An experimental plot of ln(k) vs. 1/T is obtained in lab for a reaction. The slope of the best-fit line for the graph is -2610 K. What is the value of the activation energy for the reaction in kJ/mol?

Answer 1

This represents an Arrhenius plot. The linear equation is as follows:

ln k = -Ea/R (1/T) + ln A in the form of y = mx + b (straight line)

slope = -Ea/R

-Ea/R = -2610K

Ea/R = 2610K

Ea = 2610K * 8.314 J/mol-K

Ea = 21,700 J/mol-K = 21.70 kJ/mol