[confusedstud]

ΔG=ΔG°+RTlnQ is the equation used when non-standard conditions are used. However, I'm not very sure if the Q refers to Qc whereby the concentrations in M is used or Qp whereby the partial pressures are used.

Because if we were to substitute Qp or Qc into the above equation we would get different results.

Thanks

[mjc123]

Haven't we been here before? http://www.chemicalforums.com/index.php?topic=81176.0
The quantities you insert into Q must be made dimensionless by dividing by the standard value e.g. P/P°. If you are consistent with your standard values, you should get the same answer whatever units you use. Remember that for gases, if the standard pressure is 1 atm, the standard concentration is not 1M, but 1/24 M, or generally (n/V)° = P°/RT°.

[Old_Doc48]

Haven't we been here before? http://www.chemicalforums.com/index.php?topic=81176.0
The quantities you insert into Q must be made dimensionless by dividing by the standard value e.g. P/P°. If you are consistent with your standard values, you should get the same answer whatever units you use. Remember that for gases, if the standard pressure is 1 atm, the standard concentration is not 1M, but 1/24 M, or generally (n/V)° = P°/RT°.

... 'Because if we were to substitute Qp or Qc into the above equation we would get different results'...

If you are referring to working only with Q_p => 'Atm' ratios, or Q_c => 'Molar' ratios, you would be correct, units would cancel and the term would have no units. However, (and forgive if I've totally misunderstood your comment) Q_p and Q_c are not interchangeable in the Nernst Equation. K_p = K_c(RT)^Δn under standard conditions and  Q_p = Q_c(RT)^Δn non-standard conditions. The only time K_p and K_c are equal is when Σmolar volumes of gas reactants = Σmolar volumes of gas products;i.e., Δn = 0 => K_p = K_c(RT)^{Δn = 0} => K_p = K_c(1) and Q_p = Q_c(RT)^{Δn = 0} => Q_p = Q_c(1). Again, forgive if I've misunderstood your post.
 

[confusedstud]

Haven't we been here before? http://www.chemicalforums.com/index.php?topic=81176.0
The quantities you insert into Q must be made dimensionless by dividing by the standard value e.g. P/P°. If you are consistent with your standard values, you should get the same answer whatever units you use. Remember that for gases, if the standard pressure is 1 atm, the standard concentration is not 1M, but 1/24 M, or generally (n/V)° = P°/RT°.

For the concentration variation of Q (Qc) it would be c/c° right? so if we had an equation A+B C+D Qc would be equals to ([C][D]/[A][B ])/(1/24M)?

But why would it give us the same Qc and Qp if we divided it by 1/24M?

(mod edit to negate board formatting, sjb)

[Yggdrasil]

Q_c is the correct quantity to use.  The ΔG=ΔG°+RTlnQ relation can be derived from statistical mechanics, and you arrive with Q_c at the end, not Q_p.

[mjc123]

If you are referring to working only with Qp => 'Atm' ratios, or Qc => 'Molar' ratios, you would be correct, units would cancel and the term would have no units. However, (and forgive if I've totally misunderstood your comment) Qp and Qc are not interchangeable in the Nernst Equation. Kp = Kc(RT)Δn under standard conditions and  Qp = Qc(RT)Δn non-standard conditions. The only time Kp and Kc are equal is when Σmolar volumes of gas reactants = Σmolar volumes of gas products;i.e., Δn = 0 => Kp = Kc(RT)Δn = 0 => Kp = Kc(1) and Qp = Qc(RT)Δn = 0 => Qp = Qc(1). Again, forgive if I've misunderstood your post.

I'm not sure I get you. My point (made at more length in the post I linked to) is that Q must be a dimensionless quantity because you can only take the log of a number, not a physical quantity. Neither a Q_p in atm^Δn nor a Q_c in M^Δn will do. For this purpose you divide each P (or C) by the standard value to give a dimensionless number. Provided the standard condition is the same (e.g. 1 atm ≡ 1/24M), P/P° = C/C°, and Q is numerically the same whichever basis you use.
The Q_p to use is not (e.g.) P_CP_D.../P_AP_B..., but (P_C/P°)(P_D/P°).../(P_A/P°)(P_B/P°)..., which will be the same numerically if P° = 1 atm and P is in atm.
Q_c gets difficult to type, but replace (P_C/P°) by [C]/[C]° = (P_C/RT)/(P°/RT) = (P_C/P°)
Of course the strict proportionality of P and C assumes ideal gases, but I don't think your objection was based on non-ideality.

[Corribus]

must be a dimensionless quantity because you can only take the log of a number, not a physical quantity.

And yet, we do it all the time when we calculate pH. Of course this doesn't make it right - only that as chemists our bookkeeping can get sloppy. I've read that for pH we should actually be using dimensionless proton activities and not concentrations. But how many chemists do this, or at least, how many chemists recognize that every time they're making a pH calculation using a proton concentration, it may actually be a bit of mathematical nonsense?

