Question

Given the temperature coefficient for the hydrolysis of ethyl acetate by NaOH is 1.75. Calculate the activation energy of the reaction.

Answer 1

Temperature coefficient = `k_308/k_298` = 1.75

According to the Arrhenius equation,

`log_10  k_2/k_1 = E_a/(2.303 R) [1/T_1 - 1/T_2]`

Putting T₁ = 298 K and T₂ = 308 K, we have

`log_10  k_308/k_298 = E_a/(2.303 xx 8.314) [1/298 - 1/308]`

or, log₁₀ 1.75 = `E_a/(2.303 xx 8.314) xx 20/(298 xx 308)`

or, E_a = `(2.303 xx 8.314 xx 298 xx 308 xx log_10 1.75)/10`

= 42711.5 J mol⁻¹ 

= 42.7 kJ mol⁻¹