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Chapter 19
Entropy and Free Energy
Objectives:
1) Define entropy and spontaneity.
2) Predict whether a process will be
spontaneous.
3) Describe free energy.
4) Describe the relationship
between DG, K, and product
favorability.
Introduction
How to predict if a reaction can
occur, given enough time?
THERMODYNAMICS
How far will it proceed? (K)
How to predict if a
reaction can occur at a
reasonable rate?
KINETICS (Ea)
Thermodynamics
• Energy relationships
• 1st law of thermodynamics
– Energy is conserved: energy is not created nor
destroyed
– DE = q + w
– qp = DH Enthalpy: _______ transferred in a
process at constant pressure
• 2nd law of thermodynamics
– DS Entropy: increases in ________
processes
– Direction of the process
• Lighter reaction:
C4H8 + 6 O2
non spontaneous
Burning candle
Drop a pen
Gas expansion
Heat transfer
spontaneous
•
•
•
•
4 CO2 + 4 H2O
Spontaneous
Processes
• Proceed on its own without
outside assistance.
• Occurs in a define _______
• Leads to ___________
• Maybe determined by T, P
• The reverse process is
____________________
Identify spontaneous processes
Predict whether the following are a) spontaneous as described, b)
spontaneous in the reverse direction, c) in equilibrium
1)
When a piece of metal heated to 150oC is added to water at 40oC,
the water gets hotter.
2)
Benzene vapor, C6H6(g), at P= 1atm, condenses to liquid benzene
at the normal boiling point of benzene, 80.1oC
3)
Water at room temperature decomposes into H2(g) and O2(g)
4)
AgCl(s)
Ag+(aq) + Cl- (aq) ; K = 1.8 x 10-10
Reversible vs Irreversible Processes
• Sadi Carnot (French engineer -1824)
– Efficiency of heat to work (steam engines)
– Significant amount of heat is lost to surroundings
• Rudolph Clausius (German physicist ~ 1924)
– Ratio of heat in an ideal engine and temperature at
which is delivered (q/T)
– Entropy
– Amount of work extracted from spontaneous
processes depends on the manner in which the
process is carried (pathway)
Reversible vs Irreversible Processes
• Reversible: a process which reverse direction
whenever an infinitesimal change is made in some
property of the system. Often at equilibrium.
– Here entropy can be obtained at any T by measuring
the heat required rise the temperature from 0K, with the
slow addition of heat in very small amounts.
– DS = qrev/T
• Any spontaneous process is Irreversible, they
often involve non equilibrium conditions; in order
to reverse the surroundings must do some work
on the system (so the surroundings change).
Entropy
• Associated with the ___________ in a
system
• Associated with the extent to which energy
is ___________among the various
motions of molecules of the system.
• Related to heat transfer and __________.
Entropy
• S, is a ________ function (like internal energy,
E, and enthalpy, H).
• The value of S is characteristic of the state of a
system (and a property of the bulk matter).
• DS, change in entropy, depends only on the
initial and final states of the system, and not in
the path taken from one state to the other:
DS = Sfinal – Sinitial
• For isothermal processes: DS = qrev/T
– q: heat absorbed/released
– T: temperature in K
Adding entropy changes = total entropy
DS for phase changes
• Melting of a substance at its m.p. and
vaporization of a substance at its b.p. are
isothermal processes.
• Change can be achieved by adding heat to/from
the system to/from surroundings.
• qrev = DH fusion (melting)
• T = 273 K (normal m.p. 1 atm and 273K).
• DS fusion = qrev/T = DH fusion /T
Calculate DS
• Calculate DSsystem and the DSsurroundings when 1 mol of ice
(~size of an ice cube) melts in your hand.
– Process is not reversible (different T’s).
– DS can be calculated whether rev or irrev.
– DHfusion H2O = 6.01 kJ/mol (melting is endothermic processes,
DH is positive).
DSsystem = qrev/T
=
=
DSsurroundings = qrev/T
(heat gained = - heat lost and T~ 37oC)
=
=
DStotal = DSsystem + DSsurroundings
=
Spontaneous or non spontaneous?
If T’s were about the same, process would be reversible –
overall DS = 0
2nd Law of Thermodynamics
• Any irreversible process results in an overall
(increase/decrease) ___________ in entropy,
whereas a reversible process results in no overall
change in entropy.
• The sum of entropy of a system + entropy of the
surroundings = total entropy change = DS universe.
