Chemical Kinetics Question 203
Question: The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is
Options:
A) $ 59.2kJ,mo{l^{-1}} $
B) $ 39.2\text{ kJ }mo{l^{-1}} $
C) $ 52.9,kJ,mo{l^{-1}} $
D) $ 29.5,kJ,mo{l^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Activation energy can be calculated from the equation. $ \frac{{logK_2}}{{logK_1}}=\frac{-E_{a}}{2.303R}( \frac{1}{T_2}-\frac{1}{T_1} ) $ Given $ \frac{\log K_2}{\log K_1}=2T_2=308K;T_1=298K $
$ \therefore \log 2=\frac{-E_{a}}{2.303\times 8.314}( \frac{1}{308}-\frac{1}{298} ) $
$ E_{a}=52.9,kJmo{l^{-1}} $
BETA
