Question -
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be: (R=8.314JK−1mol−1andlog2=0.301) (IIT-JEE-2013)
Options:
A .  48.6kJmol−1
B .  58.5kJmol−1
C .  60.5kJmol−1
D .  53.6kJmol−1
Answer: Option D : D As per Arrhenius equation: logK2K1=Ea2.3R[T2−T1T2T1] 2.3log2=Ea8.314[10300×310] ∴Ea=53.6kJmol−1
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