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CHAPTER 14
Entropy and Free Energy
Review of Thermodynamics
The following are general concepts from thermodynamics as
applied to chemical systems.
1) First law of thermodynamics: U = q + w
w = work ( = - pV for mechanical work for a system at constant
pressure)
q = heat
U = Uf - Ui = change in internal energy
Note that U is a state function, while q and w are not.
Sign convention
w > 0 work done on system
w < 0 work done by system
(volume decreases)
(volume increases)
q > 0 heat flows into system
(endothermic)
q < 0 heat flows out of system
(exothermic)
2) Enthalpy and heat
H (enthalpy) is defined as H = U + pV (and so a state function)
At constant volume U = qV
At constant pressure H = qp
3) Calculating the enthalpy change for a chemical reaction
Hrxn = [  Hf(products) ] - [  Hf(reactants) ]
The above is in terms of the enthalpy of formation. Recall that by
definition the formation reaction is the reaction that produces one mole
of a single product out of elements in their standard (most stable) state.
For example, the formation reaction for C2H5Cl(g) is
2 C(s) + 5/2 H2(g) + 1/2 Cl2(g)  C2H5Cl(g)
Hf(C2H5Cl(g)), the enthalpy of formation for C2H5Cl(g), is the value
for H when one mole of C2H5Cl(g) is formed by the above reaction at
standard conditions.
4) The standard state conditions for a substance are defined as follows:
gas
1 atm pressure + ideal behavior
liquid pure liquid (so solvents are approximately in their
standard state)
solid
pure solid
solute 1 mol/L + ideal behavior
Standard states can also be defined for other cases, but the above is
sufficient for our purposes.
We will use the symbol “  ” to indicate the standard state. For
example:
Hf(C2H5Cl(g)) enthalpy of formation for chloroethane for
standard conditions
2 C(s) + 5/2 H2(g) + 1/2 Cl2(g)  C2H5Cl(g)
Spontaneous Processes
Consider the following two processes. Why does only one occur?
Entropy
To determine which processes consistent with the first law of
thermodynamics actually occur we must first define a new thermodynamic function, called entropy (S).
Entropy measures the amount of “disorder” in a system (in a
precise mathematical sense)*. Entropy is a state function. We may talk
about the entropy change for the system (Ssyst) and the entropy change
for the surroundings (Ssurr). The entropy change for the universe
(Suniv) is then given by the expression
Suniv = Ssyst + Ssurr
We will discuss ways to find values for S for specific types of
processes. Note that at constant temperature S = (q)rev/T *
* In fact, S = if (đq)rev/T
The Second Law of Thermodynamics
The importance of entropy lies in the fact that it can be used to
decide which processes that are consistent with the first law of
thermodynamics will occur, and which will not. To decide, we need a
new thermodynamic law, called the second law of thermodynamics
Suniv  0 for any allowed process
In words, what the above means is the amount of randomness or
disorder in the universe (as measured by entropy) must always increase
or remain constant.
Comments Concerning the Second Law
1) The units for entropy are J/K (or J/mol.K)
2) There are three types of processes according to the second law
Suniv > 0 , spontaneous, will eventually occur
Suniv = 0 , process where equilibrium has been established
Suniv < 0 , process will not occur
3) The second law of thermodynamics is based on observation.
4) Since Suniv = Ssyst + Ssurr, it is possible for Ssyst or Ssurr to
be negative and still have a spontaneous process (though both cannot be
negative for the same process).
5) There is a connection between entropy and the way in which
states can be arranged on a molecular level, though we will not explore
this (this is studied in statistical thermodynamics).
Spontaneous Processes
A process will occur spontaneously if Suniv > 0. If Suniv < 0,
then a process will not occur. Suniv = 0 indicates we are at equilibrium.
