arrhenius equation

[Dylan Do]

Why are there different forms of the arrhenius equation? (refer to images) And do I need to have the ln(k/k) one memorized?

[Sydney]

I am not exactly sure why there are different versions of the equation either, but I think that it is just two different representations of the same information.

[liamtran]

The first form allows you to determine k, Ea, or T, given all other variables in the equation. The second form just allows us to calculate k2 given k1, T1, T2, or any of these variables given the other 3. In short, the second form emphasizes a change in conditions; both forms can be derived from the other though, so you only have to remember one.

[Varun Sekar 2E]

I believe the second equation is showing what happens to the rate of the reaction when we only change the temperature. I don't think you need to memorize because one can derive the formula of the second equation by utilizing the first equation. You do so by dividing the final rate (k2) by the initial rate (k1). You should get k2/k1=Ae^(-Ea/(RT2))/Ae^(-Ea/(RT1)). You can cancel the Frequency Factor (A) and should get k2/k1=e^(-Ea/(RT2))/e^(-Ea/(RT1). All you have to do now is find the natural log (ln) of each side. You should get ln(k2/k1) = ln(e^(-Ea/(RT2))/e^(-Ea/(RT1)). One of the properties of the natural log is that ln(x/y)=ln(x)-ln(y), so you can do the same thing here. You should now get ln(k2/k1) = ln(e^(-Ea/(RT2))) - ln(e^(-Ea/(RT1))). Now since ln(e^x)=x, you now get ln(k2/k1)=-Ea/(RT2)+Ea/(RT1). Since activation energy and R are the same (we are looking at the same reaction but with different temperatures), we can simplify to ln(k2/k1) = Ea/R (1/T1-1/T2).