Question

A reaction has a rate constant of 7.98 x 10 -8 L mol -1 min -1 at 25 °C, and a rate constant of 3.30 x 10 -7 L mol -1 min -1 at 30 °C. What is the activation energy for the reaction? Give your answer to 3 significant figures.

Answer 1

Let, we have given

rate constants of a reaction,

k₁ = 7.98 x 10^-8 L mol^-1 min^-1

T₁ = 25^oC+ 273.15 =298.15K

k₂ = 3.30 x 10^-7 L mol^-1 min^-1

T2 = 30^oC+ 273.15 =303.15K​

Now, The formula is,

ln[k₂/k₁] = Ea/R [ 1/T₁ - 1/T₂]

substituting the known values in the above formula, to solve for Ea,

ln(3.30 x10^-7/7.98x10^-8) = [Ea/8.314 J /mol.K] x [ 1/298.15 - 1/303.15]

1.42 =  [Ea/8.314 J /mol.K] x [5.53x10^-5]

[Ea/8.314 J /mol.K] = 25661.32

Ea = 213348.2 J x(1 KJ/1000J)

Ea = 213 kJ

Thus, the activation energy for the reaction is 213 kJ.