Download - Solubility products -Thermochemistry

- Solubility products
-Thermochemistry
-Thermodynamics
Dr. G. Birungi
1
Solubility
• Definition
– The amount of a substance that dissolves in a given
quantity of solvent at a given temperature to form a
saturated solution (g/L)
– Question: What happens when we add a solution of a
salt to a solution of another salt?
AgNO3(aq) + NaCl(aq)  ??
KNO3(aq) + NaCl(aq)  ??
• Consider quantitative predictions
– A reaction will only happen between salts in solution if
ions are removed from solution
• Ions removed from solution by formation of a precipitate
2
Solubility- Rules
– Salts of group 1 metals, NH4+, nitrates and acetates
are soluble
– Chlorides, bromides and iodides are soluble except
those of Ag+, Pb2+, Hg22+.
– Sulfates are soluble except those of Pb2+, Hg22+, Sr2+
and Ba2+. Ag2SO4 and CaSO4 are only slightly soluble.
– Carbonates, phosphates and sulphites are insoluble
(except those of group 1 and NH4+)
– Sulfides (S2-) are insoluble except those of groups 1
and 2 and NH4+.
– Hydroxides are insoluble except those of groups 1 2
and NH4+. Hydroxides of group 2(Ca2+, Sr2+ or Ba2+) are
slightly soluble
3
Formation and Dissolution of Precipitates
• Solubility: maximum amount of solute that will dissolve in a given amount of
solvent (depends on solvent, temperature and pressure)
No compound is infinitely soluble and
no compound is perfectly insoluble.
Solute Solubility
(g solute/100 g solvent)
Qualitative Solubility
Description
Less than 0.1
Insoluble
0.1 – 1
Slightly soluble
1 – 10
Soluble
Greater than 10
Very soluble
1–4
Formation and Dissolution of Precipitates
• Saturated solution
– Contains maximum concentration of solute
• Equilibrium between undissolved and dissolved solute.
• Unsaturated solution
– Contains less than the maximum solute
• Supersaturated solution
– Contains more than maximum amount of solute
• Often reached by heating followed by cooling
• Solutes (even those called “soluble”) have a limited
solubility in a particular solvent.
• Slightly soluble (often called “insoluble”) ionic compounds
have a relatively low solubility
– Reach equilibrium with little solute dissolved
– Heterogeneous equilibrium
5
Solubility curve
• Explore how environmental factors affect
solubility (amount of solute present in
solutions)
– Example how does temperature affect solubility?
– How does pressure affect solubility?
6
Solubility Curve
7
Any solution can be made saturated, unsaturated,
or supersaturated by changing the temperature
8
Solubility curve - Gas
9
Effect of temperature and pressure summary
• Solids and liquids
– Solubility increases with temperature (decreases
with a decrease in temp)
– Liquids and solids exhibit practically no change of
solubility with changes in pressure.
• Gases
– As the temperature increases, the solubility of a
gas decrease as shown by the downward trend
– There is an increase in solubility with an increase
in pressure. (Henry's Law)
10
Why is this Important?
Dissolving and Precipitation occurs around us:
– Tooth enamel dissolves in acidic solution (tooth decay)
– Precipitation of certain salts in kidneys causes kidney stones
– Waters of Earth contains dissolved salts as water passes
over and through the ground
– Precipitation of CaCO3 from groundwater is responsible for
cave formation.
Let’s look at the factors that affect solubility!
11
Example
• Use the solubility chart on
the left to answer the
following questions:
– At 15 0C, which of the salts
is:
• Most soluble
• Least soluble
– What mass of solute will
dissolve in 100mL of water
at the following
temperatures.
• KNO3at 70°C
• NaCl at 100°C
• NH4Cl at 90°C
12
Solubility-Product Constant (Ksp)
• Solubility-product constant (Ksp): equilibrium constant
for equilibrium between slightly soluble ionic solid and a
solution of its ions
– Indicates how soluble the solid is in water
• Solubility: quantity that dissolves to form a saturated
solution (g/L)
• Molar solubility: number of moles of solute that
dissolves in forming a liter of saturated solution of solute
(mol/L)
• Solubility depends on concentrations of other ions and
pH but Ksp is a constant.
13
Solubility-Product Constant (Ksp)
Practice: Write an ionic equation for the dissolution, and
the equation for the solubility product for:
(a) Calcium carbonate
(b) Magnesium hydroxide
(c) Ag3PO4
•
Magnitude of Ksp is a measure of how far to the right
dissolution proceeds at equilibrium (saturation).
