Asked by Heidi

Find the equilibrium constant for the following reaction:

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

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

Would I just use the ΔG° =-RTlnK?
Answered by Heidi
Temperature is at 25C, I forgot to add that!
Answered by DrBob222
Yes, but first you must find dGo for the reaction. That is dGorxn = (n*dGo products) - (n*dGo reactants). Remember to use 8.314 for R.
Answered by Heidi
d as in derivative? How would I factor in the 2L though? That's also throwing me off.
Answered by Heidi
d=difference. Wow, okay- makes more sense now.
Answered by Heidi
I was thinking this entire time 2L meant two liters. My professor just didn't put a space between the 2 and the L. L has nothing to do with the two, it has to do with the compound itself.
Answered by DrBob222
The 2 is n in the work I provided; i.e.,
dGorxn = (1*dGo products) - (2*dGo reactants. And yes, I don't know how to write a delta sign so dGo means delta Go. The capital D refers D glucose and capital L to L lactose.
Answered by Heidi
Thank you! I would've been able to handle this problem without any help if I had realized it 2=n instead of 2L.
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