Chapter 19: Entropy and Free Energy
You are expected to be familiar with the terms enthalpy,
endothermic and exothermic. If you are not fully comfortable
with these concepts, please review Chapter 6 of Kotz.
Chemical Thermodynamics
Thermodynamics is defined as the study of energy. It reaches
across the different physical sciences, and you will no doubt
encounter thermodynamics in physics as well. We will be
focusing on thermodynamics from the chemist’s perspective.
Chemical thermodynamics is the study of how far reactions will
progress. Equilibrium is a key aspect of thermodynamics – the
“measuring how far” part. In Chapter 19, we will be looking at
the “explaining why the equilibrium lies where it does” part of
thermodynamics. We do this by comparing the energy of the
reactants with the energy of the products (“lower is better”). In
Chemistry 1000, we focused our study of energy on enthalpy.
In Chemistry 2000, we will also factor in entropy, allowing us to
calculate and compare free energy at different temperatures.
The term spontaneous is used to indicate the direction in which
a reaction is favoured. A product-favoured reaction is called
“spontaneous” while a reactant-favoured reaction is called
“nonspontaneous”. ***This term gives no information about a
reaction’s speed/kinetics; it just means the reaction can occur
without a net input of outside energy.***
e.g. Formation of green copper salts on brass statues/roofs is a
spontaneous process by the chemical definition – but it can
take months! Reaction between NaOH and HCl in a
titration is also spontaneous – but much faster.
Heat Flow and Calorimetry
Calorimetry is the science of heat measurement. There are two
different types of calorimetry:
• constant-pressure calorimetry (in “coffee cup calorimeter”)
• constant-volume calorimetry (in “bomb calorimeter”)
In both types, the heat produced/consumed by a reaction is
determined by measuring temperature change. The temperature
change can be related to heat produced/consumed using the
specific heat capacity of the heat-absorbing medium:
q = m C ∆T
where q is heat (in kJ), m is mass of absorbing medium (in kg),
C is specific heat capacity (in kJ·kg-1·K-1) and ∆T is temperature
change (in K).
Specific heat capacity is a measure of how easily a compound
changes temperature. A compound with a high heat capacity
must absorb a lot of energy to increase its temperature by 1 K
(and therefore lose a lot of energy to decrease by 1 K). A
compound with a low heat capacity need only absorb a small
amount of energy to increase its temperature by 1 K (and
therefore lose a small amount of energy to decrease by 1 K).
Heat capacities are known for many common substances:
e.g. copper 0.385 kJ·kg-1·K-1
glass
0.8 kJ·kg-1·K-1
air
~1.0 kJ·kg-1·K-1
water
4.184 kJ·kg-1·K-1
Alternately, it is sometimes easier to calibrate a calorimeter to
find its heat capacity (and use that value for m·C in calculations).
By convention, we view energy changes from the perspective of
the system.
• If a system gains energy from its surroundings, ∆E > 0
• If a system loses energy to its surroundings, ∆E < 0
In calorimetry, the system is self-contained, so there is no
transfer of energy to the surroundings. As such:
q absorbed by calorimeter + q released by reaction = 0
q absorbed by calorimeter = - q released by reaction
• If the calorimeter absorbs heat (q absorbed > 0), the reaction
inside was exothermic (q released < 0).
• If the calorimeter releases heat (q absorbed < 0), the reaction
inside was endothermic (q released > 0).
If there are multiple substances absorbing heat, the heat
absorbed by each must be calculated. Adding them all together
gives q absorbed.
e.g. Sulfur (S8(s), 2.56 g) is burned in a bomb calorimeter with
excess oxygen (O2(g)). The temperature increases from
21.25 ˚C to 26.72 ˚C. The bomb has a heat capacity of 923
J/K, and it contains 815 g water. Calculate the heat
evolved per mole of SO2(g) formed (“heat of formation”!).
In constant volume calorimetry, all of the energy produced by
a reaction is transferred as heat:
∆E = qv
In constant pressure calorimetry, some of the energy
produced by a reaction is used to do work; the rest is transferred
as heat:
∆E = qp + w
Work is the application of a force over a distance. Increasing
the volume of a coffee-cup calorimeter qualifies as work
because a force moves the lid. The greater the external pressure,
the greater the force that must be applied to do this work.
w = - P∆V
Therefore:
∆E = qp - P∆V
Again, we view work from the perspective of the system’s total
energy.
