Transcript Slide 1

Chapter 15
Chemical Equilibrium (and Kinetics)
Kinetics applies to the speed of a reaction, the concentration of
product that appears (or of reactant that disappears) per unit time.
(See sec’s 2 & 12)
Equilibrium applies to the extent of a reversible reaction (sec’s 3 11), the concentration of product that has appeared after an
unlimited time, or once no further change occurs.
At equilibrium:
rateforward = ratereverse
A system at equilibrium is dynamic on the molecular level; no further
net change is observed because changes in one direction are balanced
by changes in the other.
Kinetics: Reaction Rates
Some chemical reactions proceed rapidly:
like the precipitation reactions studied in Chp 7,
where the products form practically the instant the
two solutions are mixed.
Other reactions proceed slowly:
like the rusting of an iron nail left outside.
Rate of reaction is measured in the amount of reactant
that changes into product in a given period of time, in
moles of reactant used per second.
Chemists and engineers study ways of controlling
reaction rates.
Factors Affecting Reaction Rate
Reaction rate (kinetics) is affected by the concentrations of the
reactants, the temperature, the pressure, agitation, and the
presence of catalysts.
Concentration: high concentration of reactants can make it faster, but
high concentration of products can slow it.
Temperature: usually increases because particles are moving faster
and colliding more.
Pressure: affects rate only if a gas is present.
Agitation: temporary increase in rate because collisions increase.
Catalyst: made to increase rates via various mechanisms.
Factors Affecting Reaction Rate:
Reactant Concentration
The higher the concentration of reactant
molecules, the faster the reaction will
generally go.
- increases the frequency of reactant
molecule collisions (see next 2 slides)
Since reactants are consumed as the
reaction proceeds, the speed of a reaction
generally slows over time.
Factors Affecting Reaction Rate:
Concentration & Collision Theory
In order for a reaction to take place, the reacting
molecules must collide with each other.
Once molecules collide they may react together or
they may not, depending on two factors Whether the collision has enough energy to start to
break the bonds holding reactant molecules
together; and
Whether the reacting molecules collide in the proper
orientation for new bonds to form.
Factors Affecting Reaction Rate:
Concentration & Collision Theory
Collisions in which these two conditions are met
(and therefore the reaction occurs) are called
effective collisions.
The higher the frequency of effective collisions
the faster the reaction rate.
There is a minimum energy needed for a
collision to be effective – we call this the
activation energy.
The lower the activation energy “barrier” the
faster the reaction will be.
Factors Affecting Reaction Rate:
Temperature
Increasing temperature increases the reaction rate:
- for each 10°C rise in temperature, the speed of the
reaction generally doubles.
Increasing the temperature increases the number of
molecules in the sample with enough energy so that their
collisions can overcome the activation energy.
Increasing the temperature also increases the frequency of
collisions – so the rate increases because the frequency
of effective collisions increases.
Activation Energy
The energy barrier that prevents any collision
between molecules from being an effective
collision is called the activation energy.
The larger the activation energy of a reaction, the
slower it will be at a given temperature.
Relative Potential Energy
Exothermic Reaction
Activation
Energy,
large
Activation
Energy,
small
Reactants
DHreaction
Products
Progress of Reaction
Relative Potential Energy
Endothermic Reaction
Activation
Energy
Products
DHreaction
Reactants
Progress of Reaction
Effect of Temperature on Rate
High
more molecules
Low temperatures lead to fewer
molecules with
with enough
enough
energy
frequent
energy to
to overcome
overcome the
the activation
activation energy,
energy, and
and more
less frequent
reactant
reactant collisions,
collisions, therefore
therefore aa faster
slowerreaction
reactionrate
rate
Factors Affecting Reaction Rate:
Catalysts
A catalyst is a substance that increases
the rate of a reaction, but is not
consumed in the reaction.
Catalysts lower the activation energy of a
reaction.
Catalysts work by providing a easier
pathway for the reaction.
Catalysts’ Effect on
Activation Energy
Nature’s Catalysts: Enzymes
Enzymes are protein molecules produced by living
organisms that catalyze chemical reactions.
The enzyme molecules have an active site that organic
molecules bind to.
When the organic molecule is bound to the active site,
certain bonds are weakened.
This allows a particular chemical change to occur easier
and quicker, i.e. the activation energy is lowered.
