Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: mike on October 23, 2005, 10:32:21 PM
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Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?
Answer: Is it zero?
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Yes. The boiling point of a liquid is the temperature at which the gas phase and liquid phase will be at equilibrium. At equilibrium, dG is zero.
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Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?
Answer: Is it zero?
try finding out for yourself, at constant pressure and temperature
dG=dH-TdS, or assuming that neither the standard ethalpy of vaporization or standard entropy of vaporization changes significantly for a reasonable range of temperature
deltaGstd,m=delta u std = delta Hstdvap- T delta S stdvap, plug in the boiling point temperature
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try finding out for yourself, at constant pressure and temperature
dG=dH-TdS, or assuming that neither the standard ethalpy of vaporization or standard entropy of vaporization changes significantly for a reasonable range of temperature
deltaGstd,m=delta u std = delta Hstdvap- T delta S stdvap, plug in the boiling point temperature
Well this is how I got my answer:
H2O liquid <------> vapour
Keq = avapour/aliquid
and avapour = p/p0
and: aliquid = 1
therefore: Keq = p/p0
for the normal boiling point of water p = p0 = 1
so Keq = 1
now:
dG = -RTlnKeq
and as Keq = 1 then lnKeq = 0
so therefore dG = 0
:D
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dG also equals -TdS for an isothermal process. Not quite sure what you're doin there.
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Not quite sure what you're doin there.
this shows how the change in state has zero gibbs free energy change, ie using the Keq for the liquid to vapour change for water.
I am not sure how yours shows this?
:) :)
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so are you assuming constant volume, wouldn't the hemholtz equation be more appropriate (assuming that you were to measure the vapor pressure of water in a constant volume container, water wouldn't boil; if you were to do it with a frictionless piston, the boiling point would be different). If you were to say, have this occur in an open system, how would the vapor pressure of water apply?
The vaporization process still has changes of enthalpy and entropy associated with it, the chemical potential of liquid water and its vapor are not the same.
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ok, I'm getting way ahead of myself here, so at the beginning I mentioned that
dG=dH-TdS, at constant temperature and pressure
since dS=dH/T at constant pressure
dG=dH-TdH/T=0
so yes, du seems to be zero, sorry for the confusion
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since dS=dH/T at constant pressure
dG = dH - TdS
if dG = 0
dH - TdS = 0
therefore:
dH = TdS
dS = dH/T
??? ??? ???
aren't you just making the assumption that dG =0, this doesn't prove anything! What is wrong with my proof??
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alright, I did some internet searching and actually found discussions related to free energy of vaporization....with actual values. Yes, I assumed that dS(system)=dH/T, assuming that at each stage of vaporization, the system is at equilbirum, making it a reversible process (which I thought was so). Interesting.
You may actually want to plug in actual values, for instance, find the entropy and enthalpy of vaporization (CRC or NIST, scifinder) and with the respective temperatures, plug into the dG formula.
Another possibility is that at one atm, and at the corresponding boiling point, the free energy of vaporization is zero, but at temperatures otherwise, it may not be zero, since the chemical potential of the liquid and gases differ respectively.
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actually deltaGvap=u*(gas)-u*(liquid), where u* is the chemical potential of the pure gas/liquid, you can have a deltaGvap value but at the boiling point I think it's zero since the chemical potentials are equivalent for first order phase transitions.
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Thanks GCT, well you really have been researching this haven't you ;)
Take a look at my working out further up this post and see what you think now. I think with my way you don't need to know any values and you only assume that that the vapour pressure equals atmospheric pressure at boiling (which is true). Let me know what you think.
mike
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I think that I can understand your though process, however, I usually find it more exact to work with the fundamental equations, because there are a lot of assumptions that are made from the derivations. Another thing is that I'm not quite sure if you can apply free energy to an equilibrium process, that is whatever the proportion of vapor to liquid there may be, at the temperature, equivalent chemical potentials for gas and liquid for first order transitions, free energy doesn't really apply; so it seems from the fundamental equations.
