topic 9: kinetics and equilibrium part 1 – kinetics part 2 – general equilibrium part 3 –...

TOPIC 9: KINETICS AND EQUILIBRIUM

Part 1 – KineticsPart 2 – General EquilibriumPart 3 – Solution EquilibriumPart 4 – Thermodynamics

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 1 – AIMS

What is equilibrium? What is the equilibrium expression? How can we calculate K and Kp? What are external factors that affect

equilibrium?

PART 2 – 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 3 - AIMS

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?

PART 4 – AIMS

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 IS KINETICS?

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. Nature of Reactants 2. Concentration3. Surface Area4. Pressure 5. Temperature 6. Catalyst


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)


CONCENTRATION:

Generally: Increase concentration increase rate of

reaction

(especially if volume is decreased)


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


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


TEMPERATURE:

Generally: Increasing temperature increases

kinetic energy of molecules, leading to an

increase in the amount of effective collisions –

increasing the rate of 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

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 rightCompare 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



FORWARD REACTION!!

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


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)


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?

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

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

It is added to the left side of the equation because it is an endothermic reaction in which heat is absorbed as a reactant

TABLE I – Gives us ΔH values for reactions

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

EQUILIBRIUMAIM: 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))


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))


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

EQUILIBRIUMAIM: How Can We Examine Various Systems at

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.

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.


Ex) Write the equilibrium expression for the equilibrium system of nitrogen, hydrogen, and ammonia.

___N2 (g) + ___H2 (g) <-> ___NH3 (g)


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?

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: How can we calculate K and Kp?

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. Concentration3. Pressure

AIM: What are external factors that affect a reaction at equilibrium?

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 PrincipleTEMPERATURE

Increase conc. of reactants

Favors FORWARD reactionSpeeds up the forward direction

SHIFTS RIGHT

Decrease conc. of reactants

Favors REVERSE reactionSpeeds up the REVERSE direction

SHIFTS LEFT

Increase conc. of products

Favors REVERSE reactionsSpeeds up the REVERSE direction

SHIFTS LEFT

Decrease conc. of the products

Favors FORWARD reactionSpeeds up the FORWARD direction

SHIFTS RIGHT

Le Chateliers PrincipleCONCENTRATION

ADD AWAY (increase)

TAKE TOWARDS (decrease)

Pen/pencil will go up on which ever side you shift towards

CONCENTRATION

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) 4NO(g) + 6H2O(g)

4 + 5 = 9 gas molecules 4+6 = 10 gas molecules

Increase pressure: shift to the left (more to less) 9 10Decrease pressure: shift to the right (less to more) 9 10

PRESSURE

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: What are the conditions at equilibrium?

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



Mg(NO3)2 Mg2+ + 2NO3-

Solid Positive Ion + Negative Ion

Keq = [Mg2+ ] [NO3

-]2

EX) Write out the equilibrium law expression:


BaF2(S) Ba+2(aq) + 2F-

(aq)

Ksp = [Ba+2][F-]2

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


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)

AIM: How can we estimate salt solubility from Ksp?

AgBr Ag+ + Br-

(x)(x) = 5.2 x 10-

13X2 = 5.2 x 10-13 X = 7.2 x 10-

7\

Let x = the solubility

Ex) What is the solubility of PbI2

(Ksp = 7.1 x 10-9)


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?


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?


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


Ex) Calculate solubility of CaF2 in a 0.0050M solution of NaF


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: What is enthalpy?

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:


CALORIMETRY:


TABLES:


HESS’S LAW


ΔHrxn = ΣΔHf(products) – ΣΔHf(reactants)

HESS’S LAW


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)


Ex)


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!)

SUMMARY FOR ENTHALPY

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


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)


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)


Ex)


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


Ex)


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)

SUMMARY

Spontaneous reactions: DOESN’T REQUIRE EFFORT (proceed on their own without intervention)

Ex) ice melting, nuclear reactions- natural transmutation

AIM: How can we determine if a reaction is spontaneous?

2 Conditions for a reaction to be considered spontaneous1. Tendency toward lower energy (PE) exothermic

ΔH = (-) Table I * products have lower energy and more stable


2. Tendency toward randomness ΔS = (+)Entropy measure of randomness or disorder

Physical changed and entropy :

Low Entropy High Entropy

Intermediate Entropy

Solid Gas Liquid

Table I (stability) greater exothermic reaction the more stable the products are ( lower PE)



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 kJΔH = heat of reaction

T = temperature, in KelvinΔS = entropy change (in kJ · K-

1)

topic 9: kinetics and equilibrium part 1 – kinetics part 2 – general equilibrium part 3 –...

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