T = 20. C
Suniv,melt > 0 Suniv,freeze < 0
0. C
Suniv,melt = 0 Suniv,freeze = 0
T = - 20. C
Suniv,melt < 0 Suniv,freeze > 0
T=
Molar Entropy
The standard molar entropy of a substance is the value for the entropy of
1 mole of the substance for standard pressure (1 atm) and some temperature T. We usually choose T = 25. C.
Entropy Change For a Chemical Reaction
(System)
When a chemical reaction takes place, Ssyst = Srxn.
expression for the change in entropy, for standard conditions, is
The
Srxn = [  S(products) ] - [  S(reactants) ]
Note that elements in their standard state must be included in the
above calculation.
The above method works because of the fact that entropy, like
enthalpy, is a state function.
Entropy Change For a Chemical Reaction
(Surroundings)
For the entropy change for the surroundings, we may show that
(for constant pressure and standard conditions)
Ssurr = qsurr/T = - qsyst/T = - Hrxn/T
where Hrxn is the enthalpy change for the chemical reaction.
The entropy change for the universe is then
Suniv = Ssyst + Ssurr = Srxn - (Hrxn/T)
Example
Consider the following chemical reaction
N2(g) + 3 H2(g)  2 NH3(g)
What are Ssyst, Ssurr, and Suniv for the above reaction when
carried out under standard conditions (pgas = 1.0 atm, T = 25.C)? Is the
reaction spontaneous for these conditions?
Consider the following chemical reaction
N2(g) + 3 H2(g)  2 NH3(g)
What are Ssyst, Ssurr, and Suniv for the above reaction when
carried out under standard conditions (pgas = 1.0 atm, T = 25. C)? Is the
reaction spontaneous for these conditions? (Appendix 2 for data)
Substance
H°f(kJ/mol) G°f(kJ/mol) S°(J/mol.K)
N2(g)
0.0
0.0
191.6
H2(g)
0.0
0.0
130.7
- 45.9
- 16.4
192.8
NH3(g)
Consider the following chemical reaction
N2(g) + 3 H2(g)  2 NH3(g)
What are Ssyst, Ssurr, and Suniv for the above reaction when
carried out under standard conditions (pgas = 1.0 atm, T = 25. C)? Is the
reaction spontaneous for these conditions? (Appendix 2 for data)
Substance
H°f(kJ/mol) G°f(kJ/mol) S°(J/mol.K)
N2(g)
0.0
0.0
191.6
H2(g)
0.0
0.0
130.7
- 45.9
- 16.4
192.8
NH3(g)
Srxn = [  S(products) ] - [  S(reactants) ]
= [2 (192.8)] - [(191.6) + 3(130.7)] = - 198.1 J/mol.K
Hrxn = [  Hf(products) ] - [  Hf(reactants) ]
= [(2 (-45.9)] - [ 0 ] = - 91.8 kJ/mol
Srxn = - 198.1 J/mol.K
Hrxn = - 91.8 kJ/mol
So
Ssyst = Srxn = - 198.1 J/mol.K
Ssurr = - Hrxn/T = - (- 91800. J/mol) = 307.8 J/mol.K
(298.2 K)
Suniv = Ssyst + Ssurr
= (- 198.1 J/mol.K) + (307.8 J/mol.K) = 109.7 J/mol.K
Note that even though Ssyst < 0, the reaction is still spontaneous
for standard conditions.
Entropy For Pure Substances
The more random a substance is, the larger the value for entropy.
