Question

The activation energies of two reactions are ${E_a}$ and ${E_a}’$In the order ${E_a} > {E_a}’$. If the temperature of the reacting system is increased from ${T_1}$ to ${T_2}$, Predict which alternative is correct ${K’}$ is the rate constant at a higher temperature. Assume A being same for both the reaction:
(A) $\dfrac{{{K_1}’}}{{{K_1}}} = \dfrac{{{K_2}’}}{{{K_2}}}$
(B) ${K_1} < {K_2}{\text{ }}and{\text{ }}{K_1}' < {K_2}'$
(C) ${K_1} > {K_2}{\text{ }}and{\text{ }}{K_1}' > {K_2}'$
(D) $\dfrac{{{K_1}'}}{{{K_1}}} < \dfrac{{2{K_2}’}}{{{K_2}}}$

Answer 1

Hint: To solve this question we have to build our concept for chemical kinetics. We have to know all the concepts of the activation energy and rate constant in chemical kinetics, which means we should have to know the relation between activation energy and rate constant of the reaction.

Complete answer:Properties of reaction rate constant:
1. It is a proportionality constant.
2. It is equal to the rate of the reaction when the concentration of the reactants is unity.
3. It is independent of initial concentration of the reactants. It has a constant value at fixed temperature. Its unit depends upon the order of the reaction.
Activation energy: the minimum amount of energy required by reactant molecules to participate in a reactant is called activation energy.
Activation energy = threshold energy - average kinetic energy of reacting molecules.
Threshold energy=initial potential energy of the reactant molecule + activation energy.
After defining definition and properties we will give the relation between activation energy and rate constant of the reaction.
$K = A{e^{ - {E_a}/RT}}$
The above reaction is known as ARRHENIUS EQUATION
Where, k=Rate constant or velocity constant of at temperature T
A = Arrhenius constant or frequency factor.
E=Energy of the activation.
So, from the given exponential equation we have concluded that:
More is activation energy , lesser will be rate constant or vice versa.
So, if we have ${E_a} > {E_a}'$ then,
${K_1} < {K_2}{\text{ }}and{\text{ }}{K_1}' < {K_2}'$ will be true for any chemical reaction for temperature ${T_1}$ and ${T_2}$.

Hence, the correct answer will be option number (B).

Note:The ‘energy of activation’ is defined as the excess energy that the reacting molecule acquires to attain the ‘ threshold energy’ in order to undergo chemical change.
$\Delta H = $Heat of the reaction
$\Delta H = $$({E_1} - {E_2})$=activation energy of the forward reaction – activation energy of the backward reaction.