Heat of reaction

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Hi!
In most textbooks on chemical physics/thermodymanics it is said that
under fixed pressure the heat of reaction equals change of enthalpy of the system
since [itex]dU = \delta Q - p\cdot dV[/itex], and hence [itex]d(U+pV) = \delta Q[/itex].
But my question is: why they do not write a term [itex]+\mu\cdot dN[/itex] which describes changes
in internal energy due to the change of the number of particles, which obviously
changes in course of reaction ?! ([itex]\mu[/itex] denotes chemical potential of one of components).
If I add it, I get [itex]dU = \delta Q - p \cdot dV + A\cdot d\xi[/itex] (where A stands
for reaction affinity), and hence even for p = const: [itex]dH = \delta Q + A\cdot d\xi \neq \delta Q [/itex] !
I cann't believe that so many authors can be wrong. So, where is my mistake?
Thank you in advance!
 

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  • #2
jfizzix
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Hi!
In most textbooks on chemical physics/thermodymanics it is said that
under fixed pressure the heat of reaction equals change of enthalpy of the system
since [itex]dU = \delta Q - p\cdot dV[/itex], and hence [itex]d(U+pV) = \delta Q[/itex].
But my question is: why they do not write a term [itex]+\mu\cdot dN[/itex] which describes changes
in internal energy due to the change of the number of particles, which obviously
changes in course of reaction ?! ([itex]\mu[/itex] denotes chemical potential of one of components).
If I add it, I get [itex]dU = \delta Q - p \cdot dV + A\cdot d\xi[/itex] (where A stands
for reaction affinity), and hence even for p = const: [itex]dH = \delta Q + A\cdot d\xi \neq \delta Q [/itex] !
I cann't believe that so many authors can be wrong. So, where is my mistake?
Thank you in advance!
For a chemical reaction,

[itex]dU = TdS -PdV +\sum_{i}\mu_{i} dN_{i}[/itex]

so that at constant pressure

[itex]dH = \delta Q +\sum_{i}\mu_{i} dN_{i}[/itex], as you say.

These equations are only defined for quasistatic processes. In the case of a chemical reaction, you could add the reactants arbitrarily slowly, allowing the system to never stray more than an infinitesimal amount away from equilibrium.

For Chemical equilibrium, we look at the Gibbs free energy. At constant temperature and pressure (i.e. that doesn't change appreciably for an infinitesimal amount of reaction)

[itex]dG = \sum_{i}\mu_{i} dN_{i}=0[/itex](at equilibrium).

To the extent we can say we are performing the reaction quasi-statically, we know that [itex]\sum_{i}\mu_{i} dN_{i}=0[/itex] so that the total heat of reaction is indeed given by the change of enthalpy at constant pressure.
[itex]dH = \delta Q[/itex]

I TA-ed thermodynamics for physics majors, we did only a little bit of chemistry, but I believe this is the correct reason.

Hope this helps:)
 
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  • #3
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The equation
[tex]dH=TdS+VdP[/tex]
applies to the total mass of all closed systems (for which no mass is entering or leaving). It includes closed systems in which chemical reactions are occurring.
The equation
[tex]dH=TdS+VdP+\Sigma \mu_idN_i[/tex]
applies to open systems (for which mass is entering or leaving).

For a closed system in which chemical reactions take place, and for which the temperature and the pressure are held constant, Q = ΔH. In such a system, the total enthalpy of the system is equal to the number of moles of each species times the partial molar enthalpy of that species. For an ideal gas mixture, the partial molar enthalpy of each species is equal to the molar enthalpy of the pure species at the same temperature as the pure species. The molar enthalpy of each pure species can be calculated from the heat of formation at a specified reference temperature plus the integral of the molar heat capacity at constant pressure with respect to temperature from the reference temperature to the operating temperature. In this way, one can calculate the total change of enthalpy of a mixture of ideal gases that undergoes reaction.

