Question

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k = log A \(\frac{E_a}{2.303R}\,\frac{1}{T}\).

Where E is the activation energy. When a graph is plotted for log k Vs \(\frac{1}{T}\) a straight line with a slope of 4000K is obtained. Calculate the activation energy

Answer 1

E_a = – 2.303 Rm

E_a = – 2.303 x 8.314 JK^-1 mol^-1 x (-4000k)

E_a = 76,589 J mol^-1

E_a = 76,589 kJ mol^-1