July 01, 2022, 07:27:18 PM
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Topic: Deriving K when given free energy/non-standard temp  (Read 1618 times)

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Offline cormbreb

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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.

Offline Babcock_Hall

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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°.

Offline cormbreb

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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!

Offline Babcock_Hall

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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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