Download Chemical Equilibrium

Table of Contents
Chapter 18: Chemical Equilibrium
18.1: A state of dynamic balance
18.2: Factors affecting chemical
18.3: Using equilibrium constants
Table of Contents
Chapter 18: Chemical Equilibrium
18.1: Equilibrium: A state of dynamic balance
Chemical Equilibrium: Basic Concepts
• A reversible reaction is one that can occur in
both the forward and the reverse directions.
We say this is in equilibrium
• We need to distinguish between what happens to
concentration and what happens to rate
• What happens to rate at equilibrium?
equal
• What happens to concentration at equilibrium?
constant
Chemical Equilibrium: Basic Concepts
• Chemical equilibrium occurs when reversible
reactions occur at equal rates.
Rateforward reaction = Ratereverse reaction
• The rate of a reaction depends on the
concentration of the reactants.
• At equilibrium, the concentrations of reactants
and products are constant, not equal.
Chemical Equilibrium: Basic Concepts
Constants for homogeneous equilibria
• Homogeneous equilibrium: all the reactants
and products are in the same physical state.
• Heterogeneous equilibrium: reactants and
products are in different physical states
Chemical Equilibrium: Basic Concepts
Equilibrium Expressions and Constants
• To express this mathematically, we use the
law of chemical equilibrium.
Molarity of
products
Equilibrium
constant
Molarity of
reactants
Chemical Equilibrium: Basic Concepts
Equilibrium Expressions and Constants
• This ratio is called the equilibrium constant
expression (Keq). It has no units.
• The square brackets indicate the
concentration (Molarity: mol/L) the of
reactants and products
• The value of Keq is constant only at a
specified temperature.
Chemical Equilibrium: Basic Concepts
Constants for heterogeneous equilibria
• Equilibrium positions depend upon the initial
concentrations of the reactants and products.
• In an equilibrium expression, do not include
pure solids (s) or liquids (l)
• This is because their concentration is their
density, which does not change at any given
temperature.
Chemical Equilibrium: Basic Concepts
Equilibrium Expressions and Constants
• Keq > 1: More products than reactants at
equilibrium.
• Keq < 1: More reactants than products at
equilibrium.
So, Keq tells us at a glance how much we have
of our reactants and products
Practice
• Write the equilibrium constant for these reactions:
N2(g) + 3H2(g) 2NH3(g)
2
[NH
]
3
Keq =
[N2][H2]3
2NaHCO3(s)
Na2CO3(s) + CO2(g) +H2O(g)
Keq =
[CO2] [H2O]
Practice
• Write the equilibrium constant for these reactions:
Keq =
[HI]2
[H2][I2]
Keq =
[C2H5OH]
Chemical Equilibrium: Basic Concepts
Calculating the Value
of Equilibrium Constants
• Calculate the value of Keq for the equilibrium
constant expression
[.933]2
= .399
=
3
[.533][1.600]
given concentration data at one equilibrium
position: [NH3] = 0.933 mol/L, [N2] = 0.533
mol/L, [H2] = 1.600 mol/L.
Practice
• Calculate the equilibrium constant for the following
equation:
N2O4(g)
2NO2(g)
[N2O4] = 0.0185 mol/L
[NO2] = 0.0627 mol/L
Keq = 0.213
FeO(s) + CO(g)
Fe(s) + CO2(g)
[CO] = 0.0314 mol/L
Keq = 2.97
[CO2] = 0.0934 mol/L
Chemical Equilibrium: Additional Concepts
Calculating Equilibrium Concentrations
18.3
• You can use Kc to find the concentration
of individual compounds.
Steps:
1. Write equilibrium constant expression
2. Plug in what you know
3. Solve for what you don’t know
Chemical Equilibrium: Additional Concepts
Calculating Equilibrium Concentrations
• EXAMPLE 1: Keq = 66.9 for the formation of
hydrogen iodide from its elements.
• What is the concentration of HI if [H2] =
0.0295 mol/L and [I2] = 0.0174 mol/L?
2
2
x
[HI]
= 66.9 =
Keq =
[0.0295 M] [0.0174 M]
[H2] [I2]
Chemical Equilibrium: Additional Concepts
Calculating Equilibrium Concentrations
• What is the concentration of HI?
[x]2 = 66.9 [0.0295 M] [0.0174 M] = .03434
[x]
= √(.03434)
[HI] = .185mol/L
Additional Assessment Questions
Practice
At a certain temperature, Keq = 0.118 for the
following reaction.
Calculate the concentration of [H2] in an
equilibrium mixture with [CH4] = 0.0492
mol/L and [C2H2] = 0.0755 mol/L.
