Question

On increasing the temperature from 200 K to 220 K, rate constant of reaction A increases by three times and rate constant of reaction B increases by nine times then correct relationship between activation energy of A and B is:
(a) ${ E }_{ a }=3{ E }_{ b }$
(b) ${ 3E }_{ a }={ E }_{ b }$
(c) ${ E }_{ b }=2{ E }_{ a }$
(d) ${ E }_{ a }=2{ E }_{ b }$

Answer 1

Hint: Rate of a reaction refers to the change in the concentration of a reactant or a product per unit time. In order to solve this question we have to use the Arrhenius equation which gives us the effect of temperature on the rate constant of a reaction.

Complete step by step answer:
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.
In order to solve this question we are going to use the Arrhenius equation given below:
$ k=A{ e }^{ { -E }_{ a }/RT }$...(1)
Where k is the rate constant, ${ E }_{ a }$ is the activation energy, A is the pre-exponential factor (which is a constant) , R is the gas constant and T is the temperature in K.

For reaction A:
${ T }_{ 1 }=200 K$ and ${ T }_{ 2 }=220 K$
Also $3{ (k }_{ 1 }{ ) }_{ A }={ (k }_{ 2 }{ ) }_{ A }$
Where ${ (k }_{ 1 }{ ) }_{ A }$ and ${ (k }_{ 2 }{ ) }_{ A }$ are the rate constants at ${ T }_{ 1 }$ K and ${ T }_{ 2 }$ K.
Hence $\cfrac { { (k }_{ 1 }{ ) }_{ A } }{ { (k }_{ 2 }{ ) }_{ B } } =\cfrac { 1 }{ 3 } $...(2)
Putting equation (1) in equation (2), we get:
$\cfrac { 1 }{ 3 } =\cfrac { A{ e }^{ { -E }_{A }/R{ T }_{ 1 } } }{ A{ e }^{ { -E }_{ A }/R{ T }_{ 2 } } } $
Where ${ E }_{A }$ is the activation energy for reaction A.
$\Rightarrow \cfrac { 1 }{ 3 } ={ e }^{ \cfrac { { -E }_{ A } }{ R } \times (\cfrac { 1 }{ { T }_{ 2 } } -\cfrac { 1 }{ { T }_{ 1 } } ) }$
$\Rightarrow \cfrac { 1 }{ 3 } ={ e }^{ \cfrac { { -E }_{ A} }{ R } \times (\cfrac { 1 }{ 220 } -\cfrac { 1 }{ 200 } ) }$
$\Rightarrow \log { \cfrac { 1 }{ 3 } } =\cfrac { { E }_{ A } }{ R\times 2.303 } \times \cfrac { 20 }{ 220\times 200 } $
Therefore, ${ E }_{ A }=2200\times 2.303R\log { \cfrac { 1 }{ 3 } } $....(3)
Similarly for reaction B:
${ T }_{ 1 }=200 K$ and ${ T }_{ 2 }=220 K$
Also $9{ (k }_{ 1 }{ ) }_{ B }={ (k }_{ 2 }{ ) }_{ B }$

By following the same procedure, we will get:
${ E }_{ B }=2200\times 2.303R\log { \cfrac { 1 }{ 9 } } $....(4)
Where ${ E }_{ B }$ is the activation energy for reaction B
Now dividing equation (3) by (4) we will get:
$\cfrac { { E }_{ A } }{ { E }_{ B } } =\cfrac { \log { (1/3) } }{ \log { (1/9) } } =\cfrac { 1 }{ 2 }$
Therefore $2{ E }_{ A }={ E }_{ B }$
So, the correct answer is “Option C”.

Note: The pre-exponential factor A is a constant. It is also called frequency factor since it gives the frequency of binary collisions of the reacting molecules per second per litre.