Transcript Chemical Equilibrium

Chapter 14
Chemical Equilibrium
Reaction Types
There are 2 Types of Reactions
1.Completion Reactions
and
2. Reversible Reactions
Completion Reactions

A + B
C + D
Completion reactions go in one direction.
“All” ?? reactants are converted into
products and then the reactions stops.

Usually involve to formation, and
escape, of a gas….you lose products to
the environment

Sulfur + Oxygen
S8(s) + 8O2(g)
Sulfur Dioxide
SO2(g)
Reversible Reactions
will reach Equilibrium

A “reversible” reaction is a reaction
in which the products re-form the
original reactants.

NaCl(s)
Na1+(aq ) + Cl1-(aq)
5 Facts about
Reversible Reactions
1. Reversible reactions go in both
directions.
2. Opposite reactions occur at the same rate.
Rate of Dissolving = Rate of Crystallization
Sodium Chloride
NaCl(s)
Sodium + Chlorine
Na(aq) + Cl(aq)
3.Once Equilibrium is established the forward
and reverse reactions will continue.
Reversible Reactions
4. The Amounts of the reactants and the
products will remain constant as the
reaction continues.
5. Even though the rates of reactant and
product production are equal, it is rare
for the concentrations of both the
reactant or product to be equal. There’s
usual more of one than the other.
Reading the Symbols
Completion Reaction
Reversible Reaction
Reversible Reaction with a higher concentration of products
Reversible Reaction with a higher concentration of reactants
At Equilibrium



Both forward and reverse reactions
will continue to occur.
The forward reaction rate is equal
to the reverse reaction rate.
“Dynamic” equilibrium means there
is no net change!
Examples of Equilibrium


1. Homeostasis… Living Cell Chemistry
2. Biogeochemical Cycles
* The Earth’s “Recycling” Program
- The Nitrogen Cycle
- The Carbon Cycle
- The Oxygen Cycle
Complex Ions
Complex ions can increase the solubility
of certain elements.
Some Reversible Reactions require
“Complex Ions” to establish Equilibrium.
Complex Ions have + or – charges and
usually are produced from transition
metals.
Complex Ions
Complex Ions are produced by combining
cations and “Ligands”.
Ligands are molecules, like NH3, or ions that
bond to central atoms to form Complex ions.
These complex ions improve solubility. H2O is a
common ligand.
NH3 are the
Ligands
The Equilibrium Constant
The equilibrium constant is the ratio of
products to reactants when the reaction
has reached equilibrium.
The Equilibrium Constant Keq
At equilibrium the concentration of reactants and
products can be determined mathematically,
only if the temperature is constant.
The ratio of concentration of products to reactants is
raised to its coefficient as an exponent.
aA + bB
Keq =
[C]c[D]d
[A]a[B]b
cC + dD
[ ] means Molarity
Keq is called the “equilibrium constant” expression
**Solids and liquids do not show up in the expression.
Size of Keq
[C]c[D]d
=
[A]a[B]b
At equilibrium: aA + bB
cC + dD
If Keq is about 1 the concentration of reactants and
products is about equal.
If Keq is greater than 1 there is a higher
concentration of products than reactants at
equilibrium.
(The higher value K = more products less reactants.)
If Keq is less than 1 there is a higher concentration of
reactants than products at equilibrium.
(The lower the value K = more reactants less products.)
Value of Keq
Values of Keq can only be determined by
experiment.
2 types of math problems involve Keq
1 - given [M] of all reactants and products
calculate Keq
 2 - given Keq and [M] of all reactants and
products but one, solve for its [M]

Keq does not indicate a reaction’s rate.
Determination of Keq
Keq is used to determine product and
reactant concentrations at equilibrium.
Keq =
Products
Reactants
Changes in Keq
*Changes in concentrations DO NOT change Keq
adding or removing reactant or product
[R] and [P] change but Keq remains the same


Changing the pressure of the system DOES NOT change Keq


Placing it in a bigger or smaller container
Changes in temp DO cause Keq to change

