Download Unit 9 - Kinetics and Equilibrium

TOPIC 9:
KINETICS AND EQUILIBRIUM
Part 1 – Kinetics
Part 2 – General Equilibrium
Part 3 – Solution Equilibrium
Part 4 – Thermodynamics
PART 1 – AIMS
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What is kinetics?
What factors affect the rate of chemical reactions?
How can we classify energy in chemical reactions?
How can we interpret Potential Energy Diagrams?
PART 2 – AIMS
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What is equilibrium?
What is the equilibrium expression?
How can we calculate K and Kp?
What are external factors that affect equilibrium?
PART 3 - AIMS
 How can we determine and use solubility product
constants Ksp?
 How can we estimate salt solubility from Ksp?
 How can we determine the formation of precipitates?
 What is the common ion effect?
PART 4 – AIMS
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What is entropy?
How can we calculate ΔHformation?
How can we calculate bond energies?
What is entropy?
How can we use Gibbs Free Energy to predict spontaneity?
AIM: WHAT IS KINETICS?
 Kinetics: the branch of chemistry that deals with the
rates of chemical reactions
 Collision Theory:
1. In order for a reaction to occur, reactants must collide with
each other
2. An effective collision is when reactants come together with
the correct amount of energy and in the correct position to
form a product
AIM: What Factors Affect the Rate of
a Reaction?
 As the amount of effective collisions increases, the
faster products are formed (reaction rate increases)
 How can we increase the reactions rate (increase the
# of effective collisions??
AIM: What Factors Affect the Rate of
a Reaction?
 FACTORS THAT AFFECT THE RATE OF CHEMICAL REACTIONS
1.
2.
3.
4.
5.
6.
Nature of Reactants
Concentration
Surface Area
Pressure
Temperature
Catalyst
AIM: What Factors Affect the Rate of
a Reaction?
 NATURE OF REACTANTS:
Reactions involve the breaking of old bonds and the formation
of new bonds. In general:
1. Covalent bonds are slower to react than ionic bonds
2. Breaking more bonds requires more energy than making
bonds during collisions
(Table I Reactions 1-6)
AIM: What Factors Affect the Rate of
a Reaction?
 CONCENTRATION:
Generally: Increase concentration increase rate of reaction
(especially if volume is decreased)
AIM: What Factors Affect the Rate of
a Reaction?
 SURFACE AREA:
Generally: the more surface area that is exposed the more
chances there are for collisions (effective collisions) and will
increase rate of reaction
AIM: What Factors Affect the Rate of
a Reaction?
 PRESSURE:
No effect on solids and liquids, only gases
Increasing pressure, decreases the volume, increasing the rate
of effective collisions – increases the rate of reaction
AIM: What Factors Affect the Rate of
a Reaction?
 TEMPERATURE:
Generally: Increasing temperature increases kinetic energy of
molecules, leading to an increase in the amount of effective
collisions – increasing the rate of reaction
AIM: What Factors Affect the Rate of
a Reaction?
 CATALYST
Addition of a catalyst increases the rate of the reaction by
providing a different and easier pathway for the reaction
AIM: HOW CAN WE CLASSIFY
ENERGY IN CHEMICAL REACTIONS?
3 WAYS TO
CLASSIFY
ENERGY IN
CHEMICAL
REACTIONS
1.
Look at the reactions (where is E term?)
CH4 + 2O2
CO2 + 2H2O + 890.4kJ
N2 + O2 + 66.4kJ
2NO2
2. Look at the H = heat of the reaction
(Table I)
3. Potential Energy Diagrams
H = PE products – PE reactants
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
Show how the potential energy of reactant particles changes to
chemical potential energy stored in bonds
PE diagrams keep track of PE changes during a chemical
reaction in stages
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
For Example: A + B
AB
The forward reaction is read from left to right
Compare the potential energy of the reactants to the potential
energy of the products in the forward reaction (PE diagram #1)
PE REACTANTS: 25 Joules
PE PRODUCTS: 75Joules
*Energy must have been absorbed during the reactions
PE diagrams with this pattern represent endothermic reactions
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
FORWARD REACTION!!