• Irreversible processes occur of their own accord are
_________ (spontaneous/non spontaneous).
• The total entropy of the universe increases
(DSuniverse is __________ (positive/negative)) in any
spontaneous process.
• DSsystem decreases as Fe +
O2 form rust
• DSsurroundings?
• DStotal = DSsystem + DSsurroundings
Entropy- Molecular Interpretation
• Ludwig Boltzmann (1844-1906)
• Molecules store energy
– KMT
– Higher T, higher KE and broader distribution of
molecular speeds
– 3 kinds of motion: translational, vibrational , rotational
= motional energy of the molecule
Entropy- Molecular Interpretation
– Molecules moving – snapshot – microstate
W= number of microstates of a system
S = k ln W
k= Boltzmann’s constant 1.38 x 10-23 J/K
• DS = k ln Wfinal – k lnWinitial = k ln Wfin/Winit
• Change leading to increase in number of
microstates leads to a ________
(positive/negative) value of DS.
• Related to probability.
Entropy and probability
Most often case is when energy is distributed over all
particles and to a large number of states.
Entropy and probability
Probability and the locations of gas molecules. The two molecules are colored red and blue.
a) Before the stopcock is opened, both molecules are in the left-hand flask. b) After the stopcock is opened,
there are 4 possible arrangements of the two molecules. The greater number of possible arrangements
corresponds to greater disorder in the system, in general, the probability that the molecules will stay in the
original flask is (1/2)n, where n is the number of molecules.
Dispersal of Energy
• Dispersal of matter often contributes to
energy dispersal.
W increases when:
• Change involves an:
– Increase in _______
– Increase in __________
– Increase in _______________
• Entropy change, DS sign will be ________
• The maximum entropy will be achieved at
____________ – a state in which W has
the maximum value.
Explain why melting of ice is
spontaneous process
Explain formation of solutions of
some ionic solids
• If energy and matter are both dispersed in a process, the
process is (spontaneous/non spontaneous) _______________.
• If only matter is dispersed, __________________________.
• If energy is NOT dispersed, the process will (be/never be)
______________ spontaneous.
Predict the sign of DS – each
process occurs at constant T
a) Ag+(aq) + Cl- (aq)
b) H2O(l)
AgCl(s)
H2O(g)
c) N2(g) + O2(g)
d) CO(g) + 3 H2(g)
2 NO (g)
CH4(g) + H2O(g)
3rd Law of Thermodynamics
• Lower the temperature until there is only a single
microstate:
• The entropy of a pure crystalline substance at
absolute zero is ________: S (0K) = ____
• The units of the lattice have no thermal motion.
• S= k ln W =
• As T increases, S ___________:
S solid is (larger/smaller) than S liquid is
(larger/smaller) than S gas
• Then, all substances have ________
(positive/negative) entropy values at T > 0K.
Entropy Change
S increases
slightly with T
S increases a
large amount
with phase
changes
Standard Molar Entropy Values
Standard Molar Entropy Values
Thermodynamics vs Kinetics
Diamond is
thermodynamically
(favored/not favored)
___________ to convert
to graphite, and _______
(favored/not favored)
kinetically.
Standard Entropy
• So, is the entropy gained by converting it from
a perfect crystal at 0K to standard state
conditions (1 bar, 1 molal solution).
• Units: J/Kmol
• Entropies of gases are ________ than those
for liquids, entropies of liquids are
__________than those for solids.
• Larger molecules have a __________
entropy than smaller molecules, molecules with
more complex structures have
_______entropies than simpler molecules.
Entropy
• The entropy of liquid
water is _______ than
the entropy of solid
water (ice) at 0˚ C.
S˚(H2O sol) ____ S˚(H2O liq)
Entropy of Solids
• Entropy values of solids depend on:
– Columbic attractions
So (J/K•mol)
Mg2+ & O2-
MgO
26.9
NaF
51.5
Na+ & F-
The larger coulombic attraction on MgO than NaF leads to a lower entropy.
Which sample has the higher
entropy
a) 1 mol NaCl(s) or 1 ml HCl(g) at 25oC
b) 2 mol HCl(g) or 1 mol HCl(g) at 25oC
c) 1 mol HCl(g) or 1 mol Ar(g) at 298 K
d) O2 (g) or O3 (g)
Entropy Change
• The entropy change is the sum of the
entropies of the products minus the sum of
the entropies of reactants:
DS0system = S S0 (products) – S S0 (reactants)
You will find DSo values in the Appendix L of your book.
