Chemical Systems & Equilibrium
SCH4U - Unit 4
Equilibrium

Definitions
• Equilibrium is a process in which two opposing
processes occur at the same time and at the same
rate such that there is no net change.
• The phenomena of equilibrium occurs in chemical
systems
• Such systems are said to be reversible, which
means that a process occurs in one direction but
the reverse process can also occur at the same
time and at the same rate.
At Equilibrium:
Closed system – no matter/energy/pressure
changes
 No macroscopic changes
 Reactants and products both present (and usually
in different amounts)
 [reactant] = constant, [product] = constant
 Can be approached from both sides

Rate of forward reaction = rate of reverse reactions
Dynamic Equilibrium

dynamic equilibrium = a balance between
forward and reverse processes occurring at
the same rate & this is denoted by a double
arrow
** Dynamic
equilibrium will not
occur if any of the chemicals,
reactants, or products escape or
are removed from the container.
Party Analogy


30 people at a house party
8pm:



10pm:



16 people in the kitchen
14 people in the living room
16 people in the kitchen
14 people in the living room
Different people but same number in each room
Dynamic Equilibrium
Example: Closed bottle of pop
 CO2 gas leaving dissolved state and entering gas
state
 CO2 gas ALSO, leaving gas state and entering
liquid state
 No visible change

 CO2(g)
CO2(aq)
Equilibrium Double Arrow

equilibrium is symbolized with an equation
containing a forward (→) and a reverse (←)
arrow combined into:
N2O4 (g)
2NO2 (g)
Equilibrium Double Arrow

forward reaction = in an equilibrium
equation, the left-to-right reaction

reverse reaction = in an equilibrium
equation, the right-to-left reaction
Forward
CO2(g)
CO2(aq)
Reverse
Drinking Bird Equilibrium

https://www.youtube.com/watch?v=Bzw0kWvfVkA

At rest the vapor and the liquid inside the tube are in an equilibrium

Wet head of bird with water – as the water evaporates from around the head, it takes energy
with it, head cools down = vapor inside the head cools and contracts = vacuum = pulls liquid
up
3 Types of Equilibrium
1. Solubility Equilibrium (dissolving process)
2. Phase Equilibrium (change of state)
3. Chemical Reaction Equilibrium (reactants ⇆ products)
Types of Equilibrium #1

solubility equilibrium = a dynamic
equilibrium between a solute and a solvent in a
saturated solution in a closed system
I2(s)
I2(aq)
Solubility Equilibrium

Saturated solution = a solution containing
the maximum quantity of a solute

Beyond the solubility limit, any added solute
will remain solid and not dissolve
Solubility Equilibrium
kinetic molecular theory states that particles
are always moving and colliding
 even if no changes are observed


Dissolution = the process of dissolving
(a) When the solute is first added, many more ions
dissociate from the crystal than crystallize onto it.
(b) As more ions come into solution, more ions also
crystallize.
(c) At solubility equilibrium, solute ions dissolve and
crystallize at the same rate.
Digesting a Precipitate



Allow precipitates to sit for long periods of time before
filtering
The longer you wait the more pure the crystal, also the
larger the crystal
If precipitate forms quickly, impurities maybe trapped in
the precipitate
Types of Equilibrium #2
phase equilibrium = a dynamic equilibrium
between different physical states of a pure
substance in a closed system
 closed system = a system that may exchange
energy but NOT matter with it’s surroundings

H2O(l)
H2O(g)
H2O(s)
H2O(l)
Phase Equilibrium
Types of Equilibrium #3

chemical reaction equilibrium a dynamic
equilibrium between reactants and products of
a chemical reaction in a closed system

reversible reaction = a reaction that can
achieve equilibrium in the forward or reverse
direction
Equilibrium is reached when the rate of the
forward reaction equals the rate of the
reverse reaction.
H2(g) + I2(g)
2HI(g)
(H = -13kJ/mol)
The H value always refers to the forward reaction.
13 kJ of energy is liberated for every mole of HI
formed.
For the whole reaction:
H2(g) + I2(g)
2HI(g)
(H = -26kJ)
Chemical Reaction Equilibrium In a Closed
System
 N2O4(g)
2 NO2(g)
Reaction Rate
H2 + I2
2 HI
H2 + I 2
2 HI
H2 + I 2
Time
2 HI
Reversible Reactions

