Unit 1
Kinetics and Equilibrium
Chemistry 3202
6:47 AM
Part 1: Reaction Kinetics (Chp. 12)
Reaction Kinetics is the study of the
rate of a chemical reaction
 Qualitative:
Reactions may be described as being
FAST or SLOW

Fast – burning, explosions, precipitation
 Slow – rusting, fermentation

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Reaction Kinetics

POSSIBLE
UNITS ??
Quantitative:
The rate of a reaction measures how
fast products are formed or how fast
reactants are consumed
Rate = Change in quantity
Change in time
6:47 AM
Measuring Reaction Rate
The method used to determine
reaction rate will depend on the
reaction being studied. (p. 466)
Methods:
1. monitor pH if there is an acid or
base in the equation
2. record gas volume or changes in
pressure if there is a gas in the
reaction

6:47 AM
Measuring Reaction Rate

Methods: (cont’d)
3. record changes in mass if solids are
present
4. monitor absorption of light if there is a
color change
5. changes in electrical conductivity
indicate changes in ion concentration
6:47 AM
MC: What could we use to measure the rate
of this reaction?
Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)
a) pressure
b) pH
c)
d)
gas volume
mass
Answer: d) because a solid is present
6:47 AM
MC: What could we use to measure the
rate of this reaction?
SO3(g) + H2O(l)  H2SO4(aq)
a) pressure
c) gas volume
b) pH
d) mass
Answers:
a) and c) because a gas is present
b) because an acid is being produced
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6:47 AM
What determines RATE??
All chemical reactions are bond
breaking/bond forming events
 The rate of a reaction depends on how
quickly bonds are broken and how rapidly
new bonds form.
 KMT and Collision Theory are used to
explain reaction rates.

6:47 AM
Kinetic Molecular Theory (KMT)
Matter is made of particles (atoms, ions, or
molecules) in continuous motion
 An increase in temperature:
 increases the speed of particles
 reduces the forces of attraction between
particles

6:47 AM
Kinetic Molecular Theory (KMT)

KMT is supported by:
Diffusion – particles of a gas spread to fill
their container (‘perfume in a room’)
- solids dissolve uniformly in liquids over
time.
Pressure – a balloon remains inflated
because gas particles are continuously
hitting the sides of the balloon
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Energy Distribution of Particles
25 °C
# of particles
200 °C
Kinetic Energy
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Collision Theory
reactant particles must collide with
one another for a chemical reaction to
occur
 particles must collide with proper
orientation
 collisions must have enough intensity
to break old bonds and allow new
bonds to form

6:47 AM
Collision Theory

to increase reaction rate you must
increase the number of successful
collisions between reactant molecules
VIDEO (VHS): Reaction Rates
 LASERDISK: 3 VIDEO CLIPS

6:47 AM
Factors Affecting Reaction Rate
Concentration
– an increase in the concentration of a
reactant usually increases the rate of a
chemical reaction
- the rate increases because there are:
- more particles resulting in
- more collisions between particles &
- more successful collisions.
1.
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Factors Affectingmore
Reaction
Rate
successful
collisions
Temperature
- an increase in the temperature
increases the rate of a chemical reaction
- the higher temperature results in:
- more collisions between particles
faster
rate
- more intense collisions
NOTE: A temperature increase of 10 ºC
usually causes reaction rate to double.
2.
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Factors Affecting Reaction Rate
3.
Nature of Reactants
– compounds with fewer bonds to break
will react more rapidly than
compounds with many bonds
eg. propane (C3H8) burns faster than
candlewax (C25H52) because it has
fewer bonds
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Factors Affecting Reaction Rate
Nature of Reactants
- compounds with weak bonds react
more rapidly than compounds with
strong bonds
– ions will react more rapidly than atoms
and molecules
3.
6:47 AM
Factors Affecting Reaction Rate
Surface Area
- crushing a solid to produce a powder, or
changing a substance to the gas phase,
exposes more particles for collision
if more particles are available for
collision there will be:
- more collisions
- more successful collisions
4.
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faster rate
Factors Affecting Reaction Rate
Catalysts
- a catalyst increases the reaction rate by
providing a different reaction pathway or
mechanism with a lower activation
energy
- a catalyst IS NOT consumed by a
chemical reaction.
5.
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Ea with
catalyst
# of particles
Ea without
catalyst
Kinetic Energy
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Potential Energy Diagrams (p. 473)
PE diagrams show changes in potential
energy (stored chemical energy) during
chemical reactions
 Exothermic reactions release more energy
than they absorb (eg. burning)
 Endothermic reactions absorb more
energy than they release
(eg. photosynthesis)

