Download 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 – (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
so [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=12
0&cat=chemistry
This has some of the actual equilibria we will be
investigating later this week.
Le Chatelier’s Principle – Bozeman Science
http://www.youtube.com/watch?v=PciV_Wuh9V8 7:00
good explanations with visuals and excellent discussion on how
to increase yield of a reaction
Equilibrium Disturbances – Bozeman Science
http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36
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 (remember 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
Brightstorm 4:17
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.