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Deriving K when given free energy/non-standard temp
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Topic: Deriving K when given free energy/non-standard temp (Read 2940 times)
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cormbreb
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Posts: 2
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Deriving K when given free energy/non-standard temp
«
on:
May 12, 2022, 01:55:20 PM »
I'm preparing for my gen chem 2 final, and I'm unable to get in touch with anyone who normally can help me so I figured I would ask here. I'm just stuck on what to do with this.
What is the equilibrium constant for the reaction below at 298K?
The ∆G° for the reaction below is -10.36 kJ/mol.
2A
(g)
+ 2B
(g)
⇌ C
(g)
+ 2D
(g)
I keep getting an answer choice that is
available
but isn't correct. I got it by plugging in the info given to the ∆G° = -RT(lnK) equation and solving for K. (I'm getting 1.004 as my answer.)
I used 8.314 as my R and 298K as T.
The answer key say it's actually 65.37, but I don't know how to arrive at that value.
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Babcock_Hall
Chemist
Sr. Member
Posts: 5592
Mole Snacks: +319/-22
Re: Deriving K when given free energy/non-standard temp
«
Reply #1 on:
May 12, 2022, 02:07:53 PM »
Check the units of R and ΔG°.
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cormbreb
Very New Member
Posts: 2
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Re: Deriving K when given free energy/non-standard temp
«
Reply #2 on:
May 12, 2022, 02:27:13 PM »
I finally figured it out while you were writing I think. Two key errors were made: I hadn't converted R by multiplying x 1000 and I had not switched the sign of ΔG° to a positive.
OOF. Thanks!
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Babcock_Hall
Chemist
Sr. Member
Posts: 5592
Mole Snacks: +319/-22
Re: Deriving K when given free energy/non-standard temp
«
Reply #3 on:
May 12, 2022, 03:16:43 PM »
R should be
divided
by 1000, to put it into kJ, or ΔG° can be multiplied by 1000, to put it into Joules. One way or the other, the units must match. The sign is also important, as you implied.
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Deriving K when given free energy/non-standard temp