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
Lets Label the Diagram
A. PE of Reactants
B. PE of Products
C. H (heat of reaction) = H = PE products – PE reactants
If H is positive [PE products  PE reactants] – endothermic
If H is negative [PE products  PE reactants] – exothermic
Table I – shows different chemical reactions and H for each one
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
Label the Diagram
D. PE of Activated Complex – intermediate molecule that forms when
reactants have an effective collision. It is unstable and temporary
E. Activation Energy of forward reaction – amount of energy needed
to start the reaction in the forward direction
F. Activation Energy (with catalyst) – amount of energy needed to
start the reaction if a catalyst is added
* Catalysts – speed the reaction rate by lowering the activation
energy needed to start a reactions (gives an alternate pathway)
AIM: HOW CAN WE INTERPRET
POTENTIAL ENERGY DIAGRAMS?
For Example: A + B
AB
Compare the potential energy of the reactants to the potential
energy of the products in the reverse reaction (PE diagram #2)
PE REACTANTS: 75 Joules
PE PRODUCTS: 25Joules
*Energy must have been released during the reactions
PE diagrams with this pattern represent exothermic reactions
Which diagram is endothermic?
Exothermic?
TABLE I – Gives us ΔH values for
reactions
 PRACTICE USING TABLE I:
 A negative value means that the reaction is exothermic
 If we were to rewrite the equation with the heat of reaction
shown it would looks like this
𝐂𝐇𝟒 𝐠 + 𝟐𝐎𝟐 → 𝐂𝐎𝟐 𝐠 + 𝟐𝐇𝟐 𝐎 𝐥 + 𝟖𝟗𝟎. 𝟒𝐤𝐉
It is added to the right side of the equation because it is an
exothermic reaction in which heat is released as a product
TABLE I – Gives us ΔH values for
reactions
 PRACTICE USING TABLE I:
 A positive value means that the reaction is endothermic
 If we were to rewrite the equation with the heat of reaction
shown it would looks like this
𝐍𝟐 𝐠 + 𝐎𝟐 + 𝟏𝟖𝟐. 𝟔𝐤𝐉 → 𝟐𝐍𝐎(𝐠)
It is added to the left side of the equation because it is an
endothermic reaction in which heat is absorbed as a reactant
EQUILIBRIUM
AIM: What is equilibrium?
 Equilibrium: when the forward and reverse reactions
occur at the same rate – it is a state of balance
 Dynamic Equilibrium: The motion in which the
interactions of reacting particles are balanced by the
interaction of product particles
 Reversible Equilibrium: Many reactions in equilibrium
are considered reversible. This is indicated by a
double arrow
EQUILIBRIUM
AIM: What is equilibrium?
 Physical Equilibrium: the changes that take place in
chemical reactions during physical processes such as
changes of state or dissolving.
- Phase Equilibrium: Equilibrium between phases
1. Solid and liquid phase – during melting the rate of
dissolving is equal to the rate of crystallization in a
closed container (system) (H20 (s)
H2O (l))
EQUILIBRIUM
AIM: What is equilibrium?
2. Liquid and Gas phase – during this phase the rate of
evaporation is equal to the rate of condensation in a closed
container (H20 (l)
H2O (g))
EQUILIBRIUM
AIM: How Can We Examine Various
Systems at Equilibrium?
 Solution Equilibrium:
1. Solid/Liquid Solution – saturated solutions are
examples of solid/liquid solution equilibrium in a closed
system (C12H6O11 (s) / C12H22O11 (aq)
2. Gas/Liquid Solution – in a closed system or container,
there is equilibrium between the gaseous and dissolved
state of the gas
EQUILIBRIUM
AIM: What is equilibrium?
The equilibrium position – whether the reaction lies far to the right
or to the left depends on three main factors:
1. The initial concentrations (more collisions – faster reaction)
2. Relative energies of reactants and products
(nature goes to minimum energy)
3. Degree of organization of reactants and products
(nature goes to maximum disorder)
4. Significance of K
K>1 means that the reaction favors the products at eq.
K<1 means that the reaction favors the reactants at eq.
AIM: What is the equilibrium
expression?