Ah well, a topic of conversation for another day I guess. Still, I offer the following as an interesting read:

http://pubs.acs.org/doi/abs/10.1021/ed1000476?journalCode=jceda8

[Yggdrasil]

If you are referring to working only with Qp => 'Atm' ratios, or Qc => 'Molar' ratios, you would be correct, units would cancel and the term would have no units. However, (and forgive if I've totally misunderstood your comment) Qp and Qc are not interchangeable in the Nernst Equation. Kp = Kc(RT)Δn under standard conditions and  Qp = Qc(RT)Δn non-standard conditions. The only time Kp and Kc are equal is when Σmolar volumes of gas reactants = Σmolar volumes of gas products;i.e., Δn = 0 => Kp = Kc(RT)Δn = 0 => Kp = Kc(1) and Qp = Qc(RT)Δn = 0 => Qp = Qc(1). Again, forgive if I've misunderstood your post.

I'm not sure I get you. My point (made at more length in the post I linked to) is that Q must be a dimensionless quantity because you can only take the log of a number, not a physical quantity. Neither a Q_p in atm^Δn nor a Q_c in M^Δn will do. For this purpose you divide each P (or C) by the standard value to give a dimensionless number. Provided the standard condition is the same (e.g. 1 atm ≡ 1/24M), P/P° = C/C°, and Q is numerically the same whichever basis you use.
The Q_p to use is not (e.g.) P_CP_D.../P_AP_B..., but (P_C/P°)(P_D/P°).../(P_A/P°)(P_B/P°)..., which will be the same numerically if P° = 1 atm and P is in atm.
Q_c gets difficult to type, but replace (P_C/P°) by [C]/[C]° = (P_C/RT)/(P°/RT) = (P_C/P°)
Of course the strict proportionality of P and C assumes ideal gases, but I don't think your objection was based on non-ideality.

Because the standard pressure used for Q_p and the standard concentration for Q_c are not equivalent, ΔG° will differ depending on what you consider your standard concentrations (1M or 1 atm).  Most ΔG° values will be given based on using 1M as the standard concentration, so Q_c is the appropriate value to use.  If the ΔG° was determined relative to a 1 atm standard, Q_p would be the more appropriate quantity to use.

[confusedstud]

If you are referring to working only with Qp => 'Atm' ratios, or Qc => 'Molar' ratios, you would be correct, units would cancel and the term would have no units. However, (and forgive if I've totally misunderstood your comment) Qp and Qc are not interchangeable in the Nernst Equation. Kp = Kc(RT)Δn under standard conditions and  Qp = Qc(RT)Δn non-standard conditions. The only time Kp and Kc are equal is when Σmolar volumes of gas reactants = Σmolar volumes of gas products;i.e., Δn = 0 => Kp = Kc(RT)Δn = 0 => Kp = Kc(1) and Qp = Qc(RT)Δn = 0 => Qp = Qc(1). Again, forgive if I've misunderstood your post.

I'm not sure I get you. My point (made at more length in the post I linked to) is that Q must be a dimensionless quantity because you can only take the log of a number, not a physical quantity. Neither a Q_p in atm^Δn nor a Q_c in M^Δn will do. For this purpose you divide each P (or C) by the standard value to give a dimensionless number. Provided the standard condition is the same (e.g. 1 atm ≡ 1/24M), P/P° = C/C°, and Q is numerically the same whichever basis you use.
The Q_p to use is not (e.g.) P_CP_D.../P_AP_B..., but (P_C/P°)(P_D/P°).../(P_A/P°)(P_B/P°)..., which will be the same numerically if P° = 1 atm and P is in atm.
Q_c gets difficult to type, but replace (P_C/P°) by [C]/[C]° = (P_C/RT)/(P°/RT) = (P_C/P°)
Of course the strict proportionality of P and C assumes ideal gases, but I don't think your objection was based on non-ideality.

Because the standard pressure used for Q_p and the standard concentration for Q_c are not equivalent, ΔG° will differ depending on what you consider your standard concentrations (1M or 1 atm).  Most ΔG° values will be given based on using 1M as the standard concentration, so Q_c is the appropriate value to use.  If the ΔG° was determined relative to a 1 atm standard, Q_p would be the more appropriate quantity to use.

This makes a lot more sense to me. Thank you

[mjc123]

ΔG° will differ depending on what you consider your standard concentrations (1M or 1 atm). 

Quite right, and therefore I said, in the other thread I referred to, that the standard state must be that used in the definition of ΔG°.

Because the standard pressure used for Qp and the standard concentration for Qc are not equivalent,

But they must be, for precisely that reason.
Customarily, the standard state for a solution is 1M concentration, and the standard state for a gas is 1 atm pressure. This is what you will typically find for quoted values of standard thermodynamic functions.
Assuming we are talking about gases (otherwise the question of Q_c vs. Q_p will not arise), the standard state is 1 atm pressure or equivalently 1/24 M concentration. It is not "1 atm if we're talking pressure but 1 M if we're talking concentration" - these would have, as you say different G°.
In that case the normalised pressure and concentration are the same - if P_A, say, is 2 atm, then the pressure of A is twice the standard pressure, and its concentration is twice the standard concentration. Either way, the number 2 goes into the expression for Q.