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Chapter 16
Spontaneity, Entropy and Free Energy
16.1 Spontaneity and Entropy
 A reaction that will occur without outside
intervention. We can’t determine how fast.
 We need both thermodynamics and kinetics to
describe a reaction completely.
 Thermodynamics compares initial and final
states but not how long it will take.
Kinetics
(YDVD)
describes the pathway between.
Thermodynamics
 1st Law - the energy of the universe is
constant. (Cannot create or destroy)
 Keeping track of thermodynamics
doesn’t correctly predict spontaneity.
 Entropy (S) is disorder or randomness.
 2nd Law - the entropy of the universe
increases.
Entropy
 The driving force for a spontaneous
process is an increase in the entropy of
the universe.
 Defined in terms of probability.
 Substances take the arrangement that is
most likely.
 The most likely is the most random.
Think about this.
 It would seem like reactions with a negative DH value
would proceed spontaneously.
 But NH3 = O2  NO + H2O will not occur
spontaneously even though DH = -909kJ/mol
 It would also seem like reactions with a positive DH
value would not occur spontaneously.
 But KCl(s)  K+(aq) + Cl-(aq) will occur
spontaneously even though DH =+19.2 kJ/mol
 Activation energy is involved in both reactions but
there must be something else involved.
 Entropy is that driving force!
Positional Entropy
 A gas expands into a vacuum because the
expanded state has the highest positional
probability of states available to the system.
 Ssolid < Sliquid << Sgas
 There are many more ways for the molecules
to be arranged as a liquid than a solid.
 Gases have a huge number of positions
possible.
Practice
For each of the following pairs, choose the
substance with the higher positional entropy
(per mole) at a given temperature.
• solid CO2 and gaseous CO2
• N2 gas at 1 atm and N2 gas at 1.0 x 10-2 atm
Predict the sign of the entropy change for
each of the following processes.
• Solid sugar is added to water to form a solution.
• Iodine vapor condenses on a cold surface to form
crystals.
Proof?
 The change in entropy (DS) in a reaction can be found by
subtracting the absolute values of the entropy of the
reactants from the absolute entropies of the products
 DS reaction = DS products - DS reactants
 For the reaction below, the absolute entropies of the
chemicals involved are 193 J/K mol forNH3 (g), 192 J/K
mol-1 for N2 (g) and 131 J/K mol-1 for H2 (g).
 N2(g) + 3H2(g)  2NH3(g)
 DS reaction = (2 x 193) – (192 + (3 x 131)) = -199 J/K
 This makes sense since the negative value implies that the
system has become more ordered
 (four gas molecules are converted to two gas molecules).
16.2 Entropy
 Solutions form because there are many more
possible arrangements of dissolved pieces
than if they stay separate.
 2nd Law - in any spontaneous process there is
always an increase in the entropy of the
universe.
 DSuniv = DSsys + DSsurr
 If DSuniv is positive the process is
spontaneous.
 If DSuniv is negative the process is
spontaneous in the opposite direction.
 If DSuniv is 0 the system is at equilibrium.
 For exothermic processes DSsurr is
positive.
 For endothermic processes DSsurr is
negative.
 Consider this process
H2O(l) H2O(g)
 DSsys is positive
 DSsurr is negative
 DSuniv depends on temperature.
16.3 Temperature and
Spontaneity
 Entropy changes in the surroundings
are determined by the heat flow.
 An exothermic process is favored
because by giving up heat the entropy
of the surroundings increases.
 The size of DSsurr depends on
temperature.
 DSsurr = -DH/T
Practice
 In the metallurgy of antimony, the pure metal
is recovered by two different reactions. Ex16.4
 Sb2S3(s) + 3Fe(s)
2Sb(s) +3FeS(s) DH = -125 kJ
 Sb4O6(s) + 6C(s)
4Sb(s) + 6CO(g) DH = 778 kJ
 Calculate the DSsurr for each of these reactions at
25°C and 1 atm.
16.4 Gibb's Free Energy
 DG=DH-TDS at constant temperature
 Dividing by -T and substituting DH/T for DSsurr we’ve shown that:
 -DG/T = DSuniv (at constant T and P)
 This is very important! It means if
the process is carried out at constant
T and P, the process will be
spontaneous only if DG is negative.
 -DG means +DSuniv
Let’s Check





For the reaction H2O(s) H2O(l)
DSº = 22.1 J/K mol DHº =6030 J/mol
Calculate DG at 10ºC and -10ºC
Look at the equation DG=DH-TDS
Spontaneity can be predicted from the
sign of DH and DS.
DG=DH-TDS
DS
DH
+
-
At all Temperatures
+
At high temperatures,
“entropy driven”
-
At low temperatures,
“enthalpy driven”
+
Not at any temperature,
Reverse is spontaneous
+
-
Spontaneous?
DG and Spontaneity
 At what temperature is the following
process spontaneous at 1 atm?
