Question

Two reactions $R{}_1$ and $R{}_2$ have identical pre-exponential factors. Activation energy of $R{}_1$ exceeds that of $R{}_2$ by $10kJmol{}^{ - 1}$ . If $k{}_1$ and $k{}_2$ are rate constant for reactions $R{}_1$ and $R{}_2$ respectively at $300K$ then ln $\left( {\dfrac{{k{}_2}}{{k{}_1}}} \right)$ is equal to $(R = 8.314Jmol{}^{ - 1}K{}^{ - 1})$
a) $12$
b) $6$
c) $4$
d) $8$

Answer 1

Hint : The equation given by Arrhenius will solve the above problem. It is given for temperature dependence of the rate of reactions. This equation is used to calculate the activation of energy.

Complete Step By Step Answer:
The ARRHENIUS equation gives the dependence of rate constant of chemical reaction on absolute temperature.
We are going to use the equation given by Arrhenius,
$K = Ae{}^{\dfrac{{ - Ea}}{{RT}}}$
Were k= rate constant
A= pre exponential factor
-Ea= activation energy
R= universal gas constant
T= absolute temperature
Taking log on both sides
$\ln k = \ln A - \dfrac{{Ea}}{{RT}}$
In the given equation $R{}_1$ and $R{}_2$ are two reactions having the same exponential factor, meaning the value of A is the same but having different activation energy. The given between them is that $R{}_1$ exceeds $R{}_2$ by $10kJmol{}^{ - 1}$ .
Therefore,
$Ea{}_1 - Ea{}_2 = 10kJmol{}^{ - 1}$
Reaction for $R{}_1$ is written as, $\ln k{}_1 = \ln A - \dfrac{{Ea{}_1}}{{RT}}$
Similarly the reaction for $R{}_2$ is written as, $\ln k{}_2 = \ln A - \dfrac{{Ea{}_2}}{{RT}}$
Subtracting the above equation,
$\ln k{}_2 - \ln k{}_1 = - \dfrac{{Ea{}_2}}{{RT}} + \dfrac{{Ea{}_1}}{{RT}}$
$\ln \left( {\dfrac{{k{}_2}}{{k{}_1}}} \right) = \dfrac{1}{{RT}}\left[ {Ea{}_1 - Ea{}_2} \right]$
Let us substitute the given values in the above equation,
= \dfrac{1}{{8.314 \times 300}} \times 10 \times 10{}^3 \\\ = \dfrac{{100}}{{8.314 \times 3}} \\\ = \dfrac{{100}}{{24.402}} \\\ = 4.097 \\\
From the given options, option C matches with the answer we found so the correct answer is Option C.
If $k{}_1$ and $k{}_2$ are rate constant for reactions $R{}_1$ and $R{}_2$ respectively at $300K$ then ln $\left( {\dfrac{{k{}_2}}{{k{}_1}}} \right)$ is equal to $4.097$

Note :
Arrhenius said that for a reactant to transform to a product, they should have a minimum amount of energy known as activation energy. The fraction of molecules that possess kinetic energy greater than activation energy at absolute temperature T ,is calculated from statistical mechanics.