A change of phase, under equilibrium conditions, will be at constant temperature and constant pressure.. When an ideal gas expands at constant temperature from V to V , the final state i
Trang 11 1
2 1 1
1
dV dV
dS dV dV
dS
(13)
0 1 2 2 1 1 ≥ − = dV T P T P dS (14)
T
H
S = ∆
∆
K J/mol 96 108 K
15 373
J/mol 657 , 40
⋅
=
=
∆S vap
It is assumed that T = T and thus the total entropy is unchanged by a transfer of thermal
5
energy between the two parts The constant-energy restriction of equation 12 is satisfied because any work done by one part on the other can be compensated by such a transfer of thermal energy
Again consider a system constrained to constant total volume, and to have a uniform temperature Then, as before, we can write the total change in entropy that might result from a change in
volume of either part in the form
(T, V)
and, because total volume is constant, dV = - dV Substitution of equation 12 gives1 2 5
This proves that two bodies in equilibrium must have not only the same temperature but also the same pressure It also proves that, when two bodies in contact, at the same temperature, have different pressures, the body at the higher pressure will tend to expand and compress the body at the lower pressure
ISOTHERMAL ENTROPY CHANGES Entropy changes at a single temperature follow directly from equation 5 (Calculations involving a temperature difference or temperature change will be treated later.)
A change of phase, under equilibrium conditions, will be at constant temperature and
constant pressure The thermal energy transfer, Q = Q , will be ∆H Therefore, rev
For example, the entropy change for the melting of ice was found to be 1.22 J/g·K or 22.0
J/mol·K The entropy of vaporization of water at 100 C iso
More typical liquids, not involving such strong intermolecular forces, have ∆S values of about vap
92 J/mol·K at their normal boiling points This is known as Trouton’s rule.
The work done in a reversible, isothermal expansion of an ideal gas was found to be
(equation 10, Chapter 1) W = - nRT ln V /V For an ideal gas at constant temperature ∆E = 0 rev 2 1 and therefore Q = -W The entropy change is thus rev rev
Trang 2( ,I.G.,rev.) nRln V2/V1 (15)
T
Q S
Diabatic, like its better-known twin diabetic, implies that something “passes through”
6
(thermal energy or sugar) Thus adiabatic tells us there is no thermal energy transfer; Q = 0 In
practice we achieve adiabatic conditions by insulating the system, by maintaining system and surroundings at the same temperature, or by carrying out the process very rapidly Most gas expansions are fast enough to be nearly adiabatic
One sometimes sees the statement that ∆S = Q/T for a reversible process This is not
7
really wrong, but it is doubly misleading It de-emphasizes the need to find Q (which is often rev different from Q) and it provides no clue to finding ∆S for an irreversible process Equation 5, ∆S
= Q /T, covers both irreversible and reversible processes rev
If the gas expands through a pinhole or stopcock (an effusion-controlled leak) into an
evacuated container (Figure 2), no work is done because the gas exerts forces only on the
immovable walls of the container By insulating the containers we can ensure that Q = 0 The
energy is unchanged and therefore, if the gas is ideal, the temperature remains constant This serves as an experimental demonstration (performed by Joule, in 1844) of the dependence of energy only on the temperature, although it is not as sensitive as a later method by Joule and Thomson The entropy change of an ideal gas in such an irreversible, adiabatic expansion can be6 calculated as follows
Entropy is a state function, so the entropy change depends only on the initial state and the final state Entropy changes are independent of the path taken between the initial and final states
When an ideal gas expands at constant temperature from V to V , the final state is the same1 2
whether the process is reversible or irreversible Therefore, the entropy change for the adiabatic expansion of the ideal gas into a vacuum is
(I.G., T) ∆S = nR ln V /V (16)2 1
even though Q = -W = 0 in this process.
