Question

(i)

Write the mathematical expression relating to the variation of the rate constant of a reaction with temperature. 
(ii)

How can you graphically find the activation energy of the reaction from the above expression?
(iii)

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

(i)

`"k" = "Ae"^(- ("E"_"a")/("RT"))`

k = rate constant

A = collision factor

E_a = activation energy

(ii)

Slope = -`"E"_"a"/(2.303"R")`

(iii)

∴ Ea = – 2.303 × slope × R
= – 2.303 × – 5841 × 8.314
= 111838.4 J mol^-1
= 111.8 KJ mol^-1