Question

The rate of reaction becomes four times when the temperature changes from $293K$ to $313K$. Calculate the energy of activation ${{E}_{2}}$ of the reaction assuming that it does not change with temperature.
$\left[ R=8.314J/Kmo{{l}^{-1}},\log 4=0.6021 \right]$

Answer 1

Arrhenius equation, numerical articulation that depicts the impact of temperature on the speed of a chemical reaction, the premise of all prescient articulations utilized for figuring reaction- rate constants.

Complete step by step answer:
Given:

$${{T}_{1}}=293K \\\ {{T}_{2}}=313K \\\ R=8.314J{{k}^{-1}}mo{{l}^{-1}} \\\ k_2 =4{{k}_{1}} \\\ {{E}_{a}}=?$$

The formula used here is, $\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left( \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right)$
Substituting values,

$$\Rightarrow \log \dfrac{4{{k}_{1}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303\times \left( 8.314J{{K}^{-1}}mo{{l}^{-1}} \right)}\left( \dfrac{1}{293K}-\dfrac{1}{313K} \right) \\\ \Rightarrow \log 4=\dfrac{{{E}_{a}}}{2.303\times \left( 8.314J{{K}^{-1}}mo{{l}^{-1}} \right)}\left( \dfrac{313K-293K}{293K\times 313K} \right) \\\ \Rightarrow \dfrac{0.6021\times 2.303\times 8.314J{{K}^{-1}}mo{{l}^{-1}}\times 293K\times 313K}{20K}={{E}_{a}} \\\ \Rightarrow {{E}_{a}} =\dfrac{1057266.674}{20}=52863.33\,Jmo{{l}^{-1}} \\\ \Rightarrow {{E}_{a}} =52.86\, kJmo{{l}^{-1}} \\\$$

**Hence, the answer is $52.86\, kJmo{{l}^{-1}}$

Additional Information: **
The sign 'A' in the Arrhenius condition indicates the pre-exponential factor or the recurrence factor. This factor manages the crashes among atoms and can be thought of as the recurrence of accurately arranged impacts between particles that have adequate energy to start a compound reaction.

Note:
$k=$ Rate constant, the estimation of the rate constant is temperature subordinate. An enormous estimation of the rate constant implies that the response is generally quick, while a little estimation of the rate steady implies that the response is moderately moderate.
${{E}_{a} }=$Activation energy, the activation energy is the base energy needed for a response to happen. This implies that the reactant particles have enough dynamic energy to impact effectively and defeat the shock brought about by external electrons.
$R=$ Universal gas constant, the gas steady is meant by the image R or R. It is identical to the Boltzmann constant, however communicated in units of energy per temperature increase per mole.
${{T}_{{}}}=$Temperature, expanding the temperature of a response by and large accelerates the cycle in light of the fact that the rate consistent expands as per the Arrhenius Equation.