Example of an Enzyme: Sucrase
Reaction Dynamics
If the products of a reaction are removed
from the system as they are made, then a
chemical reaction will proceed until the
limiting reactants are used up.
However, if the products are allowed to
accumulate; they will start reacting
together to form the original reactants.
This is called the reverse reaction.
Reaction Dynamics
The forward reaction slows down as the amounts of
reactants decreases.
At the same time the reverse reaction speeds up as the
concentration of the products increases.
Eventually the forward reaction is using reactants and
making products as fast as the reverse reaction is using
products and making reactants. This is called chemical
equilibrium.
Dynamic equilibrium is reached when the rates of two
opposite processes are the same.
Introduction to Equilibrium
At equilibrium, the [reactants] and [products] remain
constant over time! (Remember, [ ] mean concentration
of a substance in Molarity.)
Equilibrium is approachable from any direction!
Q = reaction quotient
At equilibrium Q = Keq
Keq = equilibrium constant
Introduction to Equilibrium
Q = Reaction Quotient = [Products]m/[Reactants]n
Given a generic reaction: aA + bB  cC + dD
Q = [C]c[D]d
[A]a[B]b
If at equilibrium, Q = Keq
Example: given the reaction N2O4(g)  2 NO2(g), we would
write the Keq expression:
Keq = [NO2]2/[N2O4]
Table: Initial and Equilibrium Concentration Ratios for the
N2O4-NO2 System at 200.0C (473 K)
Ratio(Q)
Initial
Experiment [N2O4] [NO2]
[NO2]2
[N2O4]
Value of Keq
Equilibrium
[N2O4]eq [NO2]eq
[NO2]eq2
[N2O4]eq
1
0.1000
0.0000
0.0000
3.57x10-3
0.193
10.4
2
0.0000
0.1000
∞
9.24x10-4
9.83x10-3
10.4
3
0.0500
0.0500
0.0500
2.04x10-3
0.146
10.4
4
0.0750
0.0250
0.00833
2.75x10-3
0.170
_____
The change in Q during the N2O4-NO2 reaction.
The initial part is covered by
Kinetics; the Q = K part is
covered by Equilibrium!
Figure Reaction direction and the relative
sizes of Q and Keq.
Reaction
Progress
Reaction
Progress
reactants
products
Equilibrium:
no net change
Keq = Kc for many reactions
reactants
products
The range of equilibrium constants
small K
large K
intermediate K
WRITE AN EQUILIBRIUM
EXPRESSION FOR:
CaCO3(s)  CaO(s) + CO2(g)
Keq = [CaO][CO2]/[CaCO3]
But solids don’t change in concentration (moles/liter).
Keq = CCaO[CO2]/CCaCO3
Rearrange to Keq' = [CO2]
NO solids or solvents are written into the equilibrium
constant expression or reaction quotient.
The reaction quotient for a heterogeneous system.
solids do not
change their
concentrations
PRACTICE WRITING Keq (Kc) FOR
OTHER REACTION SYSTEMS:
1.
2.
3.
4.
5.
C(s) + H2O(g)  CO(g) + H2(g)
Keq = [CO][H2]/[H2O]
HNO2(aq) + H2O(l)  H3O+(aq) + NO2-(aq)
Ka = [H3O+][NO2-]/[HNO2] (Use subscript a for acid)
CaCO3(s)  Ca2+(aq) + CO32-(aq)
Ksp = [Ca2+][CO32-]
(Use subscript sp for insol cmpd)
2 N2O5(g)  4 NO2(g) + O2(g)
Keq = [NO2]4[O2]/[N2O5]2
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Kb = [NH4+][OH-]/[NH3]
(Use subscript b for base)
Calculating Keq
The value of the equilibrium constant may be
determined by measuring the concentrations of all
the reactants and products in the mixture after the
reaction reaches equilibrium, then substituting in the
expression for Keq.
Even though you may have different amounts of
reactants and products in the equilibrium mixture, the
value of Keq will always be the same:
- the value of Keq depends only on the temperature
- the value of Keq does not depend on the amounts
of reactants or products you start with
Calculating the value of Keq
Example: A mixture of CH4, C2H2 and H2 is allowed to
come to equilibrium at 1700°C. The measured
equilibrium concentrations are [CH4] = 0.0203 M, [C2H2]
= 0.0451 M, and [H2] = 0.112 M. What is the value of the
equilibrium constant at this temperature?
