Question

The slope of the line in the graph of log k (k = rate constant) versus `1/T` for a reaction is −5400 K. Calculate the energy of activation for this reaction. (R = 8.314 J K⁻¹ mol⁻¹).

Answer 1

Given:

Slope of the graph of log ⁡k vs `1/T` = –5400 K

R = 8.314 J K⁻¹ mol⁻¹

Use the Arrhenius equation in logarithmic form:

`log k = log A - E_a/(2.303 R) * 1/T`

This is in the form y = mx + c, where:

slope m = `- E_a/(2.303 R)`

⇒ `-5400 = -E_a/(2.303 * 8.314)`

∴ E_a = 5400 × 2.303 × 8.314

= 5400 × 19.147

= 103,393.8 J/mol

≈ 103.4 kJ/mol