 Br2(l) Br2(g)
 DkJ/mol DS° = 93.0 J/K•mol
 Hint: Use DG=DH-TDS and set DG to 0
16.5 Entropy Changes in Chemical
Reactions
Reactions involving gaseous molecules:
 The change in positional entropy is
dominated by the relative number of
molecules of gaseous reactants and
products.
 2C2H6 + 7O2
4CO2 + 6H2O
 DS increases because there are more
molecules on the product side than
the reactant side.
Practice
Predict the sign of DS° for each. Ex16.6
 The thermal decomposition of solid
calcium carbonate:
 CaCO3(s)
CaO(s) + CO2(g)
 The oxidation of SO2 in air:
 2SO2(g) + O2(g)
2SO3(g)
Predict the sign of ΔS｡ for each of the following
changes.
(a)Na(s) + 1/2 Cl2(g)
NaCl(s)
(b) N2(g) + 3 H2(g)
2 NH3(g)
(c) NaCl(s)
Na+(aq) + Cl -(aq)
(d) NaCl(s)
NaCl(l)
Third Law of Thermodynamics
 The entropy of a pure crystal at 0 K is 0.
 No disorder. This gives us a starting
point.
 All others must be > 0.
 Standard Entropies Sº ( at 298 K and 1
atm) of substances are listed.
 Products - reactants to find DSº (a
state function).
 More complex molecules higher Sº.
Practice
1. Predict then calculate DS° at 25°C for the
reaction. (use the DS° values in appendix
A21) Ex 16.7
2NiS(s) + 3O2(g)
2SO2(g) + 2NiO(s)
2. Predict then calculate DS° for the
reduction of aluminum oxide by hydrogen
gas. Ex 16.8
Al2O3(s) + 3H2(g)
2Al(s) + 3H2O(g)
Practice
1. For the reaction #35
C2H2(s) + 4F2(g)
2CF4(g) + H2(g)
DS° is equal to - 358J/K. Use this value and
data from the appendix to calculate the S° for
CF4(g)
Consider the reaction #39
1.
2O(g)
O2(g)
aPredict the signs of DH and DS.
b. Would the reaction be more spontaneous at
high or low temperatures?
From data in Appendix 4, calculate ΔH°, ΔS°,
and ΔG° for the following reaction at 25｡C. #41
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
16.6 Free Energy in Reactions
 DGº = standard free energy change.
 Free energy change that will occur if
reactants in their standard state turn to
products in their standard state.
 Can’t be measured directly, can be
calculated from other measurements.
 Method #1 DGº=DHº-TDSº
 Method #2 Use Hess’s Law with known
reactions.
Free Energy in Reactions
 There are tables of DGºf .(A21)
 Method #3 DG = npDGf(products) nrDGf(reactants)
 Products - reactants because it is a state
function.
 The more negative G, the further to the right
the reaction will proceed to reach
equilibrium.
 The standard free energy of formation for
any element in its standard state is 0.
 Remember - Spontaneity tells us nothing
about rate.
Practice
1.Consider the reaction:
2SO2(g) + O2(g)
2SO3(g)
Carried out at 25°C and 1 atm. Calculate DH°
& DS° using the data in A21. Find DG° using
DG°=DH°-T DS°
2.Using the following data (at 25°C)
Cdiamond(s) + O2(g)
CO2(g) DG° = -397 kJ
Cgraphite(s) + O2(g)
CO2(g) DG° = -394 kJ
calculate DG° for the reaction:
Cdiamond(s)
Cgraphite(s)
3. Calculate DG° for the reaction using data on
A21
2CH3OH(g) + 3O2(g)
2CO2(g) + 4H2O(g)
16.7 Free energy and Pressure
 DG = DGº +RTln(Q) where Q is the
reaction quotients (P of the products /P
of the reactants).
 CO(g) + 2H2(g)  CH3OH(l)
 Would the reaction be spontaneous at
25ºC with the H2 pressure of 5.0 atm and
the CO pressure of 3.0 atm?
 DGºf CH3OH(l) = -166 kJ
 DGºf CO(g) = -137 kJ DGºf H2(g) = 0 kJ
16.8 Free Energy and Equilibrium
 DG tells us spontaneity at current
conditions. When will it stop?
 It will go to the lowest possible free energy
which may be an equilibrium.
 At equilibrium DG = 0, Q = K
 DGº = -RTlnK from [DG = DGº + RTlnK]
DGº
=0
<0
>0
K
=1
>1
<1
Temperature Dependence of K
 DGº= -RTlnK = DHº - TDSº
 ln(K) = DHº/R(1/T)+ DSº/R
 A straight line of lnK vs 1/T
16.9 Free Energy and Work
 “Free energy” is energy “free” to do work.
 The maximum amount of work possible at a
given temperature and pressure.
 Never really achieved because some of the
free energy is changed to heat during a
change, so it can’t be used to do work.