A reaction has a standard free-energy change of –10.30 kJ mol–1 (–2.462 kcal mol–1). Calculate the equilibrium constant at 25ºC.?
This is what I've done so far..
∆Gº= -2.3 RT + log (K)
-10.31 kJ = -2.3 (.0083145)(298) + log K
-10.31 kJ = -2.3 (2.48) log K
log K = 10.31kJ / -5.704
log K = -1.808
K = 10^ -1.808
K = .0155
Apparently this is incorrect and I'm unsure of how to continue. I've tried using e (-∆Gº/RT) and that hasn't worked either.. and when I type in the number I've got it tells me "Incorrect. The standard free energy change is negative." I changed my answer to negative and it's still incorrect..
"A reaction has a standard free-energy change of –10.30 kJ mol–1 (–2.462 kcal mol–1). Calculate the equilibrium constant for the reaction at 25 °C."
I have tried using both e and the equation above (which may be incorrect for this problem, though according to the application it's the one they want me to use.
hahaha wrong pogger
I think your problem is the wrong formula. I believe that's dG = -2.303RT*log K
I don't know what you're doing, and it would help if you showed your work so we could find the error, but I get something like 60 or so for K.
Are you using dG = 10310 J? You may be using kJ or cal or Kcal.
Are you using 8.314 for R?
Excuse me. I re-read everything and you have all of that info at the beginning.
If you use dG = -2.303RTlnK and your numbers you come out with about 60 which is what I obtained with dG = -RTlnK.
You must be pushing the wrong buttons somewhere.
I think I tried that. Let me show u my notes:
-10.30 kJ = -(.0083145)(298) lnK
-10.30 kJ = -2.478 lnK
-10.30/-2.478 = lnK
4.157 = lnK
e^(4.157) = 63.88 = Incorrect.
I'll even try the inverse
-2.478/-10.33 =lnK
.2399 = lnK
e^(.2399)= 1.27 = incorrect
I'm using .0083145 for R because the -10.31 is in kJ
I understand. I leave those intermediate answers in the calculator; writing them down and re-entering often leads to rounding errors. I think 63.8985 which I would round to 64.0 is the correct answer.
I COULD KISS YOU RIGHT NOW
To bad I missed out on that (or those). :-) And while I'm here let me correct an error above in one of my posts. In one of them I gave dG = -2.303RTlnK as the correct formula and it isn't. The correct formula is either
dG = -2.303RTlogK OR
dG = -RTlnK