So Sgas >> Sliquid > Ssolid
S°(Ca(s)) = 41.6 J/mol.K
S°(H2O(l)) = 69.9 J/mol.K
S°(Ca(g)) = 154.9 J/mol.K
S°(H2O(g)) = 188.7 J/mol.K
Predictions For Entropy Changes For
Chemical Reactions
Since Sgas >> Sliquid > Ssolid
gases are far more disordered than solids and liquids. Based on that, we
would predict the following
If ngas > 0 then Srxn > 0
If ngas = 0 then Srxn ~ 0
If ngas < 0 then Srxn < 0
Example: Predict the value for Srxn for the following reactions
CaCO3(s)  CaO(s) + CO2(g)
C(s) + 2 H2(g)  CH4(g)
2 HCl(g) + Br2(g)  2 HBr(g) + Cl2(g)
Example: Predict the value for Srxn for the following reactions
CaCO3(s)  CaO(s) + CO2(g)
ng = +1, so S > 0
(S = + 160.2 J/mol.K)
C(s) + 2 H2(g)  CH4(g)
ng = -1, so S < 0
(S = - 80.8 J/mol.K)
2 HCl(g) + Br2(g)  2 HBr(g) + Cl2(g)
ng = 0, so S ~ 0
(S = + 1.2 J/mol.K)
Other Trends in Entropy
1) Entropy generally increases as the number of atoms per
molecule increases.
S°(CH4(g)) = 186.3 J/mol.K
S°(C2H6(g)) = 229.2 J/mol.K
S°(C3H8(g)) = 270.3 J/mol.K
2) Entropy generally increases when molecular solids or liquids
are dissolved in solvents. For ionic compounds, entropy usually (but
does not always) also increases.
S°(NaCl(s)) = 72.1 J/mol.K
S°(NaCl(aq)) = 115.0 J/mol.K
Third Law of Thermodynamics
The entropy of a pure substance increases as temperature
increases. This is because the degree of randomness in the substance
increases. This is particularly true if a phase transition occurs.
Based on low temperature experiments, it appears that the
entropy of every pure substance approaches the same value as T  0. K.
Third law of thermodynamics: The absolute entropy (S) of a
perfect crystal of any pure substance at absolute zero is 0.0 J/mol.K.
Because there are standard ways of find the change in entropy for
a pure substance as we change the temperature of the substance at
constant pressure, the third law of thermodynamics allows us to assign
values for entropy for pure substances at any temperature.
Standard molar entropy (S) – The value for the entropy of one
mole of a pure substance at p = 1.00 atm and a particular temperature.
We usually choose T = 25. C = 298. K.
Note - The above does not apply to ions in aqueous solution. In
that case, the entropy of H+ ion at T = 25. C is defined as 0.0 J/mol.K.
Free Energy - Definition
While the second law of thermodynamics is true for all processes,
it is difficult to use because it requires calculations on both the system
and the surroundings.
To help us around this problem we define a new thermodynamic
function, called the free energy (G).
G = H - TS (a state function)
Note that for a process taking place at a particular fixed
temperature
G = H - (TS) = H - TS
Free Energy and the Second Law
For a process taking place at a particular fixed temperature
G = H - (TS) = H - TS
If we divide both sides of the above equation by –T, we get
- (G)/T = - (H)/T + S = Ssurr + Ssyst = Suniv
Since the second law requires that Suniv  0, and since we know
that T > 0, it follows that the sign for G for a process where pressure
and temperature do not change tells us whether or not a process is
consistent with the second law of thermodynamics.
Since we require Suniv  0, it follows that for processes at a fixed
temperature and pressure, G  0.
Free Energy and the Second Law
For a process where temperature and pressure are constant, we
may say the following:
G < 0 process is spontaneous
G = 0 system is at equilibrium
G > 0 process is not spontaneous (will not occur)
Notice this means we need only do a calculation for the system,
but only if the process occurs at constant pressure and temperature.
Finding Gsyst For a Chemical Reaction
For a chemical reaction Gsyst = Grxn is given by the expression
Grxn = [  Gf(products) ] - [  Gf(reactants) ]
Note Gf = 0 for an element in its standard state.
Example: What are Hrxn, Srxn, and Grxn for the following process
4 Fe3O4(s) + O2(g)  6 Fe2O3(s)
Is the reaction spontaneous for standard conditions?
Example: What are Hrxn, Srxn, and Grxn for the following process
4 Fe3O4(s) + O2(g)  6 Fe2O3(s)
Is the reaction spontaneous for standard conditions?