–
Used to compare solubility
14
Ksp of Selected Ionic Compounds (25 °C)
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
3 x 10-34
Cobalt(II) carbonate, CoCO3
1.0 x 10-10
Iron(II) hydroxide, Fe(OH)2
4.1 x 10-15
Lead(II) fluoride, PbF2
3.6 x 10-8
Lead(II) sulfate, PbSO4
Mercury(I) iodide, Hg2I2
Silver sulfide, Ag2S
Zinc iodate, Zn(IO3)2
1.6 x 10-8
4.7 x 10-29
8 x 10-48
3.9 x 10-6
See Appendix D in your book for a much more extensive list.
1–15
Example
Which of the following compounds will
have the greatest molar solubility in water
A) AgCl
Ksp = 1.8 x 10-10
B) AgBr
Ksp = 5.0 x 10-13
C) AgI
Ksp = 8.3 x 10-17
Give an explanation/reason for your answer
16
Solubilities and Solubility Products
• Ksp for a slightly soluble solid can be determined from
its solubility
– as long as there is no other reaction
• Example 1:
– The aqueous solubility for compound PbI2 is 0.54
grams/100 ml at 25 degrees Celsius. What is the Ksp of PbI2
at 25 oC ?
17
• Convert 0.54 grams of PbI2 to mols
–
0.54
461.0
=0.001167137 mols of PbI2
• Covert ml to L
–
100
1000
= 0.1
• Find molarity
–
0.001167137
0.1
= 0.011714𝑀 𝑜𝑓 𝑃𝑏𝐼2
• 𝑃𝑏𝐼2 ⇋ 𝑃𝑏 2+ + 2𝐼−
• 𝐾𝑠𝑝 = 𝑃𝑏 2+ [𝐼 − ]2
• 𝐾𝑠𝑝 = 0.011714 × (0.023427)2 = 6.4 x 10−6
18
Solubilities and Solubility Products
• Example 2:
– Determine the molar solubility of MgF2 from its solubility
product (Ksp = 6.4 x 10-9).
19
Example 3
• Calculate the molar solubility of calcium
fluoride, CaF2 (Ksp = 3.9 x 10-11)
20
Example 4
• The Ksp for LaF3 is 2.0  10-19. What is the
solubility of LaF+ in water in moles per litre?
21
Example 5
• Calculate the molar solubility of AgCl at 250C,
given the relevant Ksp value is 1.82  10 -10
22
Factors that Affect Solubility: Common Ion
Effect
• The presence of a common ion decreases the
solubility of a slightly soluble ionic compound.
• The shift in equilibrium that occurs because of
the addition of an ion already involved in the
equilibrium reaction.
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
adding




NaCl( aq ) shifts equilibrium position
23
The effect of a common ion on solubility
CrO42- added
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
24
Example
Calculate the molar solubility of MgF2 in 0.10
mol dm-3 MgCl2 at 25 C.
Ksp of MgF2 = 7.4 x 10-11
25
Example
The solubility of Ca(OH)2 in water is 0.012 mol
dm-3. What is its solubility in 0.10 mol dm-3
Ca(NO3)2? Ksp of Ca(OH)2 is 6.5 x 10-6
26
Le Châtelier’s Principle
Determine the effects of solubility when each of
the following is added to a mixture of the
slightly soluble solid NiCO3 and water at
equilibrium:
(a)Ni(NO3)2
(b)KClO4
(c) K2CO3
(d) HNO3
27
Effect of pH on solubility
• [H3O+] can have a profound effect on the solubility
of an ionic compound.
– Solubility of slightly soluble salts containing basic anions
increases as [H+] increases
• More basic anion…more solubility is influenced by pH
• For example -The molar solubility of a sparingly
soluble hydroxide increases as the pH is decreased
28
Effect of pH on solubility
• Example
𝑀𝑔(𝑂𝐻)2 (𝑠) ⇌ 𝑀𝑔2+ (𝑎𝑞) + 2𝑂𝐻 − (𝑎𝑞)
– If we add an acidic buffer, what will happen to the
solubility of Mg(OH)2 ? Explain
• By adding an acid we increase [H+] and consequently
decrease [OH-] as Kw must be maintained
• If we decrease the amount of OH- in the system, the
position of equilibrium will shift to the right
• If the position of equilibrium moves to the right then
the concentration of Mg2+ increases
• As [Mg(OH)2] = [Mg2+], the molar solubility of
magnesium hydroxide increases as the pH is lowered.
