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Topic: standard gibbs free energy of Octane  (Read 7293 times)

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Offline Liz

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standard gibbs free energy of Octane
« on: December 13, 2008, 09:21:25 PM »
Hi

I have a question, I think there is something missing there but I am not really sure.

the standard gibbs free energy of octane (l) is 6.4 JK/mol at 298 K. (this is the only given info)
would we have to increase or decrease the temperature if we want to shift the equilibrium to the right, forming more octane.
 8C(graphite) + 9H2 <=> C8H18 (l)

I used dG = dH - TdS
dG is positive.
dS is negative, because we are consuming H2 gas and we are going to less ordered system.
-TdS is therefore postive.
now I got stuck in determining the sign of dH. Because dG is + and -TdS is +, then dH can only be positive number or small negative number.

I cannot use any other info other than what is given, and I should not do any calculation, only thinking. but I need the sign of dH in order to determine whether we have to increase or decrease the temperature in order to produce more of liquid octane.



the other part is asking about whether we need to increase or decrease the pressure so that H2 is in equilibrium with both graphite and octane.

I am not really sure how to do this part but here is what I did.

dG= dGo + RT ln (1/P9H2)

at equilibrium,

0 = dGo + RT ln (1/P9H2)
0 = 6400 + 8.3145 * 298 * ln (1/P9H2)
rearrange, PH2 = 1.33 bar.
which means that we have to increase the pressure from 1 bar (standard pressure) to 1.33  bar.
I am not really sure if this is right.

Can someone help me, or tell me if I am missing something?

Thanks,

don

Offline Rabn

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Re: standard gibbs free energy of Octane
« Reply #1 on: December 17, 2008, 01:06:01 AM »
1) You should reconsider your approach.  You are asked to consider a shift in equilibrium, is there a way that dG relates to equilibrium?

2) Method looks good, didn't check the numbers.

Offline Liz

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Re: standard gibbs free energy of Octane
« Reply #2 on: December 17, 2008, 08:17:21 AM »
Thanks Rabn.

for 1) I found K to be 0.0755 from -dGo=RTlnKeq
        but how is it related to increasing or decreasing T in order to produce more Octane(l)?
         And K is related to T, by the vant hoff equation
         ln(K2/K1) = - :delta:Ho/R *(1/T2-1/T1)

        the dH term is still present :S

D

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