As before, we need a table of thermodynamic data top do this
problem. Such data are given in Appendix 2 of the textbook.
Substance
H°f(kJ/mol) G°f(kJ/mol) S°(J/mol.K)
Fe3O4(s)
- 1118.4
- 1015.4
146.4
0.0
0.0
205.2
- 824.2
- 742.2
87.4
O2(g)
Fe2O3(s)
Example: What are Hrxn, Srxn, and Grxn for the following process
4 Fe3O4(s) + O2(g)  6 Fe2O3(s)
Hrxn = [6 Hf(Fe2O3(s))] - [4 Hf(Fe3O4(s))]
= [6 (- 824.2 kJ/mol)] - [4 (- 1118.4 kJ/mol)] = - 471.6. kJ/mol
Srxn = [6 S(Fe2O3(s))] - [4 S(Fe3O4(s)) + S(O2(g))]
= [6 (87.4 J/mol.K)] - [4 (146.4 J/mol.K) + (205.2 J/mol.K)]
= - 266.4 J/mol.K
Grxn = [6 Gf(Fe2O3(s))] - [4 Gf(Fe3O4(s))]
= [6 (- 742.2 kJ/mol)] - [4 (- 1015.4 kJ/mol)] = - 391.6 kJ/mol
As a check, Grxn = Hrxn - TSrxn
= - 471.6 kJ/mol - (298.2 K) (- 0.2664 kJ/mol.K)
= - 392.2 kJ/mol (process is spontaneous)
Finding When Reactions Are Spontaneous
Since
G = H - TS,
it follows that at a particular temperature
G =  H - T S
To a first approximation the values for Hrxn and Srxn (but not
the value for Grxn) are independent of temperature. We can use this to
find the approximate value for temperature at which equilibrium occurs
for standard conditions of concentration.
Grxn = 0 = Hrxn - Teq Srxn
Teq  Hrxn /Srxn
Example: At what temperature will H2O(l) and H2O(g) exist at
equilibrium? (Note this will correspond to the normal boiling point of
water).
Example: At what temperature will H2O(l) and H2O(g) exist at
equilibrium? (Note this will correspond to the normal boiling point of
water).
H2O(l)  H2O(g)
Hrxn = + 44.0 kJ/mol
Srxn = 118.8 J/mol.K
Teq  Hrxn /Srxn = (44000. J/mol)/(118.8 J/mol.K) = 370.4 K
Since
Grxn = Hrxn - T Srxn
it follows that if T > 370.4 K, Grxn < 0, and boiling is spontaneous.
Of course, the true value for the normal boiling point for water is
373.2 K. The small difference between this and the value calculated
above is due to the assumption that Hrxn and Srxn are independent of
temperature, which is only approximately correct.
Deciding When Reactions Will Be Spontaneous
Recall that
Grxn = Hrxn - T Srxn
Grxn < 0 for a spontaneous process
There are four possible combinations of Hrxn and Srxn.
Hrxn
Srxn
behavior
positive
positive
spontaneous if T > Teq
positive
negative
reaction is never spontaneous
negative
positive
reaction is always spontaneous
negative
negative
spontaneous if T < Teq
End of Chapter 14
“…even though the average disorder in the universe has been
increasing ever since the beginning, this increasing disorder is not
occurring at every particular place in the universe. You can have local
order created in certain locations at the expense of greater disorder
elsewhere. Take your refrigerator as an example. You have ice cubes in
the freezer that are certainly a manifestation of a great deal of order, but
if you go around to the back of the refrigerator there’s a lot of hot air
coming out indicating that the ‘cost’ of the order of the ice cubes is paid
by the greater amount of disorder somewhere else.” Murray Gell-Mann
“[Thermodynamics] is the only physical theory of universal
content which, within the framework of the applicability of its basic
concepts, I am convinced will never be overthrown.” Albert Einstein