29
Effect of pH on solubility
• Predict the effect on solubility of the following
if a strong acid is added.
– CaCO3(s) ⇌ Ca2+(aq) + CO32-(aq)
– AgCl(s) ⇌ Ag+(aq) + Clˉ(aq)
– Ca(OH)2(s) ⇌ Ca2+(aq) + 2OHˉ(aq)
30
If limestone (CaCO3) deposit is near surface…sinkhole
31
If limestone (CaCO3) deposit is well below the surface….caves
A view inside Carlsbad Caverns, New Mexico
32
Cango Caves, South Africa
1–33
Sudwala Caves, South Africa
1–34
Predicting the Formation of a
Precipitate
Compare Qsp to Ksp to predict if a precipitate will form
and, if not, what concentrations of ions will cause it to
do so.
Qsp = Ksp
Qsp > Ksp
Qsp < Ksp
soln is saturated & no changes occur
ppt forms until soln is saturated
soln is unsaturated & no ppt forms
35
Practice
• Example 1: Determine whether CaHPO4 will
precipitate from a solution with [Ca2+] = 0.0001 mol
dm-3 and [HPO42-] = 0.001 mol dm-3.
• Example 2: Does silver chloride precipitate when
equal volumes of a 2 x 10-4 mol dm-3 solution of
AgNO3 and a 2 x 10-4 mol dm-3 solution of NaCl are
mixed?
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
36
Practice
Will a precipitate form when 0.10 dm3 of
8.0 x 10-3 mol dm-3 Pb(NO3)2 is added to 0.40
dm3 of 5.0 x 10-3 mol dm-3 Na2SO4?
Ksp for PbSO4 = 6.3 x 10-7
37
Concentration Necessary to Form a
Ppt
• We can also determine the concentration of an ion
necessary for precipitation to begin.
• Assume that precipitation begins when Qsp = Ksp
• Example: If a solution contains 0.0020 mol CrO42- per
dm3, what concentration of Ag+ ion must be added as
AgNO3 before Ag2CrO4 begins to precipitate. (Neglect
any increase in volume upon adding the solid silver
nitrate.)
38
Thermochemistry
• Thermochemistry is the
study of energy changes
that occur during
chemical reactions
• Focus is on heat and
matter transfer
between the system
and the surroundings
39
Energy
• The ability to do work or transfer heat.
– Work is done when the energy applied causes
an object that has mass to move.
• Kinetic energy - energy of motion
• Potential energy
– energy due to position
– energy associated with forces
of attraction and repulsion
between objects
40
Units of Energy
• The SI unit of energy is the joule (J).
kg m2
1 J = 1 
s2
• An older, non-SI unit is still in widespread use
called the calorie (cal).
1 cal = 4.184 J
41
Conservation of Energy
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is a
constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
∆ E = Efinal - Einitial
42
• The energy of a system depends only on its present
state, not on the path by which the system arrived at
that state.
• So, E depends only on Einitial and Efinal.
– In the system below, the water could have
reached room temperature from either direction.
43
Exchange of Heat - System and Surroundings
• When heat is released by
the system to the
surroundings, the process is
exothermic
• When heat is absorbed by
the system from the
surroundings, the process is
endothermic.
44
Endothermicity & Exothermicity
• Process is exothermic
when H is negative.
• Process is endothermic
when H is positive.
45
Atoms & Molecules Possess Energy
• Kinetic energy
– proportional to the absolute temperature
• Potential energy
– chemical bonds hold atoms together, forces of
attraction give rise to a compound’s potential
energy (chemical energy)
bonds break - increase PE of system
bonds made - lower PE of system
Changes in chemical energy occur during chemical reactions
46
Heat Capacity & Specific Heat Capacity
• Heat capacity - the amount of energy required
to raise the temperature of a substance by 1 K
(1C).
• Specific heat capacity - the amount of energy
required to raise the temperature of 1 g of a
substance by 1 K.
