Ionic Solubilities
Chapter 16:
Enthalpy (H) →
Thermodynamics
and Equilibria
Enthalpy (H) →
(enthalpy of dissolution)
Chem 111
Dr. Gentry
Gibbs Free Energy (G) & Dissolution
NaCl(s)
Na+ (aq)
+ Cl‒ (aq)
Enthalpy is Endothermic
∆H = + 4kJ
system requires addition of
heat to dissolve NaCl …
G. Free Energy is Exergonic
∆G = ‒ 9kJ
NaCl(s)
Na+ (aq)
+ Cl‒ (aq)
… but system releases
excess Gibbs free energy
Review of Thermodynamic Terms
• Chemical Energy (E)
Free Energy (G) →
?
Free Energy (G) →
NaCl(s)
NaCl (s)
+ Cl‒ (aq)
Na+ (aq) + Cl‒ (aq)
Endothermic
∆H = + 4kJ
AgCl = “Insoluble”
Ag+ (aq)
+ Cl‒ (aq)
AgCl (s)
Ag+ (aq) + Cl‒ (aq)
Endothermic
∆H = + 65kJ
AgCl(s)
NaCl:
Ksp = 38
NaCl(s)
Na+ (aq)
+ Cl‒ (aq)
Ag+ (aq)
+ Cl‒ (aq)
NaCl (s)
Na+ (aq) + Cl‒ (aq)
AgCl: Ksp = 1.6 × 10-10
AgCl (s)
Ag+ (aq) + Cl‒ (aq)
AgCl(s)
Chemical Energy = energy involved in
bonds, intermolecular forces, etc.
• Work (W)
• Heat (Q)
2 H2 (g) + O2 (g) → 2H2O (l)
• Enthalpy (H)
• Entropy (S)
• Gibbs Free Energy (G)
• Standard State
Terms were in Chapter 8 (end of 1st semester)
but will review as if new for Chapter 16
Underlying thrust of Chapter 16 will be in
applying concepts to 2nd semester topics
Energy →
Enthalpy (H) →
Free Energy (G) →
Na+ (aq) + Cl‒ (aq)
NaCl = “Soluble”
Na+ (aq)
But… Why equilibrium?
Why not go all the way downhill?
NaCl = Soluble
NaCl (s)
?
Change
in Energy,
∆E = ‒ 563kJ
Reactants
Product
Reaction Timeline →
1
Chemical Energy
Bond Dissociation Energy
2H2 (g) + O2 (g) → 2H2O (g) → 2H2O (l)
Reaction requires:
1st: Break two H2 bonds and one O2 bond
∆E = + 1370 kJ
2nd: Form four new O-H bonds for 2 H2O’s
∆E = – 1840 kJ
• Energy required to break a covalent bond in an
isolated gaseous molecule.
3d: Condensation of H2O from gas to liquid ∆E = – 88 kJ
Total energy for reaction: add steps together
∆E ~ – 558 kJ
actual number:
∆E = ‒ 563 kJ
One Small Problem
Energy of system may change due to chemical rxn
But conservation of total energy requires system to
transfer energy to or from surroundings
• In open lab experiments
∆E (chemical energy) for 2 H2O formation is: – 563 kJ/mol
But observed heat flow is:
– 570 kJ/mol
Explanation requires discussion of:
“Energy”, “Heat”, and “Work”
Heat (q)
Work (w)
transfer of thermal energy to raise temperature
all other transfers of energy
System
Energy
of
Chemical
System
Change in Enthalpy, ∆H
Enthalpy is what we observe in lab
∆E
Heat
Work
Paraffin + O2 →
CO2 + H2O + energy
Exothermic Reaction
Enthalpy →
∆H = ∆E + P∆V
(at constant pressure)
Heat + Work (Light)
Surroundings
Reactants
Release energy,
∆H = negative
Products
= change in chemical energy
P∆V = any work required
Air Pressure
∆H = ∆Eice→liquid + P∆Vwork
P∆V work
Endothermic Reaction
= total change in energy
Enthalpy →
∆H
Products
Absorb energy,
∆H = positive
Reactants
2
Spontaneous Processes
• Energy and Enthalpy are important
… BUT … not the only things that are important
Entropy (S)
• Entropy is a measure of the randomness in a
system.
Enthalpy (H)
- Sometimes called degree of disorder
- More accurately is number of available options
NH4
++
- Has units of J/K (Joules per Kelvin).
Cl–
Endothermic
∆H = +
But Still
Spontaneous
NH4Cl(s)
• The important question is how the entropy changes
during a reaction
∆S = Sfinal – Sinitial
“cold packs” from pharmacy
∆S = Positive value:
• Must also consider change in Entropy of system
increased # available options
(good)
∆S = Negative value: decreased # available options
(bad)
Random Gas Expansion
More Accurately….
The greater the number of choices available for
arranging the system ⇒ the greater the entropy
7 people in a
phone booth
7 people in a
dorm lounge
Start with gas
on just one side
Low Entropy
Open valve
Which is better (preferred) ?