 Mathematical expression that shows the relationship of
reactants and products in a system at equilibrium is called the
equilibrium expression
 It is a fraction with the concentrations of reactants and
products expressed in moles per liter
 Each concentration is then raised to the power of its
coefficient in a balanced equation.
 The expression equals a value called the equilibrium constant
Keq, which remains the same for a particular reaction at a
specified temperature
AIM: What is the equilibrium
expression?
To write an equilibrium expression follow these steps:
 Write a balanced equation for the system.
 Place the products as factors in the numerator of a fraction and
the reactants as factors in the denominator,
 Place a square bracket around each formula. The square bracket
means molar concentration.
 Write the coefficient of each substance as the power of its
concentration. The resulting expression is the equilibrium
expression, which should be set equal to the Keq for that reaction.
AIM: What is the equilibrium
expression?
AIM: What is the equilibrium
expression?
Ex) Write the equilibrium expression for the equilibrium
system of nitrogen, hydrogen, and ammonia.
 ___N2 (g) + ___H2 (g) <-> ___NH3 (g)
AIM: What is the equilibrium
expression?
AIM: What is the equilibrium
expression?
 The equilibrium constant is a specific numerical value for a
given system at a specified temperature
 Changes in concentrations will not cause a change in the
value of Keq, nor will the addition of a catalyst.
 Only temperature will change the value
PURE SOLIDS AND LIQUIDS ARE NOT INCLUDED IN THE
EQUILIBRIUM EXPRESSION (g) and (aq) are.
AIM: How can we calculate K and Kp?
AIM: How can we calculate K and Kp?
AIM: How can we calculate K and Kp?
AIM: How can we calculate K and Kp?
Can you…
 … write an equilibrium constant expression?
 … tell how to find K for a summary equation?
 … explain what K is telling you about a reaction?
AIM: What are external factors that
affect a reaction at equilibrium?
 If a stress is applied to a system at equilibrium the
equilibrium will shift to release the effects of the stress
Stressors:
1. Temperature
2. Concentration
3. Pressure
Le Chateliers Principle
TEMPERATURE
Increase in temperature favors the endothermic
reaction
Decrease in temperature will favor the exothermic
reaction
* *The side you shift towards will increase and the side
you shift away from will decrease**
Le Chateliers Principle
CONCENTRATION
Increase conc. of
reactants
Favors FORWARD reaction
Speeds up the forward direction
SHIFTS RIGHT
Decrease conc. of
reactants
Favors REVERSE reaction
Speeds up the REVERSE direction
SHIFTS LEFT
Increase conc. of products Favors REVERSE reactions
Speeds up the REVERSE direction
SHIFTS LEFT
Decrease conc. of the
products
SHIFTS RIGHT
Favors FORWARD reaction
Speeds up the FORWARD direction
CONCENTRATION
 ADD AWAY (increase)
 TAKE TOWARDS (decrease)
 Pen/pencil will go up on which ever side you shift
towards
PRESSURE
Need to know how many gas molecules are on the reactant side
and on the product side
Equal # of gas molecules – pressure will have no effect!!
Increase pressure – shift from more gas molecules  towards less
gas molecules
Decrease pressure – shift from less  toward more
PRESSURE
Ex)
4NH (g) + 5O2(g)
4 + 5 = 9 gas molecules
4NO(g) + 6H2O(g)
4+6 = 10 gas molecules
Increase pressure: shift to the left (more to less) 9
Decrease pressure: shift to the right (less to more) 9
10
10
AIM: What are the conditions at
equilibrium?