Chet
 
  • #4
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The equation
[tex]dH=TdS+VdP[/tex]
applies to the total mass of all closed systems (for which no mass is entering or leaving). It includes closed systems in which chemical reactions are occurring.
The equation
[tex]dH=TdS+VdP+\Sigma \mu_idN_i[/tex]
applies to open systems (for which mass is entering or leaving).
It is difficult for me to accept this, since we can imagine a box with impenetrable wall
which splits the total volume into two parts, each containing a gas of particles A and B
respectively. Such system as a whole is clearly closed one: no "external" particle can
enter or leave it. However, it can be easily verified that the entropy and other
thermodynamical functions will depend on whether the wall is present or absent
(i.e., whether two gases are separated or mixed).
Hence, even when the total number of particles is fixed, internal energy can depend
on additional variables
(not only S and V) if the system contains more than one type
of particles. It seems likely that in the case of chemical reactions, when the system
is clearly multi-component, we can not say that U depends on two variables only...
 
  • #5
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In the case of a chemical reaction, you could add the reactants arbitrarily slowly, allowing the system to never stray more than an infinitesimal amount away from equilibrium.

For Chemical equilibrium, we look at the Gibbs free energy. At constant temperature and pressure (i.e. that doesn't change appreciably for an infinitesimal amount of reaction)

[itex]dG = \sum_{i}\mu_{i} dN_{i}=0[/itex](at equilibrium).

To the extent we can say we are performing the reaction quasi-statically, we know that [itex]\sum_{i}\mu_{i} dN_{i}=0[/itex] so that the total heat of reaction is indeed given by the change of enthalpy at constant pressure.
[itex]dH = \delta Q[/itex]
So, as I right that your suggestion is that in each moment of the reaction (if we consider
it as being a very-very slow one) its affinity equals zero, i.e., reactants and products are
in chemical equilibrium ?
 
  • #6
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It is difficult for me to accept this, since we can imagine a box with impenetrable wall
which splits the total volume into two parts, each containing a gas of particles A and B
respectively. Such system as a whole is clearly closed one: no "external" particle can
enter or leave it. However, it can be easily verified that the entropy and other
thermodynamical functions will depend on whether the wall is present or absent
(i.e., whether two gases are separated or mixed).
Hence, even when the total number of particles is fixed, internal energy can depend
on additional variables
(not only S and V) if the system contains more than one type
of particles. It seems likely that in the case of chemical reactions, when the system
is clearly multi-component, we can not say that U depends on two variables only...
I'm sorry (for your sake) that it is so difficult for you to accept this.

Here is a problem for you to attempt (if you are willing to try it) to illustrate what I am trying to say. Let's do your "box" problem:

You have a box containing two different ideal gases, with a barrier between them. Each gas occupies half the volume, and there is one mole of each gas in each half of the container. The pressure of each gas is P and the temperature of each gas is T. At a certain time, you remove the barrier between the two sides of the container, and let the gases diffuse into one another until the system reaches thermodynamic equilibrium. The container is adiabatic and rigid, so that no heat enters or leaves, and no total volume change occurs.

So, State 1: Barrier present. Gas in each chamber uniform.
State 2: Barrier removed. Gases throughout container well mixed with each other.

Determine
(a) The change in internal energy of the system (Hint: the heat of mixing of ideal gases is zero)
(b) The change in temperature
(c) The change in pressure
(d) the change in entropy
(e) The change in gibbs free energy
(f) the change in chemical potential of each species.

Chet

P.S., Are you familiar with Gibbs Theorem?: A total extensive property of an ideal gas mixture is the sum of the total properties of the individual species, each evaluated at the mixture temperature but at its own partial pressure.
 
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  • #7
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Let's do your "box" problem:

You have a box containing two different ideal gases, with a barrier between them. Each gas occupies half the volume, and there is one mole of each gas in each half of the container. The pressure of each gas is P and the temperature of each gas is T. At a certain time, you remove the barrier between the two sides of the container, and let the gases diffuse into one another until the system reaches thermodynamic equilibrium. The container is adiabatic and rigid, so that no heat enters or leaves, and no total volume change occurs.

