Question

In Arrhenius equation for a certain reaction the values of $A$ and ${E_a}$ are $4 \times {10^{13}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$ and $98.6{\text{kJmo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively. At what temperature the reaction will have specified constant $1.1 \times {10^{ - 3}}\,{\text{se}}{{\text{c}}^{{\text{ - 1}}}}$.

Answer 1

To answer this question, you should recall the concept of the Arrhenius equation. We shall substitute the values in the formula given to calculate the temperature.

Formula used:
The Arrhenius equation $k = A{e^{ - {E_a}/RT}}$,
where $A$: The frequency or pre-exponential factor, $T$: temperature, ${E_a}$: Activation energy and $R$: Universal gas constant.

Complete step by step answer:
We know that ${e^{ - {E_a}/RT}}$ in the Arrhenius equation is the fraction of collisions that have enough energy to react i.e., have energy greater than or equal to the activation energy${E_a}$. This equation is used to study the dependence of a reaction on temperature.

As we know, the formula for the Arrhenius equation is given as follows: $k = A{e^{ - {E_a}/RT}}$
By taking logarithmic function on both sides the equation can be modified as:
$\Rightarrow lo{g_e}k = lo{g_e}A - \dfrac{{{E_{a}}}}{{RT}}lo{g_{e}}e$.
This will be equal to
$\Rightarrow \;2.303lo{g_{10}}k = 2.303lo{g_{10}}A - \dfrac{{{E_{a}}}}{{RT}}$.
Substituting the correct values in the equation:
$\Rightarrow 2.303log(1.1 \times {10^{ - 3}}) = 2.303log(4 \times {10^{13}}) - \dfrac{{98.6 \times {{10}^3}}}{{8.314 \times T}}$.
After solving we can get the value of temperature:

$T = \dfrac{{98.6 \times {{10}^3}}}{{8.314 \times 2.303 \times 16.56}}{\text{K}} = 310.96{\text{K}}$.

Note:
The Arrhenius equation is not only simple but a remarkably accurate formula too. Not only it is used to study reaction rates but also to model the temperature variance of permeation, diffusion and solubility coefficients, and other chemical processes over moderate temperature ranges. You should know the importance of the Arrhenius equation. $RT$ is the average kinetic energy, and the exponent is just the ratio of the activation energy ${E_a}$ to the average kinetic energy. Larger this ratio, the smaller the rate. It can be concluded that high temperature and low activation energy favour larger rate constants, and thus speed up the reaction.