[R] gets bigger and [P] gets smaller
 Or the other way around
Determination of Keq
H2CO3(aq) + H2O(l) <--> H30+(aq) + HC03-(aq)
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
Determine the value of Keq for the
reaction above if
H2CO3(aq) = [3.3 x 10-2]
H30+(aq) = [1.2 x 10-4]
HC03-(aq) = [1.2 x 10-4]
Solution
Keq =
[C]c[D]d
[A]a
[H3O+][HCO-3]
Keq =
H2CO3
[1.2 x 10-4] [1.2 x 10-4]
-7
4.4
x
10
=
Keq =
[3.3 x 10-2]
Given Keq:Determination of Concentration
The reaction;
N2O4(g)
2NO2(g)
Has a value of K is 1.50.
Calculate the concentration of the N2O4(g)
present in the system when [NO2] = 0.91 M
Solution
Keq =
1.5 =
Products
Reactants
[NO2]2
[x]
x = 0.55
=
[0.91]2
[x]
The Solubility product
constant...Ksp
Three parts to this section:
 Learn how to write the solubility
constant expression Ksp.
 Learn how to determine solubility
from Ksp.
 Learn how to determine Ksp from
solubility.
The Solubility Product Constant, Ksp
You can dissolve salt in water until the
solution becomes saturated. When a
solution reaches its saturation point it
starts to reform solid reactants.
A saturated solution is at
“Dynamic Equilibrium”.
Many ionic compounds are only “slightly”
soluble in water.
Example of Using Ksp
If you try to dissolve AgI in water you’ll
find it’s only “slightly” soluble and you’ll
end up with virtually no Ag+ or I- ions.
So…how can you compare the solubility of
AgI(s) to (Ag+(aq) or I-(aq) )?
Easy! Eliminate the reactant and determine
the Solubility of the Products.
Using Ksp
You can compare the solubility of “slightly”
soluble salts using Ksp.
The solubility product constant, Ksp, is the
product of the concentration of the ions
produced, each raised to their coefficient as an
exponent.
The Higher the Ksp, the more soluble the salt will be!
The Lower the Ksp, the less soluble the salt will be!
Solubility Product Constant
Written for the equilibrium of the solution
of slightly soluble salts.

AaBb(s)



ksp = [A+]a [B-]b
CaCO3(s)

Ca2+(aq ) + CO32-(aq)
ksp = [Ca2+] [CO32-]
Ag2CO3(s)

aA+(aq) + bB-(aq)
2Ag1+(aq) + CO32-(aq)
ksp = [Ag+1]2 [CO32-]
Solid and liquids do not show up in the ksp expression
This will make you or Break You!
For every NaF that breaks apart 1 Na+ and 1 Fare formed NaF(s)
Na+(aq) + F-(aq)
s
s
Ksp = [Na+][F-]
or Ksp = [s][s]
or Ksp = s2
For every Ag2CO3 that breaks apart 2 Ag+ and 1
CO32- are formed
Ag2CO3(s)
2Ag+(aq) + CO32-(aq)
2s
s
Ksp = [Ag+]2[CO32-] or Ksp = [2s]2[s] or Ksp = 4s3
Write ksp expression for
each of the following:
AgI(s) 
Ag2S(s) 
PbI2(s) 
MgCO3(s) 
Ca3(PO4)2(s) 
Write ksp expression for
each of the following:
AgI(s)  Ag1+ (aq) + I1- (aq)
Ag2S(s)  2Ag1+ (aq) + S2- (aq)
PbI2(s)  Pb2+ (aq) + 2I1- (aq)
MgCO3(s)  Mg2+(aq) + CO32-(aq)
Ca3(PO4)2(s)  3Ca2+(aq) + 2PO43- (aq)
Solubility Product Constant = ksp