So, State 1: Barrier present. Gas in each chamber uniform.
State 2: Barrier removed. Gases throughout container well mixed with each other.

Determine
(a) The change in internal energy of the system (Hint: the heat of mixing of ideal gases is zero)
(b) The change in temperature
(c) The change in pressure
(d) the change in entropy
(e) The change in gibbs free energy
(f) the change in chemical potential of each species.

Chet

P.S., Are you familiar with Gibbs Theorem?: A total extensive property of an ideal gas mixture is the sum of the total properties of the individual species, each evaluated at the mixture temperature but at its own partial pressure.
Ok. In canonical ensemble statistical sum for the mixture will be
[itex]
Z_{mix} = \frac{1}
{{N_A !}}\frac{1}
{{N_B !}}\left( {z_A } \right)^{N_A } \left( {z_B } \right)^{N_B }
[/itex]
where [itex]z_i = V/\lambda _i ^3 [/itex] with [itex]\lambda _i = \frac{h}
{{\sqrt {2\pi m_i kT} }}
[/itex]
The free energy is then
[itex]
\frac{{F_{mixture} }}
{{kT}} = - \ln Z_{mix} = \ln N_A ! + \ln N_B ! - N_A \ln \left( {\frac{V}
{{h^3 }}\left( {2\pi m_A kT} \right)^{3/2} } \right) - N_B \ln \left( {\frac{V}
{{h^3 }}\left( {2\pi m_A kT} \right)^{3/2} } \right)
[/itex]
or
[itex]
F_{mixture} = F_A (V_{1/2} ) + kTN_A \ln \left( {\frac{{V_{1/2} }}
{V}} \right) + F_B (V_{1/2} ) + kTN_B \ln \left( {\frac{{V_{1/2} }}
{V}} \right)
[/itex]
Hence, it clearly changes in isothermal mixing.
The change in entropy is
[itex]
S_{mixture} - S_A (V_{1/2} ) - S_B (V_{1/2} ) = - kN_A \ln \left( {\frac{{V_{1/2} }}
{V}} \right) - kN_B \ln \left( {\frac{{V_{1/2} }}
{V}} \right)
[/itex]
In fact, this illustrates what I've meant writing
"entropy and other thermodynamical functions will depend on whether the wall is
present or absent (i.e., whether two gases are separated or mixed)."...

For ideal gases the change the change in internal energy will be zero.
However, for non-ideal systems we have
[itex]
\mu _i \left( {T,p} \right) = \mu _i ^{pure} \left( {T,p} \right) + kT \cdot \ln \left( {\gamma _i \left( {T,p} \right)\frac{{N_i }}
{{N_{total} }}} \right)
[/itex]
so that the Gibbs energy of the system is
[itex]
G = \sum\limits_i^{} {\mu _i N_i } = G_1 ^{pure} + G_2 ^{pure} + kT\left( {N_1 \ln \left( {\gamma _1 \frac{{N_1 }}
{{N_{total} }}} \right) + N_2 \ln \left( {\gamma _2 \frac{{N_2 }}
{{N_{total} }}} \right)} \right)
[/itex]
Now I can get internal energy as
[itex]
U = G + TS - pV = G - T\frac{{\partial G}}
{{\partial T}} + p\frac{{\partial G}}
{{\partial p}}
[/itex]
which gives
[itex]
U = U_1 ^{pure} + U_2 ^{pure} - kT\left( {N_1 \frac{T}
{{\gamma _1 }}\frac{{\partial \gamma _1 }}
{{\partial T}} + N_2 \frac{T}
{{\gamma _2 }}\frac{{\partial \gamma _2 }}
{{\partial T}}} \right) + kT\left( {N_1 \frac{p}
{{\gamma _1 }}\frac{{\partial \gamma _1 }}
{{\partial p}} + N_2 \frac{p}
{{\gamma _2 }}\frac{{\partial \gamma _2 }}
{{\partial p}}} \right)
[/itex]
and is not zero in general case.
 
