Question

A certain reaction is $50\%$ complete in $20$ min at $300K$ and the same reaction is again $50\%$ complete in $5\min$ at $350K$ . The activation energy is , if it is first order reaction :
A. $23.091kjmo{l^{ - 1}}$
B. $24.093kjmo{l^{ - 1}}$
C. $21.059kjmo{l^{ - 1}}$
D. None of these

Answer 1

As we know that activation energy is the energy which must be supplied to compounds to result in a reaction. It is measured in joules per mole which is derived from $(Ea)$ .
First order reaction is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting medium.

Complete answer: or Complete step by step answer:
As we know that, in chemistry, Activation energy is the minimum quantity of energy which to activate the atoms or molecules for a condition in which they can go through chemical changes or physical transport . It is also defined as the energy required to break the bonds of reactive molecules .This energy is provided to compounds to result in reaction .
Some-times, the rate of the reaction decreases with increasing temperature. Activation energy is required for all types of chemical reaction.
The source of activation energy is typically heat, with reactant molecules absorbing thermal energy from their environment. This thermal energy speeds up the motion of reactant molecules.
Once a reactant molecule absorbs enough energy to reach the transition state, it can proceed through the reaction .
The value of the slope (M) is equal to $- Ea/R$ where R is equal to $8.314J/mol - K$. The activation energy can also be found algebraically by the converting to rate constants $({K_1},{K_2})$ and two related reaction temperatures $({T_1},{T_2})$ into Arrhenius equation$(2)$.
So, according to the definition of activation energy;
$K = A{e^{}}\dfrac{{ - Ea}}{{RT}}$
Taking log on both sides;
$I{n^{}}K = I{n^{}}{A^{}} - \left( {\dfrac{{Ea}}{{RT}}} \right)I{n^{}}e$
$2.303\log k = 2.303\log A - \left( {Ea/RT} \right)$
$\log k = \log A - (Ea/2.303RT)$
Where, K = Rate constant
A = Arrhenius constant
$Ea$= Activation energy
R = Gas constant = $8.34J/K/mol$
According to the question, we use half time equation for value of K;
${t_{1/2}} = \dfrac{{0.693}}{k}$
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
For, first reaction;
${k_1} = \dfrac{{0.693}}{{20\min }}at300K = 0.03465{\min ^{ - 1}}$
For, same reaction again;
${k_2} = \dfrac{{0.693}}{{5\min }}at350K = 0.1386{\min ^{ - 1}}$
According to the formula of activation energy;
$\dfrac{{{k_1}}}{{{k_2}}} = - \dfrac{{Ea}}{{2.3R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
$\therefore \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.3R}}[\dfrac{{350 - 300}}{{350 \times 300}}\,]$
$\log (\dfrac{{0.693/5}}{{0.693/20}}) = \dfrac{{{E_a}}}{{2.3 \times 8.314J{k^{ - 1}}mo{l^{ - 1}}}}(\dfrac{{50}}{{350 \times 300}})K$
$\log 4 = \dfrac{{{E_a}}}{{40156.6}}$
${E_a} \approx 40156.6 \times 2 \times 0.3 \approx 24.093kjmo{l^{ - 1}}$

Note:
The activation energy of a chemical reaction is kind of the same as the jump you have to make to get yourself out of bed. Even energy released reactions require some amount of energy input to get going before they proceed their energy releasing steps.