Question

For the reaction, ${ H }_{ 2 }+{ I }_{ 2 }\xrightarrow [ { k }_{ 2 } ]{ { k }_{ 1 } } 2HI$ the rate law expression is:
(a) $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] ={ k }_{ 1 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] $
(b) $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] =\cfrac { { k }_{ 1 }\left[ HI \right] ^{ 2 } }{ { k }_{ 2 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] } $
(c) $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] ={ k }_{ 1 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] -{ k }_{ 2 }\left[ HI \right] ^{ 2 }$
(d) $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] ={ k }_{ 1 }{ k }_{ 2 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] $

Answer 1

Hint: The reaction given in the question is a reversible reaction where ${ k }_{ 1 }$ is the rate constant for the forward reaction and ${ k }_{ 2 }$ is the rate constant for the backward reaction. The overall rate for reversible reaction can be written such that you would have to subtract the rate of the backward reaction from the rate of the forward reaction.

Complete step by step solution:
First let us understand the meaning of the rate of a reaction. It refers to the change in the concentration of a reactant or a product per unit time. As a reaction progresses, there will be a decrease in the concentration of the reactants and an increase in the concentration of the products. Therefore the rate equation will be:
Rate of a reaction =$ -\cfrac { \Delta \left[ R \right] }{ \Delta t } =\cfrac { \Delta \left[ P \right] }{ \Delta t } $
Where $ \Delta \left[ R \right]$ is the change in the concentration of the reactant and $ \Delta \left[ P \right]$ is the change in the concentration of the product for a particular time interval.
But, the above formula for the rate of the reaction represents the average rate of the reaction. In order to know the actual rate of the reaction we use instantaneous rate of reaction. Its formula is given below:
Instantaneous rate of reaction = $ -\cfrac { d\left[ R \right] }{ dt } =\cfrac { d\left[ P \right] }{ dt } $
For a particular reaction, when we calculate the rate of the reaction in terms of any reactant or the product, then the rate needs to be divided by the respective stoichiometric coefficients of the reactants or the products.
Hence for a reaction: $ aA+bB\rightarrow xX+yY$; the rate of the reaction will be:
Rate of the reaction = $-\cfrac { 1 }{ a } \cfrac { d\left[ A \right] }{ dt } =-\cfrac { 1 }{ b } \cfrac { d\left[ B \right] }{ dt } =\cfrac { 1 }{ x } \cfrac { d\left[ X \right] }{ dt } =\cfrac { 1 }{ y } \cfrac { d\left[ Y \right] }{ dt } $
Now, we can also express the rate of a reaction using the rate law. According to the rate law, the rate of a reaction can be expressed in terms of the molar concentrations of the reactants with each term raised to a power which may or may not be equal to the stoichiometric coefficients of the respective reactants. The sum of these powers to which the molar concentrations of the reactants are raised in the rate law equation is called the order of the reaction.
The reaction given in the question is a reversible reaction where ${ k }_{ 1 }$ is the rate constant for the forward reaction and ${ k }_{ 2 }$ is the rate constant for the backward reaction. Experimentally it has been found that the rate of the forward and the backward reaction are of second order. Hence the rate equations are:
Rate of the forward reaction= ${ k }_{ 1 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] $
Rate of the backward reaction = ${ k }_{ 2 }\left[ HI \right] ^{ 2 }$
The overall rate for reversible reaction can be written such that you would have to subtract the rate of the backward reaction from the rate of the forward reaction. So, the rate for the reaction: ${ H }_{ 2 }+{ I }_{ 2 }\xrightarrow [ { k }_{ 2 } ]{ { k }_{ 1 } } 2HI$ will be:
Rate of the reaction = $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] ={ k }_{ 1 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] -{ k }_{ 2 }\left[ HI \right] ^{ 2 }$

Hence (c) $\left[ -\cfrac { 1 }{ 2 } \cfrac { d\left[ HI \right] }{ dt } \right] ={ k }_{ 1 }\left[ { H }_{ 2 } \right] \left[ { I }_{ 2 } \right] -{ k }_{ 2 }\left[ HI \right] ^{ 2 }$ is the correct answer.

Note: Please note that the order of a reaction cannot be found out theoretically. It has to be found out experimentally. Also the order for elementary reactions is generally equal to the molecularity while the order of complex reactions is generally equal to the molecularity of the rate determining step. But still, we should confirm the order of the reaction experimentally for both types of reactions.