Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Acut on June 07, 2012, 07:29:20 PM
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Hi!
I am having some trouble in figuring out how units can be used when writing reaction quotients/ equilibrium constants / Nernst Equation. This discussion started in another forum (http://www.physicsforums.com/showthread.php?t=610792) but it seems I've reached a dead end there, so any help is welcome.
Essentially, I'm puzzled by how we can use apparently arbitrary units when writing the reaction quotient. Suppose we have a very simple reaction, so that its ΔG can be written as
ΔG = - RT ln p0 (equation 1)
So far so good. But what if we change the units? Suppose instead of using atm to measure pressure, we choose another unit, mta, such that 1 mta = 2atm. Well, this means that the new value of pressure p will be p = p0/2. But R also changes when going from atms to mtas, and to fix it, we may use the ideal gas law. Then we get that the new value for R, that I will call R' is half of the original R.
Now, lets restate equation one using our new set of units:
ΔG = - R'T ln p
And, by changing the back to the commonly using atm we get another expression of ΔG:
ΔG = - 0.5 R ln(p0/2) (equation 2)
Now, we are still talking about the same reaction, so the value of ΔG should NOT change regardless of what units we choose. But clearly, equations 1 and 2 give different values. Why?
(if you check the link above, you'll see that the discussion started somewhat differently there. I began by inquiring how we could simply mix pressures and concentrations in the same reaction quotient when using Nernst's equation, without doing some rescaling. It appears bizarre to me that one can use mol/L and atmospheres in the same reaction quotient and get the correct result. If you do some math using the ideal gas law to find the "concentration" of gas molecules in an ideal gas under 1 atm, you find that it is FAR from being 1 mol/L so it doesn't make sense that one can plug values in such different units into the same equation and still get meaningful results).
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The Gibbs energy depends on pressure. At 1 bar (the usual reference pressure), the quantity is called the standard Gibbs energy, G°. The difference in standard molar Gibbs energies of the products and the reactants is the standard reaction Gibbs energy, ΔrG°. Accordingly, we can write ΔrG° = −RT ln K°.
The standard (or thermodynamic) equilibrium constant is in your simple case K° = p/p°, where p° = 1 bar. To get a unitless K°, you need to give p in bars. Using, for example, p° = 1 atm means you have a different "standard" Gibbs energy.
The gas constant does not really change. Consider R = 0.08206 L atm K−1 mol−1. With your mta unit it is R = 0.5 × 0.08206 L mta K−1 mol−1. The numerical value does change, but it does not imply that "Rmta = Ratm/2". (I would prefer R = 8.314 J K−1 mol−1 because ΔG is usually given in kilojoules per mole.)
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Note that when you write a logarithm, lnx, or lognx, or exp(x), that x must be dimensionless. x never has units. So when you write
lnP
then really it means ln(P/Po) where Po is some standard. So if the standard has units of Pascals, atmospheres, torr, mm of Hg, etc. then you must use the same units for P in lnP. Usually Po is understood. When the distinction is important, then it is written
lnPμ (just to remind us it is really P/Po)
Note that you come across the Boltzmann energy a lot:
exp(-E/RT)
RT is say, Joules mol-1 and so the energy E has the same units.
BTW, you cannot mix concentration and pressure units in equilibrium expressions or the Nernst equation. I would need to see the example to understand what is going on.
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BTW, you cannot mix concentration and pressure units in equilibrium expressions or the Nernst equation.
He refers to things like hydrogen electrode potential, where both pressure and concentration are present.
[tex]E = \frac {RT} {2F} \ln \frac {[H^+]^2}{P_{H_2}}[/tex]
(E0 omitted for brevity, as it is 0)
What you wrote about a using ratio to some standard state (IOW activity) still holds.
It occurred to me right now that we can start with concentration of dissolved hydrogen, replace concentration with kPH2 (Henry's law) and rearrange moving [itex]\frac{RT}{2F}\ln k[/itex] into E0 - so both equations (using pressure and using concentration) are equivalent, they just differ by the E0 value.
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Hi!
I am having some trouble in figuring out how units can be used when writing reaction quotients/ equilibrium constants / Nernst Equation. This discussion started in another forum (http://www.physicsforums.com/showthread.php?t=610792) but it seems I've reached a dead end there, so any help is welcome.
Essentially, I'm puzzled by how we can use apparently arbitrary units when writing the reaction quotient. Suppose we have a very simple reaction, so that its ΔG can be written as
ΔG = - RT ln p0 (equation 1)
So far so good. But what if we change the units? Suppose instead of using atm to measure pressure, we choose another unit, mta, such that 1 mta = 2atm. Well, this means that the new value of pressure p will be p = p0/2. But R also changes when going from atms to mtas, and to fix it, we may use the ideal gas law. Then we get that the new value for R, that I will call R' is half of the original R.
Now, lets restate equation one using our new set of units:
ΔG = - R'T ln p
And, by changing the back to the commonly using atm we get another expression of ΔG:
ΔG = - 0.5 R ln(p0/2) (equation 2)
Now, we are still talking about the same reaction, so the value of ΔG should NOT change regardless of what units we choose. But clearly, equations 1 and 2 give different values. Why?
(if you check the link above, you'll see that the discussion started somewhat differently there. I began by inquiring how we could simply mix pressures and concentrations in the same reaction quotient when using Nernst's equation, without doing some rescaling. It appears bizarre to me that one can use mol/L and atmospheres in the same reaction quotient and get the correct result. If you do some math using the ideal gas law to find the "concentration" of gas molecules in an ideal gas under 1 atm, you find that it is FAR from being 1 mol/L so it doesn't make sense that one can plug values in such different units into the same equation and still get meaningful results).
The argument of ln cannot have units, because what is the meaning of ln of an atmosphere?
I.e. the equation is really
ΔG = - RT ln {p}
where the braces denote the quantity without the units. E.g. if p = 2 atm then {p}=2. As Prof. Sanctuary points {p0} can be also written as p/p0 so what units cancel inside the ln.