Chemical Kinetics - Result Question 49
####50. What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^{\circ} C$ to $35^{\circ} C$ ? $(R=8.314 J mol^{-1} K^{-1})$
[NEET 2013]
(a) $269 kJ mol^{-1}$
(b) $34.7 kJ mol^{-1}$
(c) $15.1 kJ mol^{-1}$
(d) $342 kJ mol^{-1}$
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Solution:
(b) $\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}(\frac{1}{T_1}-\frac{1}{T_2})$
$ \begin{aligned} & \log 2=\frac{E_a}{2.303 \times 8.314}[\frac{1}{293}-\frac{1}{308}] \\ & 0.3=\frac{E_a}{2.303 \times 8.314} \times \frac{15}{293 \times 308} \\ & E_a=\frac{0.3 \times 2.303 \times 8.314 \times 293 \times 308}{15} . \\ & =34673 J mol^{-1}=34.7 kJ mol^{-1} \end{aligned} $
