Chemistry
Unit 10: Kinetics and Equilibrium
Chemistry
Learning Objectives Kinetics and Equilibrium
Essential knowledge and skills:




Interpret reaction rate diagrams.
Identify and explain the effect the following factors have on the rate of a chemical reaction: catalyst,
temperature, concentration, size of particles.
Distinguish between irreversible reactions and those at equilibrium.
Predict the shift in equilibrium when a system is subjected to a stress (Le Chatelier’s Principle) and
identify the factors that can cause a shift in equilibrium (temperature, pressure, and concentration.)
Essential understandings:



Kinetics is the study of reaction rates. Several factors affect reaction rates, including temperature,
concentration, surface area, and the presence of a catalyst.
Reaction rates/kinetics are affected by activation energy, catalysis, and the degree of randomness
(entropy). Catalysts decrease the amount of activation energy needed.
Reactions occurring in both forward and reverse directions are reversible. Reversible reactions can reach
a state of equilibrium, where the reaction rates of both the forward and reverse reactions are constant. Le
Chatelier’s Principle indicates the qualitative prediction of direction of change with temperature,
pressure, and concentration.
Some reactions take place instantly, but most are much slower and it is possible to measure how long these
reactions take to reach a certain stage. As a chemical reaction proceeds, the concentration of the reactants
decreases and the concentration of the products increases. The decrease in the concentration of reactants per
unit time, or the increase in the concentration of products per unit time, is known as the rate of reaction.
The study of rates of reaction is known as kinetics.
Collision Theory
Collision Theory is a theory used to account for the process by which a reaction proceeds (at the molecular
level).
In order for a chemical reaction to occur, effective collisions between reacting particles must take place. An
effective collision is one that actually gives product molecules. Thus not every collision between reactant
molecules actually results in a chemical change.
Two conditions determine the “effectiveness” of a chemical reaction. An effective collision results when:
a) a favorable geometry must occur between the colliding particles
b) a threshold or minimum energy must be met and/or overcome by the colliding particles. This energy is
known as the activation energy (Ea)
The basic postulate of collision theory is that the reaction rate is proportional to the number of effective
collisions per second among reactant molecules. Conditions that increase the frequency of effective collisions
should, therefore, increase the rate.
2
FACTORS AFFECTING THE RATE OF REACTION
The rate of a chemical reaction can be changed in a number of ways:
-
by changing the concentration of the reacting particles
by changing the pressure of the system (if some of the reacting particles are in the gas phase)
by changing the particle size (if some of the reacting particles are in the solid state)
by changing the temperature of the system
by adding a catalyst
SUMMARY
Effect:
On collision
frequency
On collision
energy
On activation
energy
On rate
No effect
On fraction of
successful
collisions
No effect
Decrease particle size
(solids only)
Increases
No effect
Increase concentration
(liquids and gases)
Increases
No effect
No effect
No effect
Increases
Increase pressure
(gases)
Increases
No effect
No effect
No effect
Increases
3
Increases
Increase temperature
Increases
Increases
No effect
Increases
Increases
Add a catalyst
No effect
No effect
Decreases
Increases
Increases
Reaction rates and graphs
Jack investigated how the rate of reaction between hydrochloric acid and marble chips varied with
temperature. He added 5 g of marble chips to 50 cm3 of hydrochloric acid at 20 °C, and timed how long it
took to collect 30 cm3 of carbon dioxide. He then repeated the experiment with different temperatures of
hydrochloric acid and completed a table of results.
Time for 30 cm3 of carbon
dioxide to be collected (s)
35
15
11
11
7
7
Temperature of
hydrochloric acid (°C)
20
30
40
50
60
70
4
Draw an appropriate graph using the data in the table, then answer the questions on the page.
a) Jack’s results contained one piece of anomalous data (an outlier). Which one is it?
b) How long would you expect it to take to collect 30 cm3 of carbon dioxide when the temperature of
hydrochloric acid is 25 °C? Indicate on your graph how you obtained this value.
c) Describe the relationship between temperature and rate of reaction of hydrochloric acid and marble
chips. What conclusions can you draw from Jack’s data?
Worksheet
1. In your own words, describe what effect cooling has on the frequency at which particles of reactants can
collide. Provide a real life example as to how we use temperature to alter reactions for our benefit
5
2. In your own words, describe why an increase in concentration can result in a change in the rate of a
reaction. Provide a real life example as to how we adjust concentration to adjust a reaction for our
benefit
3. Complete the following table by indicating whether each of the following scenarios would either
increase or decrease the rate of reaction. The first one has been done for you.
Scenario
Increase or Decrease
Adding heat.