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NEW SLIDE

ΔH written in the equation and outside the
equation (Thermochemical Equation)
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Potential Energy Diagrams
∆H represents the heat of reaction or
enthalpy of reaction
 ∆H is the difference between the PE of
the reactants and the PE of the
products
 the minimum energy needed for a
chemical reaction to occur is the
activation energy

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Potential Energy Diagrams

the activated complex for a reaction
is a temporary, unstable, intermediate
species that quickly decomposes to
products
eg.
H2 + I2 → H2I2 → 2 HI
activated
complex
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ENDOTHERMIC
site of the
activated complex
Products
Eareverse
PE
∆H
(positive)
Reactants
activation energy
(Ea forward)
Reaction Progress
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EXOTHERMIC
site of AC
Ea reverse
PE Reactants
∆H
(negative)
Ea forward
Products
Reaction Progress
6:47 AM
Formula: (OPTIONAL)
Eaforward - Eareverse = ΔH
This formula is NOT necessary if you prefer
using the PE diagram.
Animation
6:47 AM
Ea fwd
Ea rev
25
ΔH
Endothermic
or
Exothermic
-30
50
20
150
250
65
28
Sketch a PE diagram for each reaction
Photosynthesis
Earev
C6H12O6 + O2
PE
Eafwd
CO2 + H2O
Reaction Progress
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∆H
Respiration
Earev
Eafwd
PE
C6H12O6 + O2
∆H
CO2 + H2O
Reaction Progress
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∆H
p. 474
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∆H
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∆H
p. 475
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Effect of a catalyst
PE
catalyzed
no catalyst
Reaction Progress
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EXOTHERMIC
no catalyst
PE
Reaction Progress
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 Sample
problem: p. 475
 Questions:
p. 476; #’s 1 – 4
p. 484; #’s 1 – 4
p. 486; #’s 1,2, 4, 6, & 7
p. 468; # 4
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Reaction Mechanisms (pp. 477 – 485)
reaction mechanism – the steps that
occur in a chemical reaction
elementary reaction - each step in a
reaction mechanism
reaction intermediate – a molecule,
atom or ion formed in one step and
consumed in a later step
NOTE: reaction intermediates are NOT
included in the overall equation
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Reaction Mechanisms
eg. #1
Step #1
Step #2
NO(g) + O2(g)  NO3(g)
NO3(g) + NO(g)  2 NO2(g)
Overall Equation:
2 NO(g) + O2(g)  2 NO2(g)
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HBr + O2 → HOOBr
fast
HOOBr + HBr → 2 HOBr
slow
2 HOBr + 2 HBr → 2 H2O + 2 Br2 fast
p. 478 #’s 5 – 8
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Reaction Mechanisms
rate-determining step (RDS)
- the slowest step in a reaction mechanism
- to increase the rate of a reaction you must
speed up the RDS
- increasing the concentration of a reactant
will increase the rate ONLY IF the reactant
is in the RDS
6:47 AM
Reaction Mechanisms
PE diagrams
- every step in a reaction mechanism
has an activation energy which can be
drawn on a PE diagram
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Reaction Mechanisms
3-step mechanism
#2
PE
#1
RDS ??
#3

