## Activation energy and k

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

Olivia Young 1A
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### Activation energy and k

Why is it that when the activation energy increases, k decreases?

harshitasarambale4I
Posts: 58
Joined: Fri Sep 28, 2018 12:26 am

### Re: Activation energy and k

We can determine this trend by looking at the Arrhenius equation:
ln k = ln A - Ea/RT

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

### Re: Activation energy and k

Looking at the equation: k= A * e^(-Ea/RT) = A / e^(Ea/RT)
When activation energy (Ea) increases, the term "e^(Ea/RT) increases as well. Since that term is in the denominator, the bigger the denominator, the smaller the overall rate (k).

The other factors that affect the rate, k, are the frequency factor, A, and the temperature, T.
When the frequency factor (A) increases, the numerator increases, so the overall rate (k) increases as well.
When the temperature increases, the term "e^(Ea/RT)" becomes smaller, and with a smaller denominator, the overall rate (k) increases.