Question

For a reaction when log k was
plotted against 1/T, a straight line
with a slope of - 6000 was
obtained. What is the activation
energy of the reactian?​

Answer 1

Answer:

Ea​=6008.314

Explanation:

In order to solve the problem given in question statement, we would be using Arrhenius equation

This equation is written as

K = Ae−Ea/RT

When we take log on both sides

Logk = RT−Ea + log

Comparing this equation with straight line equation, we get

m=R−Ea

Here m represents the slope of line, which is given as -6000

−6000=8.314−Ea

Ea=6008.314