Activation Energy

Question

Calculate the activation energy of a reaction if a temperature increase from 25.0 °C to 55.0 °C triples the rate constant. R = 8.3145 ×10⁻³ kJ/mol･K.

Answer 1

As a follow up to the previous answer, you might be more familiar with these calculations in the conventional form of the Arrhenius equation.

Arrhenius Equation:

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

k1 = initial rate constant = 1

k2 = final rate constant = 3

Ea = energy of activation = ?

R = constant = 8.314 J/Kmol

T1 = initial temperature = 25C + 273 = 298K

T2 = final temperature = 55C + 273 = 328K

Plug in values and solve for Ea:

ln (3/1) = -Ea/8.314 (1/328 - 1/273)

1.0986 = -Ea/8.314 (-0.000614)

1.0986 = 7.385x10^-5 Ea

Ea = 14,876 J/mol

Ea = 14.9 kJ/mol

Answer 2

r = A₀e^-E/RT

r(T₂)/r(T₁) = exp^((-E_A/R)*(1/T₂-1/T₁)

Take ln of both sides and substitute

ln(3) = (E_A/(8.314)J/moleK(1/298K - 1/328 K) (1/T's reversed and changed sign)

Solve for E_A in Joules. The actual rate was irrelevant.

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