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  • #8
BruceW
Homework Helper
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It is difficult for me to accept this, since we can imagine a box with impenetrable wall
which splits the total volume into two parts, each containing a gas of particles A and B
respectively. Such system as a whole is clearly closed one: no "external" particle can
enter or leave it. However, it can be easily verified that the entropy and other
thermodynamical functions will depend on whether the wall is present or absent
(i.e., whether two gases are separated or mixed).
Hence, even when the total number of particles is fixed, internal energy can depend
on additional variables
(not only S and V) if the system contains more than one type
of particles. It seems likely that in the case of chemical reactions, when the system
is clearly multi-component, we can not say that U depends on two variables only...
the term ##\sum_i \mu_i \ dN_i## is only for an open system. Your box is a closed system, therefore there is no ##\sum_i \mu_i \ dN_i## term in the equation for the system of the box.

Now, if instead you talk about sub-systems inside the box (i.e. left-side and right-side) and if there is no separating wall, then they are open with respect to each other, so you do need a ##\sum_i \mu_i \ dN_i## term for these sub-systems. But still, for the box as a whole, you do not need a ##\sum_i \mu_i \ dN_i## term.
 
  • #9
BruceW
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you might think that the box system needs a ##\sum_i \mu_i \ dN_i## term, since the properties of the system could depend on if there is a wall inside. It is true that the properties depend on whether there is a wall or not, but this is taken into account when you define the system. There is no need to add a ##\sum_i \mu_i \ dN_i## term.

edit: more explanation: If you do have a term ##\sum_i \mu_i \ dN_i## in your equations, then it is because you are considering sub-systems inside the box. So yes, it can be done this way. But you need to remember that it refers to exchange of particles from different places inside the box.
 
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  • #10
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But still, for the box as a whole, you do not need a ##\sum_i \mu_i \ dN_i## term.
I'm afraid that this is not obvious for me.
##
G = \sum\limits_i^{} {G_i }
##
since
##
G = \sum\limits_i^{} {\mu _i N_i }
##
This means that if number of particles in subsystems change, this may affect total Gibbs free energy of the system.
(unless affinity is non-zero)

Furthermore, if there was no need in ##\sum_i \mu_i \ dN_i## term for the total Gibbs free energy,
it would be a function of T and p only. Now, how would the change in total Gibbs
free energy be possible under fixed T and p in this case ?
But we know that if there are chemical reactions such a changes of G under fixed T and p are possible.
 
  • #11
BruceW
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yeah, what I really meant was that we don't need a ##\sum_i \mu_i \ dN_i## term for the swapping of particles with the world outside the box. And if we could somehow incorporate the inhomogeneity of our system without using the concept of subsystems, then we wouldn't need any ##\sum_i \mu_i \ dN_i## terms at all. But then of course, this is equivalent to considering subsystems and making use of ##\sum_i \mu_i \ dN_i## terms for our subsystems. So what I've been saying has not really added much to the conversation. sorry about that :(

I've read back to the start of the thread now. It is an interesting question. possibly the answer is that the change in thermodynamic quantities due to mixing is typically negligible compared to the effects caused by the actual reaction. For example, if you mix water and orange juice, it is not really going to give off a lot of heat just due to the mixing, right? (hehe, I'm resorting to common sense, which is maybe not a very scientific thing to do).
 
  • #12
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yeah, what I really meant was that we don't need a ##\sum_i \mu_i \ dN_i## term for the swapping of particles with the world outside the box. And if we could somehow incorporate the inhomogeneity of our system without using the concept of subsystems, then we wouldn't need any ##\sum_i \mu_i \ dN_i## terms at all. But then of course, this is equivalent to considering subsystems and making use of ##\sum_i \mu_i \ dN_i## terms for our subsystems. So what I've been saying has not really added much to the conversation. sorry about that :(

I've read back to the start of the thread now. It is an interesting question. possibly the answer is that the change in thermodynamic quantities due to mixing is typically negligible compared to the effects caused by the actual reaction. For example, if you mix water and orange juice, it is not really going to give off a lot of heat just due to the mixing, right? (hehe, I'm resorting to common sense, which is maybe not a very scientific thing to do).
Even in this case, the change in enthalpy for the system as a whole is zero. What you said in your previous posts were correct.