0.3357 mol/L
End 18.1
Table of Contents
Chapter 18: Chemical Equilibrium
18.2: Factors affecting chemical equilibrium
Chemical Equilibrium: Basic Concepts
Factors Affecting Chemical Equilibrium
• Le Châtelier’s principle: If a system at
equilibrium is stressed, it will shift to reduce
the stress.
• Rule of Thumb: “Same side does the
opposite, opposite side does the same”
• Equilibrium shifts in the direction that
increases in the end.
Chemical Equilibrium: Basic Concepts
Factors Affecting Chemical Equilibrium
• Equilibrium can be disturbed by 3 factors:
1. Changing concentration: adding or removing
a reactant or product.
2. Changing temperature
3. Changing volume & pressure
Chemical Equilibrium: Basic Concepts
Factors Affecting Chemical Equilibrium
1. Changing Concentration: Removing a Reactant
or a Product
N2(g) + 3H2(g) 2 NH3(g)
• Adding a reactant makes it form more products ,
causes the equilibrium to shift to the right
• Adding the product makes it form more reactant,
causes a shift to the left
Chemical Equilibrium: Basic Concepts
Factors Affecting Chemical Equilibrium
1. Removing a Reactant or a Product
N2(g) + 3H2(g) 2 NH3(g)
• Removing/diluting a reactant makes it form more
reactants, causes the equilibrium to shift to the left
• Removing/diluting the product makes it form
more product, causes a shift to the right
Practice adding and removing:
• How will it effect equilibrium if you…
 Add
NO?
shifts to the right, forms more products
 Add
Br2?
shifts to the right, forms more products
 Add
NOBr?
shifts to the left, forms more reactants
 Remove
NOBr? shifts to the right, forms more products
 Remove
Br2?
 Remove
NO? shifts to the left, forms more reactants
shifts to the left, forms more reactants
Chemical Equilibrium: Basic Concepts
Factors Affecting Chemical Equilibrium
clear
in H2O
I- + I2
purple
I3-
brown/yellow
in H2O
in hexane
Keq =
[I3-]
[I-] [I2]
Solutions: I- (KI) + I2(s) → I3-(let KI limit – leave some I2 on the bottom)
mix I- & I2; add I2; add I3-; dilute with hexane; pour off I2; dilute with water
Chemical Equilibrium: Basic Concepts
2. Changes in temperature
• Changes in temperature, causes a shift in
equilibrium by changing the value of Keq.
• In contrast, changes in concentration or
pressure cause a shift in equilibrium without
changing Keq
Chemical Equilibrium: Basic Concepts
2. Changes in temperature
• Treat heat as any other reactant or product
• Exothermic = heat is a product
Reactants ↔ Products + heat
• Endothermic = heat is a reactant
Reactants + heat ↔ Products
Chemical Equilibrium: Basic Concepts
2. Changes in temperature
• Increase in temp
shift to left, Keq decreases
• Decrease in temp
shift to right, Keq increases
Practice changes in temp:
heat + 6CO2(g) + H2O(l) = C6H12O6(aq) + 6O2(g)
• How will it effect equilibrium if you…
 Increase
temperature
 Decrease
temperature
shift to the right, Keq increases
shift to the left, Keq decreases
Chemical Equilibrium: Basic Concepts
Gas tubes
NO2(g)
N2O4(g) + heat
brown
clear
one tube in ice and water; heat one tube
Chemical Equilibrium: Basic Concepts
3. Changes in volume & pressure
• In a gaseous system, volume and pressure are
interconnected.
• ↓ volume, ↑ pressure Shift to side with
fewest moles
• ↑ volume, ↓ pressure Shift to side with
most moles
Chemical Equilibrium: Basic Concepts
3. Changes in volume & pressure
• How will this reaction respond to the
following changes:
N2(g) + 3H2(g) 2 NH3(g)
• Increase in pressure
shift to right
• Increase in volume
shift to left
Practice changes in vol &
press:
• How would increasing the volume of the reaction
vessel impact each of these equilibria?
shift to the left
shift to the right
Practice changes in vol &
press:
• How would decreasing the volume of the reaction
vessel impact each of these equilibria?
shift to the right
shift to the left
Chemical Equilibrium: Basic Concepts
3. Changes in volume & pressure
When molesreactants = molesproducts
volume and pressure have no effect on equilibrium.
Chemical Equilibrium: Basic Concepts
Rubber bands: final revision
unstretched
stretched + heat
Is stretching a rubber band endothermic or
exothermic?
So…if you heat the rubber bands, will the wrench
go down or up?
