Question

At 27°C and 37°C the rates of a reaction are given as 1.6*10^-2 mol L^-1s^-1 and 3.2*10^-2 mol L^-1s^-1.Calculate the energy of activation for the give reaction.

Answer 1

Given :

1. T1 = 27 + 273 = 300 K
2. T2 = 37 + 273 = 310 K
3. k1 = k and k2 = 2 k

Equation to find activation energy using temperature condition is given as :
⇒Log (k2 / k1) = Ea / 2.303 R [(T2 – T1) / T1 T2]
⇒Log (2 k / k) = Ea / 2.303 x 8.314 {(310 – 300) / 300 x 310}
⇒Log 2 = Ea / 2.303 x 8.314 x (10 / 300 x 310)
⇒Ea = 53598.6 J mol-1
⇒Ea = 53.6 kJ mol-1
Thus activation energy for the 10 degree rise in temperature is given as is 53.6 kJ mol-1