Download 2/22 Lecture Slides

Chem. 31 – 2/22 Lecture
Announcements I
• Upcoming Assignments
–
–
–
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Exam 1 – Monday, March 6th (Monday after next)
AP2 – Due March 1st (next Wednesday)
Next quiz will be from Homework Set 2 (after Exam 1)
Lab reports (glassware resubmission – due 2/27; Cl lab
report – due 3/8)
• Chemistry Seminar – This Friday
– Gallo Winery (mentioned because they also have an
internship program mid-summer to late fall)
Announcements II
• Today’s Lecture – Chapter 6 (Chemical Equilibria
and Thermodynamics)
– Thermodynamics
– Le Châtelier’s Principle
– Solubility Equilibria
Thermodynamics
ΔG = Change in Gibbs free energy
This tells us if a process is spontaneous
(expected to happen) or non-spontaneous
ΔG < 0 process is spontaneous (favored)
ΔG = ΔH - TΔS (T is absolute temperature)
processes that are exothermic (Δ H < 0) and
increase disorder (Δ S > 0) are favored at all T
processes that have Δ H > 0 and Δ S > 0 are
favored at high T
Example question
The reaction N2(g) + O2 (g) ↔ 2NO(g) has a
positive DH. Under what conditions is this
process spontaneous?
- all temperatures
- low temperatures
- high temperatures
- never
Thermodynamics
• Example Question:
The ΔG° for the reaction
Ca2+ + 2OH- => Ca(OH)2(s) is -52 kJ/mol
Determine K at T = 20.°C for Ca(OH)2(s) =>
Ca2+ + 2OH-
Le Châtelier’s Principle
Intuitive Method
Mathematical Method
– Addition to one side results
in switch to other side
DG  DG   RT ln Q
– Example:
Q
DG  RT ln  
K
reaction shifts to
reactants (more AgCl(s))
AgCl(s) ↔ Ag+ + ClAddition of Ag+
When Q>K, ΔG>0 (toward
reactants)
When Q<K, ΔG<0 (toward
products)
Example: Q = [Ag+][Cl-]
As Ag+ increases, Q>K
Le Châtelier’s Principle
Stress Number 1 Reactant/Products:
Addition of reactant: shifts toward product
Removal of reactant: shifts toward reactant
Addition of product: shifts toward reactant
Removal of product: shifts toward product
Le Châtelier’s Principle
Stress Number 1 Example:
CaCO3(s) + 2HC2H3O2(aq) ↔ Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
1. Add HC2H3O2(aq)
2. Remove CO2(g)
3. Add Ca(C2H3O2)2(aq)
4. Add CaCO3(s)
No effect because (s)
Le Châtelier’s Principle
Stess Number Two: Dilution
Side with more moles is favored at
lower concentrations
Example: HNO2(aq) ↔ H+ + NO2If solution is diluted, reaction goes
to products
If diluted to 2X the volume:
K

H NO 



2
HNO2 
  
1  1
H
NO2
2
Q 2
1
HNO2 
2

1
Q K
2
So Q<K, products favored
Le Châtelier’s Principle
Stess Number Two: Dilution – Molecular Scale View
Concentrated Solution
H+ NO2-
H+ NO2-
Diluted Solution – dissociation allows
ions to fill more space
H+ NO2-
H+ NO2-
H+
H+
H+ NO2H+ NO2NO2-
H+ NO2-
H+ NO2NO2-
Le Châtelier’s Principle
Stress Number 3: Temperature
If ΔH>0, as T increases, products favored
If ΔH<0, as T increases, reactants favored
Easiest to remember by considering heat a
reactant or product
Example:
OH- + H+ ↔ H2O(l) + heat
Increase in T
Some Le Chatelier’s Principle Examples
•
Looking at the reaction below, that is initially
at equilibrium,
AgCl(s) ↔ Ag+(aq) + Cl-(aq) (ΔH°>0)
determine the direction (toward products or
reactants) each of the following changes will
result in
a)
b)
c)
d)
increasing the temperature
addition of water (dilution)
addition of AgCl(s)
addition of NaCl
Ch. 6 – Solubility Problems
•Why Solubility is Important
• Use in gravimetric analysis (predict if precipitation is
complete enough)
• Use in precipitation titrations (in next chapter)
• Use in separations (e.g. separation of Mg2+ from Ca2+ in tap
water for separate analysis)
• Understand phase in which analytes will exist
•Problem Overview
• Dissolution of sparingly soluble salts in water
• Dissolution of sparingly soluble salts in common ion
• Precipitation problems (and selective precipitation problems)
Solubility Product Problems
- Solubility in Water
Example: solubility of Mg(OH)2 in water
Solubility defined as mol Mg(OH)2 dissolved/L sol’n
or g Mg(OH)2 dissolved/L sol’n or other units
Use ICE approach:
Mg(OH)2(s) ↔ Mg2+ + 2OHInitial
0
0
Change
+x
+2x
Equilibrium
x
2x
Note: x = [Mg2+] = solubility
Solubility Product Problems
- Solubility of Mg(OH)2 in water
Equilibrium Equation: Ksp = [Mg2+][OH-]2
Ksp = 7.1 x 10-12 = x(2x)2 = 4x3 (see
Appendix F for Ksp)
x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 M
Solubility = 1.2 x 10-4 M = [Mg2+]
Conc. [OH-] = 2x = 2.4 x 10-4 M
Solubility Product Problems
- Solubility of Mg(OH)2 in Common Ion
If we dissolve Mg(OH)2 in a common ion
(OH- or Mg2+), from Le Châtelier’s
principle, we know the solubility will be
reduced
Example 1) What is the solubility of
Mg(OH)2 in a pH = 11.0 buffer?
No ICE table needed because, from pH, we
know [OH-]eq and buffer means dissolution
of Mg(OH)2 doesn’t affect pH.
Solubility Product Problems
- Solubility of Mg(OH)2 at pH 11 – cont.
[H+] = 10-pH = 10-11 M
and [OH-] = Kw/[H+] = 10-3 M
Ksp = [Mg2+][OH-]2
Moles Mg(OH)2 dissolved = moles Mg2+
[Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2
[Mg2+] = 7 x 10-6 M
Solubility Product Problems
- Solubility of Mg(OH)2 in Common Ion
Example 2) Solubility of Mg(OH)2 in 5.0 x
10-3 M MgCl2.