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Author Topic: Understanding Enthalpy, Work, and State Functions  (Read 2472 times)

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Il Divo

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Understanding Enthalpy, Work, and State Functions
« on: June 14, 2012, 04:23:37 AM »

So recently I decided to go back and attempt to make sense of enthalpy (intuitively) as a concept, something which I'd always used mathematically but never bothered to fully understand, to my great regret. I was hoping someone could double-check my logic and reasoning to make sure I have this all together and to answer some questions.

From what I've gathered, enthalpy is an arbitrary definition which under certain conditions allows us to observe how much heat (q) is either released or absorbed by any particular chemical reaction.

Enthalpy (H) is defined H = U + PV. Because it is composed of state functions, enthalpy must also be a state function, therefore ΔH = ΔU + ΔPV. By employing the definition of change in internal energy and constant pressure conditions, we can reduce this to show that ΔH = qp. Because enthalpy remains a state function, this shows us that under conditions of constant pressure, how much heat is absorbed or released by a chemical reaction will always be the same. Meaning, when pressure is constant, enthalpy is simply heat content.

My issue is this:

1) Does Enthalpy, as a concept, have any value in situations where pressure is not being held constant? Under these conditions heat (q) will not be a state function, meaning that it will be path-dependent for any chemical reaction.

2) Why is work not considered a state function? The difficulty I'm having with this is that when I first saw enthalpy defined as H = U + PV it was explained that because all the variables contained in the definition were state functions that enthalpy would also be a state function. Intuitively, this makes sense; a state function will always have the same values if returned froma  final to an initial state, therefore mathematically enthalpy will have the same values between any two states. However, work is composed of both pressure and volume, both state functions, hence shouldn't it also be a state function?

3)  Under conditions of constant pressure, is work also considered a state function? Under these conditions internal energy will still be ΔU = qp + w. Because internal energy and heat are both state functions, mathematically the only way to keep this consistent is for work to also be a state function.

Thank you to everyone in advance and sorry for the long post.
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juanrga

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Re: Understanding Enthalpy, Work, and State Functions
« Reply #1 on: June 16, 2012, 08:01:33 AM »

From what I've gathered, enthalpy is an arbitrary definition which under certain conditions allows us to observe how much heat (q) is either released or absorbed by any particular chemical reaction.

Enthalpy (H) is defined H = U + PV. Because it is composed of state functions, enthalpy must also be a state function, therefore ΔH = ΔU + ΔPV. By employing the definition of change in internal energy and constant pressure conditions, we can reduce this to show that ΔH = qp. Because enthalpy remains a state function, this shows us that under conditions of constant pressure, how much heat is absorbed or released by a chemical reaction will always be the same. Meaning, when pressure is constant, enthalpy is simply heat content.

My issue is this:

1) Does Enthalpy, as a concept, have any value in situations where pressure is not being held constant? Under these conditions heat (q) will not be a state function, meaning that it will be path-dependent for any chemical reaction.

Enthalpy is the thermodynamic potential that plays the role of internal energy U when the thermodynamic variable V is changed by pressure p. When we apply a Legendre transform V :rarrow: P to U = U(S,V,N), to obtain the new potential in the new thermodynamic variables, the result is H = H(S,P,N) when H is given by H ≡ U + PV.

When pressure is not constant enthalpy is not a thermodynamic potential and its utility is inferior.

2) Why is work not considered a state function? The difficulty I'm having with this is that when I first saw enthalpy defined as H = U + PV it was explained that because all the variables contained in the definition were state functions that enthalpy would also be a state function. Intuitively, this makes sense; a state function will always have the same values if returned froma  final to an initial state, therefore mathematically enthalpy will have the same values between any two states. However, work is composed of both pressure and volume, both state functions, hence shouldn't it also be a state function?

3)  Under conditions of constant pressure, is work also considered a state function? Under these conditions internal energy will still be ΔU = qp + w. Because internal energy and heat are both state functions, mathematically the only way to keep this consistent is for work to also be a state function.

Mechanical work is defined as ##W \equiv - \int PdV ## and this quantity is in general path dependant. Of course in some situations it can be path-independant. For instance if P is constant ##W = - P \Delta V## and this only depends on initial and final states, but still work is not a state function because is not defined using a state but two (initial and final). That work is neither a property of the initial state nor of the final.

The same about heat, it is not a state function, even when P is constant. As you correctly wrote it is ΔH = Qp not ΔH = ΔQp.
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Zeppos10

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Re: Understanding Enthalpy, Work, and State Functions
« Reply #2 on: June 19, 2012, 04:00:15 AM »

So recently I decided to go back and attempt to make sense of enthalpy (intuitively) as a concept, something which I'd always used mathematically but never bothered to fully understand, to my great regret.
[
you are not alone in the predicament.
 
From what I've gathered, enthalpy is an arbitrary definition which under certain
conditions allows us to observe how much heat (q) is either released or absorbed by any particular chemical reaction.
this statement is correct, except that the definition is not arbitrary at all, and without specifying the special conditions you are still in the dark.

Enthalpy (H) is defined H = U + pV. By employing the definition of change in internal energy and constant pressure conditions, we can reduce this to show that ΔH = qp. Because enthalpy remains a state function, this shows us that under conditions of constant pressure, how much heat is absorbed or released by a chemical reaction will always be the same. Meaning, when pressure is constant, enthalpy is simply heat content.
no it is not: The question is: how do you connect ∆H=Q with H=U+pV ??We know ∆U = Q + W = Q + W' + W" where W" = useful work = shaft work = technical work. W' is work done against the atmosphere = -p'∆V, if p' is the external=atmospheric pressure. Now define H=U+p'V (!!!!) and the first law becomes: ∆H= Q+W" .(this is the equation to remember, it is as general as ∆U = Q + W).If, and only if, no useful work is done (W"=0: this not the case when you are charging a battery) and the external pressure remains constant, it follows: ∆H=Q.Note how important the 'constant pressure' condition is: it makes p=p', if you didn't do that at the beginning. (also: solids have enthalpy as property, but not internal pressure).

My issue is this:
1) Does Enthalpy, as a concept, have any value in situations where pressure is not being held constant? Under these conditions heat (q) will not be a state function, meaning that it will be path-dependent for any chemical reaction.
This question is answered above: the answer is yes.

2) Why is work not considered a state function? Work is composed of both pressure and volume, both state functions, hence shouldn't it also be a state function?
The pV term in H=U+pV is called 'energy of  displacement' or 'pressure energy' and this is a state function, but it depends (through p) on the state of the environment and (through V) on the state of the system. Note that from H=U+pV follows ∆H= ∆U+p∆V whether H is a function of state or not: the only condition is ∆p=0.

3)  Under conditions of constant pressure, is work also considered a state function? Under these conditions internal energy will still be ΔU = qp + w. Because internal energy and heat are both state functions, mathematically the only way to keep this consistent is for work to also be a state function.
You are confusing enthalpy H with heat Q, because you ignore useful work.

 ref: Mannaerts SHWM, Energy-balance of the Joule-Thomson experiment: Enthalpy change at decompression. npt-procestechnologie. 2010; 17(4)18-22.

see also:  http://www.physicsforums.com/showthread.php?t=612087

for discussion of Legendre transforms in thermodynamics see: http://www.physicsforums.com/showthread.php?t=313535

   

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