Question

The rate of a reaction triples when temperature changes from 20 to 50°C. Calculate the energy of activation for such a reaction.

Answer 1

Given:

Rate triples → `k_2/k_1` = 3

T₁ = 20°C = 293 K

T₂ = 50°C = 323 K

R = 8.314 J mol⁻¹ K⁻¹

To calculate the activation energy (E_a), we use the Arrhenius equation in the following logarithmic form:

`ln(k_2/k_1) = E_a/R ((T_2 - T_1)/(T_1T_2))`

`ln(k_2/k_1) = ln(3) = 1.0986`

Apply the values into the equation:

`1.0986 = E_a/8.314 ((323 - 293)/(293 xx 323))`

= `E_a/8.314 * 30/94639`

= `E_a/8.314 xx 3.17 xx 10^-4`

∴ `1.0986 = (E_a xx 3.17 xx 10^-4)/8.314`

⇒ `E_a = (1.0986 xx 8.314)/(3.17 xx 10^-4)`

= `9.132/(3.17 xx 10^-4)`

= 28800 J/mol

= 28.8 kJ mol⁻¹