Here is what Smith and Van Ness have to say on the subject:

"Eqns. (6.4) through (6.6) have the same range of applicability as Eq. (6.1). All are written for the entire mass of any closed system."

Eqn. (6.1) is dU = TdS-PdV
Eqn. (6.6) is dH=TdS-PdV

The emphasis is mine.
 
  • #13
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Ok. In canonical ensemble statistical sum for the mixture will be
[itex]
Z_{mix} = \frac{1}
{{N_A !}}\frac{1}
{{N_B !}}\left( {z_A } \right)^{N_A } \left( {z_B } \right)^{N_B }
[/itex]
where [itex]z_i = V/\lambda _i ^3 [/itex] with [itex]\lambda _i = \frac{h}
{{\sqrt {2\pi m_i kT} }}
[/itex]
The free energy is then
[itex]
\frac{{F_{mixture} }}
{{kT}} = - \ln Z_{mix} = \ln N_A ! + \ln N_B ! - N_A \ln \left( {\frac{V}
{{h^3 }}\left( {2\pi m_A kT} \right)^{3/2} } \right) - N_B \ln \left( {\frac{V}
{{h^3 }}\left( {2\pi m_A kT} \right)^{3/2} } \right)
[/itex]
or
[itex]
F_{mixture} = F_A (V_{1/2} ) + kTN_A \ln \left( {\frac{{V_{1/2} }}
{V}} \right) + F_B (V_{1/2} ) + kTN_B \ln \left( {\frac{{V_{1/2} }}
{V}} \right)
[/itex]
Hence, it clearly changes in isothermal mixing.
The change in entropy is
[itex]
S_{mixture} - S_A (V_{1/2} ) - S_B (V_{1/2} ) = - kN_A \ln \left( {\frac{{V_{1/2} }}
{V}} \right) - kN_B \ln \left( {\frac{{V_{1/2} }}
{V}} \right)
[/itex]
In fact, this illustrates what I've meant writing
"entropy and other thermodynamical functions will depend on whether the wall is
present or absent (i.e., whether two gases are separated or mixed)."...

For ideal gases the change the change in internal energy will be zero.
However, for non-ideal systems we have
[itex]
\mu _i \left( {T,p} \right) = \mu _i ^{pure} \left( {T,p} \right) + kT \cdot \ln \left( {\gamma _i \left( {T,p} \right)\frac{{N_i }}
{{N_{total} }}} \right)
[/itex]
so that the Gibbs energy of the system is
[itex]
G = \sum\limits_i^{} {\mu _i N_i } = G_1 ^{pure} + G_2 ^{pure} + kT\left( {N_1 \ln \left( {\gamma _1 \frac{{N_1 }}
{{N_{total} }}} \right) + N_2 \ln \left( {\gamma _2 \frac{{N_2 }}
{{N_{total} }}} \right)} \right)
[/itex]
Now I can get internal energy as
[itex]
U = G + TS - pV = G - T\frac{{\partial G}}
{{\partial T}} + p\frac{{\partial G}}
{{\partial p}}
[/itex]
which gives
[itex]
U = U_1 ^{pure} + U_2 ^{pure} - kT\left( {N_1 \frac{T}
{{\gamma _1 }}\frac{{\partial \gamma _1 }}
{{\partial T}} + N_2 \frac{T}
{{\gamma _2 }}\frac{{\partial \gamma _2 }}
{{\partial T}}} \right) + kT\left( {N_1 \frac{p}
{{\gamma _1 }}\frac{{\partial \gamma _1 }}
{{\partial p}} + N_2 \frac{p}
{{\gamma _2 }}\frac{{\partial \gamma _2 }}
{{\partial p}}} \right)
[/itex]
and is not zero in general case.
In the last equation, if you did the math correctly (which I'm guessing you did), all you've shown is that, since the change in U is going to be zero, a certain constraining relationship must exist between the partials of the γ's with respect to temperature and pressure.

