Question

What is the activation energy for a reaction of its rate double when the temperature is raised from ${20^ \circ }C - {35^ \circ }C$ ? Given, $R = 8.314\,J\,mo{l^{ - 1}}{K^{ - 1}}$.

Answer 1

Hint: We know that, Activation energy can be calculated using the Arrhenius equation,
The Arrhenius equation is $K = A{e^{ - Ea/RT}}$
Where, K is the rate constant
The activation energy is ${E_a}$
R is the gas constant
T is temperature in Kelvin
A is frequency factor or Arrhenius factor which explains the rate of collision and the fraction of collisions with the proper orientation for the reaction to occur.

Complete step by step answer: We know that Activation energy is the difference between energy level of reactants and the transition state for the reaction to happen reactants need to cross the transition state energy barrier and hence lower activation energy faster will be the reaction.
Given,
The initial temperature is ${20^ \circ }C$.
The final temperature is ${35^ \circ }C$.
The gas constant value R is$8.314J\,mo{l^{ - 1}}\,{K^{ - 1}}$.
Since rate of the reaction become doubled on raising the temperature, Hence the rate of the reaction is,
${r_2} = 2{r_1}$
Taking the natural log of both sides of Arrhenius equation gives,
\[lnk = lnA - Ea/RT\]
The activation energy can also be calculated if two known temperatures are directly given and a rate constant at each temperature also be given. We calculate the activation energy using Equation,
$\log \,\dfrac{{{K_2}}}{{{K_1}}} = - \dfrac{{{E_a}}}{{2.303\,R}}\left[ {\dfrac{{{T_1} - {T_2}}}{{{T_1}{T_2}}}} \right]$
Where \[{T_{1{\text{ }}}}and{\text{ }}{T_2}\] two different temperatures, \[{k_1}{\text{ }}and{\text{ }}{k_2}\] are reaction rate constants.
We know that the rate of the reactions and the reactions rate are directly proportional so,
$\dfrac{{{K_2}}}{{{K_1}}} = 2$
Substituting the know values in the equation,
$\log \,2 = - \dfrac{{{E_a}}}{{2.303\,\left( {8.314} \right)}}\left[ {\dfrac{{{{293}_1} - 308}}{{293 \times 308}}} \right]$
Substituting the log value of 2 we get,
$0.3010 = - \dfrac{{{E_a}}}{{2.303\,\left( {8.314} \right)}}\left[ {\dfrac{{ - 15}}{{293 \times 308}}} \right]$
${E_a} = \dfrac{{0.3010 \times 2.303 \times 8.314 \times 293 \times 308}}{{15}}$
Multiplying the all values and divided by 15 we get,
${E_a} = 34.67\,kJ\,mo{l^{ - 1}}$
The activation energy for a reaction is$34.67\,kJ\,mo{l^{ - 1}}$.

Note: We must remember, to convert the temperature in Celsius to Kelvin using the conversion,
${0^ \circ }C = 273\,K$
To convert the values in joule divide the value by $1000$.
$1J = 0.001\,kJ$