 At equilibrium the rate of the forward
reaction is equal to the rate of the
reverse reaction
 Concentrations are constant not equal
 Adding a catalyst would speed up the
forward and reverse reactions to the
same extent
AIM: How can we determine and use
solubility product constants? Ksp
 SOLUBILITY PRODUCT CONSTANT
 A special case of equilibrium involving
dissolving
 Because the constant is a product of a
solubility, we call it the solubility
product constant Ksp
AIM: How can we determine and use
solubility product constants? Ksp
 SOLUBILITY PRODUCT PROBLEMS:
 Given Ksp, find solubility
 Given solubility, find Ksp
 Predicting precipitation
 Find solubility in a solution with a
common ion
AIM: How can we determine and use
solubility product constants? Ksp
Solid  Positive Ion + Negative Ion
Mg(NO3)2 
Keq =
Mg2+ + 2NO3-
2+
[Mg
] [NO3
-]2
AIM: How can we determine and use
solubility product constants? Ksp
EX) Write out the equilibrium law expression:
BaF2(S)  Ba+2(aq) + 2F-(aq)
Ksp = [Ba+2][F-]2
AIM: How can we determine and use
solubility product constants? Ksp
SOLUBILITY GENERALIZATIONS (TABLE F)
 All nitrates are soluble
 All compounds of the alkali metals are
soluble (Li, Na, K, etc.)
 All compounds of the ammonium (NH4+) are
soluble
AIM: How can we estimate salt
solubility from Ksp?
Need to set up RICE tables
 R- Reaction
 I – initial concentration
 C – Change in concentration
 E – Concentration at equilibrium
AIM: How can we estimate salt
solubility from Ksp?
Ex) What is the solubility of silver bromide
(Ksp = 5.2 x 10-13)
AgBr  Ag+ + Br-
Let x = the solubility
(x)(x) = 5.2 x 10-13
X2 = 5.2 x 10-13
X = 7.2 x 10-7\
AIM: How can we determine and use
solubility product constants? Ksp
Ex) What is the solubility of PbI2
(Ksp = 7.1 x 10-9)
AIM: How can we determine the
formation of precipitates?
 Step 1: Write out the dissolving equations
 Step 2: Determine the most likely precipitate and write out its
equation
 Step 3: Determine the molar concentrations and calculate the
reaction quotient (Q)
 The reaction quotient (Q) - The product of the Ksp equation
using the ion concentration before any reaction interaction
 If Q > Ksp then a precipitate will form
AIM: How can we determine the
formation of precipitates?
Ex) A student mixes 0.010 mole
Ca(NO3)2 in 2 liters of 0.10M Na2CO3
solution. Will a precipitate form?
AIM: How can we determine the
formation of precipitates?
Ex) 0.15 moles of AgNO3 is mixed with
5 liters of .02M NaCl solution. What is
the most likely precipitate and will it
form?
AIM: What is the common ion effect?
 The common ion effect is used to describe the
effect on an equilibrium involving a substance that
adds an ion that is a part of the equilibrium.
 The common ion effect is responsible for the
reduction in the solubility of an ionic precipitate
when a soluble compound is combining one of the
ions of the precipitate is added to the solution in
equilibrium with the precipitate.
AIM: What is the common ion effect?
 When AgNO3 is added to a salt solution of AgCl
it is described as a source of a common ion, Ag+
ion.
 Common ion – ion that enters the solution from
2 different sources
 Common ion effect can be used to make an
“insoluble” salt even less soluble
AIM: What is the common ion effect?
Ex) Calculate solubility of CaF2 in a 0.0050M
solution of NaF
AIM: What is enthalpy?
 Enthalpy (H) – flow of energy (heat exchange)
at constant pressure when two systems are in
contact
 Measure only the change in enthalpy ΔH (the
difference between the potential energies of
the products and reactants)
 Exothermic reactions are favored ΔH=  Δ H = q at constant pressure - open container
AIM: How can we calculate ΔH ?
Δ H can be calculated from several sources
including:
Stoichiometry
Calorimetry q=mcΔT
From tables of standard values
Hess’s Law
Bond Energies
AIM: How can we calculate ΔH ?
STOICHIOMETRY:
AIM: How can we calculate ΔH ?
CALORIMETRY:
AIM: How can we calculate ΔH ?
CALORIMETRY:
AIM: How can we calculate ΔH ?
TABLES:
AIM: How can we calculate ΔH ?
HESS’S LAW
ΔHrxn = ΣΔHf(products) – ΣΔHf(reactants)
AIM: How can we calculate ΔH ?
HESS’S LAW
AIM: How can we calculate ΔH ?
HESS’S LAW
AIM: How can we calculate bond
energies ?