In my earlier post, I showed that, at least in the case of an ideal gas, your objection to what I originally said was not warrented. I showed this for the exact example that you felt it could not be done for. That example included the important feature of a mixture of gases. If you would like, I can also show how the same equation applies to a mixture of ideal gases undergoing chemical reaction between states 1 and 2, by again applying Gibbs Theorem. I invite you to try this yourself, however, and see what you get. Once we've dispensed with your discomfort for cases involving ideal gases, we can move on to non-ideal gases.

Chet
 
  • #14
jfizzix
Science Advisor
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So, as I right that your suggestion is that in each moment of the reaction (if we consider
it as being a very-very slow one) its affinity equals zero, i.e., reactants and products are
in chemical equilibrium ?
In thermodynamic equilibrium, the reaction has already taken place.

If you have a closed system with a mixture of reactants and products, thermodynamics only accurately describes the equilibrium state, when they are thoroughly mixed and the reaction has reached a balance.

In keeping with this, thermodynamics also accurately describes how new equilibrium states are related to old equilibrium states after an infinitesimal change has taken place.

If you have an otherwise closed system at constant pressure with some already equilibrated mixture of reactants and products, and you add an infinitesimal amount of a reactant (increasing the total mass, yes), the amount of reactants and products will change an infinitesimal amount until equilibrium is reestablished.

The internal energy will change as
[itex]dU = -PdV +TdS +\sum_{i}\mu_{i} dN_{i}[/itex]
The enthalpy will change (at constant pressure) as
[itex]dH = TdS +\sum_{i}\mu_{i} dN_{i}=\delta Q + \sum_{i}\mu_{i} dN_{i}[/itex]
Usually, we can also assume the environment is also at constant temperature, in which case, the change in Gibbs free energy is given by
[itex]dG = \sum_{i}\mu_{i} dN_{i}[/itex]

However, before you added the extra reactant, the system was already at equilibrium under constant temperature and pressure. Under these conditions, the Gibbs free energy is at a local minimum, which means there is no first order change in the Gibbs free energy, i.e., [itex]dG=0[/itex].

This is almost certainly why [itex]dH =\delta Q[/itex] at constant pressure and temperature in chemical reactions. This business about first order changes in Gibbs free energy may sound like hand waving, but in the thermodynamic limit, noticeable spontaneous deviations from equilibrium are very improbable.
 
  • #15
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Dear timntimn,

In response #13, I replied to your response #7 in which I suggested that, as a result of a certain constraint, the last two terms in parenthesis in your final equation are zero. You had claimed that the the presence of these terms is an indication that the change in internal energy is not zero in the general case (of our sample problem). The constraint I was referring to is the gibbs-duhem equation:

[tex]N_1dμ_1+N_2dμ_2=0[/tex]

If follows from the gibbs-duhem equation that the two terms in parenthesis are indeed zero. Consequently, the total change in internal energy for the general case is zero, even though irreversible mixing has occurred within the container in the interim between the initial and final equilibrium states.

If you'd like, I can now continue by discussing how Q=ΔHreaction between the initial and final states of a system held at constant temperature and pressure that undergoes an irreversible reaction.

Chet
 
  • #16
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Dear timntimn,

In response #13, I replied to your response #7 in which I suggested that, as a result of a certain constraint, the last two terms in parenthesis in your final equation are zero. You had claimed that the the presence of these terms is an indication that the change in internal energy is not zero in the general case (of our sample problem). The constraint I was referring to is the gibbs-duhem equation:

[tex]N_1dμ_1+N_2dμ_2=0[/tex]

If follows from the gibbs-duhem equation that the two terms in parenthesis are indeed zero. Consequently, the total change in internal energy for the general case is zero, even though irreversible mixing has occurred within the container in the interim between the initial and final equilibrium states.

If you'd like, I can now continue by discussing how Q=ΔHreaction between the initial and final states of a system held at constant temperature and pressure that undergoes an irreversible reaction.