Reaction Progress
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Reaction Mechanisms
eg:
Step #1
H2CO2 + H+  H2CO2H+
fast
Step#2
H2CO2H+  HCO+ + H2O
slow
Step #3
HCO+  CO + H+
fast
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Reaction Mechanisms
eg:
Overall
H2CO2  H2O + CO
Omit H+ - catalyst
Omit H2CO2H+ & HCO+ - reaction intermediates
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Reaction Mechanisms
PE
H2CO2 + H+
Reaction Progress
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CO + H+
p. 476; #’s 1 – 4
p. 484; #’s 1 – 4
p. 486; #’s 1,2, 4, 6, & 7
p. 468; # 4
p. 478; #’s 5 - 8
 p. 484 #’s 5 – 9
 p. 485 #’s 10, 12
 p. 486 #’s 8, 10, 11
 p. 487 #’s 14, 17
 p. 829 #’s 128,129, 131, 132
6:47 AM
p. 829 # 128
Step 1
H2(g) + NO(g) → H2O(g) + N(g)
Step 2
Step 3
H2(g) + O(g) → H2O(g)
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
Part 2: Chemical Equilibrium
Equilibrium
A balancing Act!
Text Ch 13: p 488 - 541
Part 2: Chemical Equilibrium
All reactions we have done have shown
reactants being converted 100% to
products
 Many reactions are reversible with some
products being converted back to reactants

6:47 AM
Part 2: Chemical Equilibrium
Dynamic equilibrium occurs when 2
opposing processes occur at the same rate
 A chemical equilibrium occurs when two
opposing chemical reactions occur at equal
rates.
 Demo: p. 491

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Types of Equilibria
1.
Phase Equilibria
An equilibrium may be established
between different phases of a compound
in a sealed container
eg.
H2O(l) in a sealed container
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Types of Equilibria
Initially: H2O(l) changes to H2O(g)
H2O(l) → H2O(g)
Gradually:
H2O(g) changes to H2O(l)
H2O(l) ← H2O(g)
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Types of Equilibria
Using equilibrium notation:
H2O(l)
⇌
H2O(g)
Temperature changes?
Closed vs. open system?
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Types of Equilibria
2.


Solubility Equilibria
occur in saturated solutions
when NaCl(s) is placed in water, the
initial rate of dissolving is fast
NaCl(s)
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NaCl(aq)
Types of Equilibria

as more solid dissolves, the rate of
dissolving slows and recrystallization
begins.
eg. NaCl(s)
NaCl(aq)
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Types of Equilibria

when the solution is saturated there are
NO VISIBLE CHANGES

At equilibrium, the RATE of dissolving
and the RATE of recrystallization are
EQUAL.
equilibrium
eg. NaCl(s)
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⇌
NaCl(aq)
Types of Equilibria
 Temperature
change??
 Open vs. Closed ??
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Types of Equilibria
3.

Chemical Equilibrium
Chemical reactions that are reversible
usually result in chemical equilibrium
eg.
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NO2 gas changing to N2O4
Types of Equilibria

Initially the forward rate is high
eg. 2 NO2(g)
N2O4(g)

as more product forms, the reverse
reaction begins and increases in rate.
eg. 2 NO2(g)
N2O4(g)
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Types of Equilibria

eventually the forward rate slows and
the reverse rate increases such that
the FORWARD AND REVERSE
RATES ARE EQUAL
eg. 2 NO2(g)
⇌
N2O4(g)
http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm
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Conditions for Equilibrium (p. 492)
1.
2.
3.
Macroscopic properties are constant
ie. NO OBSERVABLE CHANGE
Forward and reverse rates must be equal
A CLOSED SYSTEM is required for
equilibrium
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Conditions for Equilibrium
4.
Equilibrium may occur from either
direction
eg. 2 NO2(g) ⇌ N2O4(g)
OR
N2O4(g) ⇌ 2 NO2(g)

p. 493;
#’s 1 – 6
Kinetics & Equilibrium #4
6:47 AM
Shifts in Equilibrium


Equilibrium occurs when the forward
rate equals the reverse rate.
Changes in concentration,
temperature and pressure/volume can
cause the forward or reverse reaction
rate to change.
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Shifts in Equilibrium
Eventually, a new equilibrium will be
established with different reactant and
product concentrations
 Le Châtelier’s Principle is used to
predict changes in concentrations
when a stress is applied to a system at
equilibrium.