Increase
Removing heat
Adding a catalyst
Diluting a solution
Removing an enzyme
Lowering the temperature
Decreasing the surface area
Increasing the concentration of a solution
Breaking a reactant down into smaller pieces
4. Complete the following table by indicating which factor would have the greatest impact on the rate of
reaction. Choose from concentration, temperature, surface area or catalyst.
Scenario
Factor that has the greatest
impact on the rate of reaction.
Blowing air on a campfire to help get it going.
Raw carrots are cut into thin slices for cooking.
Protein is broken down in the stomach by the enzyme
pepsin.
A Woolly Mammoth is found, perfectly preserved, near the
Arctic circle.
More bubbles appear when a concentrated solution of
hydrochloric acid is added to a magnesium strip than when a
dilute solution of the acid is added.
Exhaust from a car engine passes through a catalytic
converter changing most of the poisonous nitrogen oxides
into nitrogen gas and oxygen gas.
A dust explosion occurs in a saw mill.
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5. Use the following vocabulary terms to correctly fill in the blanks.
Vocabulary
catalyst
energy
catalytic converter
heat
collisions
rate of reaction
concentration
surface area
dilute
temperature
1. A freshly exposed surface of metallic sodium tarnishes almost instantly if exposed to air and moisture,
while iron will slowly turn to rust under the same conditions. In these two situations, the
__________________ refers to how quickly or slowly reactants turn into products.
2. Adding _________________________ will increase the rate of reaction because this causes the
particles of the reactants to move more quickly, resulting in more collisions and more
______________________.
3. Removing heat will lower the ____________________, causing the particles of the reactants to slow
down, resulting in less frequent collisions.
4. ___________________ refers to how much solute is dissolved in a solution. If there is a greater
concentration of reactant particles present, there is a greater chance that __________________ among
them will occur. More collisions mean a higher rate of reaction.
5. A concentrated acid solution will react more quickly than a _______________ acid solution because
there are more molecules present, increasing the chance of collisions.
6. Grains of sugar have a greater ______________________ than a solid cube of sugar of the same mass,
and therefore will dissolve quicker in water.
7. A ______________________, for example an enzyme, is used to speed up a chemical reaction but is not
used up in the reaction itself.
8. A ______________________ in car has metallic catalysts where several reactions occur. Carbon
monoxide, which was produced in the combustion of gasoline, is changed into carbon dioxide and water
in the presence of these metallic catalysts.
7
Entropy (S) – a measure of the disorder or randomness of the particles that make up a system
Law of Disorder – Spontaneous processes ALWAYS proceed in a way that the entropy of the universe increases
(increase disorder)
Entropy changes associated with state of matter:
Solids – molecules are tightly packed and cannot move (more order)
Liquids – molecules have some freedom to move (some order)
Gas – unrestricted movements (very little order)
Entropy increases as a substance changes from solidliquidgas
Ex: As water (l) vaporizes (g) – entropy increases
Ex: As a metal (s) melts to liquid (l) – entropy increases
1. 2KClO3(s) → 2KCl(s) + 2O2(g)
__________
2. H2O(l) → H2O(s)
__________
3. N2(g) + 3H2(g) → 2NH3(g)
__________
4. NaCl(s) → Na+(aq) + Cl-(aq)
__________
5. KCl(s) → KCl(l)
__________
6. CO2(s) → CO2(g)
__________
7. H+(aq) + C2H3O2-(aq) → HC2H3O2(l)
__________
8. C(s) + O2(g) → CO2(g)
__________
9. H2(g) + Cl2(g) → 2HCl(g)
__________
10. Ag+ + Cl-(aq) → AgCl(s)
__________
11. 2N2O5(g) →4NO2(g) + O2(g)
__________
12. 2Al(s) + 3I2(s) → 2AlI3(s)
__________
13. H+(aq) + OH-(aq) → H2O(l)
__________
14. 2NO(g) →N2(g) + O2(g)
__________
15. H2O(g) → H2O(l)
__________
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Exothermic reactions– Reactions that release/lose energy (examples: burning, a beaker getting warm)


the products have less energy than the initial reactants
The ∆ H rxn is ALWAYS negative
Ex: 4Fe + 3O2  2 Fe2O3 ∆H rxn = -1625 kJ
Endothermic reactions – Reactions that gain/absorb energy (ex: cooking, decomposition of water)


The products have more energy than the initial reactants.
The ∆ H rxn is ALWAYS positive for endothermic reactions!
Ex: NH4NO3 (s)  NH4+ (aq) + NO3-(aq) ∆H rxn = +27 kJ
9
Exothermic and Endothermic Energy Profile Worksheet
1.
2.
Answer the following questions based on the potential energy diagram shown here:
a.