47
Heat Capacity & Specific Heat Capacity
To calculate a quantity of heat (q)
heat = specific heat  mass  temperature change
q
=
J
=
m

T
J g-1 K-1  g

K
s

48
Specific Heats…
q = s  m  T
If ∆T > 0, then q > 0 and heat is gained by the
system
• If ∆T < 0, then q < 0 and heat is lost by the system
• Molar heat capacity is the product of specific
heat times the molar mass of a substance (units
are J mol–1 K–1)
49
Examples
• Pg 140
• How much heat is needed to warm 250 g of
water from 220C (about room temperature) to
near its boiling point, 980C, given the specific
heat capacity of water is 4.18 Jg-1K-1?
– 7.9  104 J
• What is the molar heat capacity of water?
– 75.2 Jmol-1K-1
50
Calorimetry
• Calorimetry is a technique used to
measure heat exchange in chemical
reactions
• A calorimeter is the device used to
make heat measurements.
• Calorimetry is based on the law of
conservation of energy.
• qwater = - q surroundings
51
Coffee Cup Calorimeter
• We indirectly measure the
heat change for the system
by measuring the heat
change for the water in the
calorimeter
• The specific heat for water is
well known (4.184 J/g/K),
• We can measure H for the
reaction with this equation:
q = m  s  T
52
Calorimetry example
• How much heat is given off when an 869 g iron bar cools
from 94 °C to 5 °C?
(specific heat of Fe = 0.444 J g-1 K-1)


?
q = m x s x Dt
Dt
= tfinal – tinitial
= 5°C – 94°C
= -89°C
q = 869 g x 0.444 J/g/K x -89 K
= -34,000 J
(q < 0, heat lost!)
53
Example
In a neutralization reaction between a strong acid
and a strong base, the following reaction forms
water.
H+(aq) + OH-(aq)
H2O(l)
If 25.00 mL of 2.50 M HCl and 25.00 mL of 2.50 M
NaOH, both at 21.1 0C react and the final
temperature is 37.8 0C, determine the heat of the
neutralization reaction expressed per mole of H2O
formed.
54
Solution:
• Assume that 50.00 mL of water forms which
absorbs all the heat.
Heat of reaction = qneutr = -qcalorim
qcalorim = m x s x ∆T
= (50.00 mL x 1.00 g/mL) x 4.18 J/g/K
x (37.8 – 21.1) K
= 3.5 x 103 J
qneutr = -qcalorim = -3.5 kJ
55
n(H+) = C x V
= 2.50 mol dm-3 x 25.00 x 10-3 dm3
= 0.0625 mol H+
H+(aq) + OH-(aq) → H2O(l)
n(H+) = n(OH-) = n(H2O) = 0.0625 mol
Amount of heat produced per mole of H2O is
qneutr = - 3.5kJ/0.0625 mol
= - 56 kJ/mol H20
(Since q < 0, heat is lost, that is neutralisation reaction is exothermic!)
[Do example 4.6 in textbook]
56
Enthalpy
• Enthalpy (H) is used to quantify the heat flow into or out
of a system in a process that occurs at constant pressure
• Standard enthalpy of reaction (Hrxn): the enthalpy
change for the transformation of reactants in their
standard states to products in their standard states.
• There are many types of heats of reaction, e.g.
– heat of formation,
– heat of combustion,
– heat of hydrogenation.
57
Standard State
• Standard state is the reference state for the material's
thermodynamic state properties.
• Required for comparison purposes.
• Standard state of any substance - the physical state at
which it is most stable at 1 bar and 298 K.
• When a reaction occurs at constant pressure, the heat
of reaction is equal to the enthalpy change.
58
Enthalpies of Reaction
• The change in enthalpy, H, is
the enthalpy of the products
minus the enthalpy of the
reactants:
H = Hproducts − Hreactants
59
Hess’s Law
• Hess’s law states that “If
a reaction is carried out
in a series of steps, H
for the overall reaction
will be equal to the sum
of the enthalpy changes
for the individual
steps.”
60
Hess’s Law
• The total enthalpy
change depends only
on the initial state of
the reactants and the
final state of the
products.
61
Calculation of H
We can use Hess’s law in this way:
Horxn= nHof(products) - mHof(reactants)
where n and m are the stoichiometric
coefficients.
62
Heat of Combustion
• The heat of combustion is the heat released in the
reaction of one mole of a substance in its standard
state with oxygen.
• The equation for the combustion of a compound must
show the reaction of one mole of the compound with
sufficient oxygen to convert all of the carbon and
hydrogen present to gaseous CO2 and liquid H2O.