End 18.2
Table of Contents
Chapter 18: Chemical Equilibrium
18.3: Using equilibrium constants
The Plan
• Tuesday – individual work:
o
o
o
o
o
WebAssign Ch 18 (a big one!)
Ch 18 practice problems
Preview notes for ch.19
WebAssign CRT Review (a bigger one!)
The sub has permission to kick you off of computers if you
are misusing time
• Wednesday – group work:
o Ch 18 practice problems (due Thursday)
o Ch 18 Review – due with the 21 May
o The sub has permission to separate groups if you are not
using your time well
Chemical Equilibrium: Basic Concepts
Equilibrium Expressions and Constants
• What is Keq?
• Why do we use Keq?
• Write Keq for the following reaction:
Chemical Equilibrium: Additional Concepts
Solubility equilibria
• The solubility product constant (Ksp) is
an equilibrium constant for the dissolving
of a sparingly soluble ionic compound in
water.
Equilibrium of a
dissolved substance
Tells us if we have more
solid or more ions
Chemical Equilibrium: Additional Concepts
Solubility equilibria
When you know Ksp, you can…
1. calculate the molar solubility of a
sparingly soluble ionic compound
2. calculate moles per liter of a saturated
substance.
3. calculate ion concentrations in a saturated
solution.
These all use the same process!
Chemical Equilibrium: Additional Concepts
1. Calculating Molar Solubility
• EXAMPLE 2: What is the molar solubility
of copper (II) hydroxide?
1. Write the equilibrium equation
2. Look up Ksp
3. Replace Molarites with x
2.2 x 10-20 = [Cu2+][OH-]2
2.2 x 10-20 = [x][2x]2
Chemical Equilibrium: Additional Concepts
1. Calculating Molar Solubility
• EXAMPLE 2: what is the solubility of
copper (II) hydroxide?
4. Solve for x
2.2 x 10-20 = [x][2x]2 = x4x2 = 4x3
[x]3 = 5.5 x 10-21
[x] = 1.8 x 10-7 M
The solubility of Cu(OH)2 is = 1.8 x 10-7 mol/L
Chemical Equilibrium: Additional Concepts
2. Ion Concentration from Ksp
WE ALSO KNOW…
The moles per liter of the saturated substance is
1.8x10-7 mol/L
Ion concentrations
[Cu2+] = 1.8x10-7 M
x = [Cu2+]; 2x = [OH-]
[OH-] = 2x = 2(1.8 x 10-7 M)
[OH-] = 3.6 x 10-7 M
Practice
1. How many moles per liter of silver chloride will be in
a saturated solution of AgCl? Ksp = 1.8x10-10
AgCl(s) Ag+(aq) + Cl-(aq)
1.3x10-5 mol/L
1.3e-05 mol/L
2. Calculate the molar solubility of strontium
chromate (SrCrO4)in water if Ksp = 3.7x10-5
0.00608 mol/L
Chemical Equilibrium: Additional Concepts
Predicting precipitates
• Ksp can also be used to predict whether a
precipitate will form
• Use the ion product, Qsp.
• Ksp = Qsp at only at equilibrium
• They are calculated the same way
• Instantaneous concentrations
• The relationship between K and Q allows
us to predict the direction of a rxn
Chemical Equilibrium: Additional Concepts
Predicting precipitates
• If Qsp < Ksp, no precipitate forms
• If Qsp > Ksp, a precipitate will form
• If Qsp = Ksp, no change will occur
Chemical Equilibrium: Additional Concepts
Common ion effect
• Common ion effect: The solubility of a
substance is reduced when the substance is
dissolved in a solution containing a common ion.
• For example, PbI2 is less soluble in an
aqueous solution of NaI than in pure water.
• Because the common ion I– is already
present in the NaI solution. It reduces the
maximum possible concentration of Pb2+ and
thus reduces the solubility of PbI2.
Homework
• WebAssign ch 18: Due Wednesday, 11 pm
o Don’t do #19 or #21
o You will not be graded on those problems
• Ch 18 Practice Problems: due Thursday, 10
May
• Ch 18 review: due 21 May
How will adding H2
effect eq?
N2(g) + 3H2(g) 2 NH3(g)
Practice changes in temp:
6CO2(g) + H2O(l) = C6H12O6(aq) + 6O2(g) + heat
• How will it effect equilibrium if you…
 Increase
temperature
 Decrease
temperature
shift to the left, Keq decreases
shift to the right, Keq increases
Practice
• Use Ksp values to predict whether a precipitate will
form when equal volumes of the following aqueous
solutions are mixed:
• .1M Pb(NO3)2 and .03M NaF
• .25 M K2SO4 and .01 AgNO3
A precipitate
of PbF2 forms
No precipitate
forms