The same dynamic equilibrium composition is
reached whether we start from pure N2O4(g),
pure NO2(g), or a mixture of the two, provided
that environment, system and total mass
remain the same.
Calculating the Equilibrium Constant



The equilbrium constant, Keq, is the ratio of
equilibrium concentrations at a particular temp
Kc for solution-phase systems or Kp for gasphase systems
Keq = [C]c[D]d for the eqn
[A]a[B]b
aA+bB
cC+dD
Note: The equilibrium constant depends ONLY on the
concentration of gases (not liquids/solids)
Questions: Equilibrium Law Expression
1. Write the equilibrium law expression for the following:
[N2O4(g)]
a) 2NO2(g) ↔ N2O4(g)
K=
[NO2(g) ]2
b) 2HI(g) ↔ H2(g) + I2(g)
K=
[H2(g)] [I2(g) ]
[HI(g) ]2
2. A reaction vessel contains NH3, N2 and H2 gas at equilibrium at a certain
temperature. The equilibrium concentrations are [NH3] = 0.25mol/L, [N2] =
0.11mol/L and [H2] = 1.91 mol/L. Calculate the equilibrium constant for the
decomposition of ammonia.
2NH3(g) ↔ N2(g) + 3H2(g)
K=
[N2(g)] [H2(g) ]3
[NH3(g) ]2
[0.11] [1.91 ]3
K=
[0.25 ]2
K = 12.3
Questions: Equilibrium Law Expression
3. Nitryl chloride gas, NO2Cl, is in equilibrium at a certain
temperature in a closed container with NO2 and Cl2 gases. At
equilibrium, [NO2Cl] = 0.00106mol/L and [NO2] = 0.0108mol/L.
If K = 0.558, what is the equilibrium concentration of Cl2?
4. Write a balanced equation for the reaction with the following
equilibrium law expression:
[NO2(g)]2
K=
[NO (g) ]2 [O2 (g) ]
Heterogeneous Equilibria

homogeneous equilibria = equilibria in
which all entities are in the same phase


Reactants and products are all gas or all aqueous
heterogeneous equilibria = equilibria in
which reactants and products are in more than
one phase

Reactants and products are in different phases
Homogenous equilibrium applies to reactions in
which all reacting species are in the same
phase.
N2O4 (g)
[NO2]2
Kc =
[N2O4]
2NO2 (g)
Kp =
2
PNO
2
PN2O4
Heterogenous equilibrium applies to
reactions in which reactants and products
are in different phases.
CaCO3 (s)
[CaO(s) ][CO2(g)]
Kc =
[CaCO3(s)]
Kc
[CaO(s)]
[CaCO3(s)]
= [CO2(g)]
CaO (s) + CO2 (g)
[CaCO3(s)] = constant
[CaO(s) ] = constant
Kc = [CO2(g)]
The concentration of solids and pure liquids are
considered to be constant and are not included in the
expression for the equilibrium constant.
CaCO3 (s)
CaO (s) + CO2 (g)
PCO2 = Kp
PCO does not depend on the amount of CaCO3 or CaO
2
N2O4 (g)
2NO2 (g)
equilibrium
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2
& N2O4
Equilibrium favors the reactant side
CHECKPOINT
The reaction at 200C between ethanol and ethanoic acid
produces ___________________ and
__________________.
1. Write the equation for this reaction
2. Determine the equilibrium constant expression for the
reaction
Calculating Equilibrium Concentrations
(when given one concentration)
Sample Problem:
When ammonia is heated it decomposes:
2NH3(g)↔ N2(g) + 3H2(g)
When 4.0 mol of ammonia is introduced in a 2.0L container and heated. The
equilibrium amount of ammonia is 2 0 mol. Determine the equilibrium concentrations of
the other two entities.
STEP 1: Determine the concentration (initial and equilibrium) for known values
STEP 2: Setup an ICE Table
STEP 3: Determine the value of X
STEP 4: Use x value to determine the other quantities
Determine the concentrations
[NH3]initial = 4.0mol/2.0L = 2.0mol/L
[NH3]equilibrium = 2.0mol/2.0L = 1.0mol/L
Setup ICE Table
Determine the value of X
[NH3](g)equil = 2 0mol / L - 2x
[NH3](g)equil = 1.0mol/L (from calculations in Step 1)
2.0mol/L – 2x = 1.0mol/L
-2x = - 1.0mol/L
x = 0.5mol/L
Use X to determine other quantities
constant
Reversible Reactions
For a given overall system composition, the
same equilibrium concentrations are reached
whether equilibrium is approached in the
forward or the reverse direction
 What about Keq will it be the same in fwd/rev?

