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Gibbs Free Energy of Formation: 1C or 25C? (And other exciting questions.)

by Danny.Boy
Tags: 25c, energy, exciting, formation, free, gibbs
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Danny.Boy
#1
Dec21-12, 10:02 AM
P: 3
Hi there:

When using ΔG=ΔG+RT ln Q to calculate the energy yield of a reaction, does it matter if I use ΔG calculated at 1C or 25C? Also, why are there two choices and when are they each applicable? Finally, I have also seen ΔG' written (note the prime). What does this mean and how does it differ from ΔG conceptually and numerically?

I realise that these are very basic questions, so if you want to point me towards some elementary reading material I understand.

Thanks in advance,
Danny.Boy
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DrDu
#2
Dec21-12, 11:22 AM
Sci Advisor
P: 3,571
Quote Quote by Danny.Boy View Post
When using ΔG=ΔG+RT ln Q to calculate the energy yield of a reaction, does it matter if I use ΔG calculated at 1C or 25C?
Yes, it matters. If you use Delta G at 1 deg C, then you will also have to use a corresponding T and then equilibrium constant will be that for this temperature. The same way using Delta G at 25 deg Celsius yields Q for this very temperature.
The change of Q with temperature can be calculated with the van't Hoff equation:
http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation
Danny.Boy
#3
Feb4-13, 12:18 PM
P: 3
Thanks for your reply DrDru, but I'm afraid I don't really follow. Perhaps a concrete example would help me understand. For example, consider this reaction at 50C (i.e., 323.15K):

[A]+[B]→[C]+[D]

Using ΔG=ΔG+RT ln Q, I get something like this:

ΔG=ΔG+R323.15ln (([C][D])/([A][B]))

but what is the value of ΔG that I should use? The value at 1C or 25C?

DrDu
#4
Feb4-13, 02:38 PM
Sci Advisor
P: 3,571
Gibbs Free Energy of Formation: 1C or 25C? (And other exciting questions.)

Quote Quote by Danny.Boy View Post
but what is the value of ΔG that I should use? The value at 1C or 25C?
Neither of the two but the value at 50 deg. Celsius.
If you have both the values at 1 and at 50 degrees, you could linearly interpolate as a first approximation.
But, as I said, it would be more exact to calculate the value of Delta G0 at 50 degs from the van't Hoff or Gibbs-Helmholtz equation, see:
http://en.wikipedia.org/wiki/Gibbs-Helmholtz_equation

E.g. ##T_1=1^\circ##C, ##T_2=25^\circ##C and ##T_3=50^\circ##C,
then
##\Delta G^0(T_1)/T_1-\Delta G^0(T_2)/T_2=\Delta H ^0(1/T_1-1/T_2)##.
Solve this for ##\Delta H^0## and then solve
##\Delta G^0(T_1)/T_1-\Delta G^0(T_3)/T_3=\Delta H ^0(1/T_1-1/T_3)##
for ##\Delta G^0(T_3)##.
Danny.Boy
#5
Feb4-13, 02:41 PM
P: 3
Aha! Makes sense. Thanks for explaining that.


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