Energy must be added/absorbed to BREAK
bonds (endothermic) in order to
overcome the attractive forces between
each nuclei and the shared electrons
Energy is released when bonds are
FORMED (exothermic) because resulting
attractive forced between the bonded
atoms lowers potential energy causing a
release.
AIM: How can we calculate bond
energies ?
**BARF**
ΔH = Σ Bond Energies (broken) – Σ Bond Energies (formed)
AIM: How can we calculate bond
energies ?
Ex)
SUMMARY FOR ENTHALPY
What does ΔH tell you about the changes in
energy regarding a chemical reaction?
ΔH = + reaction is endothermic and heat
energy is added into the system
ΔH = - reaction is exothermic and heat
energy is lost from the system
(Nature tends to favor the lowest possible
energy state!)
AIM: What is entropy?
ENTROPY ΔS: disorder or randomness of
the matter and energy of a system (more
disordered/dispersal is favored)

Nature favors CHAOS (high entropy low
energy)
AIM: What is entropy?
Thermodynamically favored processes or
reactions are those that involve a
decrease in internal energy of the
components (ΔH<0) and increase in
entropy (ΔS >0)
These are spontaneous or
thermodynamically favored
AIM: What is entropy?
ΔS is + when dispersal/disorder increases
(favored)
ΔS is – when dispersal/disorder decreases
NOTE: Units are usually J/(molrxn • K)
(not kJ!)
ΔSrxn = ΣΔSf(products) – ΣΔSf(reactants)
AIM: What is entropy?
Ex)Predict which has the largest increase in
entropy:
CO2(s)  CO2(g)
H2(g) + Cl2(g)  2HCl(g)
KNO3(s)  KNO3(l)
C(diamond)  C(graphite)
AIM: What is entropy?
Ex)
AIM: How can we use Gibbs free
energy to predict spontaneity?
 The calculation of Gibbs free energy, ΔG is what
ultimately decides whether a reaction is
thermodynamically favored or not

 A NEGATIVE sign on ΔG indicates that a reaction is
thermodynamically favored (spontaneous)
 Several ways to calculate ΔG that links
thermochemistry, entropy, equilibrium, and
electrochemistry together!
AIM: How can we use Gibbs free
energy to predict spontaneity?
ΔG = ΔH – TΔS
AIM: How can we use Gibbs free
energy to predict spontaneity?
Ex)
SUMMARY
 Ex) If ΔG is NEGATIVE, the reaction is
thermodynamically favorable
 If ΔG is ZERO, the reaction is at equilibrium and
 If ΔG is POSITIVE, the reaction is NOT
thermodynamically favorable
SUMMARY
Ex)
Spontaneous reactions: DOESN’T
REQUIRE EFFORT (proceed on
their own without intervention)
Ex) ice melting, nuclear reactionsnatural transmutation
AIM: How can we determine if a reaction is spontaneous?
AIM: How can we determine if a reaction is spontaneous?
2 Conditions for a reaction to be considered
spontaneous
1. Tendency toward lower energy (PE)
exothermic
ΔH = (-) Table I
* products have lower energy and more
stable
AIM: How can we determine if a reaction is spontaneous?
2. Tendency toward randomness ΔS = (+)
Entropy  measure of randomness or disorder
Physical changed and entropy :
Solid
Liquid
Low Entropy
Intermediate
Entropy
Gas
High Entropy
Table I (stability) greater exothermic reaction the more stable the products are ( lower PE)
AIM: How can we determine if a
reaction is spontaneous?
Chemical changes and Entropy:
- Free elements (ex. O2, Na, Fe) high entropy
- Compounds (ex. H2O, NH3) lower entropy
- ***NATURE FAVORS REACTIONS THAT HAVE
LOW ENERGY AND HIGH ENTROPY***
Free Energy
ΔG = Lets us know if a reaction is spontaneous
ΔH  (negative, exothermic)
ΔS  (positive, high entropy)
**always have ΔG = (-) SPONTANEOUS
Gibbs Free Energy Equation
ΔG = ΔH - TΔS
where
ΔG = Gibbs Free Energy, in kJ
ΔH = heat of reaction
T = temperature, in Kelvin
ΔS = entropy change (in kJ · K-1)