Does the graph represent an endothermic or
exothermic reaction?
b.
Label the postion of the reactants, products,
and activated complex.
c.
Determine the heat of reaction, ΔH,
(enthalpy change) for this reaction.
d.
Determine the activation energy, Ea for this
reaction.
e.
How much energy is released or absorbed
during the reaction?
f.
How much energy is required for this reaction to occur?
Sketch a potential energy curve that is represented by the following values of ΔH and Ea. You may make up
appropriate values for the y-axis (potential energy).
ΔH = -100 kJ and Ea = 20 kJ
Is this an endothermic or exothermic reaction?
3.
In this unit we will be discussing reactions that are reversible, and can go in either the forward or reverse directions.
For example, hydrogen gas and oxygen gas react to form water, but water can also be broken down into hydrogen and
oxygen gas.
We typically write a reaction that can be reversed this way, using the double arrow symbol (
2 H 2 + O2
2H O
2
This reaction is exothermic in the forward direction:
2 H2 + O2  2 H2O + 285 kJ
but endothermic in the reverse direction:
2 H2O + 285 kJ  2 H2 + O2
10
or ↔ or
):
Consider a general reversible reaction such as:
A+B
C+D
Given the following potential energy diagram for this
reaction, determine ΔH and Ea for both the forward and
reverse directions. Is the forward reaction endothermic or
exothermic?
4.
Sketch a potential energy diagram for a general reaction A + B
Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ
11
C+D
Factors Influencing Reaction Rate - Activation Energy
1.
Answers
Answer the following questions based on the potential energy diagram shown here:
g.
Does the graph represent an endothermic or exothermic
reaction?
h.
Label the postion of the reactants, products, and
activated complex.
i.
Determine the heat of reaction, ΔH, (enthalpy change)
for this reaction.
j.
Determine the activation energy, Ea for this reaction.
k.
How much energy is released or absorbed during the reaction?
l.
How much energy is required for this reaction to occur?
Solution
a.
The graph represents an endothermic reaction
b.
the reactants, products, and activated complex are shown
c.
ΔH = +50 kJ.
Since this in an endothermic reaction, ΔH will have a positive
value.
ΔH is the difference in energy between the energy levels of the initial reactants (50 kJ) and the final products
(100 kJ), and does not depend on the actual pathway (remember Hess's Law?)
d.
Ea = +200 kJ.
Activation energy is the amount of energy required to go from the energy level of the reactants
(50 kJ) to the highest energy point on the graph, the activated complex (250 kJ).
e.
50 kJ of energy are absorbed during this endothermic reaction (this is the value of ΔH)
f.
200 kJ of energy are required for this reaction to occur (E a).
Even though ΔH is only 50 kJ, enough energy must be supplied to reach the activated complex, or the reaction
will not occur.
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2.
Sketch a potential energy curve that is represented by the following values of ΔH and Ea. You may make up
appropriate values for the y-axis (potential energy).
ΔH = -100 kJ and Ea = 20 kJ
Is this an endothermic or exothermic reaction?
Solution:
Since ΔH is a negative number, we know that the reaction is exothermic.
Therefore, begin by sketching an exothermic potential energy graph:
Next, assign values to the y-axis that will satisfy our requirements,
namely
ΔH = -20 kJ and Ea = +60kJ
Remember:
ΔH is the difference between energy levels of the reactants and
products - this difference will equal -20 kJ.
Ea is the difference between energy levels of the initial reactants and the peak of the curve, the activated complex.
3.
In the next unit we will be discussing reactions that are reversible, and can go in either the forward or reverse
directions. For example, hydrogen gas and oxygen gas react to form water, but water can also be broken down into
hydrogen and oxygen gas.
We typically write a reaction that can be reversed this way, using the double arrow symbol (
or ↔ or ):
2 H2 + O2  2 H2O
This reaction is exothermic in the forward direction:
2 H2 + O2  2 H2O + 285 kJ
but endothermic in the reverse direction:
2 H2O + 285 kJ  2 H2 + O2
13
Consider a general reversible reaction such as:
A+BC+D
Given the following potential energy diagram for this reaction, determine ΔH and Ea for both the forward and reverse
directions. Is the forward reaction endothermic or exothermic?
Solution:
You should immediately see that the forward reaction is exothermic - the products (C + D) are at a lower energy
level than the reactants.
4.
ΔHforward = -20 kJ
the difference in energy level between the reactants and products.
Ea forward = +60 kJ
amount of energy required to go from the initial reactants A + B to the activated
complex, the peak of the graph.
ΔHreverse = +20 kJ
difference in energy from C + D and A + B. Notice the value of ΔH does not change,
only the sign.