6:47 AM
Le Châtelier’s Principle (p.520)
When a stress is applied to a system at
equilibrium, the system will adjust or shift
WHAT???
to relieve the stress.
A change(stress) is applied to a system
at equilibrium
 Forward or reverse rate will change
 New equilibrium established

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Le Châtelier’s Principle
Changes in concentration
An increase in concentration on one
side of an equation favors or drives the
reaction to the opposite side.
eg. What will happen if CO is added to
this system at equilibrium?
1.
CO(g) + 2 H2(g)
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⇌
CH3OH(g)
forward rate increases;
reverse rate catches up
Le Châtelier’s Principle
- the system will ‘shift’ to the right to use
the CO and produce more CH3OH
- some H2 will be used
[CH3OH] will increase
[H2] will decrease
[CO] ‘spikes’ and then drops to a value
higher than it was before the change
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?? possible graph ??
CO
mol/L
CH3OH
H2
time
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Le Châtelier’s Principle
An equilibrium shifts away from a
substance that increases in
concentration or toward a
substance that decreases in
concentration.
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Le Châtelier’s Principle
WHY???
- rates at equilibrium ???
- increasing the [CO] & forward rate ??
- reverse rate??
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Le Châtelier’s Principle
- [H2] ??
- new equilibrium concentrations ??
** NO CHANGE in Keq **
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Le Châtelier’s Principle
CO(g) + 2 H2(g)
⇌
What happens if we:
decrease [CO]
decrease [CH3OH]
increase [H2]
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CH3OH(g)
Le Châtelier’s Principle
IMPORTANT NOTE:
- adding a solid or liquid does not change
molar concentration
- changing the amount of a solid or liquid in
an equilibrium will NOT cause a shift
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Le Châtelier’s Principle
eg.
CaCO3(s) ⇌ CaO(s) + CO2(g)
What happens if we:
add CaCO3(s) ??
add CO2(g) ??
6:47 AM
Le Châtelier’s Principle
2.
(ΔH & equations)
Temperature
- Raising the temperature of an
exothermic equilibrium favors the
formation of reactants.
eg. CO(g) + 2 H2(g)
⇌
CH3OH(g) + 65 kJ
What happens if we increase
temperature in this equilibrium?
6:47 AM
Le Châtelier’s Principle
- Raising the temperature of an
endothermic equilibrium favors formation
of products.
eg. CaCO3(s) + heat ⇌ CaO(s) + CO2(g)
What happens if we increase temperature
in this equilibrium?
6:47 AM
Le Châtelier’s Principle
OR
- Raising the temperature shifts the
equilibrium away from the energy term.
- Decreasing the temperature shifts the
equilibrium toward the energy term.
CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
CaCO3(s) + heat ⇌ CaO(s) + CO2(g)
6:47 AM
Le Châtelier’s Principle
eg. How would an increase in temperature
affect these equilibria?
N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat
2 SO3(g) ⇌ 2 SO2(g) + O2(g) ΔH = +197 kJ
2 SO3(g) + 197 kJ ⇌ 2 SO2(g) + O2(g)
**A change in temperature changes Keq**
6:47 AM
Le Châtelier’s Principle
3.

Pressure/Volume
An increase in pressure of a system at
equilibrium has the same effect as a
decrease in the volume of the system.
(inverse relationship)
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Le Châtelier’s Principle
Increasing the pressure of a system at
equilibrium by reducing volume causes
the equilibrium to shift in the direction
that reduces pressure
ie. shift to the side with fewer molecules
of GAS!!

