Question

A first order reaction is $ 50\% $ completed in $ 20 $ minutes at $ 27^\circ C $ and in $ 5 $ minutes at $ 47^\circ C $ . The energy of activation of the reaction is $ \_\_\_\_\_\_\_\_ $ (in $ KJ/mol $ ).

Answer 1

Hint :Activation energy of a reaction is the energy that is going to be provided to the compounds to start the desired chemical reaction. This can be measured in kilojoules per mole, joules per mole or in kilocalories per mole.
Arrhenius equation formula:
 $ k = A{e^{\dfrac{{ - {E_a}}}{{RT}}}} $
Where, $ k $ is rate constant of the reaction, $ A $ is pre exponential factor, $ {E_a} $ is the activation energy, is universal gas constant and $ T $ is absolute temperature.
Half – life of a reaction:
 $ k = \dfrac{{\ln 2}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} $
Where, $ {t_{\dfrac{1}{2}}} $ is the time at which the half of the reaction is completed and $ k $ is rate constant of the reaction.
Relation between the temperature and the rate constants:
 $ \ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = - \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right) $
Where, $ {E_a} $ is the activation energy, $ R $ is universal gas constant, $ {k_1} $ and $ {k_2} $ are rate constants at $ {T_1} $ and $ {T_2} $ respectively and $ {T_1} $ and $ {T_2} $ are absolute temperature.

Complete Step By Step Answer:
Given, the half-life time of the reactions, $ {t_{1/2}} = 20 min $
${t_{1/2}}' = 5 min $
The temperature (in kelvin), $ {T_1} = 27^\circ C = (27 + 273)K = 300K $
The temperature (in kelvin), $ {T_2} = 47^\circ C = (47 + 273)K = 320K $
Universal gas constant, $ R = 8.314 $
From the half-life time of the reactions, $ {t_{1/2}} = 20 min $
${t_{1/2}}' = 5 min $ ,
As the rate of reactions are inversely proportional to the rate constants of the reaction, we can find out the ratios of the rate of reactions,
Hence, $ \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{t_{\dfrac{1}{2}}}}}{{{t_{\dfrac{1}{2}}}'}} = \dfrac{{20}}{5} = 4 $
Relation between the temperature and the rate constants:
 $ \ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = - \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right) $
Where, $ {E_a} $ is the activation energy, $ R $ is universal gas constant, $ {k_1} $ and $ {k_2} $ are rate constants at $ {T_1} $ and $ {T_2} $ respectively and $ {T_1} $ and $ {T_2} $ are absolute temperature.
Now, putting all the values in the above equation,
 $ \ln \left( 4 \right) = - \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{1}{{320}} - \dfrac{1}{{300}}} \right) $
 $ \Rightarrow \ln \left( 4 \right) = - \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{{300 - 320}}{{96000}}} \right) $
 $ \Rightarrow 1.386 = - \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{{ - 1}}{{4800}}} \right) $
 $ \Rightarrow {E_a} = 1.386 \times 8.314 \times 4800 = 55311.4Jmo{l^{ - 1}} $
 $ \Rightarrow {E_a} = 55.32KJmo{l^{ - 1}} $
Hence, the energy of the activation of the reaction is $ 55.32KJmo{l^{ - 1}} $ .

Note :
When the temperature of the reaction increases the rate constant also increases according to the Arrhenius equation. As rate constant increases the rate of the reaction also increases.