Thermochem question

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I am having difficulty understanding this equation:

deltaG0 = -RTlnK'eq

and K'eq = [C]eq[D]eq/[A]eq/eq

but at equilibrium aren't the concentrations products equal to the concentrations of reactants? So in that case, wouldn't Keq always reduce to 1? And thus, lnKeq = 0?

Thanks

 

[Dr Gerrard]

No, at equilibrium, rate of the forward reaction = rate of the backwards reaction.

 

[Doodl3s]

I am having difficulty understanding this equation:

deltaG0 = -RTlnK'eq

and K'eq = [C]eq[D]eq/[A]eq/eq

but at equilibrium aren't the concentrations products equal to the concentrations of reactants? So in that case, wouldn't Keq always reduce to 1? And thus, lnKeq = 0?

Thanks

NO, absolutely not... or else the need for Keq wouldn't be there! At equilibrium every reaction has a different set of concentrations.

EDIT:ADD: Gerrard got in first. But he actually may have hit on what you are confusing.. that the rates ARE THE SAME at equilib. NOT THE conc.

 

[sv3]

I am having difficulty understanding this equation:

deltaG0 = -RTlnK'eq

and K'eq = [C]eq[D]eq/[A]eq/eq

but at equilibrium aren't the concentrations products equal to the concentrations of reactants? So in that case, wouldn't Keq always reduce to 1? And thus, lnKeq = 0?

Thanks

This gave me fits before. At equilibrium, the concentrations are what the Keq dictates them to be. So let's say Keq = 2 for a certain reaction, at equilibrium, there will be twice as much product than reactant in terms of concentration - just like the Keq formula. Each reaction has its own equilibrium constant, which will dictate the concentrations of the products and reactants at equilibrium.

As far as deltaG0 = -RTlnK'eq is concerned, that's for standard state conditions. If the Keq is 1, then yes delta G would be 0. IF greater, than you get a negative G which indicates a spontaneous reaction. This makes sense if your Keq is greater than 1 since there would be more product than reactant in a reaction, indicating that the reaction does occur and does so in favour of the products. The opposite applies if Keq<1 since that would give you a positive G.

hope it helps, it's confusing at first but wrap your head around.....something related to this is bound to pop up on the mcat if just in a discrete.

 

[Dr Gerrard]

I am having difficulty understanding this equation:

deltaG0 = -RTlnK'eq

and K'eq = [C]eq[D]eq/[A]eq/eq

but at equilibrium aren't the concentrations products equal to the concentrations of reactants? So in that case, wouldn't Keq always reduce to 1? And thus, lnKeq = 0?

Thanks

Yeah sorry for the lack of explanation.

But if the concentration of products and reactants were always equal at equilibrium, then you are very right is saying Keq = 1, and the right side of the equation would equal 0.

But this makes sense. deltaG0 is deltaG for standard conditions. One stipulation of standard conditions is the concentration of products and reactants initially are all equal to 1M. If they are all equal to 1M, then the ratio would be zero, and Q = K.

Since this signifies the reaction is already at equilibrium, then deltaG would be zero, which is always the case at equilibrium (note, I said delta G, not delta G standard)