Transcript Chemical Equilibrium

Forward and reverse reactions taking
place at equal rates
It is a dynamic state - reactions are
constantly occurring
(a) Start: 10 goldfish in the left tank and 10 guppies in the right.
(b) Equilibrium state. with 5 of each kind of fish in each tank. The equilibrium is dynamic;
an averaged state and not a static condition . The fish do not stop swimming when
they have become evenly mixed.
(c) If we were to observe one single fish (here a guppy among goldfish). we would find
that it spends half its time in each tank .
Brightstorm videos
Chemical Equilibrium Definition 5:01
http://www.youtube.com/watch?v=FYc_SoW2M40&list=PL06C3C4E3F
84C6A24&index=42
Crash course chemistry
http://www.youtube.com/watch?v=g5wNg_dKsYY 10:56 Equilibrium
http://www.youtube.com/watch?v=DP-vWN1yXrY 9:28 Equilibrium
equations
You don’t need to know how to do the RICE table starting at 4:40
Isaacs Teach http://www.youtube.com/watch?v=g4TKRInLdPA 10:09
Equilibrium
Good basic explanation!
http://www.youtube.com/watch?v=4z4_rc6nsKU 12:46 What is the
equilibrium constant, Keq? Also very good explanation
Equilibrium constant expressions
aA + bB  cC + dD
c
d
Keq = [C] [D]
a
b
[A] [B]
General information about the Keq
expression
• Equilibrium [ ] of products are placed in the
numerator.
• Equilibrium [ ] of reactants are placed in the
denominator.
• Each [ ] term is raised to an exponent equal to its
coefficient in the balanced equation.
• If there is more than 1 product or reactant, the terms
are multiplied.
• Solids and liquids (pure substances) are not
included in the Keq expression. This is because their
[ ] are their densities. The density of a substance
does not change with changing temperatures.
• Keq is constant for a given reaction at a given
temperature. There are no units associated with
the value of Keq.
• The value of Keq is independent of the:
– individual [ ] of reactants and products
– original [ ] of reactants and products
– volume of the container.
• The value of Keq is dependent on temperature.
• What does the value of Keq tell you about a
reaction?
Keq >1: more products than reactants at equilibrium
Keq < 1: more reactants than products at equilibrium
Using equilibrium constants
Calculating equilibrium concentrations:
Example: At 1405 K, hydrogen sulfide, also
called rotten egg gas (because of its bad
odor), decomposes to form hydrogen and a
diatomic sulfur molecule,S2. Keq = 2.27 x 10-3.
(a) Write the balanced equation for the
reaction described above. Write out the Keq
expression.
(b) Calculate the concentration of hydrogen
gas if [S2] = 0.0540 M and [H2S] = 0.184 M.
Solving the problem – part (a)
2H2S (g)  2H2 (g) + S2 (g)
Keq =
2
[H2] [S2]
2
[H2S]
Solution – part (b)
[H2]2= Keq [H2S]2 =
[S2]
(2.27 x 10-3)(0.184 M)2 =
0.0540 M
[H2]2= 1.42 x 10-3 M
 [H2] = 3.77 x 10-2 M
Le Châtelier’s principle:
1884 - Henri Le Châtelier
When a stress is
applied to a system
at equilibrium, the
system shifts in the
direction that
relieves the stress.
Δ in concentration
• Adding more of a reactant or product: the
reaction will shift in the direction to consume
a portion of what was added.
– more reactant  shifts right
– more product  shifts left
• Removing some of a reactant or product:
the reaction will shift in a direction to restore
part of what was removed.
– reactants removed  reaction shifts left (i.e. the
reverse reaction)
– products removed  reaction shifts right (i.e. the
forward reaction).
Δ in volume
Relevant when discussing gaseous equilibria and when
the number of moles of gaseous reactants differ from the
number of moles of gaseous products. The change in
volume is a result of a change in pressure of the gaseous
system.
• When P↓, the reaction will shift in a direction to↑ number of
moles of gas.
PCl5 (g)  PCl3 (g) + Cl2 (g)
1 mol  2 mol
2NH3(g)  N2(g) + 3H2(g)
2 mol  4 mol
• When P↑, the reaction will shift in a direction to ↓ number of
moles gas.
PCl5 (g)  PCl3 (g) + Cl2 (g)
1 mol  2 mol
2NH3(g)  N2(g) + 3H2(g)
2 mol  4 mol
Δ in temperature
View changes in temperature as reactants or products.
When the temperature of an equilibrium system is ↑ the
reaction that is endothermic (ΔH>0) will take place.
* forward rxn is endothermic  more product (shifts to the
right).
* reverse rxn is endothermic  less product (shifts to the
left)
When the temperature of an equilibrium system is ↓, the rxn
which is exothermic (ΔH<0) will take place.
* forward rxn is exothermic – more product (shifts to the right).
* reverse rxn is exothermic – less product (shifts to the left)
General rule: if the forward rxn is endothermic,↑K.
If the forward rxn is exothermic ↓K.
Animation demonstration
http://www.learnerstv.com/animation/animation.php?ani=
120&cat=chemistry
http://www.youtube.com/watch?v=PciV_Wuh9V8
7:00 Le Chatelier’s Principle; good explanations with visuals
and excellent discussion on how to increase yield of a reaction
http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36
Equilibrium disturbances
This one will help you with the lab we’re doing. He also
discusses the effect of disturbances (changes) in an
equilibrium system and how they affect the value of K (the
equilibrium constant)
Reactions that go to completion
Formation of a gas
H2CO3 (aq)  H2O (l) + CO2 (g)
Formation of precipitate (Double
displacement reactions)
Formation of a slightly ionized product;
often times H2O (i.e. in a neutralization
reaction)
H3O+ + OH-  2H2O (l)
Solubility equilibria
http://www.youtube.com/watch?v=YJdyEtB66A&feature=topics
The Solubility Product Constant, Ksp
• Many important ionic compounds are only
slightly soluble in water and equations are
written to represent the equilibrium between
the compound and the ions present in a
saturated aqueous solution.
