Question
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.
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.
Answers
I think your problem is the wrong formula. I believe that's dG = -2.303RT*log K
Nope. That gives me .0156621, and when I input that I got the response "Incorrect. The standard free energy change is negative."
I changed it to a negative and it still didn't work.
I changed it to a negative and it still didn't work.
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?
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.
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.
-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
-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
dG = -2.303RTlogK OR
dG = -RTlnK
hahaha wrong pogger
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