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

[Danny.Boy]

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 1°C or 25°C? 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 realize 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

 

[DrDu]

When using ΔG=ΔG°+RT ln Q to calculate the energy yield of a reaction, does it matter if I use ΔG° calculated at 1°C or 25°C?

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_'t_Hoff_equation

 

[Danny.Boy]

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 50ºC (i.e., 323.15K):

[A]+→[C]+[D]

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

ΔG=ΔG°+R×323.15×ln (([C][D])/([A]))

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

 

[DrDu]

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

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]

Aha! Makes sense. Thanks for explaining that.