Ea reverse = +80 kJ
amount of energy required to go from C + D to the activated complex.
Sketch a potential energy diagram for a general reaction A + B  C + D
Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ
Solution:
Begin by determining whether the forward reaction (A + B →
C + D) is endothermic or exothermic:
Since ΔHreverse = -10 kJ you can determine that ΔHforward = +10
kJ (same value, just change the sign).
Since ΔHforward is positive you know that the forward direction
of the reaction is endothermic. Begin by sketching a general
endothermic potential energy curve that is endothermic. Don't
worry about drawing to scale at this time:
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At what energy level should you put the activated complex, the top of the peak? Since E a forward = +40 kJ, you
know that from the initial reactants (A + B) you need to go up 40 energy units. Pick any value for a starting energy
level (let's use a starting point of 10 here), then add 40 to find the peak of your graph (10 + 40 = 50).
Return to ΔH to determine the energy level of the products. Since ΔH forward = +10 kJ, add 10 to our starting value
to determine the energy level of the products:
Exothermic and Endothermic Profiles Worksheet 2
1. The graph below is a potential energy diagram for the hypothetical reaction:
A + B  C + D
Is the forward reaction endothermic or exothermic? Calculate the value of ΔH for this
reaction.
Is the reverse reaction endothermic or exothermic? Calculate the value of ΔH for this
reaction.
What is the value of the potential energy of the activated complex?
Calculate the activation energy for the forward reaction.
Draw a dashed line to show the effect of adding a catalyst to the system
15
2.
On the graph below, draw a potential
energy diagram for the following
reaction:
Q + R  S + T
given the following information:
the potential energy of Q + R is 150 kJ
the potential energy of S + T is 250 kJ
the potential energy of the activated complex is
375 kJ
Is the forward reaction endothermic
or exothermic? Calculate the value
of ΔH for this reaction.
What is the activation energy of the
reverse reaction?
Calculate the activation energy for
the forward reaction.
Draw a dashed line to show the
effect of adding a catalyst to the
system.
What is the total potential energy
content of the activated complex?
Equilibrium Notes
Equilibrium is established when the rate of the forward reaction and the rate of the reverse reaction are
constant.
2 H2O  H3O+ + OHThe double arrow , indicates that the reaction is reversible
Reversible reaction:
 Reaction that can be reversed so that products react to form reactants
Example: 2H2(g) + O2(g)  2H2O(g)

Forward: 2H2(g) + O2(g) 2H2O(g)
o
o Reverse: 2H2O(g)  2H2(g) + O2(g)
Complete reaction
 Reactions that go to completion
 Solid is formed and precipitates out of solution
 Gas is formed and escapes from solution
Dynamic Equilibrium
 Particles in constant motion
 Reactants collide to form products
16

At some point amount of products has increased and they collide to form reactants
Equilibrium Constant (Keq or K)
 Gives a measure of the percent yield of a reaction
 Used to determine
o If a reaction is at equilibrium
o The stability of the reactants compared to the products
o If reactants or products are favored
 If Keq > 1, products are favored
 If Keq < 1, reactants are favored
 If Keq = 1, than neither side is favored
o What will happen if concentrations are changed
 Symbol = Keq
 Keq is only good for one temperature. If the reaction is conducted at a different temperature, a
new Keq is needed.
How to write equilibrium expressions:

Given the following reaction aA + bB  cC + dD
[C ]c [ D]d
Keq 
[ A]a [ B]b

Given the following reaction: 2H2O(g)  2H2(g) + O2(g)
Keq 
[ H 2 ]2 [O2 ]
[ H 2O]2

Pure solids and liquids are not included in equilibrium expressions

Given the concentrations of reactants and products at equilibrium (these are NOT the initial
starting concentrations), you can put them into the equation above and calculate the same number
– this is what makes it a constant.
The set of concentrations at equilibrium are said to be the position of equilibrium

17
Example:
[H2]
[O2]
[H2O]
Keq



(Example 1)
Concentration of
substance after
equilibrium is reached
10.0 M,
5.0 M
2.0 M
(Example 2)
Concentration of
substance after
equilibrium is reached
20.0 M
1.25 M
2.0 M
102 5
Keq  2 =125
2
202 1.25
=125
Keq 
22
This example is showing
that even when the
concentrations at
equilibrium have
changed (because
starting amounts were
changed), the value of
Keq is the same
You can also use Keq to find out of a reaction is at equilibrium
Let’s say that you start a reaction by mixing H2 and I2 gas. You need to know when the reaction
has reached equilibrium so you periodically measure the concentration of H2, I2, and HI that are
present
After 1 minute the concentrations are as follows
[H2]
8.0 M,
[O2]
4.0 M
[H2O]
1.0 M
Keq
Keq=(82)(4)/12 = 256
o The reaction is not at equilibrium yet because the concentrations do not show an
equilibrium constant of 125.