Equilibrium
Tubes
Heat + N2O4 (g)
2NO2 (g)
Brown
Colourless
ENDOTHERMIC
Rxn
The effects of temperature on equilibrium
Very Cold
Cold
Hot
NO2 is one of the chemicals in smog!

In the summer on hot, windless days an orange haze is
seen over the horizon, this is NO2
N2O4 (g)
Colourless

2NO2 (g)
Brown
In the winter, the smog doesn't go away, it is just less
noticeable. The cooler temperatures lead to more N2O4
and less NO2 which we can't see as well!
Qualitative Changes in Equilibrium Systems
You should be familiar with your own body’s attempt at
maintaining equilibrium or “homeostasis”:




If body T too high  sweat, surface blood vessels dilate
If body T too low  shiver, surface blood vessels
constrict
If blood CO2 levels ↑  breathe deeper & faster
If blood sugar levels ↑  insulin released to remove
excess glucose
Le Châtelier’s Principle

When a chemical system at equilibrium is
disturbed by a change in a property, the
system adjusts in a way that opposes the
change.

In other words: If an external stress is applied
to a system at equilibrium, the system adjusts
in such a way that the stress is partially offset
as the system reaches a new equilibrium
position.
Le Châtelier’s Principle
Le Chatelier’s Principle: if you disturb an
equilibrium, it will shift to undo the
disturbance.
equilibrium shift = movement of a system at
equilibrium, resulting in a change in the
concentrations of reactants and products

https://www.youtube.com/watch?v=dIDgPFEucFM
Le Châtelier’s Principle
1.
2.
System starts at equilibrium.
A change/stress is then made to
system at equilibrium.



3.
Change in concentration
Change in temperature
Change in volume/pressure
System responds by shifting to
reactant or product side to restore
equilibrium.
Le Châtelier’s Principle
Change in Reactant or Product
Concentrations
•
•
•
Adding a reactant or product shifts the
equilibrium away from the increase.
Removing a reactant or product shifts the
equilibrium towards the decrease.
To optimize the amount of product at equilibrium,
we need to flood the reaction vessel with reactant
and continuously remove product.
Le Châtelier’s Principle
Change in Reactant or Product
Concentrations
N2 (g) +
•
•
•
•
3H2 (g) ↔
2NH3 (g)
If H2 is added while the system is at equilibrium, the
system must respond to counteract the added H2
That is, the system must consume the H2 and produce
products until a new equilibrium is established.
Equilibrium shifts to the right.
Therefore, [H2] and [N2] will decrease and [NH3]
increases.
Change in Reactant or Product
Concentrations
N2 (g) + 3H2 (g)
Equilibrium
shifts left to
offset
stress
2NH3 (g)
Add
NH3
Change in Reactant or Product
Concentrations
aA + bB
Change
cC + dD
Shifts the Equilibrium
Increase concentration of product(s)
Decrease concentration of product(s)
Increase concentration of reactant(s)
Decrease concentration of reactant(s)
left
right
right
left
Le Châtelier’s Principle
Effect of Temperature Changes
•
•
•
The equilibrium constant is temperature
dependent.
For an endothermic reaction, H > 0 and heat
can be considered as a reactant.
For an exothermic reaction, H < 0 and heat can
be considered as a product.
Effect of Temperature Changes
Effect of Temperature Changes
Adding heat (i.e. heating the vessel) favors away from the
increase:
– if H = + (Endothermic), adding heat favors the
forward reaction,
– if H = - (Exothermic), adding heat favors the
reverse reaction.
Removing heat (i.e. cooling the vessel), favors towards the
decrease:
– if H = + (Endothermic), cooling favors the reverse
reaction,
– if H = - , (Exoothermic), cooling favors the forward
reaction.
Gas Law – Boyle’s Law
Relationship: Pressure & Volume

As pressure on a gas increases, the volume of
the gas decreases
Le Châtelier’s Principle
Effects of Volume and Pressure
•
•
•
•
•

