Ok, let me rephrase, because I see what I wrote was confusing.
What I meant was: Now, how do you relate each of the quantities on the right hand side of the equation in terms of the starting pressure of the reactant at time zero and the pressure of the reactant at time t? Then you can worry about substitution for f.
(Don't worry about moles - that's just confusing things. Everything can be done with pressures, which are your concentrations.)
I'm not sure. I could manage this relation in terms of moles but I don't get how to draw the relation in terms of pressures (since the reaction equation occurs in terms of moles).
My thoughts are still like this: n
t=n
0*(1+(3/2)f). I can't help but feel this means P
t=P
0*(1+(3/2)f), but the question is how to quantitatively prove that the n relationship is equivalent to the P relationship? If it's harder then we can leave that aside and focus on the pressures for now, but I'm not really sure where to start with them.
Edit: I see. What I should have mentioned before is that this reaction occurs in a closed-volume container.
With this in mind, any relation I can find between moles also applies to pressures (as I can replace n in any situation with PV/(RT), assuming ideal gases, and then divide through by V/(RT) to leave an expression where P has simply replaced n, as V is constant). So P
t=P
0*(1+(3/2)f) is therefore proven.
It's also quite easy to show that if we start with n
0[N2O5], n
0[O2] and n
0[NO2], then: (subscript 0 means initial, subscript t means at time t, [total] means total) n
t[total]=n
0[N
2O
5]*(1+(3/2)f)+n
0[NO2]+n
0[O2]. This translates into P
t[total]=P
0[N
2O
5]*(1+(3/2)f)+P
0[NO2]+P
0[O2]. If we then substitute P
0[component]=P
0[total]*n
0[Component]/n
0[total], we end up with:
P
t[total]=
(P
0[total]*((n
0[N
2O
5]*(1+(3/2)f))+n
0[NO
2]+n
0[O
2])
)/(n
0[N
2O
5]+n
0[NO
2]+n
0[O
2]). I should think this formula is correct particularly since it reduces to my last one when n
0[NO
2]=n
0[O
2]=0.
Thanks for the help. Can you think of any extension questions I might use to go further?