
#1
Dec2112, 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 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 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 



#2
Dec2112, 11:22 AM

Sci Advisor
P: 3,370

The change of Q with temperature can be calculated with the van't Hoff equation: http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation 



#3
Feb413, 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 50ºC (i.e., 323.15K):
[A]+[B]→[C]+[D] Using ΔG=ΔG°+RT ln Q, I get something like this: ΔG=ΔG°+R×323.15×ln (([C][D])/([A][B])) but what is the value of ΔG° that I should use? The value at 1ºC or 25ºC? 



#4
Feb413, 02:38 PM

Sci Advisor
P: 3,370

Gibbs Free Energy of Formation: 1°C or 25°C? (And other exciting questions.)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 GibbsHelmholtz equation, see: http://en.wikipedia.org/wiki/GibbsHelmholtz_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_11/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_11/T_3)## for ##\Delta G^0(T_3)##. 



#5
Feb413, 02:41 PM

P: 3

Aha! Makes sense. Thanks for explaining that.



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