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
aditya kashyap , 11 Years ago
Grade upto college level
2 Answers
Sunil Kumar FP
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
Last Activity: 11 Years ago
Yash Chourasiya
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
Last Activity: 5 Years ago
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