{4}}}\text{(}g)$ $\text{CC}{{\text{l}}_{\text{3}}}\text{(}g)\,\,+\,\,\text{Cl}(g)\,\,\to \,\,\text{CC}{{\text{l}}_{\text{4}}}\text{(}g)$
In this reaction are first order in $\text{CHC}{{\text{l}}_{\text{3}}}(g)$ and half order in\[\text{C}{{\text{l}}_{\text{2}}}(g)\], which step about the relative rates of step 1, 2 and 3 is correct?
(A)Step first is the slowest
(B)Step 1 and step 2 must both be slow
(C)Step 2 must be slower than step 1
(D)Step 3 must be the slowest

Answer 1

Hint: The rate of a complex chemical reaction is determined by the slowest step. The rate law from the slowest step in a mechanism should agree with experimental rate law.

Complete answer: In chemical kinetics, reactions which occur in a single step are termed as simple or elementary reactions and those reactions which take place in two or more steps are termed as complex reactions. In complex reactions the order of the slowest elementary reaction gives the order of the chemical reaction.
In this question the given rate of the complex reaction represents the experimental rate law of the reaction.
\[\text{rate}\,\,\text{=k }\!\![\!\!\text{ CHC}{{\text{l}}_{\text{3}}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ C}{{\text{l}}_{\text{2}}}{{\text{ }\!\!]\!\!\text{ }}^{{}^{\text{1}}/{}_{\text{2}}}}\] Where, k = rate constant.
So this rate law will be possible if step second will be slower than step first because in the second step intermediate is formed in the presence of chlorine atoms. A transition state $\text{CC}{{\text{l}}_{\text{3}}}$ is formed in the second who determine the rate of formation of product in equation (III). We can say that the rate law of equation (I) and equation (III) could not be itself an elementary reaction.

So, option (C) will be the correct option.

Note: A reaction intermediates are formed in between the reaction which is consumed in the later steps. So intermediates are the rate determining steps in most of the multistep reactions.