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#### Simons2

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##### Equilibrium Constant?
« on: October 12, 2017, 02:08:07 PM »

Find the equilibrium constant for the following reaction at 25C (298K):

D-Glucose (aq) <-> 2L (+) lactic acid (aq)

ΔG ° for D-Glucose= -914.5kJ/mol
ΔG ° for lactic acid = -538.8kJ/mol

The 2L (+) is really throwing me off, I'm not sure how I would factor that into the equation; ΔG°=-RTlnK.

I know I subtract products and reactants and -RTlnK =-(8.314)(298)lnK.
« Last Edit: October 12, 2017, 02:59:14 PM by Simons2 »
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#### XeLa.

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##### Re: Equilibrium Constant?
« Reply #1 on: October 12, 2017, 02:57:08 PM »

The L-(+)-lactic acid represents an optical (geometric) isomer of lactic acid. Just think of the equation as:

D-Glucose(aq) 2 Lactic acid(aq)

I'm not sure how I would factor that into the equation; ΔG°=-RTlnK.

That equation looks right to me - now how are you going to use it?

EDIT

XeLa
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#### Simons2

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##### Re: Equilibrium Constant?
« Reply #2 on: October 12, 2017, 03:01:20 PM »

The L-(+)-lactic acid represents an optical (geometric) isomer of lactic acid. Just think of the equation as:

D-Glucose 2 Lactic acid

(the other information isn't involved in your calculations). That equation looks right to me - now how are you going to use it?

EDIT

XeLa

I was thinking this whole time 2L meant 2 liters. It's 2 L(+)? So my final equation would be (2*(-538.8 )-(-914.5)=-(8.314)(298)lnK and K=1.07. Spaces really would've been helpful in the original problem.
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