As volume is decreased pressure increases.
The system shifts to decrease pressure.
An increase in pressure favors the direction that
has fewer moles of gas.
Decreasing the number of molecules in a
container reduces the pressure.
In a reaction with the same number of product
and reactant moles of gas, pressure has no effect.
Only a factor with gases.
Effects of Volume and Pressure
A (g) + B (g)
C (g)
Change
Shifts the Equilibrium
Increase pressure
Decrease pressure
Increase volume
Decrease volume
Side with fewest moles of gas
Side with most moles of gas
Side with most moles of gas
Side with fewest moles of gas
Le Châtelier’s Principle
Adding a Catalyst
•
•
does not shift the position of an equilibrium
system
system will reach equilibrium sooner
Le Châtelier’s Principle
Adding a Catalyst

lowers the activation energy for both forward
and reverse reactions by an equal amount, so
the equilibrium establishes much more rapidly
but at the same position as it would without
the catalyst
Adding a Catalyst
Le Châtelier’s Principle
Adding Inert Gases
pressure of a gaseous system at equilibrium
can be changed by adding a gas while keeping
the volume constant
 If the gas is inert in the system, for example, if
it is a noble gas or if it cannot react with the
entities in the system, the equilibrium position
of the system will not change

Le Châtelier’s Principle Summary
No Sweat!






Chickens cannot perspire.
When a chicken gets hot, it pants like a dog.
Farmers have known for a long time that chickens
lay eggs with thin shells in hot weather.
These fragile eggs are easily damaged.
Eggshell is primarily composed of calcium
carbonate, CaCO3(s).
The source of the carbonate portion of this
chalky material is carbon dioxide, CO2, produced
as a waste product of cellular respiration.
No Sweat!

The carbon dioxide dissolves in body fluids
forming the following equilibrium system:
No Sweat!



When chickens pant, blood carbon dioxide
concentrations are reduced, causing a shift
through all four equilibria to the left and a
reduction in the amount of calcium
carbonate available for making eggshells.
Solution: Give the chickens carbonated
water to drink in the summer.
This shifts the equilibria to the right,
compensating for the leftward shift caused
by panting.
Le Chatelier’s Principle: Warm-up

Page 459 # 4 & 6
AgNO3 – Hint: check solubility table
Graph for Question #4
Le Chatelier’s Principle: Warm-up

Page 459
A = ↑ volume
B = ↑ temperature
C = ↑ [C2H6]
D = catalyst/inert gas
E = ↓ [C2H4]
The Equilibrium Law Expression &
The Equilibrium Constant, K

At constant temperature, regardless of initial
concentrations the concentrations of reactants and
products always give a constant value K
aA + bB
[C]c[D]d
K=
[A]a[B]b
cC + dD
Products
Reactants
Equilibrium Law Expression



The molar concentrations of the products are
always multiplied by one another and written in
the numerator, and the molar concentrations of
the reactants are always multiplied by one
another and written in the denominator.
The coefficients in the balanced chemical equation
are equal to the exponents of the equilibrium law
expression.
The concentrations in the equilibrium law
expression are the molar concentrations of the
entities at equilibrium.
Recall: Equilibrium achieved from any
combination of reactants and products
Regardless of initial concentrations, at a given temperature, the
relationship of the equilibrium concentrations of reactants and
products always yields a CONSTANT value, K
N2O4 (g)
[NO2]2
K=
[N2O4]
2NO2 (g)
= 4.6 x 10-3
constant
Ways Different States of Matter Can
Appear in the Equilibrium Constant, K
Molarity
Partial Pressure
gas, (g)
YES
YES
aqueous, (aq)
YES
---
liquid, (l)
---
---
solid, (s)
---
---
Questions: Equilibrium Law Expressions
Write the equilibrium law expressions for the following
reactions:
1. NH4Cl(s) ↔ NH3(g) + HCl(g)
K = [NH3 (g) ] [HCl(g)]
2.
2H2O(l) ↔ 2H2(g) + O2 (g)
3.
2NaHCO2(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)
K = [H2 (g) ]2 [O2(g) ]
K = [H2O (g) ] [CO2(g) ]
Equilibrium Law Expression

Note: equilibrium constants give no
information about the rate of a reaction; they
provide only a measure of the equilibrium
position of the reaction

K is independent of the initial concentration of
the reactants and products, but on the
concentrations at the equilibrium
What Does the Value of K Mean?

If K >> 1, the reaction is product-favoured;
product predominates at equilibrium.