Calculate the standard entropy changes for the evaporation
of 1.0 mol of liquid ethanol to ethanol vapor.
C2H5OH(l)  C2H5OH(g)
Calculate the standard entropy change for forming
2.0 mol of NH3(g) from N2(g) and H2(g)
N2(g) + 3 H2(g)  2 NH3 (g)
Using standard absolute entropies at 298K, calculate the
entropy change for the system when 2.35 moles of NO(g)
react at standard conditions.
2 NO(g) + O2(g)  2 NO2(g)
Calculate the standard entropy change for the
oxidation of ethanol vapor (CH2H5OH (g)).
Show that DS0univ is positive (>0) for
dissolving NaCl in water
DS0univ = DS0sys + DS0surr
1) Determine DS0sys
2) Determine DS0surr
NaCl(s) NaCl (aq)
Classify the following as one of the
four types of Table 19.2
CH4 (g) + 2 O2 (g)
2 H2O (l) + CO2 (g)
2 FeO3(s) + 3 C (graphite)
4 Fe(s) + 3 CO2 (g)
DH0 (kJ)
-890
+467
DS0 (J/K)
-242.8
+560.7
Calculate the entropy change of the UNIVERSE
when 1.890 moles of CO2(g) react under standard
conditions at 298.15 K.
Consider the reaction
6 CO2(g) + 6 H2O(l)  C6H12O6 + 6O2(g)
for which DHo = 2801 kJ and DSo = -259.0 J/K at 298.15 K.
• Is this reaction reactant or product favored under standard conditions?
Gibbs Free Energy
DSuniv = DSsurr + DSsys
DSsurr = -DHsys/T
DSuniv = -DHsys/T + DSsys
Multiply equation by –T
-T DSuniv = DHsys –TDSsys
J. Willard Gibbs (1839-1903)
DGsys = -T DSuniv
DGsys = DHsys –TDSsys
DGsys < 0, a reaction is spontaneous
DGsys = 0, a reaction is at equilibrium
DGsys > 0, the reaction is not spontaneous
Gibbs Free Energy and
Spontaneity
• J. Willard Gibbs (1839-1903)
• Gibbs free energy, G, “free energy”, a thermodynamic
function associated with the ________________.
G = H –TS
H- Enthalpy
T- Kelvin temperature
S- Entropy
• Changes during a process: DG
• Use to determine whether a reaction is spontaneous.
DG is ___________related to the value of the
equilibrium constant K, and hence to product
favorability.
“Free” Energy
DG = w max
• The free energy represents the maximum energy
____________________________.
Example: C(graphite) + 2 H2 (g)
CH4 (g)
DH0rx = -74.9 kJ; DS0rx = -80.7 J/K
DG0rx = DH0 – TDS0
= -74.9 kJ – (298)(-80.7)/1000 kJ
= -74.9 kJ + 24.05 kJ
DG0rx = - 50.85 kJ
• Some of the energy liberated by the reaction is needed
to “order” the system. The energy left is energy
available energy to do work, “free” energy.
DG < 0, the reaction is _______________.
Calculate DGo for the reaction below at 25.0 C.
P4(s) + 6 H2O(l) → 4 H3PO4(l)
Species DH (kJ/mol) S (J/K·mol)
P4(s)
0
22.80
H2O(l) -285.8
69.95
H3PO4(l) -1279.0
110.5
f
f
DG0rx = DH0 – TDS0
Standard Molar Free Energy of
Formation
• The standard free energy of formation of a compound, DG0f, is the
free energy change when forming one mole of the compound from
the component elements, with products and reactants in their
standard states.
• Then, DG0f of an element in its standard states is _________.
Gibbs Free Energy
DG0rxn is the increase or decrease in free
energy as the reactants in their standard
states are converted completely to the
products in their standard states.
* Complete reaction is not always
observed.
* Reactions reach an ______________.
DG0system = S G0 (products) – S G0 (reactants)
Calculating DG0rxn from DG0f
DG0system = S G0 (products) – S G0 (reactants)
Calculate the standard free energy change for the
oxidation of 1.0 mol of SO2 (g) to form SO3 (g).
DGf0 (kJ/)
SO2(g) -300.13
SO3(g) -371.04
DG0system =
Free Energy and Temperature
• G = H – TS
• G is a function of T, DG will change as T
changes.