• C6H6(l) + 7½O2(g)  6CO2(g) + 3H2O(l) H° = -3274 kJ
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H° = -890.2 kJ
63
Example
The heat of combustion of ethane gas, C2H6(g), is
–1560.1 kJ mol-1. If H for CO2(g) and H2O(l) are
–393.5 kJ mol-1 and –285.8 kJ mol-1 respectively, calculate
the heat of formation of ethane.
combustion equation: C2H6(g) + 3½O2(g)  2CO2(g) + 3H2O(l)
Horxn
= nHof (products) - mHof (reactants)
= [(2 mol)Hof (CO2) + (3 mol)Hof (H2O)] [(1 mol)Hof (C2H6) + (3½ mol)Hof (O2)]
-1560.1 kJ
= (2 x -393.5) + (3 x -285.8) -Hof (C2H6) - 0 kJ
 Hof (C2H6) = -84.3 kJ mol-1
64
Enthalpy of Formation
• The standard heat of formation of a
compound (H°f): the heat change when one
mole of the compound in its standard state is
formed from its elements in their standard
states at a specified temperature.
• The standard enthalpy of formation of an
element in its standard state is taken to be
zero.
65
Enthalpies of Formation
66
Standard Enthalpies of Formation
Standard enthalpies of formation, H0f, are
measured under standard conditions (25°C and 1.00
atm pressure).
67
Calculate the standard enthalpy of formation of CS2
(l) given that:
C(graphite) + O2 (g)
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
CO2 (g)
H0rxn = -393.5 kJ
SO2 (g)
H0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g) H0rxn = -1072 kJ
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
68
2. Add the given rxns so that the result is the desired rxn.
(molar ratios!)
C(graphite) + O2 (g)
CO2 (g)
H0rxn = -393.5 kJ
2S(rhombic) + 2O2 (g)
2SO2 (g)
H0rxn = -296.1x2 kJ
0
+ CO2(g) + 2SO2 (g)
CS2 (l) + 3O2 (g) Hrxn
= +1072 kJ
C(graphite) + 2S(rhombic)
CS2 (l)
H0rxn = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
69
Benzene (C6H6) burns in air to produce carbon dioxide
and liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of
formation of benzene is 49.04 kJ/mol.
2 C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
Hrxn
= nH0f (products) -mH0 f (reactants)
0
Hrxn
= [12H0f (CO2) +6H0 f(H2O)] - [2H0f (C6H6)]
0
Hrxn
= [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ
-6535 kJ
= - 3267 kJ/mol C6H6
2 mol
70
Application of Hess’s Law
Find H of NO2(g) given that
½N2(g) + ½O2(g)  NO(g)
2NO2(g)  2NO(g) + O2(g)
Desired equation:
½ N2(g) + O2(g)  NO2(g)
Hf = 90.25 kJ …..(1)
Hf = 114.14 kJ …(2)
H = ?
2NO(g) + O2(g)  2NO2(g)
NO(g) + ½ O2(g)  NO2(g)
H = -114.14 kJ
H = -57.07 kJ..(-2)/2
+ ½O2(g)  NO(g)
NO(g) + ½O2(g)  NO2(g)
½N2(g) + O2(g)  NO2(g)
H = 90.25 kJ
H = -57.07 kJ
H = 33.18 kJ
½N2(g)
71
Bond Energies
• bond breaking - endothermic
• bond formation – exothermic
• Bond energies indicate the amount of
energy required to break a particular bond
and are therefore positive.
H = [sum of bond energies of all bonds in
reactants (bonds broken)] - [sum of bond
energies of all bonds in products (bonds
formed)]
72
Example - hydrogenation reaction
C2H4(g) + H2(g)  C2H6(g)
Bond energies of reactants = EC=C + 4EC-H + EH-H
= 606.1 + 4(409.64) + 430.5 kJ
= 2675.16 kJ
Bond energies of products = EC-C + 6EC-H
= 334.4 + 6(409.64) kJ
= 2792.24 kJ
H
= 2675.16 - 2792.24 kJ
= -117.08 kJ mol-1
73
Thermodynamics
Spontaneous Change:
Entropy and Free Energy
74
• In previous chapters we have studied:
– How fast does the change occur
– How is rate affected by concentration and
temperature
– How much product will be present
• Why does a change occur in the first place?
75
Chemical Thermodynamics…What are the
Driving Forces?
