Question

A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.
[Given: log 2 =0.3010, log 4 = 0.6021, R = 8.314 J/K mol]

Answer 1

Activation energy is the energy that should be provided to compounds to make the changes of reactants to products in a chemical reaction. If activation energy is zero the reaction won’t happen means there are no products are going to form. Formula to calculate activation energy is as follows.
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right]$
Where ${{k}_{1}},{{k}_{2}}$= chemical reactions rates at different time intervals
${{E}_{a}}$ = Activation Energy
R = Gas constant
${{T}_{1}},{{T}_{2}}$= Temperature in Kelvin at respective time intervals.

Complete step by step answer:
In the question it is given that the chemical reaction is first order and 50% of the reaction completed in 40 minutes at 300 K and in 20 minutes at 320 K.
From the question ${{T}_{1}}$ = 300, ${{T}_{2}}$ = 320

We know the formula to calculate activation energy.
$\log \dfrac{{{k}_{2}}}{{{k}_{1}}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}} \right]$
Where ${{k}_{1}},{{k}_{2}}$= chemical reactions rates at different time intervals
${{E}_{a}}$ = Activation Energy
R = Gas constant
${{T}_{1}},{{T}_{2}}$= Temperature in Kelvin at respective time intervals.

We know that,

& {{k}_{1}}=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}=\text{ }\dfrac{0.693}{40} \\\ & \text{ } \\\ & {{k}_{2}}\text{ }=\text{ }\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}=\dfrac{0.693}{20} \\\ \end{aligned}$$ Now substitute $${{k}_{1}},{{k}_{2}}$$values in the main equation. $$\begin{aligned} & \log \dfrac{\dfrac{1.693}{20}}{\dfrac{1.693}{40}}=\dfrac{{{E}_{a}}}{2.303R}\left[ \dfrac{1}{300}-\dfrac{1}{320} \right] \\\ & \log 2=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{320-300}{300\times 320} \right] \\\ & 0.3010=\dfrac{{{E}_{a}}}{2.303\times 8.314}\left[ \dfrac{20}{300\times 320} \right] \\\ & {{E}_{a}}=27663.8J/mol \\\ \end{aligned}$$ **Note:** Don’t be confused between first order and second order reaction. First order chemical reaction: In a chemical reaction, in rate determining step only one reactant is going to play a role. Then the reaction is called a first order reaction. Second order reaction: In a chemical reaction, in rate determining step both the reactants are going to play a role. Then the reaction is called a second order reaction. In the above formula $${{t}_{{}^{1}/{}_{2}}}$$ means half-life of the reaction. Half-life: The time taken by the reactants to fall to half of its original concentration is called half-life of the reaction.