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.
Slide 3
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
Slide 4
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.
Slide 5
Party Analogy 30 people at a house party 8pm: 16 people in the
kitchen 14 people in the living room 10pm: 16 people in the kitchen
14 people in the living room Different people but same number in
each room
Slide 6
Dynamic Equilibrium Example: Closed bottle of pop CO 2 gas
leaving dissolved state and entering gas state CO 2 gas ALSO,
leaving gas state and entering liquid state No visible change CO
2(g) CO 2(aq)
Slide 7
Equilibrium Double Arrow equilibrium is symbolized with an
equation containing a forward ( ) and a reverse ( ) arrow combined
into: N2O4 (g)N2O4 (g) 2NO 2 (g)
Slide 8
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 CO 2(g) CO 2(aq)
Forward Reverse
Slide 9
Drinking Bird Equilibrium
https://www.youtube.com/watch?v=Bzw0kWvfVkA
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
Slide 10
3 Types of Equilibrium 1. Solubility Equilibrium (dissolving
process) 2. Phase Equilibrium (change of state) 3. Chemical
Reaction Equilibrium (reactants products)
Slide 11
Types of Equilibrium #1 solubility equilibrium = a dynamic
equilibrium between a solute and a solvent in a saturated solution
in a closed system I 2(s) I 2(aq)
Slide 12
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
Slide 13
Solubility Equilibrium kinetic molecular theory states that
particles are always moving and colliding even if no changes are
observed Dissolution = the process of dissolving
Slide 14
(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.
Slide 15
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
Slide 16
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 its surroundings H 2 O (l) H 2 O (g) H 2
O (s) H 2 O (l)
Slide 17
Phase Equilibrium
Slide 18
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
Slide 19
H 2 (g) + I 2 (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. H 2 (g) + I 2 (g) 2HI(g) ( H = -26kJ) For
the whole reaction: Equilibrium is reached when the rate of the
forward reaction equals the rate of the reverse reaction.
Slide 20
Chemical Reaction Equilibrium In a Closed System N 2 O 4(g) 2
NO 2(g)
Slide 21
Reaction Rate Time H 2 + I 2 2 HI H 2 + I 2 2 HI
Slide 22
Reversible Reactions The same dynamic equilibrium composition
is reached whether we start from pure N 2 O 4(g), pure NO 2(g), or
a mixture of the two, provided that environment, system and total
mass remain the same.
Slide 23
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 gas- phase systems K
eq = [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)
Slide 24
Questions: Equilibrium Law Expression 1. Write the equilibrium
law expression for the following: a) 2NO 2(g) N 2 O 4(g) b) 2HI (g)
H 2(g) + I 2(g) 2. A reaction vessel contains NH 3, N 2 and H 2 gas
at equilibrium at a certain temperature. The equilibrium
concentrations are [NH 3 ] = 0.25mol/L, [N 2 ] = 0.11mol/L and [H 2
] = 1.91 mol/L. Calculate the equilibrium constant for the
decomposition of ammonia. K = [N 2 O 4(g) ] [NO 2(g) ] 2 K = [H
2(g) ] [I 2(g) ] [HI (g) ] 2 K = [N 2(g) ] [H 2(g) ] 3 [NH 3(g) ] 2
K = [0.11] [1.91 ] 3 [0.25 ] 2 K = 12.3 2NH 3(g) N 2(g) + 3H
2(g)
Slide 25
Questions: Equilibrium Law Expression 3. Nitryl chloride gas,
NO 2 Cl, is in equilibrium at a certain temperature in a closed
container with NO 2 and Cl 2 gases. At equilibrium, [NO 2 Cl] =
0.00106mol/L and [NO 2 ] = 0.0108mol/L. If K = 0.558, what is the
equilibrium concentration of Cl 2 ? 4. Write a balanced equation
for the reaction with the following equilibrium law expression: K =
[NO 2(g) ] 2 [NO (g) ] 2 [O 2 (g) ]
Slide 26
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
Slide 27
Homogenous equilibrium applies to reactions in which all
reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c =
[NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P
Slide 28
Heterogenous equilibrium applies to reactions in which
reactants and products are in different phases. CaCO 3 (s) CaO (s)
+ CO 2 (g) K c = [CaO (s) ][CO 2(g) ] [CaCO 3(s) ] [CaCO 3(s) ] =