The example just cited illustrates a general method for evaluating entropy changes in
irreversible processes Having determined what are the initial and final states, a path is found
between those states that will be entirely reversible Then equation 5 tells us ∆S = Q /T rev 7
It is of interest to calculate the entropy change for the surroundings in the reversible and
irreversible expansions above In the reversible expansion, Q = Q = - Q rev surr = -(Q ) rev surr
Trang 3(rev) ∆S surr = - nR ln V /V2 1
The adiabatic expansion into a vacuum produces no change at all in the surroundings, and
therefore
∆S surr = 0
Adding together the entropy changes for system and surroundings, we obtain, for the reversible expansion,
(∆S)system + surroundings = 0 and for the irreversible expansion, into a vacuum,
(∆S)system + surroundings = nR ln V /V > 02 1
This result agrees with the requirements of the second law
The particular irreversible path chosen is an extreme case If the gas had been allowed to expand against a constant external pressure (as in Figure 1, Chapter 1) the entropy change for system plus surroundings would have been positive, but less than that for the expansion into a vacuum The magnitude of the total entropy change (system plus surroundings) can be taken as a measure of the degree of thermodynamic irreversibility of any process
INTERPRETATION OF ENTROPY We have seen that we can measure the increase in entropy by
measuring the amount of energy transferred to the system, as thermal energy transfer, Q, divided
by the (absolute) temperature, provided we measure Q along a path between initial and final
states that is completely reversible To know whether a process will actually occur, we must make a similar measurement for changes in the surroundings, as we will emphasize below
Without getting involved in the detailed calculations of statistical mechanics, it is important
to look more carefully at the meaning of entropy as a spread function In particular, there are two
ways of measuring the spreading, or the randomness, of a state We have seen that entropy is appropriately called the spread function because it is a measure of the amount of spreading of
energy among the molecules of a substance However, in addition to considering where the packets of energy are located we should look also at where the molecules are (like the yellow and
white rice molecules we mixed earlier) Remarkably, the same measurement of Q rev /T gives
information on this second type of spreading
Consider a sample of an ideal gas, occupying an initial volume V (e.g., 1 m ) Assume, quite i 3 arbitrarily for the moment, that each molecule requires some very small volume, V (e.g., 2 Å ,o 3 most of which volume elements are unoccupied at any given instant) Then we can say there are
spaces for V /V molecules (5 x 10 in our example) If we let the gas expand to a larger volume, i o 29
V , then by the same reasoning there are now spaces for V /V molecules The important thing is f f o
Trang 4that we have increased the number of available spaces by the ratio V /V But we already know f i that if we expand an ideal gas, the entropy increases by ∆S = nR ln V /V In other words, when f i
we increased the number of available, or accessible, spaces by the ratio V /V , the entropy f i
increased by the logarithm of this same ratio
We could equally well make the same kind of argument for molecules in a crystal, where the number of available spaces is more readily counted If some number of “stray” molecules are introduced and given an opportunity to diffuse throughout the crystal, we would obtain a
substantial increase in entropy per molecule Or we may simply drop a crystal of salt into water When we return, very likely the salt molecules will have diffused throughout the water The salt molecules, or their constituent ions, are attracted to each other and roughly equally attracted to
water molecules, so there is negligible energy difference, but there is a very high probability of
the salt diffusing into the water, never to return to the crystal unless external conditions are
changed It is like a drop of water on the pavement Even though energetically the water drop would prefer to remain with its neighbors, the lure of wide open spaces is enough to make the drop evaporate The process is driven by an increase in entropy, or in probability
The crystal is highly ordered; the solution is highly disordered, or random There are more states accessible to the salt when it is dissolved in water, or accessible to gas molecules when a larger volume is available to the molecules More available, or accessible, states can be equated
to a greater probability Thus entropy is variously described as measuring disorder, or
randomness, or probability We can always organize molecules, forming crystals or condensing a
gas to a liquid or arranging conditions such that molecules will form living structures To do so
always has a cost involved, specifically the cost of increased randomness in the surroundings
The second law of thermodynamics does not tell us we cannot decrease entropy in any given sample It tells us only that, for the universe as a whole (system plus surroundings), entropy will always increase
A study of how signals may be communicated from one place to another gave rise to a new
understanding of probability under the heading of information theory Information theory turned
out to be of major importance in understanding the meaning of entropy as well as in discovering better ways to organize the storage and distribution of information over telephone lines, as radio
or television signals, or in computers
Gibbs Free Energy