• If a system does work on its surroundings, its energy
decreases: ∆E < 0 therefore w < 0
• If a system has work done on it by its surroundings, its
energy increases: ∆E > 0 therefore w > 0
By definition, heat change at a constant pressure is enthalpy:
qp = ∆H
Therefore:
∆E = ∆H - P∆V
Or:
∆H = ∆E + P∆V
At the introductory level, we generally consider that ∆H ≈ ∆E
and don’t correct for pressure-volume work. Since the reaction
we are studying is our system:
• If ∆E < 0 then ∆H < 0 (energy released to surroundings).
This is an ____________________ reaction.
)
• If ∆E > 0 then ∆H > 0 (energy absorbed from surroundings.
This is an ____________________ reaction.
Recall Hess’ Law. Enthalpies are additive.
e.g. In a two-step reaction, step 1 has ∆H = 5 kJ/mol and step 2
has ∆H = 10 kJ/mol. Overall, ∆H = 15 kJ/mol.
e.g. In a two-step reaction, step 1 has ∆H = 5 kJ/mol and step 2
has ∆H = - 10 kJ/mol. Overall, ∆H = - 5 kJ/mol.
We can use Hess’ Law to calculate the enthalpy change for any
reaction in which we know the standard enthalpies of formation
for all of the reactants and products.
Note that standard conditions for thermodynamics are 25 ˚C
and 1 bar (0.9869 atm). This is different from the Ideal Gas
Law which has 0 ˚C as its standard temperature.
e.g. The reactions below are two of the key reactions in the
processing of uranium for use as fuel in nuclear power
plants:
UO2(s) + 4 HF(g) → UF4(s) + 2 H2O(g)
UF4(s) + F2(g) → UF6(g)
.
(a) Calculate the enthalpy change for each reaction.
(b) Calculate the enthalpy change for the overall process:
UO2(s) + 4 HF(g) + F2(g) → UF6(g)
+ 2 H2O(g)
compound
UO2(s)
UF4(s)
UF6(g)
HF
H2O(g)
∆Hf˚
(kJ/mol)
-1085
-1914
-2147
-271.1
-241.818
Laws of Thermodynamics
1. The total energy of the universe is constant.
2. The total entropy of the universe always increases for a
spontaneous process.
3. The entropy of a pure, perfectly formed crystal at 0 K is
zero.
Entropy
Entropy is a measure of order. To be more precise, it is a
measure of lack-of-order, or chaos. The second law of
thermodynamics states that entropy will increase for any
spontaneous process – this can occur via dispersal of matter or
via dispersal of energy.
Dispersal of Matter:
When methanol is dissolved in water, the solution gets cold.
This is clearly an endothermic process – yet it is spontaneous!
This is because the entropy of the system increases since the
water and methanol molecules are less organized when mixed
than when separate.
Dispersal of Energy:
When bromomethane (CH3Br) and magnesium metal (Mg)
react, they make CH3MgBr. In this case, the entropy of the
system has decreased (two molecules bond to make one more
organized molecule), but the reaction is still spontaneous. Why?
This reaction is very exothermic, dispersing heat energy out into
the surroundings. Since this heat energy is now more spread out
(rather than being tied up in the molecules), the entropy of the
universe has increased.
Mathematically, we can express this as:
∆S universe = ∆S system + ∆S surroundings
The only way a reaction can affect the entropy of its
surroundings is by absorbing or emitting heat. If a reaction
absorbs heat (∆H > 0), it reduces the entropy of its surroundings
(∆S surroundings < 0). If a reaction emits heat (∆H < 0), it increases
the entropy of its surroundings (∆S surroundings > 0). This effect is
also inversely proportional to temperature:
∆S surroundings = - ∆H system
T
Therefore:
∆S universe = ∆S system - ∆H system
T
Therefore:
T ∆S universe = T ∆S system - ∆H system
Or:
- T ∆S universe = ∆H system - T ∆S system
Finally, we define –T∆S universe as Gibbs Free Energy (∆G), the
effect a reaction has on the order of the universe:
∆G system = ∆H system - T ∆S system
If ∆G < 0, the reaction makes the universe less ordered and is
therefore product-favoured (spontaneous in forward direction).
If ∆G > 0, the reaction makes the universe more ordered and is
therefore reactant-favoured (spontaneous in reverse direction).
∆G can only be used for reactions at constant pressure. Why?
