Question

The rate constants of a reaction at 500K and 700K are $0.02{{s}^{-1}}$and $0.07{{s}^{-1}}$respectively. Calculate the values of ${{E}_{a}}$ and A

Answer 1

Hint: To solve this question, we must recall the concept of Arrhenius Equation. The mathematical equation for the same is given below which we shall utilize to solve the given problem.
$k=A{{e}^{-{}^{{{E}_{a}}}/{}_{RT}}}$

Complete step by step solution:
Arrhenius had combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry which gives us the temperature dependence of the rate of the reaction. In the equation, the A symbolizes the frequency and the exponential term represents the fraction of collisions that have enough energy to react at temperature T.

We shall use the non-exponential form of the equation to make our calculations easier.
Taking logarithm on both sides we get,

$\ln \,k=\ln (A{{e}^{^{^{-}{}^{{{E}_{a}}}/{}_{RT}}}})=\ln A+\ln ({{e}^{^{^{-}{}^{{{E}_{a}}}/{}_{RT}}}})$
$\ln \,k=\left( \dfrac{-{{E}_{a}}}{R} \right)\left( \dfrac{1}{T} \right)+\ln A$
But when we are comparing between two rate constants and temperatures, the equation will take the form:
$\ln \left( \dfrac{{{k}_{2}}}{{{k}_{1}}} \right)=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right]$
In the question given,
${{k}_{1}}=0.02{{s}^{-1}}$
${{k}_{2}}=0.07{{s}^{-1}}$
${{T}_{1}}=500K$
${{T}_{2}}=700K$

Substituting these values in the above reaction, we get:

$\ln \left( \dfrac{0.07}{0.02} \right)=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{1}{500}-\dfrac{1}{700} \right]$

Or, $\ln \left( \dfrac{7}{2} \right)=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{2}{3500} \right]$

Or, $0.5441=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{2}{3500} \right]$

Or, ${{E}_{a}}=\dfrac{0.5441\times 2.303\times 8.314\times 3500}{2}$

Or, ${{E}_{a}}=18231.4J$

Now to calculate A we will use the formula by substituting the value we obtained for Activation energy:

$\ln A=\ln k+\dfrac{{{E}_{a}}}{2.303RT}$

We get,

$\ln A=\ln (0.02)+\dfrac{18321.4}{2.303\times 8.314\times 500}\sim 1.585$

Hence, the correct answer is ${{E}_{a}}=18231.4J$and A = 1.585.

Note:
We must always remember that chemical reactions occur more rapidly at higher temperatures. This is because thermal energy correlates direction to motion at the molecular level. As the temperature rises, molecules move faster and they have more chances of colliding more vigorously which would in turn greatly increase the likelihood of bond cleavages and rearrangements.