o The reaction will proceed in a direction that increases the amount of reactants (reverse
direction)
Keq Expressions and Simple Calculations
Write the equilibrium expressions for the following reactions:
a) 2ICl(g) ↔ I2(g) + Cl2(g)
b) 3O2(g) ↔ 2O3(g)
c) 2Bi3+(aq) + 3H2S(g) ↔ Bi2S3(s) + 6H+(aq)
d) CaC2(s) + 2H2O(l) ↔ C2H2(g) + Ca(OH)2(s)
e) 2C6H6(l) + Br2(l) ↔ 2C6H5Br(l) + 2HBr(g)
f) 2H2(g) + O2(g) ↔ 2H2O(l)
g) SO2(g) + 1/2O2(g) ↔ SO3(g)
18
Consider the following equilibrium. Which one(s) will favor the products, which will favor the reactants?
a)
b)
c)
d)
2NO2(g)  N2O4(g)
Cu2+(aq) + 2Ag(s)  Cu(s) + 2Ag2+(aq)
Pb2+(aq) + 2Cl-(aq)  PbCl2(s)
SO2(g) + 1/2O2(g)  SO3(g)
Keq = 2.2
Keq = 1 x10-15
Keq = 6.3 x 104
Keq = 110
The Reaction Quotient (Q)
Consider the reaction
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
KC = 4.0
If 0.020 mole CO, 0.020 mole H2, 0.010 mole CH4 and 0.010 mol H2O are placed in a 1L reaction flask
which direction would the reaction go towards the reactants or the products
To determine this use the reaction quotient Q
The reaction quotient is the same as the equilibrium constant but the concentration values are not
necessarily those at equilibrium
For the reaction above, calculate Q
Compare Qc to Kc to determine which way the reaction will proceed in order to reach equilibrium
 If Qc > Kc the reaction will go towards the _____________
 If Qc = Kc the reaction is at equilibrium
 If Qc < Kc the reaction will go towards the _____________
Example
Consider the reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
Kc = 0.0500
A 50.0L reaction vessel contains 1.00 mol N2, 3.00 mol H2 and 0.500 mol NH3. Which way will the
reaction proceed in order to reach equilibrium?
19
Solve the following calculations, SHOW all your work
1. A reaction vessel with a capacity of 1.0 L, in which the following reaction:
SO2(g) + NO2(g) ↔ SO3(g) + NO(g)
Had reached a state of equilibrium, was found to contain 0.40 mol of SO3, 0.30 mol of NO, 0.15 mol of
NO2, and 0.20 mol of SO2. Calculate the Keq for this reaction.
2. Given the equilibrium: 2A(g) + B(g) ↔ C(g) + 3D(g) and the equilibrium concentrations ; [A]
= 2.0 M, [B] = 0.50 M, [C] = 0.25 M and [D] = 1.5 M. Find the Keq.
3. Consider the following equation: H2(g) + I2(g) ↔ 2HI(g). Calculate the Keq if at 300 K the
concentrations are [H2] = 0.40 M, [I2] = 0.45 M and [HI] = 0.30 M
4. The equilibrium constant for the formation of ammonia by the Haber process is 2.0 at a certain
temperature. If the equilibrium concentration of N2 in a mixture is 5.0 M and H2 is 2.0 M,
determine the concentration of ammonia.
20
5. The equilibrium constant for the formation of CO2 by the reaction
2CO(g) + O2(g) ↔ 2CO2(g)
is 1.0 x 104. If at equilibrium, the concentration of CO2 is 3.0 M and the concentration of O2 is
9.0 M, calculate the concentration of CO
6.
At 100°C, Keq = 0.135 for the reaction N2(g) + 3H2(g) ↔ 2NH3(g). In a reaction mixture at
equilibrium [NH3] = 0.030 M and [N2] = 0.50 M. What is the [H2]?
7. At a certain temperature the reaction CO(g) + 2H2(g) ↔ CH3OH(g) has a Keq = 0.500. If a
reaction mixture at equilibrium contains 0.180 M of CO and 0.200 M of H2, what is the
concentration of CH3OH?
Le Chatelier’s Principle


A reaction at equilibrium will proceed in a direction that relieves the stress put upon it.
The equilibrium position changes but the equilibrium constant does not change unless
temperature changes
Stress #1: Change of Concentration
1)
2 CO2(g) ↔ 2 CO(g) + O2(g)
by adding more CO2(g) you shift the
reaction towards the products
2)
2 CO2(g) ↔ 2 CO(g) + O2(g)
by removing CO(g) you shift the reaction
towards the products.
