CHEMISTRY COMMUNITY

According to a person on Chemistry community, you solve this problem by deriving an equation using the Arrhenius equation and getting rid of A using this method:

k/k' = (e^-Ea/RT)/(e^-Ea/RT')
k/k' = e^(-Ea/RT + Ea/RT')
ln (k/k') = -Ea/RT + Ea/RT'
ln (k/k') = (Ea/R)(1/T' - 1/T)

Why are we able to combine Ea? Isn't the activation energy of the reverse reaction completely different form that of the forward reaction?

15.61 states: "The rate constant of the rst-order reaction 2 N2O(g) + S --> 2 N2(g) + O2(g) is 0.76 s 1 at 1000. K and 0.87 s 1 at 1030. K.

Recall that the rate constant changes with different temperatures. We are still doing the same reaction, but simply at different temperatures with different corresponding rate constants.

This problem isn't asking for the activation of the reverse reaction though, it's asking for the activation of the forward reaction. Activation energy doesn't change with temperature. (you're right that the activation energy WOULD be different for the reverse reaction though, that's just not what the question is asking)

I agree with Lindsay. The question doesn't really need us to think about the activation energy for the reverse reaction. It just asks us to calculate the activation energy for the forward reaction, which doesn't change at different temperatures. I think we derived the equation we use under the assumption that the activation energy remains the same (ex. we'll use it to calculate the Ea for a forward reaction at different temperatures or to calculate the Ea for a reverse reaction at different temperatures).

It might be confusing because the answer book uses k and k', which we usually associate with the forward and reverse rate constants. It might be easier just to think about it as k1 and k2, like the book does on page 642.