Question

Consider this reaction data: T(k) 325, 825 and k(s-1) 0.389, 0.735

If you were going to graphically determine the activation energy of this reaction, what points would you plot? A). Point 1 x, y and Point 2 x, y? B). Rise, Run, and Slope? C). What is the activation energy of this reaction? J/mol?

Answer 1

Solution:

Arrhenius equation

Ln k = - Ea/R x ( 1/T)

Here k is rate constant , T is temperature , R is gas constant and Ea is activation energy.

Here ln k is the function of 1/T . So we should plot ln k Vs 1/T

We are given temperature in K. and value of rate constant k as well.

A.

Point 1 :

y = ln K = ln ( 0.389) = -0.94418

x = 1/T = 1/325 = 0.003077

Point 2 :

y ln K = ln (0.735 ) = -0.30788

x = 1/T=1/825 = 0.001212

B.

Rise = y2-y1 =(0.30788 – (-0.94418))=0.636291

Run = x2-x1= 0.001212 – 0.003077= -0.00186

Slope = rise / run = 0.63291 / - 0.00186 = -341.211

c.

Activation energy calculation :

We use Arrhenius equation in the form of y =mx+c

Ln k = - (Ea/R) 1/T

Here y = ln k

Slope = - Ea/R

R is gas constant and its value is 8.314 J per K per mol

Therefore Ea = - slope x R

=-(-341.211) x 8.314

= 2836.83 J /mol

Ea = 2836.8 J /mol