When investigating equilibrium or reversibility, it was necessary to find the change for the system but also for the surroundings Usually we prefer to ignore the surroundings most of the time The free energy functions provide a means of doing just that
We began with the variables P, V, and T, to which we added energy, E, and subsequently have added enthalpy, H, entropy, S, and now the two free energy functions, F and G, as shown in
equations 17-19 and in Figure 3
H = E + PV (17)
F = E - TS (18)
G = H - TS (19)
Trang 5
V P G T = ∂ ∂ (23)
S T G P − = ∂ ∂ For many years the Gibbs free energy and the Helmholtz free energy were not clearly 8 distinguished, and thus both were designated by the symbol F In recent decades the symbol F has been selected for Helmholtz free energy and the symbol G for Gibbs free energy Unless specifically indicated otherwise, the term “free energy” will always mean the Gibbs free energy in this book [Fig 3: A becomes F.] G = F + PV (19a)
G = E + PV - TS (19b)
We introduced the Helmholtz free energy from the relationship (eqn 4) ∆A = ∆F = W and8 rev the observation that a change in F represents the minimum amount of work that must be done on the system, at constant temperature, to get from the initial to the final state For example, you cannot put a box on a shelf without doing some work on the box (and gravitational field), nor can you compress a gas without doing work on the gas There is good reason for picking the correct function Recall that we found enthalpy, H, to be much more convenient than energy, E, when measuring thermal energy transfers under conditions of constant pressure For much the same reasons, we will find G to be generally more convenient than F The Helmholtz free energy is more appropriate at constant volume; Gibbs free energy is better for constant pressure FREE ENERGY AND EQUILIBRIUM Assume for the moment that the only work done is work of expansion or compression, and that the composition of the system does not change Then, from equation 19b, dG = dE + d(PV) - d(TS) As we have seen, because E is a state function, dE is independent of path, so dE = q + w = q + w rev rev regardless of path We may therefore substitute q + w for dE, to obtain rev rev dG = q + w + P dV + V dP - T dS - S dT (20) rev rev But q = T dS and w = - P dV Therefore, quite generally, rev rev dG = V dP - S dT (21)
We effectively split this equation into two parts, by holding constant first the temperature, then the pressure:
Trang 6We introduced the assumption that w # w by considering an expansion or compression
9
rev
of an ideal gas, and from that and the independence of dE on path, we concluded that q = dE - rev
w rev $ q Now we may work backward from the second law, which is one of our basic postulates,
(dS)syst + surr = q /T + (q ) rev rev surr /T surr$ 0 This inequality must be valid for any process occurring in
the system, including if q surr = (q ) rev surr, which may therefore be assumed without loss of generality
But we know that q surr = - q (by combining the first law with the first-law equation), and because
we are assuming temperature is constant, we may set T surr = T Therefore (dS)syst + surr = q /T - rev q/T $ 0 and thus q $ q It follows from this that w # w rev rev
Note, of course, that P and V naturally go together, as do
T and S.
An equally valid alternative to equation 20 is obtained
by substituting dE = q + w, giving
dG = q + w + P dV + V dP - T dS - S dT
Then, for constant temperature and pressure,
(T,P) dG = q + w + P dV - T dS
(24)
But at constant pressure we may safely assume the pressure of the system is the same as the
pressure of the surroundings (meaning the part of the surroundings that is in actual contact with,
and therefore may affect, the system), so w = - P dV, where, as usual, P is the pressure of the system Also, we know that q 9 $ q Therefore,
rev (T,P) dG = q - T dS = q - q rev# 0
Thus
(T,P) dG # 0 (25)
in general, and for the limiting case of the reversible, or equilibrium, process,
(T,P) dG = 0 (25a)
This shows that ∆G must be negative, or, in the limit of a reversible process, zero That is, G is a
minimum for an equilibrium state This is the more convenient criterion for equilibrium that was
sought Looking for the change in free energy, G, under our normal operating conditions of
Trang 7∫ ∫ = ∫ = ∫ =
∂
∂
=
=
P
G dG
G
T
constant temperature and constant pressure, we need only look at changes of the system It is not necessary to make separate calculations for the surroundings
We began this section with the assumption that the only work done was P dV work, or work
of expansion or compression Now we revisit that question Retain the conditions of constant temperature and constant pressure, but from equation 20 we obtain
(T, P) dG = P dV + w rev
The total work done on the system may be broken into expansion/compression work and other
forms (e.g., electrical work) which we label w’ For reversible processes,
w = - P dV + w’ rev rev
It follows that
(T,P) dG = w’ (26) rev
Thus, apart from providing a convenient measure of whether the system is or is not at equilibrium
(with only P dV work), the free energy tells us how much other work can be expected from a process In an electrochemical cell, w’ is the electrical energy generated by the chemical reaction Work against a gravitational field or against a magnetic field would also be included in w’.
IDEAL-GAS EXPANSIONS The change in free energy of an ideal gas in an isothermal expansion or compression is particularly important because it has been found possible to express free energy changes of all systems, whether gases or not, in this same form The broader theory will be
developed in the discussion of physical equilibria
Start with the definition of G, equation 19 At constant temperature,
(T) ∆G = ∆H - T ∆S (27)
In an isothermal, reversible expansion of an ideal gas, ∆E = 0 and ∆H = 0.
