Chapter 8 Chemical and Physical Change: Energy, Rate, and Equilibrium Denniston Topping Caret 4 th Edition Copyright The McGraw-Hill Companies, Inc.

Chapter 8

Chemical and Physical Change: Energy, Rate, and Equilibrium

Denniston Topping Caret

4th Edition

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

8.1 Thermodynamics

• Thermodynamics - the study of energy, work, and heat.– applied to chemical change– applied to physical change

• The laws of thermodynamics help us to understand why some chemical reactions occur and others do not.

8.1

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sThe Chemical Reaction and Energy

• Important points to kinetic molecular theory

– molecules and atoms in a reaction mixture are in constant, random motion;

– these molecules and atoms frequently collide with each other;

– only some collisions, those with sufficient energy, will break bonds in molecules; and

– when reactant bonds are broken, new bonds may be formed and products result.

8.1

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s • As the bonds are broken and new bonds are formed, energy is required or released.

• We can measure the change in energy during these changes.

• System - the process under study

– Usually the chemical reaction or physical change of interest.

• Surroundings - the rest of the universe.

• We will be able to measure the change in energy in the form of heat as the temperature changes.

8.1

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8.1

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sExothermic and Endothermic Reactions 1

• The first law of thermodynamics - energy of the universe is constant.

• Where does the energy come from that is released and where does the energy go when it is absorbed?

• The chemical bond is stored chemical energy.

A-B + C-D A-D + C-B

8.1

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sA-B + C-D A-D + C-B

These bonds must be broken.

This requires energy.

These bonds are formed.

This releases energy

• If the energy required to break the bonds is less then the energy released when the bonds are formed, there is a net release of energy…Exothermic reaction.

• If the energy required to break the bonds > the energy released when the bonds are formed, there will need to be an external supply of energy…Endothermic reaction.

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Exothermic Reaction

Endothermic Reaction

8.1

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Combustion of organic compoundsCH4(g) + 2O2(g)CO2(g) + 2H2O(g) + 211 kcal

exothermic reaction

Decomposition of ammonia22 kcal + 2NH3(g) N2(g) + 3H2(g)

endothermic reaction

8.1

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sEnthalpy

• Enthalpy - represents heat energy.

• Change in Enthalpy (Ho) - energy difference between the products and reactants.

• Energy released (exothermic), enthalpy change is negative

– In the combustion of CH4, Ho = -211 kcal

• Energy absorbed (endothoermic), enthalpy change is positive.

– In the decomposition of NH3, Ho=+22 kcal

2

8.1

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sSpontaneous and Nonspontaneous Reactions

• Spontaneous reaction - occurs without any external energy input.

• Often, but not always, exothermic reactions are spontaneous.

• Thermodynamics is used to help predict if a reaction will occur.

• Another factor is needed.

8.1

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sEntropy

• The second law of thermodynamics - the universe spontaneously tends toward increasing disorder or randomness.

• Entropy (So) - a measure of the randomness of a chemical system.

• High entropy - highly disordered system

• Low entropy - well organized system

• No such thing as negative entropy.

2

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s So of a reaction = So(products) - So(reactants)

• A positive So means an increase in disorder for the reaction.

• A negative So means a decrease in disorder for the reaction.

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So is positive

So is negative

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All of the above have a positive So

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• If exothermic and positive So…

SPONTANEOUS

• If endothermic and negative So…

NONSPONTANEOUS

• For any other situations, it depends on the relative size of Ho and So.

2

8.1

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sFree Energy

• Free energy (Go) - represents the combined contribution of the enthalpy and entropy values for a chemical reaction.

• Predicts spontaneity

• Negative Go…Spontaneous

• Positive Go…Nonspontaneous

Go = Ho - TSo

T in Kelvins

2

8.2 Experimental Determination of Energy Change in Reactions

• Calorimetry - the measurement of heat energy changes in a chemical reaction.

• Calorimeter - devise which measures heat changes in calories

• The change in temperature is used to measure the heat loss or gain.