• Methane burns but not the reverse
CH4 + 2O2  CO2 + 2H2O
• Steel chain rusts, but a rusty one will not
become shiny
• A cube of sugar dissolves in a cup of coffee
after a few seconds of stirring, but not the
reverse
76
Spontaneity
• How can we tell if a reaction will proceed or
not?
– Some chemical and physical changes take place by
themselves, given enough time.
• A spontaneous chemical reaction is one that,
given sufficient time, will achieve chemical
equilibrium, with an equilibrium constant
greater than 1, by reacting from left to right.
77
Spontaneous Reaction
Cu(s) + Cl2(g)  CuCl2(s)
 spontaneous 
2H2(g) + O2(g)  2H2O(g)
• spontaneous reaction but occurs only if you ignite the
mixture
78
Nonspontaneous Reaction
O3(g)  O(g) + O2(g)
nonspontaneous 
*This does not mean that it does not occur at all. It
means that, when equilibrium is achieved, not many O3
molecules have broken down into products.* That is
[O][O 2 ]
K
 0.0063  1 at 25 C
[O3 ]
79
Chemical Thermodynamics
• Thermodynamics lets us predict whether a
process will occur given enough time.
• Spontaneous process: occurs by itself without
ongoing input of energy from outside system
– Freezing of water at 1 atm and –5 °C
– Burning or falling may need a little “push” to get
started, but will continue without external aid
• Nonspontaneous process: does not occur unless
we make it happen; system must be supplied with
continuous input of energy
– A book rises only if something else supplies energy
80
Spontaneity
• Spontaneous does not mean instantaneous
and has nothing to do with the rate.
– Given enough time it will occur by itself
– Chemical reaction proceeding toward equilibrium
is an example of a spontaneous change.
If a change is spontaneous in one direction, it is
not spontaneous in the other.
81
Mechanical Systems
• Some spontaneous processes occur with loss of
energy.
• change in the direction which lowers their energy
• position of equilibrium is when the energy
“available” is minimized (energy has different forms:
thermal, potential, kinetic)
82
System and Surroundings
• System – portion of the universe we wish to study.
• Surroundings – everything else.
Universe = System + Surroundings
Example: In the chemistry lab, system is usually a flask,
beaker, etc and the surrounds are the rest of the
laboratory.
83
Chemical Systems
• The enthalpy (H) is a measure of the total
energy of a “system” (that part of the universe
we are considering – the rest is called the
“surroundings”).
• Can we use the enthalpy to predict the
direction of chemical change?
• Some scientists though that the sign of H
determined spontaneity.
84
Exothermic
Physical:
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
Chemical:
8Al(s) + 3Fe3O4(s)  4Al2O3(s) + 9Fe(l)
reactants
H
products
The enthalpy
(energy) of the
chemical system
is lowered.
85
Endothermic
Physical: H2O(s)  H2O(l)  H2O(g)
NH4Cl(s)  NH4+(aq) + Cl-(aq)
Chemical:
Ba(OH)28H2O(s) + 2NH4NO3(s) 
Ba(NO3)2(aq) + 2NH3(g) + 10H2O(l)
products
H
reactants
Some spontaneous
chemical reactions are
endothermic (enthalpy
increases).
86
Enthalpy
• H cannot be used to predict if a reaction or
process will go.
• The reason is that it represents the total
energy of the system. We need to examine
the available energy (the energy available to
do useful work). The unavailable energy per
degree kelvin is known as the entropy (S) of a
system.
87
Identifying Spontaneous Processes
Predict whether the following are spontaneous?
1. When a piece of metal heated to 150 C is added to water at 40
C, water gets hotter.
2. Water at room temperature decomposes into H2(g) and O2(g).
3. Reaction of sodium metal and chlorine gas to form sodium
chloride.
4. Reaction of nitrogen atoms to form N2 molecules at 25 C and 1
atm.
How can we tell the direction of a spontaneous change when it is
not as obvious?
88
Direction of Chemical Change
G = H – TS
– G ≡ available energy, Gibbs energy
– H ≡ total energy, enthalpy
– S ≡ unavailable energy per Kelvin, entropy
– T ≡ temperature
• The direction of chemical change is the
direction which lowers the Gibbs energy.
89
Criterion for Spontaneity
ΔG = ΔH – TΔS
A process is spontaneous in the direction in
which the Gibbs energy decreases: i.e. G is
negative.