constant [CaO (s) ] = constant K c = [CO 2(g) ] The concentration
of solids and pure liquids are considered to be constant and are
not included in the expression for the equilibrium constant. KcKc
[CaO (s) ] [CaCO 3(s) ] = [CO 2(g) ]
Slide 29
P CO2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not
depend on the amount of CaCO 3 or CaO
Slide 30
N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start
with NO 2 & N 2 O 4 equilibrium Equilibrium favors the reactant
side
Slide 31
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
Slide 32
Sample Problem: When ammonia is heated it decomposes: 2NH 3(g)
N 2(g) + 3H 2(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
Calculating Equilibrium Concentrations (when given one
concentration)
Determine the value of X [NH 3 ] (g)equil = 2 0mol / L - 2x [NH
3 ] (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
Slide 35
constant
Slide 36
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 K
eq will it be the same in fwd/rev?
Slide 37
Equilibrium Tubes The effects of temperature on equilibrium
Heat + N 2 O 4 (g) 2NO 2 (g) Colourless Brown Very ColdCold Hot
ENDOTHERMIC Rxn
Slide 38
NO 2 is one of the chemicals in smog ! In the summer on hot,
windless days an orange haze is seen over the horizon, this is NO 2
In the winter, the smog doesn't go away, it is just less
noticeable. The cooler temperatures lead to more N 2 O 4 and less
NO 2 which we can't see as well! N 2 O 4 (g) 2NO 2 (g) Colourless
Brown
Slide 39
Qualitative Changes in Equilibrium Systems You should be
familiar with your own bodys 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
CO 2 levels breathe deeper & faster If blood sugar levels
insulin released to remove excess glucose
Slide 40
Le Chteliers 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.
Slide 41
Le Chteliers Principle Le Chateliers 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
https://www.youtube.com/watch?v=dIDgPFEucFM
Slide 42
1. System starts at equilibrium. 2. A change/stress is then
made to system at equilibrium. Change in concentration Change in
temperature Change in volume/pressure 3. System responds by
shifting to reactant or product side to restore equilibrium. Le
Chteliers Principle
Slide 43
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.
Slide 44
Le Chteliers Principle Change in Reactant or Product
Concentrations If H 2 is added while the system is at equilibrium,
the system must respond to counteract the added H 2 That is, the
system must consume the H 2 and produce products until a new
equilibrium is established. Equilibrium shifts to the right.
Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases. N
2 (g) + 3H 2 (g) 2NH 3 (g)
Slide 45
Change in Reactant or Product Concentrations N 2 (g) + 3H 2 (g)
2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress
Slide 46
Change in Reactant or Product Concentrations ChangeShifts the
Equilibrium Increase concentration of product(s)left Decrease
concentration of product(s)right Decrease concentration of
reactant(s) Increase concentration of reactant(s)right left aA + bB
cC + dD
Slide 47
Le Chteliers 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.
Slide 48
Effect of Temperature Changes
Slide 49
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.
Slide 50
Gas Law Boyles Law Relationship: Pressure & Volume As
pressure on a gas increases, the volume of the gas decreases
Slide 51
Le Chteliers 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.
Slide 52
Slide 53
Effects of Volume and Pressure ChangeShifts the Equilibrium
Increase pressure Side with fewest moles of gas Decrease
pressureSide with most moles of gas Decrease volume Increase
volumeSide with most moles of gas Side with fewest moles of gas A
(g) + B (g) C (g)
Slide 54
Le Chteliers Principle Adding a Catalyst does not shift the
position of an equilibrium system system will reach equilibrium
sooner
Slide 55
Le Chteliers 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
Slide 56
Adding a Catalyst
Slide 57
Le Chteliers 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
Slide 58
Le Chteliers Principle Summary
Slide 59
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.