Ksp shows the solubility of the salt


High ksp = more soluble
Low ksp = less soluble
Table 14-3
pg 508
3 Types of problems
1. Calculating ksp given molar concentrations of the ions.
2. Calculating Molar Solubility given the ksp.
3. Calculating ksp from the Molar Solubility.
Values For Ksp At 25 oC
Large Ksp = more soluble
Small Ksp = less soluble
1. Calculating ksp given molar concentrations
of the ions.
Calculate the Solubility Product Constant, ksp for Copper(I) Bromide.
The concentration of Cu1+ ions in solution is 7.9 x 10-5 mol/L.
CuBr(s)
Cu1+(aq) + Br1-(aq)
Ksp = Cu1+(aq) + Br1-(aq)
Ksp = [7.9 x 10-5 mol/L] [7.9 x 10-5 mol/L]
= 6.2 x 10-9
2. Calculating Molar Solubility given the ksp.
Calculate the solubility of AgI in water if
Ksp = 1.8 x 10-10
AgI
Ag+ + I[s]
[s]
Ksp = [Ag+][I-] = s2
1.8 x 10-10 = s2
√
s = 1.8 x 10-10 = 1.3 x 10-5
3. Calculating ksp from the Molar Solubility.
It is found that 1.2 x 10-3 mol of lead (II)
iodide, PbI2, dissolves in 1.0 L of aqueous
solution at 25oC. What is the Ksp at this
temperature?
PbI2(s)
Pb2+(aq) + 2I1-(aq)
[s]
Ksp = [Pb2+][2I-]2
= 4s3
Ksp = 4(1.2 x 10-3)3
Ksp = 6.9 x 10-9
[2s]
LeChatelier’s Principle

When an system at equilibrium is disturbed, the
system adjusts in a way to relieve the stress….
It will “shift” to the right or to the left.
Equililibrium shifts toward the reactants or the products.
3 types of changes (stresses) affect equilibrium

D [concentration]




[ ] means molarity (mol/L)
Increase or decrease
D temperature
D pressure – for gas systems only
Using La Chatelier’s Principle
Systems at equilibrium can be forced to
completion by changing concentrations,
pressure or temperature.
Increase in Concentration
Increase in the amount of reactants


in number of collision between reactants.
Forward rx occurs more than the reverse
rx…..there’s a shift to the right.




Concentration of all r and p change
K remains the same
An increase in concentration of a substance
pushes the equilibrium away from the side
of the increase.
Same thought process for an increase in
the concentration of the products.
Decrease in Concentration
Decrease in the amount of reactants

in number of collision between reactants
 Reverse rx occurs more than the forward
rx….there’s a shift to the left.


Decreases in concentration of a substance
pulls the equilibrium towards the side of the
decrease.
Same thought process for a decrease in the
concentration of the products.


Concentration of all r and p change
K remains the same
Examples
2NO2(g)
N2O4(g)
What happens to k if the concentration of NO2
is increased?

What happens to concentration of N2O4 if the
concentration of NO2 is decreased?
 What happens to k if the concentration of
N2O4 is increased?
 What happens to concentration of NO2 if the
concentration of N2O4 is increased?

LeChatelier’s Principle
Changing the pressure on the container



Increasing the pressure... The reaction will shift
toward the side with the fewer moles of gas.
Decreasing the pressure... The reaction will shift
toward the side with more moles of gas.
Changing the pressure will make no difference if there
is an equal number of moles of gas on each side.
k does NOT change
LeChat Pressure
N2(g) + 3H2(g) <---> 2NH3(g)
 Which way will the equilibrium shift if
the pressure is increased?
 Which direction will the equilibrium shift
if the container size is increased?
Change in Temperature

Do you know the forms for endothermic
and exothermic reactions???
A + B
heat + A + B
C + D + heat
C +D
Heat, energy, or a number of KJ on reactant
side = endothermic
 Heat, energy, or a number of KJ on product
side = exothermic

Changes in Temperature
When a temperature change occurs you write
“heat”, or the number of KJ into the equation as
a reactant or product.

Which side is heat written on?

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Based on whether reaction is endothermic or exothermic
Treat “heat” as a reactant or product
in heat, push reaction away from the side “heat” is located
in heat, pull reaction towards the side “heat” is located
K will get bigger or smaller depending on which
direction the equilibrium reaction shifts.. If the shift is
to the right, K get bigger.. If the shift is to the left, K
gets smaller.

Change in Temperature
The reaction 2NO2(g)
N2O4(g)
gives off 57.2 KJ of energy. What happens
to the concentration of N2O4 if the
equilibrium system is heated?
What happens to the value of K?

The reaction is exothermic therefore:
2NO2(g)
N2O4(g) + heat (57.2 KJ)
The decrease in heat will pull reaction to the
right.
Common Ion Effect
The Common Ion Effect explains why there
is a reduction in the solubility of a salt
due to the presence of a common ion.
Example: Think about the solubility of
CaF2(s) in NaF(aq)…. The CaF2(s) can’t
breakdown as easily because there is
already F- ions in solution.