At constant temperature,
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
90
For ΔG to be negative:
ΔH
negative
ΔS
positive
negative
negative but ΔH more negative
than (-TΔS) (lower T)
positive
positive but (-TΔS) more
negative than ΔH (higher T)
91
Spontaneity
• melting of ice
–endothermic
–highly organised structure becomes less wellordered
• octane combustion
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
–exothermic
–27 molecules in the gas phase are converted to 34
molecules
92
Spontaneity
Clearly two factors should be considered
when trying to predict spontaneity:
• a decrease in energy or enthalpy
• an increase in disorder or entropy
Sometimes these effects reinforce one
another, but at other times they oppose
one another. The final outcome is
determined by their relative magnitudes.
93
Spontaneous Process
• Matter changes from a more ordered to a
less ordered state.
A change in order is a change in the number
of ways of arranging the particles, and is a
key factor in determining the direction
of a spontaneous process.
94
Disorder and Entropy
• There is a natural tendency for a system to become
disordered
– For parts of a system to have more ways of being
arranged.
• Creating order requires work
• Disorder increases when a process results in more
ways for the atoms, ions, or molecules in the system
to be arranged.
95
The number of ways to arrange a deck of playing cards
96
Entropy
• The driving force for a spontaneous process is
an increase in the entropy of the universe.
• Entropy, S, can be viewed as a measure of
randomness, or disorder.
– Number of microstates of system is number of
ways it can disperse its thermal energy among
various modes of motion of all its particles
– units of J K-1 mol-1
97
Third Law of Thermodynamics
• A perfect crystal has zero entropy at a temperature of absolute
zero
Ssys = 0 at 0 K
• “Perfect” means that all the particles are aligned flawlessly in
the crystal structure with no defects of any kind.
• At absolute zero, all particles in the crystal have their minimum
energy, and there is only one way they can be arranged.
• Because S is explicitly known (= 0) at 0 K, S values at other
temperatures can be calculated.
98
Absolute Entropies
• Elements in their standard states have an
entropy greater than zero, i.e. S° > 0.
• The entropy of any pure substance can
be measured at a given temperature.
• The absolute entropy of one mole of a
substance in its standard state is called
the standard molar entropy, S°.
99
Standard Molar Entropies
• Standard molar entropy (S°): J K-1 mol-1(Appendix D)
– Predicting relative values of S°
– Affected by temperature, physical state, dissolution,
and atomic or molecular complexity.
1. Temperature changes: S° increases as temp rises
2. Physical states and phase changes:
-when a more ordered phase changes to a
less
ordered one, S +ve
-S° increases as the substance changes from a
solid to a liquid to a gas: Ssolid < Sliquid << Sgas
100
Ssolid < Sliquid << Sgas
101
Standard Molar Entropies
3. Dissolution of a solid or liquid: the entropy of a
dissolved solid or liquid solute is greater than the
entropy of the pure solute
-Type of solute and solvent affect overall entropy
change.
4. Dissolution of a gas: a gas is so disordered to begin
with that it becomes more ordered when it dissolves
in a liquid
-Entropy of a solution of a gas in a liquid is always less
than entropy of pure gas.
5. Atomic size or molecular complexity:
S° (KCl) < S° (CaCl2) < S° (GaCl3)
102
Entropy and vibrational motion
NO
NO2
N 2 O4
103
Examples
1. Choose the member with the higher entropy:
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 1 mol of KBr(s) or 1 mol of KBr(aq)
(d) Seawater in midwinter (2 C) or midsummer (23 C)
2. Predict whether S is positive or negative for the following:
(a) H2O(l) → H2O(g)
(b) Ag+(aq) + Cl-(aq) → AgCl(s)
(c) 4Fe(s) + 3O2(g) → 2Fe2O3(s)
(d) N2(g) + O2(g) → 2NO(g)
104
Entropy changes for reversible phase
transitions
heat transferred
ΔS 
temp at which change occurs
q

T
105
Melting of ice at normal boiling point
H2O(s, 1 atm) → H2O(l, 1 atm)
q
ΔS 
T
ΔH fusion

Tmp

6.02 kJ mol -1

273.15 K
 22.0 J K -1 mol -1
What is ΔS° when one mole of water freezes?
106
Calculating the Change in Entropy of a
Rxn
• Chemists are especially interested in learning to
predict and calculate change in entropy as a
reaction occurs.