21
Key Point:
increase the amount of a reactant and you shift the equilibrium
forward (make more products)
increase the amount of a product and you shift the equilibrium backwards
(make more reactants)
decrease the amount of a product and you shift the equilibrium
forward (make more products)
decrease the amount of a reactant and you shift the equilibrium backwards
(make more reactants)
Explanation:
More reactants results in more collisions therefore producing more products until
a new equilibrium is established.
Similarly, reducing the amount of a product will decrease the number of reverse
reactions shifting the equilibrium forward towards the production of more
products
Equilibrium:
Examples:
Kc is not changed. (ratio of reactants and products remain the same)
1)
CaO(s) + H2O(l) ↔ Ca(OH)2(aq) + heat
a) What happens when CaO is added to the system?
b) What happens when CaO is added to the system?
2)
Co(H2O)62+(aq) + 2Cl-(aq) + heat ↔ Co(H2O)4Cl2(aq) + 2 H2O(l)
Pink
blue
a) What colour change occurs when Co(H2O)62+(aq) is added to the system?
b) What colour change occurs when Co(H2O)4Cl2(aq) is added to the system?
c) What colour change occurs when Cl-(aq) is added to the system?
22
Stress #2: Change of temperature (addition/removal of heat, treat heat as reactant/product)
Exothermic Reactions:
Explanation: Notice on the potential energy diagram that forward reactions only need achieve 60 kJ of
energy to overcome the activation energy “hump” while reverse reactions need 80 kJ of energy. At lower
temperatures, there is only enough energy for the A + B (forward)
particle collisions to have effective collisions. Very few C + D
(reverse) collisions occur. But at higher temperatures, the increase
in energy allows for more C + D (reverse) reactions to occur. As a
result there is a large increase in reverse effective collisions but only
a small increase in forward effective collisions. The net increase in
reverse collisions results in an equilibrium shift towards the reactant
side (and a lowering of K).
One way of remembering this is by looking at the energy term in a
thermochemical equation as a reactant or product. In an exothermic reaction, the energy term is on the
product. So, adding heat (energy) is just like adding a product. Adding a product causes reactants to be
formed (equilibrium shift to the left, K decreases).
Endothermic Reactions:
Endothermic reactions work the reverse of exothermic reactions. Less energy
is needed to overcome the reverse reaction. Hence, an increase in temperature
will increase the number of forward reactions compared to the number of
reverse reactions.
Increasing temperature is similar to increasing the number of reactants (more
products are formed).
Key Point:
for exothermic reactions:
increase the temperature and the equilibrium is shifted backwards (make more reactants). It is similar to
adding a product.
decrease the temperature and the equilibrium is shifted forwards (make more products). It is similar to
removing a product.
for endothermic reactions:
23
increase the temperature and the equilibrium is shifted forwards (make more products). It is similar to
adding a reactant.
decrease the temperature and the equilibrium is shifted backwards (make more reactants). It is similar to
removing a reactant.
Equilibrium:
for exothermic reactions: Kc decreases as temperature increases.
for endothermic reactions: Kc increases as temperature decreases.
Examples:
1)
CaO(s) + H2O(l) ↔ Ca(OH)2(aq) + heat
a) What happens when temperature is increased?
b) What happens when temperature is decreased?
2)
Co(H2O)62+(aq) + 2 Cl-(aq) + heat ↔ Co(H2O)4Cl2(aq) + 2 H2O(l)
Pink
a)
b)
blue
What colour change occurs when temperature is increased?
What colour change occurs when temperature is decreased?
Stress #3: Change in the volume of a gaseous equilibrium system:
Decrease the volume (compress the gases) for these reactions:
1)
N2(g) + 3H2(g) ↔ 2NH3(g)
equilibrium will shift towards the products as there are fewer "entities" on
equation.
2)
the right side of the
H2O(l) ↔ H2O(g)
equilibrium will shift towards the reactant as this side as H2O(l) is the lower volume entity (fewer gaseous
entities).
3)
H2(g) + I2(g) ↔ 2HI(g)
equilibrium will not shift because there are the same number of gas molecules on both sides of equation.
Key Point: a decrease in volume (compression) will shift the equilibrium towards
fewer gaseous entities.
the side with
An increase in volume (expansion) will shift the equilibrium towards the side with more gaseous entities
Explanation:
Decrease in volume, More entities on the reactant side:
24
When volume is decreased more reactions occur on the side with more entities. For example, if there are
more moles of reactants than products, then more collisions between reactants will occur and the
equilibrium will shift towards products. Eventually a new equilibrium will be created when the number
of reactants has decreased. Since equilibrium has shifted towards the products, K will increase.