q = - w = rev rev InRT dV/V = nRT ln V /V2 1
∆S = q /T = nR ln V /V rev 2 1
∆G = - T ∆S = - nRT ln V /V2 1
or
(T, I.G.) ∆G = nRT ln P /P (28)2 1
The same result could be obtained from equation 22
FREE-ENERGY CHANGES IN CHEMICAL REACTIONS A very important application of the
free-energy function is to the description of chemical reactions Because G is a state function, free
energies can be added and subtracted in the same manner as enthalpies It is not possible to
Trang 8assign meaningful absolute values to free energies, but handbook tables give values of G relative
to the elements from which compounds are formed Thus the free energy of formation of any
element is, by definition, zero From Table 1 we can see that the standard free energy of
formation of SO in its standard state (gas, at 1 atm pressure, at 25 C) is - 370.4 kJ/mol, whereas3 o
∆G (SO ) = - 300.2 kJ/mol Subtracting (and setting ∆G (O ) = 0) gives ∆G o f 2 o f 2 reaction = - 70.2 kJ/mole The negative value indicates that the reaction is spontaneous, showing that oxygen gas will combine with sulfur dioxide to form sulfur trioxide at this temperature and pressure, at least
in the presence of a suitable catalyst
Table 1 STANDARD ENTHALPIES AND FREE ENERGIES OF FORMATION*
*Values are in kJ/mol, for 25 C, 1 atm The standard states for liquids (lq) and crystalline solids (c) are the pure o materials at 1 atm pressure and the standard states for gases (g) are the pure gases at 1 atm pressure in the “ideal gas state” (that is, extrapolated from low pressures assuming ideal gas behavior) “aq” refers to the (hypothetical ideal) 1 molal solution in water.
Trang 9( )
T
reactants products
G G
P
G
∂
−
∂
=
∂
∆
∂
T reactants
T
products
G P
G P
G
∂
∂
−
∂
∂
=
∂
∆
∂
reactants products
T
V V
P
G
−
=
∂
∆
∂
(29)
V P
G T
∆
=
∂
∆
∂
Free energy values do not tell us anything about rates of reaction There may be a significant activation energy required to get a reaction to proceed, even when it is thermodynamically
allowed (That is what catalysts are good for By offering a different path, they may involve a
substantially lower activation energy and therefore a faster rate.) On the other hand, if ∆G > 0,
the reaction cannot proceed, under the specified conditions, regardless of catalysts or time
LECHATELIER’S PRINCIPLE AND EQUILIBRIUM For every chemical reaction there exists an equilibrium point at which there is no further tendency for the reaction to proceed forward or backward Although the equilibrium point of most reactions is far to one side or the other, a true equilibrium point can be calculated, and for nearly all systems the equilibrium point can be demonstrated if the experimental techniques are sufficiently sensitive
LeChatelier’s principle states that when any stress is applied to a system at equilibrium, the system will respond in a manner that will reduce the stress For example, the reaction
2 NO + O2!6 2 NO2
releases thermal energy (∆H = - 56.53 kJ/mol(NO )) but causes a decrease of volume, or2
pressure Therefore an increase of temperature will tend to drive the reaction to the left, so that the reaction will absorb thermal energy, but an increase of pressure, by compression, will tend to drive the reaction to the right to decrease the number of moles of gas and hence the pressure These relationships can be derived, quantitatively, from the second law
The change in ∆G reaction = G products - G reactants with change of pressure is
The derivative of the sum is the sum of the derivatives:
From equation 22, these derivatives are the respective volumes:
or
Trang 10(30)
S T G P ∆ − = ∂ ∆ ∂ (30a)
T H G T G P ∆ − ∆ = ∂ ∆ ∂ (31)
T
H T
G
P
∆
−
=
∂
∆
∂
It is assumed that there are no other changes that might cause a reverse effect For 10
example, changes in the relative concentrations of the reactants, or addition of an inert gas, are not within the scope of equations 29 and 30 or 31
Similarly, from equation 23,
This equation tells how ∆G (measured at some constant temperature and pressure) depends upon the temperature at which the process occurs The ∆S on the right-hand side is similarly to be
measured at a constant pressure and at a constant temperature (the temperature at which the
derivative, or the slope of ∆G vs T, is determined) We may therefore substitute for - ∆S the quantity (∆G - ∆H)/T.
To find the shift of equilibrium point we set ∆G = 0 and obtain
The first result, equation 29, says that if there is a volume increase during the process, a pressure
increase will cause an increase in ∆G, making the process less spontaneous The second equation (31) says that if the process is endothermic (∆H positive), a temperature increase will make ∆G
more negative, thus making the process more spontaneous The equations apply to phase
changes or other equilibrium processes as well as to chemical reactions The importance of the10
equations is that they give a quantitative description of how ∆G changes, and therefore of how
the equilibrium point changes
Free Energy and the Energy-Entropy Battle
For the formation of ammonia, we may write the equation
½ N +2 3/2 H2&6 NH3
∆G = - 16.4 kJ/mol, ∆H = - 45.9 kJ/mol, and ∆S = -99.2 J/mol·K The reaction clearly is
spontaneous at room temperature and pressure, but it doesn’t go The activation barrier is too great From LeChatelier’s rule, we can see that an increase in temperature, to overcome the