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The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be: (R = 8.314 JK –1 mol –1 and log 2 = 0.301) (1) 48.6 kJ mol –1 (2) 58.5 kJ mol –1 (3) 60.5 kJ mol –1 (4) 53.6 kJ mol –1



The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be: (R = 8.314 JK –1 mol –1 and log 2 = 0.301) (1) 48.6 kJ mol –1 (2) 58.5 kJ mol –1 (3) 60.5 kJ mol –1 (4) 53.6 kJ mol –1

Grade:upto college level

2 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
6 years ago

Using Arrhenius Equation

ln(k2/k1) = (Ea/R)(1/T1-1/T2)

We get,

2.303 log2 = -(Ea/8.314)(1/300-1/310)

Ea = 5359.59 Jmol-1= 53.6 kJ mol-1

Hence, the correct option is D.

thanks and regards

sunil kr

askIITian faculty

Yash Chourasiya
askIITians Faculty 256 Points
9 months ago
Dear Student

Using Arrhenius Equation

ln(k2/k1) = (Ea/R)(1/T1- 1/T2)

We get,

2.303 log 2 = -(Ea/8.314)(1/300 – 1/310)

Ea = 5359.59 Jmol-1
= 53.6 kJ mol-1

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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