For instance, say that we're talking about the situation besides the normal boiling point, another point of the equilibrium curve of the liquid-gas phase diagram, where the pressure is not 1 atm, but you're still at the boiling point for that respective pressure. What will be the result of your method?
p.s. I'm not actually on the equilibrium chapter yet, just finished activity
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consider boiling water in a close system
boiling occurs at constant temperature, so the change in molar gibb's free energy is given as: dg = dh - Tds
given that boiling occurs at saturated condition, then this process can be regarded as reversible. This implies that T.ds is therefore equivalent to the heat supplied (q = Tds)
boiling occurs at constant pressure:
du = q - p.dv
h = u + pv
dh = du + p.dv = (q - p.dv) + p.dv = q
given that q = T.ds = dh then it must be true that dg = dh - Tds = 0
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consider boiling water in a close system
boiling occurs at constant temperature, so the change in molar gibb's free energy is given as: dg = dh - Tds
given that boiling occurs at saturated condition, then this process can be regarded as reversible. This implies that T.ds is therefore equivalent to the heat supplied (q = Tds)
boiling occurs at constant pressure:
du = q - p.dv
h = u + pv
dh = du + p.dv = (q - p.dv) + p.dv = q
given that q = T.ds = dh then it must be true that dg = dh - Tds = 0
first off, I don't think that boiling pertains to closed isolated systems, a supercritical state would be relevant where you have no distinction between the gas and liquid phase.
I've already mentioned this case in one of the above posts, but it's a bit insignificant ....you say that dG=dH-TdS, at constant temperature and pressure, the dS actually pertains to the entropy of the system, unfortunately dSsys is not always defined as dq/T, though it equals dq/T when you assume an reversible process.
For instance, dS does not equal dq/T for an irreversible process.
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For instance, dS does not equal dq/T for an irreversible process.
Is assuming boiling a reversible process a bad assumption? I doubt so..
The conditions where boiling occur is such that both liquid and gas phases are at equilibrium with each other.
consider boiling in an open system:
dh = q + v.dp = q (dp = 0 at constant pressure)
q = T.ds (since boiling is a reversible process)
since q = T.ds = dh, then dg = dh - Tds = 0
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dG = dGo + RTlnK
at equilibrium dG = 0
therefore:
0 = dGo + RTlnK
dGo = -RTlnK
K = equilibtium constant for this equilibrium:
H2O(l) <----> H2O(g)
now:
K = avapour/aliquid right?
and aliquid = 1 ok?
therefore:
K = avapour = p/po
and:
po = p at the boiling point (vapour pressure equals atmospheric pressure)
so: K = 1
therefore:
dGo = -RTlnK
dGo = -RTln1
dGo = 0
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according to mike, dG = 0
so dH = TdS must be valid too.
given at constant pressure, Q = dH, then Q = TdS must be valid.
This shows that boiling is a reversible process.
GCT: what's wrong with my method?
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according to mike, dG = 0
so dH = TdS must be valid too.
given at constant pressure, Q = dH, then Q = TdS must be valid.
This shows that boiling is a reversible process.
GCT: what's wrong with my method?
I don't really think there's anything explicitly wrong with your method, but the whole point of this thread was to find the basis of why and if dG (rather du, chemical potential), free energy of vaporization, is zero for a boiling point transition. It seems to me that you have assumed that dG=0 for free energy of vaporization from the start. All in all, you're right, the whole process of boiling is a reversible process,it's when the chemical potential of the gas and liquid are equivalent for a first order transition. I though that there was a more fundamental way to show this, for instance du(gas)-du(liquid)=dG vap (note that dG vaporization is not always zero at non-equivalent chemical potentials). It's always interesting though, when through a discussion similiar to this, one discovers exceptions, I was simply trying to find any exceptions and so far I've found none for the case at boiling point temperature free energy of vaporization for a first order transition.
mike, try working with a = gX , g is equivalent to the activity coefficient. g=1 for a pure substance. a=X. K=[gas vapor pressure]/1, the vapor pressure refers to constant volume, we can equate the gas vapor pressure for different respective atmospheric pressures to find our temperature of interest, boiling temperature. Really doesn't prove anything though, if you do everything correctly though, you may find that dG=0, but this simply proves that the fundamental derivations of the equations were done correctly. You're merely checking upon the validity of the derived equation. And that's why we need to work with the fundamentals.
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I don't really think there's anything explicitly wrong with your method, but the whole point of this thread was to find the basis of why and if dG (rather du, chemical potential), free energy of vaporization, is zero for a boiling point transition.
The chemical potential of a pure substance is equivalent to the molar chemical potential.
Question: What is the standard molar Gibbs free energy of vaporisation at the normal boiling point of water?
We are indeed considering the case for pure substance - water.