Chet

Let us consider two systems, initially separated by the wall, first system of particles of type A, and of particles of type B the second. Let the interparticle potential U(1,2) be zero for ##
\left| {\vec r_1 - \vec r_2 } \right| > \sigma
##, but for ##
\left| {\vec r_1 - \vec r_2 } \right| < \sigma
## to have the form:
u(1,2) = 0, if 1=A and 2=A;
u(1,2) = u0, if 1=A and 2=B or 2=A and 1=B;
u(1,2) = 0, if 1=B and 2=B.
Now consider isothermal mixing at the temperature which is so low, that we can neglect
##TS## terms everywhere (making internal energy simply equal to mechanical energy).
Should the change in internal energy be zero now?
 
  • #17
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In thermodynamic equilibrium, the reaction has already taken place.

If you have a closed system with a mixture of reactants and products, thermodynamics only accurately describes the equilibrium state, when they are thoroughly mixed and the reaction has reached a balance.

In keeping with this, thermodynamics also accurately describes how new equilibrium states are related to old equilibrium states after an infinitesimal change has taken place.

If you have an otherwise closed system at constant pressure with some already equilibrated mixture of reactants and products, and you add an infinitesimal amount of a reactant (increasing the total mass, yes), the amount of reactants and products will change an infinitesimal amount until equilibrium is reestablished.

The internal energy will change as
[itex]dU = -PdV +TdS +\sum_{i}\mu_{i} dN_{i}[/itex]
The enthalpy will change (at constant pressure) as
[itex]dH = TdS +\sum_{i}\mu_{i} dN_{i}=\delta Q + \sum_{i}\mu_{i} dN_{i}[/itex]
Usually, we can also assume the environment is also at constant temperature, in which case, the change in Gibbs free energy is given by
[itex]dG = \sum_{i}\mu_{i} dN_{i}[/itex]

However, before you added the extra reactant, the system was already at equilibrium under constant temperature and pressure. Under these conditions, the Gibbs free energy is at a local minimum, which means there is no first order change in the Gibbs free energy, i.e., [itex]dG=0[/itex].

This is almost certainly why [itex]dH =\delta Q[/itex] at constant pressure and temperature in chemical reactions. This business about first order changes in Gibbs free energy may sound like hand waving, but in the thermodynamic limit, noticeable spontaneous deviations from equilibrium are very improbable.
But it seems to me that you obtain [itex]dG=0[/itex] only if you add a small amount
of reactants and products as well. The following makes me think so:
##
\eqalign{
& G\left( {T,p,N_i + \delta N_i } \right) = G\left( {T,p,N_i } \right) + \sum\limits_i^{} {\left( {\frac{{\partial G}}
{{\partial N_i }}} \right)_{T,p} \cdot \delta N_i } = G\left( {T,p,N_i } \right) + \sum\limits_i^{} {\mu _i \cdot \delta N_i } \cr}
##
where ##\sum\limits_i^{} {\mu _i \cdot \delta N_i } \ne 0## unless ##
{\delta N_i }## are "concerted" as ##\delta N_i = \tilde \nu _i \cdot dN## where ##
\tilde \nu _i ## are stoichiometric coefficients.
Did I miss something?
 
  • #18
20,148
4,218
Let us consider two systems, initially separated by the wall, first system of particles of type A, and of particles of type B the second. Let the interparticle potential U(1,2) be zero for ##
\left| {\vec r_1 - \vec r_2 } \right| > \sigma
##, but for ##
\left| {\vec r_1 - \vec r_2 } \right| < \sigma
## to have the form:
u(1,2) = 0, if 1=A and 2=A;
u(1,2) = u0, if 1=A and 2=B or 2=A and 1=B;
u(1,2) = 0, if 1=B and 2=B.
Now consider isothermal mixing at the temperature which is so low, that we can neglect
##TS## terms everywhere (making internal energy simply equal to mechanical energy).
Should the change in internal energy be zero now?
In the sample problem, we were not talking about isothermal mixing. We were talking about adiabatic mixing at constant volume. In that particular case, even if the temperature is very low, the change in potential energy will equal the change in kinetic energy of the molecules, and the change in internal energy will be zero (even though the temperature changes).