6:47 AM
Le Châtelier’s Principle
eg. How would an increase in pressure caused by a decrease in volume - affect
these equilibria?
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)
6:47 AM
Le Châtelier’s Principle
eg. Why would a change in pressure NOT
affect the following equilibria?
H2(g) + I2(g) ⇌ 2 HI(g)
2 Ag(s) + Zn2+(aq) ⇌ 2 Ag+(aq) + Zn(s)
6:47 AM
Le Châtelier’s Principle
pp. 529, 530 #’s 33 – 37
(omit 36 for now)
p. 533 #’s 1 – 3, 5
Answers on p. 537
Animation
Lab: Perturbing Equilibrium
6:47 AM
Le Châtelier’s Principle
4. Catalyst
- Does NOT cause a shift in equilibrium
- Increases BOTH rates equally so
equilibrium is reached faster
5. Surface Area - same as a catalyst
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Equilibrium Constant (Keq)
For any system at equilibrium, there is a
mathematical relationship between
reactant and product concentrations
(Guldberg and Waage, 1864)
P. 494
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Equilibrium Constant (K)
See p. 495
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eg. N2O4(g)
⇌
2 NO2(g)
Equilibrium Constant (K)
For the general equilibrium below:
aP + bQ ⇌ cR + dS
Keq =
c
[R]
d
[S]
a
b
[P] [Q]
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Equilibrium Constant (K)
-
-
-
a, b, c, & d are coefficients used to
balance the equation
P, Q, R, & S are the reactants and
products
Kc is sometimes used instead of Keq
when units are molar concentration
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Equilibrium Constant (K)
eg. Write the expression for Keq for:
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
2 HCl(g) ⇌
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H2(g) + Cl2(g)
Equilibrium Constant (K)
NOTE!!
Solids or liquids ARE NOT included in
the Keq or Kc expression because their
concentration is constant
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Equilibrium Constant (K)
eg. Write the expression for Kc for:
2 NaCl(s) + H2SO4(aq) ⇌ 2 HCl(g) + Na2SO4(aq)
CaCO3(s) ⇌
p. 497;
6:47 AM
CaO(s) + CO2(g)
#’s 1 - 5
Interpreting K
Large
vs
small
K values?
K much larger than 1 - products favoured
K much smaller than 1 - reactants favoured
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Changing K (p. 497)

For a given system at equilibrium, the
value of the equilibrium constant
depends only on temperature
ie. For any equilibrium, the only way
to change the actual value of K is to
change the temperature
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Calculations with K
Types of K calculations:
1. Given equilibrium concentrations, find K
2. Find a missing concentration given K and
other concentrations.
3. Given initial concentrations and
equilibrium data, find K (ICE tables)
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1.Given equilibrium concentrations, find K
eg. Calculate Kc for this equilibrium
using the equilibrium concentrations
given:
H2(g) + I2(g) ⇌ 2 HI(g)
[H2] = 0.22 mol/L
[I2] = 0.30 mol/L
[HI] = 1.56 mol/L
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Given the equilibrium concentrations;
[CO(g)] = 0.105 mol/L,
[H2(g)] = 0.250 mol/L
[CH3OH(g)] = 0.00261 mol/L,
What is the value of Keq for the equilibrium below?
CO(g) + 2 H2(g) ⇌ CH3OH(g)
(A) 0.0994
(B) 0.398
(C) 2.51
(D) 10.0
6:47 AM
Be Careful!!
The other answers are
possible if you mess up
the calculation
Given the equilibrium concentrations below,
what is the value of Keq for
N2(g) + O2(g) ⇌
2 NO(g) ?
[N2(g)] = 0.10 mol/L
[O2(g)] = 0.20 mol/L
[NO(g)] = 0.0030 mol/L
(A)
(B)
(C)
(D)
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2.2×10−4
4.5×10−4
1.5×10−1
3.0×10−1
Find Keq for the equilibrium below:
4 NH3(g) + 5 O2(g)
⇌
4 NO(g) + 6 H2O(g)
[NH3(g) ] = 0.100 mol/L
[O2(g) ] = 0.200 mol/L
[ NO(g) ] = 0.300 mol/L
[ H2O(g) ] = 0.250 mol/L
6:47 AM
2. Find a missing concentration . . .
eg. Find the [HI] in the equilibrium
below if Kc = 36.9, [H2] = 0.125 mol/L
and [I2] = 2.56 mol/L.
H2(g) + I2(g) ⇌ 2 HI(g)
p. 499; #’s 6 – 9
p. 538; # 6
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Which reaction is
faster - forward
or reverse?
3. ICE tables