• Entropy-favored and enthalpy-disfavored
• Entropy-disfavored and enthalpy-favored
Changes in DG0 with T
Consider the reaction below. What is DG0 at 341.4 K and will
this reaction be product-favored spontaneously at this T?
CaCO3(s)  CaO(s) + CO2(g)
Thermodynamic values:
DHf0 (kJ/mol) S0 (kJ/Kmol)
CaCO3(s) -1206.9
+0.0929
CaO(s)
-635.1
+ 0.0398
CO2(g)
-393.5
+ 0.2136
Estimate the temperature required to decompose
CaSO4(s) into CaO(s) and SO3(g).
CaSO4(s)
CaO(s) + SO3(g)
DH0sys = S H0 (products) – S H0 (reactants)
DH0sys = S S0 (products) – S S0 (reactants)
Thermodynamic values:
DHf0 (kJ/mol) S0 (J/Kmol)
CaSO4(s) -1434.52
+106.50
CaO(s)
-635.09
+ 38.20
SO3(g)
-395.77
+ 256.77
For the reaction: 2H2O(l)  2H2(g) + O2(g)
DGo = 460.8 kJ and DHo = 571.6 kJ at 339 K and 1 atm.
• This reaction is (reactant,product) _____________
favored under standard conditions at 339 K.
• The entropy change for the reaction of 2.44 moles of
H2O(l) at this temperature would be _________J/K.
DGorxn = DHorxn - DT Sorxn
DSo = (DHo - DGo)/T
DG0, K, and Product Favorability
• Large K – product favored
• Small K – reactant favored
• At any point along the
reaction, the reactants are
not under standard
conditions.
• To calculate DG at these
points:
DG = DG0 + RT ln Q
R – Universal gas constant
T - Temperature (kelvins)
Q - Reaction quotient
DG0, K, and Product Favorability
DG = DG0 + RT ln Q
For a A + b B
cC+dD
Q = [C]c [D]d
[A]a [B]b
DG of a mixture of reactants and products is
determined by DG0 and Q.
When DG is negative (“descending”) the reaction is
_____________ . At equilibrium (no more change in
concentrations), DG = 0.
0 = DG0 + RT ln K (at equilibrium)
DG0 to be negative, K must be larger
DG0 = - RT ln K For
than 1 and the reation is product favored.
Summary DG0 and K
• The free energy at equilibrium is ________ than the free
energy of the pure reactants and of the pure products.
DG0 rxn can be calculated from:
DG0rxn = S G0 (products) – S G0 (reactants)
DGorxn = DHorxn - DT Sorxn
DGorxn = - RT ln K
• DGrxn describes the direction in which a reaction proceeds
to reach ___________, it can be calculated from:
DGrxn = DG0rxn + RT ln Q
– When DGrxn < 0, Q < K, reaction proceeds spontaneously to convert
______________________ until equilibrium is reached.
– When DGrxn > 0, Q > K, reaction proceeds spontaneously to convert
______________________ until equilibrium is reached.
– When DGrxn = 0 , Q = K, reaction is ___________________.
The formation constant for [Ag(NH3)2]+ is
1.6 x107. Calculate DG0 for the reaction below.
Ag+ (aq) + 2 NH3 (aq)  [Ag(NH3)2]+ (aq)
DG0 = -RTlnK
The reaction below has a DG0 = -16.37 kJ/mol.
Calculate the equilibrium constant.
1/2 N2 (g) + 3/2 H2 (g)  NH3 (g)
DG0rxn = DG0f NH3 (g)
DG0 = -RTlnK
The value of Ksp for AgCl (s) at 25oC is 1.8 x 10-10.
Determine DGo for the process:
Ag+ (aq) + Cl- (aq)  AgCl (s) at 25oC.
The standard free energy change for a chemical reaction is
-18.3 kJ/mole. What is the equilibrium constant for the
reaction at 87 C? (R = 8.314 J/K·mol)
Thermodynamics
• First Law: The total energy of the universe
is a constant.
• Second Law: The total entropy of the
universe is always increasing.
• Third Law: The entropy of a pure, perfectly
formed crystalline substance at 0K is zero.
- A local decrease in entropy (the assembly of
large molecules) is offset by an increase in
entropy in the rest of the universe -.
End of Chapter
• Go over all the contents of your
textbook.
• Practice with examples and with
problems at the end of the chapter.
• Practice with OWL tutor.
• Work on your OWL assignment for
Chapter 19.