If K << 1, the reaction is reactants-favoured;
reactants predominates at equilibrium.
Products
Reactants
Question: Magnitude of K
Consider the reaction: H2(g) + I2(g) ↔ 2HI(g) + heat
At 448⁰C, K=50.5. Would you predict the value of K to be
higher or lower at 300⁰C?



At 448⁰C K >> 1 = PRODUCTS favoured
Heat lowered = rxn shifts to PRODUCT side
At 300 ⁰C K > 50.5
Equilibrium Reactions in Solution

In addition to gas-phase and heterogeneous reactions,
equilibrium reacts can also take place in solution.

It is important to write the reaction components as they
ACTUALLY EXIST IN SOLUTION -- represent ions in
solution as individual entities

Get the equilibrium law expression from the net ionic
equation
Equilibrium Reactions in Solution
Example: Write the equilibrium reaction and equilibrium law
expression for the reaction between zinc metal and copper
(II) chloride solution.
Cancel out the spectator ions
The Reaction Quotient, Q

If a chemical system begins with reactants only, it is
obvious that the reaction will initially proceed to
the right, toward products.

If, however, reactants and products are both
present, the direction in which the reaction
proceeds is usually less obvious.

In such a case, we can substitute the
concentrations into the equilibrium law expression
to produce a trial value that is called a reaction
quotient, Q
The Reaction Quotient, Q

reaction quotient, Q = a test calculation
using measured concentration values of a
system in the equilibrium expression
think of Q as being similar to K
 K is calculated using concentrations at
equilibrium
 Q may or may not be at equilibrium

The Reaction Quotient, Q
aA + bB
Q=
cC + dD
[C]c[D]d
[A]a[B]b
The Reaction Quotient, Q

Q is equal to K, and the system is at equilibrium.

Q is greater than K, and the system must shift left
(toward reactants) to reach equilibrium, because the
product-to-reactant ratio is too high.

Q is less than K, and the system must shift right
(toward products) to reach equilibrium, because the
product-to-reactant ratio is too low.
ICE Table

Initial, Change, Equilibrium
I = initial concentration of reactants and
products before reaction
 C = change in the concentrations of reactants
and productsthe start and the point at which
equilibrium is achieved
 E = concentrations of reactants and products
at equilibrium.

Solving Equilibrium Problems with ICE
Example1:
For the above reaction [N2]i = 0.32mol/L and [H2]i =
0.66mol/L. At a certain T and P, [N2]eq = 0.20mol/L. What is
the value of K under these conditions?
Example 2
At 150°C, K for the reaction I2(g) + Br2(g) ↔ 2IBr(g) is found
to be 1.20x102 . Starting with 4.00mol of each of iodine and
bromine in a 2.00L flask, calculate the equilibrium
concentrations of all reaction components.
Example 3



Unlike the previous two examples, it is not always
obvious if a system is already at equilibrium, or which way
the reaction will shift to reach equilibrium.
In these situations, it is helpful to determine the Reaction
Quotient, Q
When the reaction 2HI(g) ↔ H2(g) + I2(g) takes place at
445°C, the value of K is 0.020. If [HI]=0.20mol/L,
[H2]=0.15mol/L and [I2]=0.09mol/L, is the system at
equilibrium? If not, in which direction will it shift to reach
equilibrium?
Example 4

For the reaction H2(g) + F2(g) ↔ 2HF(g) , K is 1.5x102 at
SATP. Calculate all equilibrium concentrations if 4.00mol
of H2(g) , 4.00mol of F2(g) and 6.00mol of HF(g) are initially
placed in a 2.00L reaction vessel.
Calculations with Imperfect Squares
Our ability to square both sides of the
equilibrium law equation greatly simplified the
calculation of equilibrium concentrations.
 In the absence of perfect squares, a different
simplification technique helps us solve the
problem.

Assumption
“The 100 rule”
 if the concentration to which x is added or
from which x is subtracted is at least 100 times
greater than the value of K
initial conc. divided by K
 If # is greater 100 then drop the x in the
denominator

When the 100 Rule assumption fails

We must use the quadratic equation
Example 5
If 0.50 mol of N2O4(g) is placed in a 1.0L closed container at
150C, what will be the concentrations of N2O4(g) and NO2(g)
at equilibrium? (K = 4.50)
N2O4(g) ↔ 2NO2(g)
Homework


Read section 7.5
Questions
Remember Solubility?

