A first-order reaction has rate constants of 2.0 10-2 s-1 and 8.4 10-2 s-1 at 0°C and 20.°C, respectively.

Question

What is the value of the activation energy?

Answer 1

Ea = R ln (k2/k1) / (1/T1 - 1/T2)
Ea = 8.314 ln (0.084/0.020) / (1/273 - 1/293)
Ea = 11.93 / 0.00025 = 47720 J = 47.7 kJ

Obviously, apachowko would not have changed his answer had I not posted this. Nice to give a thumbs down to someone who pointed out where you were wrong.

Answer 2

This can be solved using Arrhenius equation which is an expression that shows the dependence of the rate constant k of chemical reactions on the temperature T (in Kelvin) and activation energy Ea, and A is the pre-exponential factor

k = A* exp(-Ea/RT)

We can write two equations
2 x 10^-2 = A*exp(-Ea/R*273)---------Equation 1

8.4 x 10^ -2 = A exp (-Ea/R*293)--------- Equation 2)

We can divide equation 1 by equation 2 to give

2 x 10^-2/ 8.4 x10^-2 = exp(-Ea/273R)/exp(-Ea/293R)

0.238095231* exp(-Ea/293R) = exp(-Ea/273R)

Taking natural logs we have

ln (0.238095231) - Ea/293R = -Ea/273R

(Ea/R)(1/293 - 1/273) = ln (0.238095231)

Ea = ln (0.238095231)*R/(1/293 - 1/273)

Ea = -1.435084525*8.314/(3.412969283 x 10^-3 -3.66300363 x 10^-3)

Ea = 47718.615J/mol.