2 CH4(g)  C2H2(g) + 3 H2(g)
Write the expression: Keq = [C2H2][H2]3/[CH4]2
Fill in the data: Keq = (0.0451)(0.112)3/(0.0203)2
= 0.154
Using Keq to find the concentration at
equilibrium
In an equilibrium mixture the concentrations of H2 and I2 are
both 0.020 M. What is the value of the equilibrium
concentration of HI?
H2(g) + I2(g)  2 HI(g)
Keq = 69 at 340°C
Write the expression: Keq = [HI]2/[H2][I2]
Put in the known data: 69 = [HI]2/(0.020)2
Rearrange and solve: [HI] = 0.166 M or 0.17 M
Using Keq and initial concentrations to find
equilibrium concentrations
We can find equilibrium concentrations if we know initial
conditions and the value of Keq.
Find the equilibrium concentrations for both the reactant
and product for the reaction below, given that you start
with 2.00 moles of the reactant in a 1.00 L container.
Initial concentration of the reactant is therefore 2.00 M
and the product is 0.
A. Set up an ICE table:
N2O4(g)  2 NO2(g) Keq = 0.20
Initial [ ]
2.00M
0.0M
(Note Q = 0)
Change
-x
+2x
Equil [ ]
(2.00-x) (2x)
Keq = [NO2]2/[N2O4]
B. Determine changes – using regular old stoichiometry.
Using Keq and initial concentrations to find
equilibrium concentrations
C. Put everything in the Keq expression:
Keq = [NO2]2/[N2O4] = 0.20
Keq =(2x)2/(2.00-x)=0.20M
Rearrange and solve for x, using the quadratic
equation.
x= 0.29, but you’re not done. Plug x into
equstions:
[N2O4] = 2.00 - 0.29 = 1.71M
[NO2] = 2x = 0.58 M
D. Plug these concentrations back into Keq and
solve, if close to 0.20M, then they are correct.
Solubility & Solubility Product, Ksp
Even “insoluble” salts dissolve somewhat in water
insoluble = less than 0.1 g per 100 g H2O.
Equilibrium constant for this process is called the
solubility product constant, Ksp.
Given a salt, AnYm, Ksp = [A+]n[Y-]m
If there is undissolved solid in equilibrium with the
solution, the solution is saturated.
Larger Ksp = more soluble salt.
Example - Determine the Ksp of PbBr2 if its
solubility is 1.44 x 10-2 M
PbBr2(s)  Pb2+(aq) + 2 Br–(aq)
I -0
0
E -0.0144
0.0288
Ksp = [Pb2+][Br–]2 = (0.0144)(0.0288)2
= 1.19 x 10-5
LeChatelier’s Principle
When a chemical system at equilibrium is
subjected to a stress, the system will
return to equilibrium by shifting to reduce
the stress.
Principle 1: The first stress we look at is a change
in concentration of reactant or product:
If the concentration increases, the system reacts to
consume some of it.
If the concentration decreases, the system reacts
to produce some of it.
Principle 1: Concentration
The effect of added Cl2 on the PCl3-Cl2-PCl5 system.
PCl3(g) + Cl2(g)  PCl5(g)
After adding more reactant
chlorine, we see both
reactants decreased and the
product increased, to
reestablish equilibrium.
Sample Problem
Predicting the Effect of a Change in Concentration
on the Equilibrium Position
To improve air quality and obtain a useful product, chemists often
remove sulfur from coal and natural gas by treating the fuel
contaminant hydrogen sulfide with O2;
2H2S(g) + O2(g)
What happens to
(a) [H2O] if O2 is added?
(c) [O2] if H2S is removed?
2S(s) + 2 H2O(g)
(b) [H2S] if O2 is added?
(d) [H2S] if sulfur is added?
PRINCIPLE 2: EFFECT OF INCREASE OR
DECREASE IN PRESSURE
Decreasing the size of the container increases the
concentration of all the gases in the container
increases their partial pressures.
If their partial pressures increase, then the total
pressure in the container will increase.
According to Le Châtelier’s Principle, the equilibrium
should shift to remove that pressure.
The way to reduce the pressure is to reduce the number
of molecules in the container.
When the volume decreases, the equilibrium shifts to
the side with fewer molecules.