Solubility = the concentration of a saturated
solution of a solute in solvent at a specific
temperature and pressure


Solubility is a specific maximum concentration
Degree of Solubility:



Unsaturated
Saturated
Supersaturated
Solubility

Unsaturated solution = a solution containing less
than maximum quantity of a solute

Saturated solution = a solution containing the
maximum quantity of a solute

Supersaturated solution = a solution whose
solute concentration exceeds the equilibrium
concentration
Solubility Curve of Solids
The Solubility Product Constant

Solubility product constant (Ksp) = the value
obtained from the equilibrium law applied to a
saturated solution
Similar to Keq
no units
At a specific temp.

Example:



AgCl(s) ↔ Ag+(aq) + Cl-(aq)
+
-10 at 25⁰C
[Ag
]
[Cl
]
=
1.8x10
(aq)
(aq)
Ksp =
Equilibrium exists between a saturated
solution and excess solute.
Dissolving
Precipitation
Saturated Solution
Excess Solute
Solubility vs. Solubility Product

Solubility = the amount of a salt that dissolves in a given
amount of solvent to give a saturated solution


mol/L or g/100mL
Solubility Product = the product of the molar
concentrations of a the ions in the saturated solution

Ksp has no units
Table of Ksp

Appendix C8 (page 802)

Usually only for low solubility
ionic compounds

High solubility compounds
form solutions that do not
tend to be saturated & no
equilibrium is established
Calculating Solubility using the Ksp Value
Example 1:
Calculate the molar solubility of cobalt (II) hydroxide at
25⁰C if Ksp = 1.1x10-15 at this temperature.

Calculating Ksp using Solubility values
Example 2:
Calculate Ksp for silver chromate (Ag2CrO4) if its solubility
is 0.29g/L at 25 ⁰ C.

Predicting Precipitation



Instead of using a solubility table…
using Q to determine whether, after mixing, the ions are
present in too high a concentration, in which case a
precipitate will form
Trial Ion Product = the reaction quotient applied to the
ion concentrations of a slightly soluble salt
Using Q to Predict Solubility
Q is greater than Ksp
Q is equal to Ksp
Q is less than Ksp
supersaturated
solution
Precipitate will from
Saturated
Precipitate will not
form
unsaturated
Precipitate will not
form
Demo: KI + Pb(NO3)2
Calculations involving the prediction of a
precipitate (using Q)
Example 3:
If 500mL of a 4.0x10-6 mol/L CaCl2 solution is mixed with
300mL of a 0.0040mol/L AgNO3 solution, will a precipitate
form?

Homework:

Page 486 #1,2,4
Page 488 #5
Worksheet: Extra Solubility Problems

Quiz on Thursday April 18  ICE problem + solubility


Common Ion Effect

Common Ion Effect = A reduction in the solubility of a
salt caused by the presence of another slat having a
common ion
Energy & Equilibrium:
The Laws of Thermodynamics
Thermodynamics

Thermodynamics = the study of energy
transformation
3 fundamental laws of thermodynamics
 Laws used to understand why certain changes
occur but others do not

First Law of Thermodynamics
“Conservation of Energy”
 The
total amount of energy in the
universe is constant. Energy can be
neither created nor destroyed, but can be
transferred from one object or place to
another, or transformed from one form
to another.
First Law of Thermodynamics
Remember:

Total energy of the universe = system +
surrounding

Hess’s Law = the value of ∆H for any reaction
that can be written in steps equals the sum of
the ∆H values for each of the individual steps
Enthalpy Changes & Spontaneity



bond energy = the minimum energy required to
break one mole of bonds between two particular
atoms; a measure of the stability of a chemical
bond
It is also equal to the amount of energy released
when a mole of a particular bond is formed.
It is measured in kilojoules (kJ) and is equal to the
minimum energy required to break the
intramolecular bonds between one mole of
molecules of a pure substance.
Bond Energy

Bond energy is measured in kilojoules (kJ) and
is equal to the minimum energy required to
break the intramolecular bonds between one
mole of molecules of a pure substance.