N2(g) + 3 H2(g) ↔ 2 NH3(g)
For the system:
Note that as volume decreases, the reaction shifts towards the product side in order to achieve a fewer
number of particles.
For the system:
N2O4(g)
Colourless
↔
2NO2(g)
orange-brown
Increase in volume, more entities on the reactant side:
When volume is increased fewer reactions occur. But more importantly, fewer reactions occur on the
side with fewer entities. For example, if there are more moles of reactants than products, then the
decreased number of collisions between products will occur and the equilibrium will shift towards
products. Eventually a new equilibrium will be created when the number of reactants has decreased.
Since equilibrium has shifted towards the products, K will increase.
Equilibrium:
Kc is not changed
Stress #4: Change in the external pressure of a gaseous system:
Key Point: this is the same as changing the volume (stress #3)
Explanation: according to Boyle's Law, the pressure of a gas is inversely proportional to its volume.
As pressure increases, volume decreases. Hence a decrease in pressure will increase volume and therefore
shift equilibrium to the side of fewer gaseous entities.
Equilibrium:
Kc is not changed.
Stress #5: Add an inert gas to a gaseous equilibrium at a constant volume (increase internal pressure)
1)
Add Argon or Neon to N2(g) + 3 H2 (g) <==> 2 NH3 (g)
Key Point: Since collisions with inert gases are ineffective, there is neither an increase nor decrease in the
number of forward or reverse collisions.
25
Explanation: Though gas pressure increases, the probability of effective collisions for both reactants
and reactants is not increased.
Equilibrium:
Kc is not changed. (ratio of reactants and products remain the same)
Stress #6:
Addition of a catalyst to a system at equilibrium:
Key Point:
adding a catalyst will not affect the equilibrium
Explanation: adding a catalyst simply decreases the amount of time it takes for a reaction to occur – in
effect it changes the rate of reaction (forward or reverse), but does not change how far a reaction will
progress in a forward or reverse direction.
Equilibrium:
Kc is not changed. (ratio of reactants and products remain the same)
Summary of Responses of Equilibrium System to Changes
Change in Condition
Concentration
Increase
products
Decrease
products
Increase
reactants
Decrease
reactants
increase
Pressure
decrease
increase
Temperature
decrease
Catalyst
added
→
Direction of shift in
equilibrium position
Effect on K
In the reverse direction
No change in K
In the forward direction
No change in K
In the forward direction
In the reverse direction
In the direction with the
least number of moles of
gas
In the direction with the
greater number of moles of
gas
In the direction of the
endothermic reaction
In the direction of the
exothermic reaction
No change in the
equilibrium position.
Equilibrium reached more
quickly (i.e. reaction rate
changes).
26
No change in K
No change in K
No change in K
No change in K
K changes
K changes
No change in K
In the problems below, for each change given in the first column of the table, use Le Chatelier's principle
to predict
• the direction of shift of the equilibrium.
• the effect on the quantity in the third column.
Example: For the following reaction
5CO(g) + I2O5(s) ↔ I2(g) + 5CO2(g)
ΔH = -1175 kJ
For each change listed, predict the equilibrium shift and the effect on the indicated quantity.
Change
Direction
of Shift
( forward, reverse or no
change)
Effect on
Quantity
(a)
decrease in volume
(b)
raise temperature
amount of CO(g)
(c)
addition of I2O5(s)
amount of CO(g)
(d)
addition of CO2(g)
amount of I2O5(s)
(e)
removal of I2(g)
amount of CO2(g)
Effect
(increase, decrease,
or no change)
Kc
1. Consider the following equilibrium system in a closed container:
Ni(s) + 4CO(g) ↔ Ni(CO)4(g)
ΔH = -161 kJ
In which direction will the equilibrium shift in response to each change, and what will be the effect on the
indicated quantity?
Change
Direction
of Shift
( left, right or no change)
Effect on
Quantity
(a)
add Ni(s)
Ni(CO)4(g)
(b)
raise temperature
(c)
add CO(g)
(d)
remove Ni(CO)4(g)
CO(g)
(e)
decrease in volume
Ni(CO)4(g)
(f)
lower temperature
CO(g)
(g)
remove CO(g)
Kc
amount of Ni(s)
Kc
27
Effect
(increase, decrease,
or no change)
2. NaOH(s)  Na+(aq) + OH-(aq) + energy
(Remember that pure solids and liquids do not affect equilibrium values.)
