At 407 K the rate constant of a chemical reaction is $\text{9}\text{.5 x 1}{{\text{0}}^{-5}}\text{ }{{\text{s}}^{-1}}$ and at 420 K, the rate constant is$\text{1}\text{.9 x 1}{{\text{0}}^{-4}}\text{ }{{\text{s}}^{-1}}$ . Calculate the frequency factor of the reaction.

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Hint: The equation called Arrhenius equation is usually written as $k=A{{e}^{-{{E}_{a}}/RT}}$ where the pre-exponential factor A is a constant and is called frequency factor and ${{E}_{a}}$ is called the activation energy, R is the gas constant and T is the temperature. The activation energy is calculated by the formula $\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}} \right]$ where ${{k}_{1}}\text{ and }{{k}_{2}}$ are rate constants at different temperatures.

Complete answer:
According to the question,
Rate constant of the first reaction is $\text{9}\text{.5 x 1}{{\text{0}}^{-5}}\text{ }{{\text{s}}^{-1}}$at 407 K.
${{k}_{1}}=\text{9}\text{.5 x 1}{{\text{0}}^{-5}}\text{ }{{\text{s}}^{-1}}$
Rate constant of the second reaction is $\text{1}\text{.9 x 1}{{\text{0}}^{-4}}\text{ }{{\text{s}}^{-1}}$ at 420 K
${{k}_{2}}=\text{1}\text{.9 x 1}{{\text{0}}^{-4}}\text{ }{{\text{s}}^{-1}}$
The value of gas constant is taken in the SI unit. The value of R = 8.314 $Jmo{{l}^{-1}}{{K}^{-1}}$
So, with all these factors we can calculate the value of activation energy.
The activation energy is calculated with the formula = $\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}} \right]$
So, putting all the values, we get
$\log \frac{1.9\text{ x 1}{{\text{0}}^{-4}}}{9.5\text{ x 1}{{\text{0}}^{-5}}}=\dfrac{{{E}_{a}}}{2.303\text{ x 8}\text{.314}}\left[ \dfrac{420-407}{420\text{ x 407}} \right]$
So, the value of activation energy is 757582.3 joule per mole

Since, we know the value of activation energy is calculated, now, the value of frequency factor can be calculated easily.
According to the Arrhenius equation,
The logarithm form of this equation will be,
$\log k=\log A-\dfrac{{{E}_{a}}}{2.303RT}$
Since, we have two reactions, we can put the value of any reaction.
Let us take the first reaction:
${{k}_{1}}=\text{9}\text{.5 x 1}{{\text{0}}^{-5}}\text{ }{{\text{s}}^{-1}}$
So, putting all the values in the equation, we get
$\log 9.5\text{ x 1}{{\text{0}}^{-5}}=\log A-\dfrac{75782.3}{2.303\text{ x 8}\text{.314 x 407}}$
$\log \dfrac{A}{9.5\text{ x 1}{{\text{0}}^{-5}}}=\dfrac{75782.3}{2.303\text{ x 8}\text{.314 x 407}}=9.7246$
A = antilog (9.7246)
$A=5.04\text{ x 1}{{\text{0}}^{5}}{{s}^{-1}}$
So, the value of frequency factor is $5.04\text{ x 1}{{\text{0}}^{5}}{{s}^{-1}}$ .

Note: To find the frequency factor it is not necessary to apply the equation on the first reaction, we can put the values of any reaction in the formula. The Arrhenius equation tells the effect of temperature on the rate of reaction.