Slide 60
No Sweat! The carbon dioxide dissolves in body fluids forming
the following equilibrium system:
Slide 61
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 Chateliers Principle: Warm-up Page 459 A = volume B =
temperature C = [C 2 H 6 ] D = catalyst/inert gas E = [C 2 H 4
]
Slide 69
At constant temperature, regardless of initial concentrations
the concentrations of reactants and products always give a constant
value K aA + bB cC + dD K = [C] c [D] d [A] a [B] b Products
Reactants The Equilibrium Law Expression & The Equilibrium
Constant, K
Slide 70
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.
Slide 71
Recall: Equilibrium achieved from any combination of reactants
and products
Slide 72
constant N 2 O 4 (g) 2NO 2 (g) = 4.6 x 10 -3 K = [NO 2 ] 2 [N 2
O 4 ] Regardless of initial concentrations, at a given temperature,
the relationship of the equilibrium concentrations of reactants and
products always yields a CONSTANT value, K
Slide 73
Ways Different States of Matter Can Appear in the Equilibrium
Constant, K MolarityPartial Pressure gas, (g) YES aqueous,
(aq)YES--- liquid, (l)--- solid, (s)---
Slide 74
Questions: Equilibrium Law Expressions Write the equilibrium
law expressions for the following reactions: 1. NH 4 Cl (s) NH 3(g)
+ HCl (g) 2. 2H 2 O (l) 2H 2(g) + O 2 (g) 3. 2NaHCO 2(s) Na 2 CO
3(s) + H 2 O (g) + CO 2(g) K = [NH 3 (g) ] [HCl (g) ] K = [H 2 (g)
] 2 [O 2(g) ] K = [H 2 O (g) ] [CO 2(g) ]
Slide 75
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
Slide 76
What Does the Value of K Mean? If K >> 1, the reaction is
product-favoured; product predominates at equilibrium. If K
Question: Magnitude of K Consider the reaction: H 2(g) + I 2(g)
2HI (g) + heat At 448C, K=50.5. Would you predict the value of K to
be higher or lower at 300C? At 448C K >> 1 = PRODUCTS
favoured Heat lowered = rxn shifts to PRODUCT side At 300 C K >
50.5
Slide 79
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
Slide 80
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
Slide 81
Slide 82
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
Slide 83
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
Slide 84
The Reaction Quotient, Q aA + bB cC + dD Q = [C] c [D] d [A] a
[B] b
Slide 85
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.
Slide 86
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.
Slide 87
Solving Equilibrium Problems with ICE Example1: For the above
reaction [N 2 ] i = 0.32mol/L and [H 2 ] i = 0.66mol/L. At a
certain T and P, [N 2 ] eq = 0.20mol/L. What is the value of K
under these conditions?
Slide 88
Example 2 At 150C, K for the reaction I 2(g) + Br 2(g) 2IBr (g)
is found to be 1.20x10 2. Starting with 4.00mol of each of iodine
and bromine in a 2.00L flask, calculate the equilibrium
concentrations of all reaction components.
Slide 89
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) H 2(g) + I 2(g) takes place at 445C, the value of K is
0.020. If [HI]=0.20mol/L, [H 2 ]=0.15mol/L and [I 2 ]=0.09mol/L, is
the system at equilibrium? If not, in which direction will it shift
to reach equilibrium?
Slide 90
Example 4 For the reaction H 2(g) + F 2(g) 2HF (g), K is 1.5x10
2 at SATP. Calculate all equilibrium concentrations if 4.00mol of H
2(g), 4.00mol of F 2(g) and 6.00mol of HF (g) are initially placed
in a 2.00L reaction vessel.
Slide 91
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.