Stess
Equilibrium
Shift
1.Add NaOH(s)
Amount of
NaOH(s)
[Na+]
[OH-]
Keq
____
2.Add NaCl (adds
Na+)
____
3.Add KOH (adds
OH-)
____
4.Add H+
(removes OH-)
____
5.Increase
temperature
6. Decrease
temperature
7. Increase
Pressure
8. Decrease
Pressure
Le Chatelier’s Principle Worksheet
For each of the following, indicate the direction the equilibrium would shift and what would happen to
the concentrations of each substance in equilibrium.
1. The following equilibrium may be established with carbon dioxide and steam.
CO(g) + H2O(g) ↔ CO2(g) + H2(g) + heat
What would be the effect of each of the following on the equilibrium and concentrations?
a.) The addition of more H2O?
b.) The removal of some H2?
c.) Raising the temperature?
28
d.) Increasing the pressure?
e.) Addition of a catalyst?
2. What would be the effect of each of the following on the equilibrium involving the synthesis of
methanol?
CO(g) + 2H2(g) ↔ CH3OH(g)
a.)
b.)
c.)
d.)
The removal of CH3OH?
An increase in pressure?
Lowering the concentration of H2?
The addition of a catalyst?
3. A small percentage of nitrogen gas and oxygen gas in the air combine at high temperatures found
in automobile engines to produce NO(g), which is an air pollutant.
N2(g) + O2(g) + heat ↔ 2NO(g)
a.) Higher engine temperatures are used to minimize carbon monoxide production. What
effect does higher engine temperatures have on the production of NO(g)? Why?
b.) What effect would high pressures have on the production of NO(g)? Why?
4. What would be the effect of each of the following on the equilibrium involving the reaction of
coke (C(s)) with steam to give CO(g) and H2(g)?
C(s) + H2O(g) ↔ CO(g) + H2(g)
a.)
b.)
c.)
d.)
The addition of steam?
An increase in pressure?
The removal of H2 as it is produced?
The addition of a catalyst?
5. The binding of oxygen to hemoglobin (abbreviated Hb), giving oxyhemoglobin (HbO2) is
partially regulated by the concentration of H+ and CO2 in the blood. Although the equilibrium is
rather complicated it can be summarized as follows:
HbO2 + H+ + CO2 ↔ CO2HbH+ + O2
According to Le Chatelier’s principle, what would be the effect of each of the following on the
equilibrium:
29
a.) The production of lactic acid (contains H+) and CO2 in a muscle during exercise?
b.) Inhaling fresh oxygen enriched air?
6. Here’s a biological example of an application of Le Chatelier’s principle:
Hemoglobin (Hb) reacts with oxygen to form HbO2, a substance that transfers oxygen to the tissues in the
body. Carbon monoxide (CO) also reacts with HbO2 by the process below.
HbO2(aq) + 4CO(aq)
↔ Hb(CO)4(aq) + O2(aq)
a) Use LeChatelier’s Principle to explain why inhaling CO can cause death.
b) Use LeChatelier’s Principle applied to this equation to suggest a treatment for a victim of CO
poisoning. Explain how your treatment affects the equilibrium above.
c)
7. 4 HCl (g)
+
2 O2 (g)
⇄
2 H2O (l)
(colourless)
+
2 Cl2 (g) +
(yellow)
Describe how the above equilibrium will shift after each stress below:
shift
Increase in temperature
Increase [HCl]
Decrease in [Cl2]
Decrease temperature
Add Ne at constant volume
30
colour change
98 kJ
8. Le Chatelier’s Principle
N2(g) + 3H2(g) ↔ 2NH3(g)
Le Chatelier’s Principle states that when a system at equilibrium is subjected to a stress, the system will
shift its equilibrium point in order to relieve the stress.
Directions: Complete the following chart by writing ← , → , or none for shift, and ↑, ↓, or remains the
same for the concentrations of reactants and products.
Stress
Equilibrium
shift
[N2]
[H2]
Add N2
Add H2
Add NH3
Remove N2
Remove H2
Remove NH3
Increase
Pressure
Decrease
pressure
31
[NH3]
K
9. N2O4(g) ⇄
2 NO2(g) ∆H = +59 kJ
Describe four ways of increasing the yield of for the reaction above.
Describe three ways to increase the rate of the above reaction.
10. The copper (II) ion and copper (II) hydroxide complex exist in equilibrium as follows:
Cu(OH)2 (aq) + 4 H2O (l)
violet
⇄
Cu(H2O)4 2+(aq) + 2 OH-(aq)
light blue
∆H = - 215 kJ
Describe how the above equilibrium will shift after each stress below:
shift
Increase in [Cu(H2O)4 2+]
Add NaOH
Increase in [Cu(OH)2]
Decrease in [Cu(H2O)4 2+]
Decrease in [Cu(OH)2]
Increase temperature
Decrease temperature
Add KCl (aq)
Add HCl (aq)
32
color change