Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)?
It isn't on the equation sheet my teacher gave our class
take a look at your own equations:
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) = ΔH - TΔS
now: try to combine those two equations
shouldn't be too hard to do...
and what do you use it to solve
with K(T) given, calculate [itex]\Delta[/itex] G (T) belonging to , for example:
K298 K = 10-14 [itex]\Delta G[/itex]298 K = - RT lnK = - R * 298 * ln (10-14)
so, you calculate [itex]\Delta G[/itex]298 K and [itex]\Delta G[/itex]283 K
... and put up your secondary equaitions of the general structure ΔG = ΔH - TΔS
for example: [itex]\Delta G[/itex]283 K = ΔH - 283 K* ΔS
two equations with two unknowns resulting thereof
regards
Ingo
I know that much but do I set TΔS = to 0 for both??
I did this:
ΔG
298.15 K = 774.95
ΔG
283.15 K = 786.25
I know how to do that no problem.
But what I DON'T know is how to set up both equations in such a way that I can find ONE of the variables I'm looking for, especially when none of them cancel out...
(786.25) = ΔH - (283.15)ΔS
-(774.95) = ΔH - (298.15)ΔS
^ when I do this, I get an extremely small number for ΔS which is completely 100% wrong.
(786.25) = ΔH - (283.15)ΔS
+(774.95) = ΔH - (298.15)ΔS
When I do this, i go nowhere...because I end up with
1561.12 = 2 ΔH - 581.3 ΔS. I STILL have two unknown variables.