The rate of an elementary reaction has been studied as a function of temperature.

The following data were obtained for the rate constant:

Exp 1: T = 300K; k = 6.47 mol.L^-1.s^-1

Exp 2: T = 500K; k = 2.84 x 10⁴ mol.L^-1.s^-1

Exp 3: T = 700K; k = 1.03 x 10⁶ mol.L^-1.s^-1

Calculate the activation energy for the reaction.

Question

The rate of an elementary reaction has been studied as a function of temperature.

The following data were obtained for the rate constant:

Exp 1: T = 300K; k = 6.47 mol.L^-1.s^-1

Exp 2: T = 500K; k = 2.84 x 10⁴ mol.L^-1.s^-1

Exp 3: T = 700K; k = 1.03 x 10⁶ mol.L^-1.s^-1

Calculate the activation energy for the reaction.

| General Chemistry 2

Accepted Answer

Arrhenius equation:

k = Ae^-Ea/RT

k₂ / k₁ = Ae^-Ea/RT2 / Ae^-Ea/RT1

⇒k₂ / k₁ = e^{-E_a/RT₂ + E_a/RT₁}

⇒ln (k₂ / k₁) = $\frac{E_{a}}{R}$ $(\frac{1}{T_{1}} - \frac{1}{T_{2}})$

⇒ E_a =

\frac{\ln (\frac{2.84 \times 10^{4}}{6.47}) \times 8.314}{\frac{1}{300} - \frac{1}{500}}

= 5.23 x 10⁴ J = 52.3 kJ

Arrhenius equation - 2 | Chemical Kinetics: Mechanisms