Equilibrium Constant

The last step in the biochemical pathway of glycolysis is:

Phosphoenolpyruvate + ADP Pyruvate + ATP

Its Standard Gibbs Free Energy, ΔGo, at 25C is -31.4 kJ.mol-1.

What is the equilibrium constant, Ke ?

(ΔGo = - RT ln{Ke}; R = Gas Constant = 8.3145 J.K-1.mol-1 ; T = absolute temperature)

If all reactants and products are at equal concentrations, in which direction will the reaction move naturally i.e. left or right ?

Answers
This is a very confusing question - where is it from? Well, I say it's confusing, but the maths part is half rearranging the equation, fairly simple...If deltaGo is -31.4, and R is 8.3145, then-31.4 = -8.3145 x (25 + 273) x ln(Ke) .......... [to convert celcius to Kelvin add 273.]-31.4 = -8.3145 x 298 x ln(Ke)ln(Ke) = -31.4 / (-8.3145 x 298)Ke = e^(-31.4 / (-8.3145 x 298))I'm not sure how to answer the last part of the question because I do not know what the reactants and products are in the reaction. You've only provided "Phosphoenolpyruvate + ADP Pyruvaate + ATP" which doesn't tell me much. I'd write out the equation for the equilibrium constant, with the reactants and products substitutes and work it out that way.Sorry I can't be more help, but hope it's been at least partly useful!
jamxmitchell
19 October 2014
Ah-hah! I see the problem, your question didn't include the equilibrium sign, so it should read:Phosphoenolpyruvate + ADP <---> Pyruvaate + ATPif "<--->" were the equilibrium sign!
jamxmitchell
19 October 2014
So, the order of each chemical involved is 1, so the expression for the equilibrium constant is:Ke = ([Pyruvate] x [ATP]) / ([Phosphoenolpyruvate] x [ADP])If the concentrations of all product and reactants are the same therefore, the Ke will be 1, as any number divided by itself is 1.So, then look at the Ke value you just calculated in the method I showed below. Is it greater than 1 or less than one?If it is greater than 1 then the top of the fraction is greater than the bottom, therefore there are more right-hand side products normally than left, so if your calculated Ke is more than 1 then the reaction MOVES to the right.If your calculated Ke is less than 1 then there are more left-side reactants, so the equilibrium will MOVE to the left.Hope this helps!
jamxmitchell
19 October 2014
Thanks for the help! I think I understand it now, it didn't help that the equilibrium sign was missing. It's meant to read Phosphoenolpyruvate + ADP  <---> Pyruvate + ATP (as you wrote it). I got an answer of 318061.5, and I think it's unit-less as they cancel but I'm not too sure? So that would make the equilibrium move to the right as Ke > 1
rachel25
22 October 2014
Hmm, I get an answer for Ke:e^(-31.4 / (-8.3145 x 298)) = 1.01And I agree, the equilibrium would move to the right.Where did you get this question from? Is it an exam question? If so there should be a mark scheme on the exam board's website for that particular paper, and you should be able to find the answer there. I'm curious myself to what the answer is!Glad to help, at least I hope I helped get the answer right, it's definitely not a typical question.
jamxmitchell
23 October 2014
Oh, and yes, Ke would be unit-less (I think) as it's a constant, and not a measurement of anything.
jamxmitchell
23 October 2014
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