PRINCIPLE 2: EFFECT OF INCREASE OR
DECREASE IN PRESSURE
When
Since
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pressure
are more
is decreased
gas
by
molecules
increasing
on the
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reactantsthe
side
position
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whenshifts
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the side
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the products.
side.
40
PRINCIPLE 2: INCREASE OR DECREASE
PRESSURE TO MAKE MORE CO?
C(s) + CO2(g)  2 CO(g)
Remember, the system will adjust to restore pressure:
If P decreases, system shifts to side with more
moles of gas.
If P increases, system shifts to side with fewer moles
of gas.
We would want to decrease the partial pressure of CO.
Sample
Problem
Predicting the Effect of a Change in Pressure on the
Equilibrium Position
PROBLEM:
How would you change the pressure (volume) of each of the
following reactions to increase the yield of the products.
(a) CaCO3(s)
CaO(s) + CO2(g)
(b) S(s) + 3F2(g)
SF6(g)
(c) Cl2(g) + I2(g)
2ICl(g)
SOLUTION: (a) CO2 is the only gas present. To increase its yield, we
should increase the volume, decrease the pressure.
(b) There are more moles of gaseous reactants than products, so we should
decrease the volume, increase the pressure, to shift the reaction to the right.
(c) There are an equal number of moles of gases on both sides of the
reaction, therefore a change in volume/pressure will have no effect.
Principle 3: The Effect of Temperature Changes on
Equilibrium for Exothermic Reactions
For an exothermic reaction, heat is a “product.”
Increasing the temperature is like adding heat.
According to Le Châtelier’s Principle, the equilibrium will
shift away from the added heat.
The concentrations of C and D will decrease and the
concentrations of A and B will increase.
The value of Keq will decrease.
How will decreasing the temperature affect the system?
aA + bB  cC + dD + Heat
K eq
c
d

C  D

a
b
A   B
Principle 3: The Effect of Temperature Changes on
Equilibrium for Endothermic Reactions
For an endothermic reaction, heat is a “reactant.”
Increasing the temperature is like adding heat.
According to Le Châtelier’s Principle, the equilibrium will
shift away from the added heat.
The concentrations of C and D will increase and the
concentrations of A and B will decrease.
The value of Keq will increase.
How will decreasing the temperature affect the system?
Heat + aA + bB  cC + dD
K eq
c
d

C  D

a
b
A   B
Practice - Predict the Effect on the Equilibrium
when the Temperature is Reduced
1.
2 CO2(g)  2 CO(g) + O2(g)
endothermic
2.
BaSO4(s)  Ba2+(aq) + SO42-(aq)
endothermic
3.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) exothermic
Sample Problem
Predicting the Effect of a Change in Temperature on
the Equilibrium Position
PROBLEM: How does an increase in temperature affect the concentration of
the underlined substance and Kc for the following reactions?
(a) CaO(s) + H2O(l)
Ca(OH)2(aq) DH0 = -82kJ
(b) CaCO3(s)
(c) SO2(g)
PLAN:
CaO(s) + CO2(g) DH0 = 178kJ
S(s) + O2(g) DH0 = 297kJ
Express the heat of reaction as a reactant or a product. Then consider
the increase in temperature and its effect on Kc.
Ca(OH)2(aq) heat
SOLUTION: (a) CaO(s) + H2O(l)
An increase in temperature will shift the reaction to the left, decrease
[Ca(OH)2], and decrease Kc.
(b) CaCO3(s) + heat
CaO(s) + CO2(g)
The reaction will shift right resulting in an increase in [CO2] and increase in Kc.
(c) SO2(g) + heat
S(s) + O2(g)
The reaction will shift right resulting in an decrease in [SO2] and increase in Kc.
ACID STRENGTH
Calculating equilibrium for weak
acids
The ICE table and the quadratic equation (if needed): to find equilibrium
concentration, we set up a table and use some algebra. Given 0.100 M
acetic acid, what is the H3O+ concentration at equilibrium? What is the pH?
Ka is 1.8 x 10-5.
Ka = [H3O+][CH3COO-]/[CH3COOH]
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
Initial:
0.100 M
--------0.0 M
0.0 M
Change:
-x
+x
+x
Equilibrium: 0.1 – x
x
x
1.8 x 10-5 = (x)(x)/(0.1-x) = x2/(0.1-x)
Solve for the value of x, which will also be [H3O+], by rearranging and
simplifying the equation. The pH will then be found.