Which allows people more choices on where to stand?
Open valve to
let gases move
Higher Entropy
After time, gas
randomly fills
both sides
2nd Law of Thermodynamics: the world prefers
having more choices available (higher entropy)
∆Suniverse > 0 for spontaneous processes
Ice
Water
Steam
Spontaneous Reactions
• To decide whether a process is spontaneous, changes in
BOTH enthalpy of system AND entropy of system must
be considered:
• Spontaneous process:
Decrease in enthalpy (–∆
∆H)
(energy goes down)
… and / or …
Solid
less random
lower entropy
Liquid
Gas
more random
higher entropy
Increase in entropy
(entropy goes up)
(+∆
∆S)
3
Gibbs Free Energy
Spontaneous Processes
• Gibbs Free Energy (∆G) combines contributions of enthalpy
and entropy to determine overall spontaneity of reaction.
• Energy (enthalpy) is important
(want to go down in energy)
Energy
• But must also consider Entropy
(want to go up in entropy – become more random)
NH4+ (aq)
+ Cl– (aq)
NH4Cl(s)
Endothermic
∆H = + (bad)
Spontaneous
Entropy Increases
∆S = + (good)
1 solid compound → 2 free ions
∆G = ∆H – T·∆S
∆G < 0
Process is spontaneous
∆G = 0
Process is at equilibrium
∆G > 0
Process is non-spontaneous
Notes:
1) Temperature (T) must be in Kelvin
2) Must convert ∆H from kJ to J
(1000 J / kJ)
∆S is J/K, so ∆H must also be in J to add them
Boiling Water at 100Cº
?
But What About Reaction Equilibrium?
liquid
gas
Change in Enthalpy - Endothermic or Exothermic?
∆Hº = + 40,660 J / mol
If have ∆G= ‒ for a spontaneous reaction,
why does reaction stop (equilibrium)
before the reaction reaches the end?
break Hydrogen bonds in liquid – no IMF’s in gas
Change in Entropy?
∆Sº
=
+ 109 J / K*mol
liquid to gas = increase in entropy
Change in Gibbs Free Energy ?
∆Gº
= ∆Hº – T∆
∆Sº = 40,660 J – (373K·109 J/K)
= 0 J / mol
Random Mixing of Gases
Entropy of Mixing, Smixing
Pure A
Start with two
gases on
separate sides
Random
mixture
of A & B
Pure B
Low Entropy
Open valve
low entropy
(less preferred)
high entropy
(more preferred)
Mixtures have greater number of ways to arrange
all the molecules than keeping them separate
… so mixtures have greater entropy – more stable
Higher Entropy
Open valve to
let gases move
around
After time, gases
will randomly
mix on both sides
4
Entropy of Mixing, Smixing
2 NO2 (g)
N2O4(g)
Gibbs Free Energy
% Conversion in Reaction ( A → B )
Pure A
low entropy
Mix A + B
Pure reactant and Pure product have lower
entropy than a mixture of the two
Entropy prefers the reaction to stop in the middle
Gibbs Free Energy
N2O4(g)
∆Gº in Joules
T
in Kelvin
R = 8.314 J/K·mol
Gtotal
Gtotal
Gmixing
all
NO2
68%
mixture
all
N2O4
Acetic acid has a pKa of 4.74 at 298K. What is the
change in Gibbs Free Energy for the acid ionization
reaction?
K a = 10 − pKa = 10 −4.74 = 1.8 × 10−5
Gmixing
98
98
∆Gº = ‒ RT ln(K)
∆Gº = ‒ RT ln(Keq)
102
Gºpure + Gmixing
Pure B
low entropy
high entropy
2 NO2 (g)
Gtotal =
102
∆Gº = ‒ (8.314 J/K·mol) · (298K) · ln (1.8 × 10-5)
∆Gº = + 27,000 J = + 27 kJ
all
NO2
68%
mixture
all
N2O4
∆Gº = ‒ RT ln(K)
The dissolution of AgCl(s) in water has the following
thermodynamic properties. What is the Ksp value for
this salt at 25ºC?
∆Hº = + 65.49 kJ
∆Sº = + 32.98 J/K
∆G° = ∆H° − T∆S° = 65.49 * 103 J − (298K)(32.98J / K)
∆G° = 55.67 * 103 J
∆ G° = − RT ln(K)
K = e
∆Gº = ‒ RT ln(Keq)
For: 2 NO2 ⇔ N2O4
What is the value of the equilibrium constant for
this reaction at 25ºC given that the change in Gibbs
Free Energy at this temperature is ∆Gº = ‒ 4.7kJ
‒ 4.7 × 103 J = ‒ (8.314 J/K·mol) * (298K) * ln(K)
+ 1.9 = ln(K)
e (1.9) = K
− 55.67*103 / (8.314*298)
K sp = 1.7 × 10−10
K =
[N2O4 ]
= 6.7
[NO2 ]2
5