Keq may be calculated given initial
concentrations and at least one
equilibrium concentration
Using the Initial concentration and the
Change in concentration we can find
the missing Equilibrium concentrations
and calculate Keq
6:47 AM
ICE tables
eg. 4.30 mol of NH3 was placed in a
1.00 L closed container to establish
this equilibrium:
2 NH3(g)
⇌
N2(g) + 3 H2(g)
Calculate Kc if the equilibrium
concentration of H2(g) = 0.500 mol/L
6:47 AM
ICE tables
2 [NH3]
I
C
E
4.30 mol/L
2x
-2x
4.30 - 2x
[N2]
3 [H2]
0
0
x
+x
3x
+3x
x
3x
0.500 mol/L
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[H2] = 0.500 mol/L
3x = 0.500
x = 0.167
[N2] = x
= 0.167 mol/L
[NH3] = 4.30 - 2x
= 3.97 mol/L
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K = (0.500)3 x (0.167)
(3.97)2
= 0.00132
ICE tables
eg. 2.00 mol of N2, 4.00 mol of H2 and
3.00 mol of NH3 were allowed to come
to equilibrium in a 1.00 L container
2 NH3(g)
⇌
N2(g) + 3 H2(g)
Calculate Kc if the equilibrium
concentration of NH3(g) = 3.50 mol/L
6:47 AM
ICE tables
I
2 [NH3]
[N2]
3 [H2]
3.00
2.00
4.00
2x
+2x
-x x
3x
-3x
C
E
3.00 + 2x
= 3.50
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2.00 - x
4.00 - 3x
[NH3] = 3.50 mol/L
3.00 + 2x = 3.50
x = 0.25
[H2] = 4.00 – 3x
= 3.25
[N2] = 2.00 - x
= 1.75 mol/L
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K = (3.25)3 x (1.75)
(3.50)2
= 4.90
ICE tables
eg. The oxidation of ammonia occurs
according to the following expression:
4 NH3(g) + 5 O2(g)
⇌
4 NO(g) + 6 H2O(g)
0.800 mol of each chemical were placed in
a 1.00 L container and there was 0.450
mol of NH3 at equilibrium. Calculate the
equilibrium concentrations and Kc
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4 NH3(g)
H2O(g)
6:47 AM
5 O2(g)
4 NO(g)
6
[NH3] = 0.450 mol/L
0.800 – 4x = 0.450
x = 0.0875
[O2] = 0.3625 mol/L
[NO] = 1.15 mol/L
[H2O] = 1.325 mol/L
6:47 AM
K = 37000
ICE tables
eg. 2.30 mol of NH3 was placed in a 2.00 L
closed container to establish this
equilibrium:
2 NH3(g)
⇌
N2(g) + 3 H2(g)
Calculate Kc if 25 % of the NH3(g) reacts.
6:47 AM
[NH3] = 2.30 mol/2.00 L = 1.15 mol/L
2 [NH3]
[N2]
3 [H2]
I
1.15
0
0
C
2x
-2x
E
1.15 - 2x
+xx
3x
+3x
x
= (0.75 x 1.15) = 0.8625 mol/L
6:47 AM
3x
[NH3] = 0.8625
1.15 - 2x = 0.8625
x = 0.1438
[N2] = x
= 0.1438 mol/L
[H2] = 3x
= 0.4314 mol/L
6:47 AM
K = 0.0155
ICE tables
eg. 0.800 mol of NH3 was placed in a
1.00 L closed container to establish
this equilibrium:
2 NH3(g)
⇌
N2(g) + 3 H2(g)
Calculate Kc if 70% of the NH3 reacts.
6:47 AM
Le Châtelier’s Principle
Chapter Review
p. 535
p. 536
p. 538
p. 539
6:47 AM
#’s 2, 3, 5-7
#’s 13, 15
#’s 7, 8, & 10 (MC)
#’s 15, 17, 19, 20, 22
ICE tables
Assignment
test
6:47 AM
Energy
PRODUCED
ΔH
& isequations

EXOTHERMIC
The energy term may be
included in a
chemical equation, or written as ΔH to the
right of the equation.
eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
OR
CO(g) + 2 H2(g) ⇌ CH3OH(g)
6:47 AM
ΔH = - 65 kJ
ENDOTHERMIC
Energy
is
REQUIRED
ΔH & equations
eg. N2(g) + O2(g) + 90 kJ ⇌ 2 NO(g)
OR
N2(g) + O2(g) ⇌
2 NO(g)
ΔH = + 90 kJ
Back to Temperature
6:47 AM
ΔH & equations

The energy term may be included in a
chemical equation, or written as ΔH to the
right of the equation.
eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ
OR
CO(g) + 2 H2(g) ⇌ CH3OH(g)
6:47 AM
ΔH = - 65 kJ