Energy is absorbed when reactant bonds break

Energy is released when product bonds form
Bond Energy
A bond that has a
higher bond energy
(i.e. Requires more
energy to break) is
more stable.
Enthalpy & Entropy Changes
Together Determine Spontaneity
Endothermic = + ∆H
 Exothermic = - ∆H


Exothermic reactions tend to proceed
spontaneously
Spontaneous Reaction

spontaneous reaction = one that, given the
necessary activation energy, proceeds without
continuous outside assistance

Example: a sparkler


Needs light from a flame for activation
Once lit, the available fuel combusts quickly and
completely, releasing large amounts of energy as
heat and light
Entropy

enthalpy is not the only factor that determines
whether a chemical or physical change occurs
spontaneously

entropy, S = a measure of the randomness or
disorder of a system, or the surroundings
Entropy

Increase entropy = increase randomness = +∆S

When entropy increases in a reaction, the entropy
of the products, Sproducts, is greater than the
entropy of the reactants, Sreactants, yielding an
overall positive change in entropy, S.
Entropy

decrease entropy = decrease randomness = -∆S

When entropy decreases in a reaction, the entropy
of the products, Sproducts, is less than the entropy of
the reactants, Sreactants, yielding an overall negative
change in entropy, S.
Increase in Entropy
Change in Volume of Gaseous Systems
Change in Temperature
Change in State
In Chemical Reactions…
Enthalpy, Entropy, and
Spontaneous Change

Changes in the enthalpy, ∆H, and entropy, ∆S,
of a system help us to predict whether a
change will occur spontaneously
Exothermic reactions (-∆H) involving an
increase in entropy (+∆S) occur spontaneously,
because both changes are favoured
 Endothermic reactions (+∆H) involving a
decrease in entropy (-∆S) are not spontaneous
because neither change is favoured

Enthalpy, Entropy, and
Spontaneous Change
But what happens in cases where the energy
change is exothermic (favoured) and the
entropy decreases (not favoured)?
 Or when the energy change is endothermic
(not favoured) but entropy increases
(favoured)?


In these situations, the temperature at which
the change occurs becomes an important
consideration as well as free energy
Free Energy

free energy (or Gibbs free energy), G =
energy that is available to do useful work

In general, a change at constant temperature
and pressure will occur spontaneously if it is
accompanied by a decrease in Gibbs free
energy, G
-∆G = spontaneous
+∆G = nonspontaneous
Second Law of Thermodynamics
“Law of Entropy”
 all
changes that occur in the universe. All
changes, whether spontaneous or not, are
accompanied by an increase in the
entropy (overall disorder) of the universe
Mathematically, Suniverse > 0
Second Law of Thermodynamics

a system’s entropy, Ssystem, can decrease (the
system becomes more ordered), so long as there
is a larger increase in the entropy of the
surroundings, Ssurroundings, so that the overall
entropy change, Suniverse, is positive.
Problem?

Living organisms seem to violate the second
law of thermodynamics.


Build highly ordered molecules such as proteins
and DNA from a random assortment of amino
acids and nucleotides dissolved in cell fluids
building highly ordered structures such as nests,
webs, and space huttles.
Not really a problem…

Living organisms obey the second law of
thermodynamics because they create order
out of chaos in a local area of the universe
while creating a greater amount of disorder in
the universe as a whole
Oh no! Thermal Death!



The second law of thermodynamics predicts that the
universe will eventually experience a final “thermal death”
in which all particles and energy move randomly
about.
Life will come to an end because there won’t be any
sources of free energy to exploit; stars will stop shining.
Waterfalls will stop falling.
All energy will have become randomized. All of the energy
that there ever was will still be there, except that it will be
uniformly distributed throughout the universe, unable to
apply an effective push or a pull on anything. According to
the second law, a state of perfect equilibrium is the
ultimate fate of the universe.
Predicting Spontaneity

The spontaneity of any reaction carried out at
constant temperature and pressure can be
predicted by calculating the value of G using the
following equation, called the Gibbs-Helmholtz
equation:
∆G, Spontaneity & Free Energy
∆G = ∆H - T∆S
 ∆G
= - = spontaneous
 ∆G = + = nonspontaneous
 Remember: K
= ºC + 273
Predicting Spontaneity
∆G = ∆H - T∆S
-∆G = spontaneous
+∆G= nonspontaneous
+∆S
-∆S
+∆H
-∆H
Spontaneity
depends on T
Spontaneous
Spontaneity
nonspontaneous depends on T
Third Law of Thermodynamics
“Law of Entropy”
 The
entropy of a perfectly ordered pure
crystalline substance is zero at absolute
zero.
 Mathematically, S
= 0 at T = 0 K
Calculating Standard Entropy
Change

standard entropy = the entropy of one mole
of a substance at STAP; units (J/molK)
Remember ∆H°

Standard enthalpy change of reaction
∆H° vs ∆S°