[Big-Daddy]

So the problem is: the activation energy for Rxn. 2 is n times that of Rxn. 1. When the temperature under which Rxn 1 is run rises from T₁ to T₂, the rate constant is increased by a factor of a. In terms of a, what is the increase (factor of multiplication) in the rate constant when Rxn 2 is taken from T₁ to T₂?

I wrote:
k₁ refers to rate constants under T₁, k₂ as under T₂

a=k₂(Rxn 1)/k₁(Rxn 1)=e^((-Ea(Rxn 1)/R)*((1/T₂)-(1/T₁))

Now we can write the Rxn 2 rate constant as k₁(Rxn 2)=A*e^((-2*Ea(Rxn 1))/(R*T₁)) which translates into (let our scale factor be s):

s=k₂(Rxn 2)/k₁(Rxn 2)=e^(((-2*Ea(Rxn 1))/R)*((1/T₂)-(1/T₁))

I clearly have just one step to take to express s in terms of a - at least I hope so - but how do I do it? I might just be missing something silly ...

[Corribus]

Seems more a math problem than a chemistry problem.  Do you have a correct answer?  I got
[tex]s = e^{n \ln a}[/tex]
If n = 1 (activation energies are equal), it reduces to s = a, so that's a good sign.

I can explain how I derived it if that is helpful.  I had to use a substitution trick.

[Big-Daddy]

Seems more a math problem than a chemistry problem.  Do you have a correct answer?  I got
[tex]s = e^{n \ln a}[/tex]
If n = 1 (activation energies are equal), it reduces to s = a, so that's a good sign.

I can explain how I derived it if that is helpful.  I had to use a substitution trick.

Well, is my working correct so far?
I should clarify that when I wrote: s=k₂(Rxn 2)/k₁(Rxn 2)=e^(((-2*Ea(Rxn 1))/R)*((1/T₂)-(1/T₁))
It should really have been: s=k₂(Rxn 2)/k₁(Rxn 2)=e^(((-n*Ea(Rxn 1))/R)*((1/T₂)-(1/T₁))

As I was working previously with a special case of n=2 but that is unnecessary I think.

I'm thinking that mathematically, we've got a=e^(-x) and s=e^(-n*x). We can then rewrite s=(e^(-x))^n and so s=a^n. Which I think is the simpler way of writing your formula 

So if n=2, s=a². Well that's nice! Thanks for all the help. Had your post not got me thinking on the right lines I would never have realized I could express it like this.

[Corribus]

Well, is my working correct so far?

Honestly, with the way you formatted your equations it's hard to tell.  If you use LaTex it's much easier to see what you're doing.  For most simple formulas it's pretty easy to learn.

I'm thinking that mathematically, we've got a=e^(-x) and s=e^(-n*x). We can then rewrite s=(e^(-x))^n and so s=a^n. Which I think is the simpler way of writing your formula 

Yes they appear to be equivalent.

[curiouscat]

Honestly, with the way you formatted your equations it's hard to tell.  If you use LaTex it's much easier to see what you're doing.  For most simple formulas it's pretty easy to learn.

Seconded.

[Big-Daddy]

Sorry I will try to do this in the future.