• The solubility product constant, Ksp, is the
product of the concentrations of the ions
involved in a solubility equilibrium, each
raised to a power equal to the stoichiometric
coefficient of that ion in the chemical equation
for the equilibrium.
The Solubility Equilibrium Equation And Ksp
CaF2 (s)
Ca2+ (aq) + 2F- (aq)
Ksp = [Ca2+][F-]2
Ksp = 5.3x10-9
As2S3 (s)
2As3+ (aq) + 3S2- (aq)
Ksp = [As3+]2[S2-]3
Ksp = 6 x 10-51
Ksp And Molar Solubility
• The solubility product constant is related to the
solubility of an ionic solute, but Ksp and molar
solubility - the molarity of a solute in a saturated
aqueous solution - are not the same thing.
• Calculating solubility equilibria fall into two
categories:
– determining a value of Ksp from experimental data
– calculating equilibrium concentrations when Ksp is
known.
Calculating Ksp From Molar Solubility
It is found that 1.2x10-3 mol of lead (II) iodide, PbI2,
dissolves in 1.0 L of aqueous solution at 25 oC.
What is the Ksp at this temperature?
Solution:
PbI2 (s)
Pb2+ (aq) + 2I- (aq)
Ksp = [Pb2+] [I-]2
Ksp = (1.2 x 10-3 M) (2 x 1.2 x 10-3 M)2
Ksp = 6.9 x10-9
Calculating Molar Solubility From Ksp
Calculate the molar solubility of silver chromate,
Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.
Solution:
Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)
Ksp = [Ag+]2 [CrO4 2-]
Ksp = (2x)2(x) = 1.1 x 10-12
4x3 = 1.1 x 10-12
X = 6.5 x 10-5 M
The Common Ion Effect In Solubility Equilibria
• The common ion effect also affects solubility
equilibria.
• Le Châtelier’s principle is followed for the shift in
concentration of products and reactants upon
addition of either more products or more reactants
to a solution.
The solubility of a slightly soluble ionic
compound is lowered when a second solute that
furnishes a common ion is added to the solution.
Solubility Equilibrium Calculation
-The Common Ion Effect
What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?
Ksp = 1.1x10-12 for Ag2CrO4.
Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)
Ksp = [Ag+]2 [CrO4 2-]
Ksp = (2x)2(0.10) = 1.1 x 10-12
x = 1.65 x 10-6 M
Comparison of solubility of Ag2CrO4
In pure water:
6.5 x 10-5 M (prior slide)
In 0.10 M K2CrO4:
1.7 x 10-6 M
The common ion effect!!
Determining Whether Precipitation Occurs
• Q is the ion product reaction quotient and is based on
initial conditions of the reaction.
• Q can then be compared to Ksp.
• To predict if a precipitation occurs:
- Precipitation should occur if Q > Ksp.
- Precipitation cannot occur if Q < Ksp.
- A solution is just saturated if Q = Ksp.
DR lab: unexpected PPT according to solubility rules!
Ca(OH)2 (s)
Ca2+ (aq) + 2OH- (aq)
Ksp = [Ca2+][OH-]2
Ksp = 6.5 x 10-6
Determining Whether Precipitation Occurs
– An Example
The concentration of calcium ion in blood plasma is
0.0025 M. If the concentration of oxalate ion is
1.0x10-7 M, do you expect calcium oxalate to
precipitate? Ksp = 2.3x10-9.
Three steps:
(1) Determine the initial concentrations of ions.
(2) Evaluate the reaction quotient Q.
(3) Compare Q with Ksp.
Solution
CaC2O4 (s)
Ca2+ (aq) + C2O42- (aq)
Ksp = [Ca2+] [C2O42-] = 2.3x10-9
Qsp = (2.5 x 10-3 M) (1.0x10-7 M) = 2.5 x 10-10
2.5 x 10-10 < 2.3x10-9
Q < Ksp therefore no ppt will be formed
Summary
• The solubility product constant, Ksp, represents
equilibrium between a slightly soluble ionic
compound and its ions in a saturated aqueous
solution.
• The common ion effect is responsible for the
reduction in solubility of a slightly soluble ionic
compound.
• The solubilities of some slightly soluble
compounds depends strongly on pH.
Equilibrium lab
Fe(OH)3 (s) Fe3+ (aq) + 3OH- (aq)
Ksp = [Fe3+][OH-]3 = 4 x 10-38
Q vs. Ksp
Q = [Fe3+][OH-]3 = (0.2M)(6.0M)3 = 43.2
Q >Ksp so a PPT forms to take the Fe3+
out of solution
Qualitative Inorganic Analysis
• Acid-base chemistry, precipitation reactions,
oxidation-reduction, and complex-ion formation all
come into sharp focus in an area of analytical
chemistry called classical qualitative inorganic
analysis.
• “Qualitative” signifies that the interest is in
determining what is present, not how much is
present.
• Although classical qualitative analysis is not as
widely used today as instrumental methods, it is
still a good vehicle for applying all the basic
concepts of equilibria in aqueous solutions.