Question

The slope of the line in the graph of log k (k = rate constant) versus `1/T` is −5841. Calculate the activation energy of the reaction.

Answer 1

Given: Slope = −5841

R = 8.314 J mol⁻¹ K⁻¹

By using 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 the slope m of the line is:

Slope = `-E_a/(2.303 R)`

⇒ −5841 = `-E_a/(2.303 xx 8.314)`

E_a = 5841 × 2.303 × 8.314

= 111838 J mol⁻¹

= 11.839 kJ/mol