Looking at the formula for Gibbs free energy, we can make
some generalizations about reaction outcomes based on enthalpy
and entropy changes:
∆Hrxn
∆Srxn
∆Grxn
K
Reaction Outcome
+
–
(enthalpy (entropy
decreases) increases)
–
–
(enthalpy (entropy
decreases) decreases)
+
+
(enthalpy (entropy
increases) increases)
–
+
(enthalpy (entropy
increases) decreases)
Note that enthalpy dominates at lower temperatures while
entropy dominates at higher temperatures.
e.g. Calculate ∆G˚ for each of the following reactions, and
indicate when (if ever) it would be product-favoured.
(a) C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ∆H˚ = -673 kJ
∆S˚ = 60.4 J/K
(b) MgO(s) + C(s, graphite) → Mg(s) + CO(g)
∆H˚ = 491.18 kJ
∆S˚ = 197.67 J/K
Like enthalpies, Gibbs free energies are additive. Hess’ Law
still applies though it’s not called that anymore. (This is also
true for entropies. Note that standard entropy of formation is S˚
– not ∆Sf˚. This is because entropies are calculated relative to
absolute zero; see the third law of thermodynamics)
Gibbs Free Energy and Equilibrium
We have seen that Gibbs free energy tells us whether a reaction
is product-favoured or reactant-favoured:
• If ∆G < 0
Reaction is product-favoured
(Q < K)
• If ∆G > 0
Reaction is reactant-favoured
We can therefore conclude that:
(Q > K)
• If ∆G = 0
Reaction is at equilibrium!
(Q = K)
We can use the reaction quotient (Q) to relate Gibbs free energy
under nonstandard conditions (∆G) to Gibbs free energy under
standard conditions (∆G˚):
∆G = ∆G˚ + R T lnQ
As Q increases, ∆G becomes more positive. This makes the
reaction less product-favoured/more reactant-favoured.
As Q decreases, ∆G becomes more negative. This makes the
reaction more product-favoured/less reactant-favoured.
If we consider the system at equilibrium (∆G = 0, Q = K), we
can relate Gibbs free energy to the equilibrium constant:
0 = ∆G˚ + R T lnK
Therefore:
∆G˚ = – R T lnK
e.g. Calculate ∆G˚ for formation of the blue complex
tetraamminecopper(II) from Cu2+ and NH3.
Cu2+(aq) + 4 NH3(aq) → [Cu(NH3)4]2+(aq)
Kf = 2 × 1014
e.g. Methanol (CH3OH, ∆Gf˚ = -166.27 kJ/mol) is widely used
as a fuel in race cars such as those in the Indianapolis 500.
The liquid fuel can be formed using the reaction:
C(s, graphite) + ½ O2(g) + 2 H2(g) → CH3OH(l)
Calculate K for the formation of methanol at 25 ˚C.
Comment on the relationship between the sign of ∆G˚ and
the magnitude of K.
Equilibrium Constants and Temperature
In Chapter 16, we noted that equilibrium constants were
temperature-dependent and noted qualitatively that increasing
the temperature would drive a reaction in the endothermic
direction while decreasing it would drive a reaction in the
exothermic direction. Now that we can relate equilibrium
constants to Gibbs free energy, we can quantitate this
relationship.
∆G˚ = – R T lnK
Therefore:
ln K = – ∆G˚ = – (∆H˚ – T∆S˚)
RT
RT .
Therefore:
ln K = – ∆H˚ + ∆S˚
RT
R .
If we assume that ∆H˚ and ∆S˚ are reasonably constant over the
range of temperatures in which we are interested, we can set up
a comparative equation, relating K to T:
∆Ho
K1
ln
= R
K2
1
1
T 1 T2
The format of this equation should be starting to look very
familiar! See similar formulae in chapters 14 and 15.
e.g. Hydrogen gas reacts with fluorine gas to make hydrogen
fluoride gas. Calculate the equilibrium constant for this
reaction at 250 ˚C.
∆Hf˚(HF(g)) = -332.63 kJ/mol
S˚(HF(g)) = 88.7 J/mol·K
Important Concepts from Chapter 19
• laws of thermodynamics
• system vs. surroundings (including sign conventions for
energy and work)
• calorimetry and specific heat capacity
• constant-volume vs. constant-pressure calorimetry
• enthalpy, entropy, and Gibbs free energy
o what they are
o use in calculations
o relationships between these three state functions
• relationship between Gibbs free energy and equilibrium
• relationship between temperature and equilibrium constants
• standard conditions for thermodynamics (25 ˚C, 1 bar)