Neet Solved Paper 2013 Question 3
Question: What is the activation energy for a reaction if its rate doubles when the temperature is raised from $ 20{}^\circ C $ to $ 34{}^\circ C $ ? ( $ \text{R=},8\text{.314},J,mo{l^{\text{-1}}},{K^{\text{-1}}} $ )
Options:
A) $ 342,kJ,mo{l^{\text{-1}}} $
B) $ 269,kJ,mo{l^{\text{-1}}} $
C) $ 34.7,kJ,mo{l^{\text{-1}}} $
D) $ 15.1,kJ,mo{l^{\text{-1}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, initial temperature, $ T _1=20+273=293K $ Final temperature $ T _2=35+273 $ $ =308K $ $ R=8.314J,mo{l^{-1}},{K^{-1}} $ Since, are becomes double on raising temperature,
$ \therefore $ $ r _2=2r _1or,\frac{r _2}{r _1}=2 $ As rate constant $ k\propto r $
$ \therefore $ $ \frac{k _2}{k _1}=2 $ From Arrhenius equation, we know that $ \log \frac{k _2}{k _1}=-\frac{E _{a}}{2.303R}[ \frac{T _1-T _2}{T _1T _2} ] $ $ \log 2=-\frac{E _{a}}{2.303\times 8.314}[ \frac{293-308}{293\times 308} ] $ $ 0.3010=-\frac{E _{a}}{2.303\times 8.314}[ \frac{-15}{293\times 308} ] $
$ \therefore $ $ E _{a}=\frac{0.3010\times 2.303\times 8.314\times 293\times 308}{15} $ $ =34673.48,J,mo{l^{-1}},=34.7,kJ,mo{l^{-1}} $