Slide 92
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
Slide 93
When the 100 Rule assumption fails We must use the quadratic
equation
Slide 94
Example 5 If 0.50 mol of N 2 O 4(g) is placed in a 1.0L closed
container at 150C, what will be the concentrations of N 2 O 4(g)
and NO 2(g) at equilibrium? (K = 4.50) N 2 O 4(g) 2NO 2(g)
Slide 95
Homework Read section 7.5 Questions
Slide 96
Slide 97
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
Slide 98
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
Slide 99
Slide 100
Solubility Curve of Solids
Slide 101
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) K sp = [Ag + (aq) ] [Cl -
(aq) ] = 1.8x10 -10 at 25 C
Slide 102
Equilibrium exists between a saturated solution and excess
solute. DissolvingPrecipitation Saturated Solution Excess
Solute
Slide 103
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
Slide 104
Table of K sp 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
Slide 105
Calculating Solubility using the Ksp Value Example 1: Calculate
the molar solubility of cobalt (II) hydroxide at 25 C if K sp =
1.1x10 -15 at this temperature.
Slide 106
Calculating Ksp using Solubility values Example 2: Calculate K
sp for silver chromate (Ag 2 CrO 4 ) if its solubility is 0.29g/L
at 25 C.
Slide 107
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
Slide 108
Using Q to Predict Solubility Q is greater than Ksp
supersaturated solution Precipitate will from Q is equal to
KspSaturated Precipitate will not form Q is less than
Kspunsaturated Precipitate will not form
Slide 109
Demo: KI + Pb(NO 3 ) 2
Slide 110
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?
Slide 111
Homework: Page 486 #1,2,4 Page 488 #5 Worksheet: Extra
Solubility Problems Quiz on Thursday April 18 ICE problem +
solubility
Slide 112
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
Slide 113
Energy & Equilibrium: The Laws of Thermodynamics
Slide 114
Thermodynamics Thermodynamics = the study of energy
transformation 3 fundamental laws of thermodynamics Laws used to
understand why certain changes occur but others do not
Slide 115
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.
Slide 116
First Law of Thermodynamics Remember: Total energy of the
universe = system + surrounding Hesss 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
Slide 117
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.
Slide 118
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
Slide 119
Bond Energy A bond that has a higher bond energy (i.e. Requires
more energy to break) is more stable.
Slide 120
Enthalpy & Entropy Changes Together Determine Spontaneity
Endothermic = + H Exothermic = - H Exothermic reactions tend to
proceed spontaneously
Slide 121
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
Slide 122
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
Slide 123
Entropy Increase entropy = increase randomness = +S When
entropy increases in a reaction, the entropy of the products, S
products, is greater than the entropy of the reactants, S
reactants, yielding an overall positive change in entropy, S.
Slide 124
Entropy decrease entropy = decrease randomness = -S When
entropy decreases in a reaction, the entropy of the products, S
products, is less than the entropy of the reactants, S reactants,
yielding an overall negative change in entropy, S.
Slide 125
Increase in Entropy
Slide 126
Change in Volume of Gaseous Systems
Slide 127
Change in Temperature
Slide 128
Change in State
Slide 129
In Chemical Reactions
Slide 130
Slide 131
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
Slide 132
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
Slide 133
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
Slide 134
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, S universe > 0
Slide 135
Second Law of Thermodynamics a systems entropy, S system, can
decrease (the system becomes more ordered), so long as there is a
larger increase in the entropy of the surroundings, S surroundings,
so that the overall entropy change, Suniverse, is positive.
Slide 136
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.
Slide 137
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
Slide 138
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 wont 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.
Slide 139
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:
Slide 140
G, Spontaneity & Free Energy G = H - TS G = - = spontaneous
G = + = nonspontaneous Remember: K = C + 273
Slide 141
Predicting Spontaneity +H-H +S Spontaneity depends on T
Spontaneous -S nonspontaneous Spontaneity depends on T G = H - TS
-G = spontaneous +G= nonspontaneous
Slide 142
Slide 143
Slide 144
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
Slide 145
Calculating Standard Entropy Change standard entropy = the
entropy of one mole of a substance at STAP; units (J/molK)