Solution:
Solution
Arrhenius equation is used to calculate the energy of activation of the reaction.
$$ log\dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{2.303R}[\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}] $$
Given that the rate constant of a first order reaction becomes 5 times when the temperature is raised from 350 K to 400 K. This means: $$ \dfrac{k_{2}}{k_{1}} = 5 , T_{1} = 350 K , T_{2} = 400 K $$
Substituting values in Arrhenius Equation to calculate activation energy:
$$ log\dfrac{k_{2}}{k_{1}} = \dfrac{E_{a}}{2.303R}[\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}}] $$
$$log5 = \dfrac{E_{a}}{2.303 \times 8.314 JK^{-1}mol^{-1}}[\dfrac{1}{350K} - \dfrac{1}{400K}] $$
$$log5 = \dfrac{E_{a}}{2.303 \times 8.314 JK^{-1}mol^{-1}}[\dfrac{400K-350K}{350K \times 400K}]$$
$$0.699 = \dfrac{E_{a}}{2.303 \times 8.314 Jmol^{-1}} [\dfrac{50K}{140000K}] $$
$$E_{a} = \dfrac{0.699 \times 2.303 \times 8.314 Jmol^{-1} \times 140000}{50} $$
$$ E_{a} = 37474.8 J mol^{-1} $$
$$ E_{a} = 37474.8 Jmol^{-1} \times \dfrac{1.00 kJ}{1000 J} = 37.5 kJ mol^{-1} $$
