Chemistry · General chemistry II · Worked example

Find Ea from two rate constants and predict a third

A reaction has k = 2.5 × 10⁻³ s⁻¹ at 298 K and k = 1.1 × 10⁻² s⁻¹ at 318 K. Find Ea, then predict k at 338 K.

ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

Set up the two-point form

Label the pairs: k₁ = 2.5 × 10⁻³ s⁻¹ at T₁ = 298 K and k₂ = 1.1 × 10⁻² s⁻¹ at T₂ = 318 K. The ratio k₂/k₁ = 4.4.

ln (4.4) = \frac{E_{a}}{8.314} (\frac{1}{298} - \frac{1}{318})

ln 4.4 = 1.482 and the bracket is 2.111 × 10⁻⁴ K⁻¹.

E_{a} = \frac{(8.314) (1.482)}{2.111 \times 10^{- 4}} = 5.84 \times 10^{4} J/mol = 58.4 kJ/mol

Use the same equation with T₂ = 338 K and the Ea just found.

k_{338} = (2.5 \times 10^{- 3}) e^{(58400 / 8.314) (1 / 298 - 1 / 338)} = 4.1 \times 10^{- 2} s^{- 1}

Ea ≈ 58.4 kJ/mol, and k ≈ 4.1 × 10⁻² s⁻¹ at 338 K (an extrapolation, with no check on curvature).

A reaction’s rate constant triples between 300 K and 320 K. What is Ea?

Show the answer and explanation

Ea = R ln 3 ÷ (1/300 − 1/320) = 8.314 × 1.0986 ÷ 2.083 × 10⁻⁴ ≈ 4.38 × 10⁴ J/mol.

E_{a} = \frac{8.314 ln 3}{\frac{1}{300} - \frac{1}{320}} \approx 43.8 kJ/mol

Enter both rate constants in the Kinetics studio and add 338 K as the prediction temperature.

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Tro, Chemistry: A Molecular Approach, 4th ed., §14.5 The Effect of Temperature on Reaction Rate, pp. 642–647 (Arrhenius plots, pp. 644–645; two-point form, p. 646)
OpenStax Chemistry 2e — Collision theory