• Calculations are similar for enthalpy changes:
– calculated from enthalpies of formation of reactants & products.
 Hrxn° = npHf(products) - nrHf(reactants)
– calculated from standard molar entropies of reactants & products.
 Srxn° = npS(products) - nrS(reactants)
107
Example
Calculate ΔS° for the reaction
SiCl4(g) + 2Mg(s) → 2MgCl2(s) + Si(s)
Substance:
SiCl4(g) Mg(s)
MgCl2(s)
Si(s)
S°/J K-1 mol-1: 330.73
32.68
89.62
18.83
A.
B.
C.
D.
E.
-254.96 J K-1
-198.02 J K-1
198.02 J K-1
254.96 J K-1
471.86 J K-1
108
Example
Calculate ΔS° for the reaction
2Cl2(g) + SO2(g) → SOCl2(g) + Cl2O(g)
Substance:
Cl2(g)
S°/J K-1 mol-1: 223.0
A.
B.
C.
D.
E.
SO2(g)
248.1
SOCl2(g)
309.77
Cl2O(g)
266.1
-118.2 J K-1
-104.8 J K-1
104.8 J K-1
118.2 J K-1
1270.0 J K-1
109
Gibbs Energy
• If a process is nonspontaneous in one
direction (G > 0), it is spontaneous in the
opposite direction (G < 0)
• Standard Gibbs Energy Change (Go)
Go = Ho – TSo
– Used frequently to find any one of these
variables when given other two
110
Example
Calculate Grxn° at 25 °C for the decomposition
of potassium chlorate, one of the common
oxidizing agents in explosives, fireworks, and
match heads.
4KClO3(s)  3KClO4(s) + KCl(s)
KClO3
KClO4
KCl
Hof
So
-397.7 kJ mol-1
143.1 J K-1 mol-1
-432.8 kJ mol-1 151.0 J K-1 mol-1
-436.7 kJ mol-1
82.59 J K-1 mol-1
111
Example
Determine the standard Gibbs energy change
at 298 K for the reaction:
2NO(g) + O2(g) → 2NO2(g)
112
Calculating Change in
Entropy/Enthalpy of Rxn
 Hrxn° = npHf(products) - nrHf(reactants)
 Srxn° = npS(products) - nrS(reactants)
• Can also calculate Gorxn using standard Gibbs
energy of formation (Gof)
  G = standard Gibbs energy change that occurs
if reactants in their standard state are converted
to products in their standard state.

Grxn° = npGf(products) - nrGf(reactants)
Use Gfvalues to calculate Grxn° for previous examples.
113
Gibbs energy of formation
•
Gof: Gibbs energy change that accompanies
the formation of exactly 1 mole of the pure
substance from free elements in their most
stable states under standard state conditions.
C(s) + O2(g) → CO2(g)
Gof = -394.4 kJ mol-1
• All elements in their standard states have
Gof = 0.
114
Gibbs Energy and Equilibrium
G = -RT ln K
•Allows us to calculate standard Gibbs energy change of
reaction (G) from its equilibrium constant, or vice versa.
• As Gbecomes more positive, K becomes smaller
– reaction reaches equilibrium with less product and
more reactant
• As Gbecomes more negative, K becomes larger
– reaction reaches equilibrium with more product
and less reactant
115
• K > 1 (lnK > 0) → ΔGrxn° < 0
• K < 1 (lnK < 0) → ΔGrxn° > 0
• K = 1 (lnK = 0) → ΔGrxn° = 0
116
A⇌ B
The Relationship Between G° and K at 25 °C
G0(kJ)
K
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
REVERSE REACTION
9x10-36
FORWARD REACTION
200
Significance
Forward reaction goes to
completion; essentially no reverse
reaction
117
Examples
1. Determine the value of the equilibrium
constant at 298.15 K for the reaction
C2H6(g) ⇌ H2(g) + C2H4(g)
G298= 101.0 kJ
2. Determine the value of K at 100 °C for
I2(g) + Cl2(g)  2ICl(g)
For this rxn, H298= -26.9 kJ & S298= 11.3 J K-1
(Assume that H298and S298do not vary with T.)
118
Example
Calculate G (kJ mol-1) for following process:
AgCl(s) ⇌ Ag+(aq) + Cl¯(aq)
Ksp = 1.6 x 10-10 at 25 C
119
Gibbs Energy and Cell Potential
ΔGrxn° = -nFEcell°
120