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

Question

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

(ii) The slope of the line in the graph of log k (k = rate constant) versus \(\frac{1}{T}\) is -5841. Calculate the activation energy of the reaction.

Answer 1

Finding activation energy graphically: Taking a log of Arrhenius equation

It is an equation for a straight line (y = mx + C) 

Therefore, on plotting a graph between different values of log10k and \(\frac{1}{T}\) at different temperature, a straight line as shown in fig. is obtained and slope of line is equal to - \(\frac{E_a}{2.303R}\)

Thus, by knowing slope, we can calculated E_a

Slope = -\(\frac{E_a}{2.303R}\)

or -E_a = -5841 x 2.303 x 8.314

E_a = 111838.45 J mol^-1 = 111.838 kJ mol^-1