Question

If a temperature increase from 12.0 ∘C to 21.0 ∘C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction

Answer 1

Answer:

53.7kJ/mol

Explanation:

Using Arrhenius equation

Given

T1 = 12°C = 273 + 12 = 285K

T2 = 21°C = 273 + 21 = 294K

k = A exp(-Ea/RT)

Where k = Rate constant

A = the pre-exponential factor

Ea = the activation energy

R = the Universal Gas Constant = 8.314J/kmol

T = the temperature

Taking logarithms of both sides of the Arrhenius Equation.

ln(k) = ln(A) - Ea/RT

If there are the rates at two different temperatures, we can derive the expression to be;

ln(k2/k1) = Ea/R(1/T1 - 1/T2)

The reaction doubles the rate constant

So, k2/k1 = 2 (Given)

Then we have

ln(2) = Ea/8.314(1/285 - 1/294)

ln(2) * 8.314 = Ea*(1/285 - 1/294)

6.9314E-1 * 8.314 = Ea*(1/285 - 1/294)

5.7628 = Ea*(1/285 - 1/294)

5.7628 = Ea*1.0741E-4

Ea = 5.7628 / 1.074E-4

Ea = 53657.35567970204J

Ea = 53.7kJ/mol