Question

The values of the rate constant of a reaction were

determined at several temperatures. A plot

logk Vs 1/T gave a straight line with slope of -2.1x10 K. What is the activation energy of the?​

Answer 1

Answer:

According to Arrhenius equation

logk=logA−

2.303

E

a

⋅

T

1

Plot of log k vs

T

1

is a straight line.

Slope = −

2.303R

E

a

Intercept = log A

solution