Question

The rate constant for a particular reaction is 1.04 x 10⁴ at 21.8°C and is 6.63 x 10⁴ at 93.5°C. What is the activation energy for the reaction, in kJ/mol? R = 8.314472 J/mol·K.

Answer 1

Use the Arrhenius equation to do so:

K = Ae^(-Ea/RT)

lnK = lnA - Ea/RT

This would be the equation to use. Now, as we are having two rates at two differents temperatures we have:

lnK1 = lnA - Ea/RT1 and lnK2 = lnA - Ea/RT2 dividing these two equations we have:

ln(K1/K2) = Ea/R (1/T2 - 1/T1) and from here we solve for Ea:

T1 = 21.8 + 273 = 294.8 K; T2 = 93.5 + 273 = 366.5 K

ln (1.04x10⁴/6.63x10⁴) = (Ea/8.3144) (1/366.5 - 1/294.8)

-1.852384 = (Ea/8.3144) (-6.6701x10^-4)

Ea = 1.852384 * 8.3144 / 6.6701x10^-4

Ea = 23088.12 J/mol

Ea = 23088.12 / 1000 = 23.09 kJ/mol

Hope this helps