## 15.61 (6th edition)

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

Schuyler_Howell_4D
Posts: 66
Joined: Fri Sep 28, 2018 12:28 am
Been upvoted: 1 time

### 15.61 (6th edition)

Can someone explain where they got the equation they use to solve for Ea:

ln(k'/k)=(Ea/R)[(1/T)-(1/T')]

Andrea Zheng 1H
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am

### Re: 15.61 (6th edition)

They got this by using the equation ln(k) = -Ea/(R*T). Because you are finding the difference in two different k values, you would use ln(k')-ln(k), which is ln(k'/k). On the other side of the equation, you would have Ea (which is constant) and R (which is constant) and T (which changes). This means you would end up with Ea/R multiplied by the difference in temperatures, which has 1/Temperature because temperature is in the denominator in ln(k) = -Ea/(R*T). Therefore, you get ln(k'/k) = (Ea/R)[(1/T)-(1/T')].