Question

Consider this reaction data:

T(k)	k(s^-1)
325	.399
775	.621

If you were going to graphically determine the activation energy of this reaction, what points would you plot?

Determine the rise, run, and slope of the line formed by these points.

What is the activation energy of this reaction?

Answer 1

The points plotted using inverse of temperature (1/K) on X -axis and natural logarithm of rate constant (ln K /s) on X axis

Activation energy from the graph :

slope, m = rise/run

rise = y2-y1 = -0.476 - (-0.919) = 0.443 and run = x2-x1 = 0.00129-0.00308 = - 0.00179

slope = rise/run = - 0.443/0.00179 = -247.4 K

the equation of straight line of a graph = y = -247.4x - 0.156

slope = -247.4 K = -Ea/R

Ea = activation energy and R = ideal gas constant = 8.314 J/mol K

247.4 K x 8.314 J/mol K= Ea

Activation energy, Ea = 2056.88 J/mol = 2.06 x 10³ J/mol = 2.06 kJ/mol =