Question

the rate of a reaction doubles when it's temperature change from 300 k to 310 k. activation energy of such a reaction will be

Answer 1

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