Question

A reaction takes place in three steps with an individual rate constant and activation energy, as given below.

	Rate constant	Activation energy
Step1	\[{k_1}\]	\[{E_{a_1}} = 180{\text{ kJ/mol}}\]
Step 2	\[{k_1}\]	\[{k_2}\]\[{E_{a_2}} = 80{\text{ kJ/mol}}\]
Step 3	\[{k_1}\]	\[{k_3}\]\[{E_{a_3}} = 50{\text{ kJ/mol}}\]

And overall rate constant \[k = {\left( {\dfrac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}}\]. The overall activation energy of the reaction will be
A )\[140{\text{ kJ/mol}}\]
B )\[150{\text{ kJ/mol}}\]
C )\[130{\text{ kJ/mol}}\]
D )\[120{\text{ kJ/mol}}\]

Answer 1

Hint: The Arrhenius equation gives the relationship between the activation energy and the rate constant. According to the Arrhenius equation, \[k = A{e^{{-E_a}/RT}}\].
Use Arrhenius equation and the expression for the overall rate constant and derive a relationship between the overall activation energy and the activation energy of individual steps.

Complete step by step answer:
The individual rate constants and activation energy for three individual steps are given. An expression for the overall rate constant is also given. You are asked to determine the overall activation energy for the reaction.
Write the Arrhenius equation for overall reaction and for three individual steps
\[k = A{e^{{-E_a}/RT}} \\
{k_1} = A{e^{{-E_{a_1}}/RT}} \\
{k_2} = A{e^{{-E_{a_2}}/RT}} \\
{k_3} = A{e^{{-E_{a_3}}/RT}} \\\]
Write the expression for the overall rate constant \[k = {\left( {\dfrac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}}\]
Substitute the values of the individual rate constants and the overall rate constant from the Arrhenius equations.
\[k = {\left( {\dfrac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}} \\
A{e^{{-E_a}/RT}} = {\left( {\dfrac{{A{e^{{-E_{a_1}}/RT}} \times A{e^{{-E_{a_2}}/RT}}}}{{A{e^{{-E_{a_3}}/RT}}}}} \right)^{2/3}} \\\]
Simplify the above equation
\[{e^{{-E_a}/RT}} = {\left( {\dfrac{{{e^{{-E_{a_1}}/RT}} \times {e^{{-E_{a_2}}/RT}}}}{{{e^{{-E_{a_3}}/RT}}}}} \right)^{2/3}} \\
{e^{{-E_a}/RT}} = {\left( {{e^{\left( { - {E_{a_1}} - {E_{a_2}} + {E_{a_3}}} \right)/RT}}} \right)^{2/3}} \\
{-E_a}/RT = \dfrac{2}{3}\left( { - {E_{a_1}} - {E_{a_2}} + {E_{a_3}}} \right)/RT \\
{-E_a} = \dfrac{2}{3}\left( { - {E_{a_1}} - {E_{a_2}} + {E_{a_3}}} \right) \\\]
Substitute values in the above equation and calculate the activation energy for the overall reaction.
\[{-E_a} = \dfrac{2}{3}\left( { - {E_{a_1}} - {E_{a_2}} + {E_{a_3}}} \right) \\
{E_a} = \dfrac{2}{3}\left[ {180 + 80 - 50} \right] \\
= 140{\text{ kJ/mol}} \\\]
The overall activation energy of the reaction will be \[140{\text{ kJ/mol}}\].

Hence, the correct option will be the option A ).

Note: You can also use the Arrhenius equation to obtain the relationship between the rate constants at two different temperatures. You can use the following equation for this purpose.
\[\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{ - {E_a}}}{R}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]\]
By using the above equation, you can calculate activation energy when rate constants at two different temperatures are given.