If we are talking about mixing at constant temperature and pressure, then heat will have to be removed from the system (if the heat of mixing is positive) in order to keep the temperature constant. In that case, for the overall system, ΔH=-Qmixing.

Chet
 
  • #19
jfizzix
Science Advisor
Insights Author
Gold Member
757
354
But it seems to me that you obtain [itex]dG=0[/itex] only if you add a small amount
of reactants and products as well. The following makes me think so:
##
\eqalign{
& G\left( {T,p,N_i + \delta N_i } \right) = G\left( {T,p,N_i } \right) + \sum\limits_i^{} {\left( {\frac{{\partial G}}
{{\partial N_i }}} \right)_{T,p} \cdot \delta N_i } = G\left( {T,p,N_i } \right) + \sum\limits_i^{} {\mu _i \cdot \delta N_i } \cr}
##
where ##\sum\limits_i^{} {\mu _i \cdot \delta N_i } \ne 0## unless ##
{\delta N_i }## are "concerted" as ##\delta N_i = \tilde \nu _i \cdot dN## where ##
\tilde \nu _i ## are stoichiometric coefficients .
Did I miss something?
This is a good question.

If you add a small amount of a reactant to the mixture, at that instant, [itex]\sum_{i}\mu_{i}dN_{i}\neq 0[/itex]. The Gibbs free energy at this instant is different than its equilibrium value, and different from what it was before you added the extra reactant. The sign of this difference will determine whether the reaction runs forward or backward toward chemical equilibrium. The amount of reactants and products will adjust until equilibrium is reached, and it is this state that thermodynamics is equipped to describe.

The amount of reactant you add will not be the same as the change in the total reactant amount between equilibrium states.

As you say, the [itex]dN_{i} [/itex] are "concerted" such that [itex]dN_{i} = \tilde \nu _{i} dN[/itex], where [itex]\tilde \nu _{i} [/itex]are stoichiometric coefficients (and [itex]dN[/itex] is the number of times the reaction takes place).



I think a lot of these troubles can be boiled down into one little paradox:

If we had a mixture in chemical equilibrium, and doubled all the atoms of all types in this mixture, we should expect the Gibbs free energy of the total system to double, since it is an extensive variable.

In principle, we should see this change whether or not we add the reactants and products one atom at a time. So what gives?

I'd be interested in hearing responses to this too.
 
  • #20
10
0
Dear colleagues!

I've asked the same question on the Russian forum at http://chemport.ru
After intense discussions, we've almost came to a solution.
Its key is that chemical reaction is non-equilibrium process (otherwise,
affinity is zero and my question "vanishes") and hence, an entropy production
takes place. Due to this fact ## T\cdot dS = \delta Q + A\cdot d\xi##, where
##\delta Q## is the amount of heat, which is given or taken to (from) the system
by its surroundings.
But in order to make me understand underlying reasoning, I would appreciate
your help in answering the following questions:
1) Is a chemical reaction and "non-equilibrium" or "irreversible" process, or are
these notions equivalent ?
2) Whatever the case in question 1, is it possible to say that for the system
with chemical reactions in it there still exist a thermodynamic potential G
(Gibbs free energy), which is a single valued function of its arguments, T, p and ##N_i## ?
3) If the answer to the previous question is "yes", what is ##S## in ##G = U-TS+pV## ?,
i.e., how should I understand an entropy of a non-equilibrium state?

Thank you for fruitful discussions :)
 
  • #21
BruceW
Homework Helper
3,611
119
Dear colleagues!

I've asked the same question on the Russian forum at http://chemport.ru
After intense discussions, we've almost came to a solution.
Its key is that chemical reaction is non-equilibrium process...
yeah. I do physics really, not chemistry. But I'm pretty sure the idea is that the chemical reaction itself happens on a much slower time scale than the usual 'equilibrium' thermodynamics. So we have a quasi-static process.
 

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