3

8.2

Cal

orim

etry

• Specific heat (S.H.) - the number of calories of heat needed to raise the temperature of 1 g of the substance 1 oC.

• S.H. for water is 1.0 cal/goC

• To determine heat released or absorbed, need:

– specific heat– total number of grams of solution– temperature change (increase or decrease)

sss SHT mQ

8.2

Cal

orim

etry 1. If 0.10 mol of HCl is mixed with 0.10 mol

of KOH in a “coffee cup” calorimeter, the temperature of 1.50 x 102 g of the solution increases from 25.0oC to 29.4oC. If the specific heat of the solution is 1.00 cal/goC, calculate the quantity of energy evolved in the reaction.

2. Is the reaction endothermic or exothermic

3. What would be the energy evolved for each mole of HCl reacted?

3

8.2

Cal

orim

etry

• Nutritional Calorie (large Calorie) = 1kilocalorie (1kcal) or 1000 calorie

• the fuel value of food

• Bomb Calorimeter is used to measure nutritional Calories

8.2

Cal

orim

etry

• One gram of a certain carbohydrate was burned in a bomb calorimeter. The temperature of 1.25 x 103 g H2O was raised from 24.5oC to 31.5oC. Calculate the fuel value of the carbohydrate (in Kcal/g).

8.3 Kinetics

• Thermodynamics determines if a reaction will occur but tells us nothing about the time it will take

• Kinetics - the study of the rate of chemical reactions– Also gives the mechanism - step-by-step

description of how reactants become products.

4

8.3

Kin

etic

s• We will look at:

– disappearance of reactants and– appearance of products.

for the reaction AB

8.3

Kin

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s

8.3

Kin

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sThe Chemical Reaction

Let’s consider the following reaction:

CH4(g) + 2O2(g) CO2(g) +2H2O(g) + 211 kcal

• C-H and O=O bonds must be broken and C=O and O-H bonds must be formed

• Energy is required to break the bonds.

– Comes from the collision of the molecules.

– Effective collision - one that leads to a chemical reaction.

8.3

Kin

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sActivation Energy and the Activated Complex

• Activation energy - the minimum amount of energy required to produce a chemical reaction. • Activated complex -

extremely unstable complex.

• Formation of activated complex requires energy. (Ea)

• Exothermic - net release of energy (Ho)

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8.3

Kin

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s• Potential energy

diagram for an endothermic reaction.

• Net absorption of energy (Ho)

68.

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Factors That Affect Reaction Rate

• structure of the reacting species,

• concentration of reactants,

• temperature of reactants,physical state of reactants, and

• presence of a catalyst

8.3

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sStructure of Reacting Species

• Oppositely charged species react more rapidly

• Ions with the same charge do not react.

• Bond strength plays a role.

• Magnitude of the activation energy is related to bond strength

• Size and shape influence the rate.

• Large molecules may block the reactive part of the molecule.

8.3

Kin

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sThe Concentration of Reactants

• Rate will generally increase as concentration increases.

• Caused by a greater number of collisions

The Temperature of Reactants

• Rate increases as the temperature increases.

• Higher temp. means higher K.E.

• Higher K.E. means higher percentage of these collisions will result in product formation.

8.3

Kin

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sThe Physical State of Reactants

• Solid state:• atoms, ions or cpds. are close but restrictive in

motion.

• Gaseous state:• particles are free to move but are far apart

causing collisions to be relatively infrequent.

• Liquid state: • particles are free to move and are close

together.

• According to the above statement, what would be the typical order of rate?

• Liquid > gas> solid

8.3

Kin

etic

sThe Presence of a Catalyst

• Catalyst - a substance that increases the reaction rate.

• Catalyst interacts with the reactants to create an alternative pathway for product production

8.3

Kin

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s

Haber Process - N2+3H2 2NH3

8.3

Kin

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s• Enzyme - a biological catalyst that

controls and speeds up thousands of essential biochemical reactions.

• Again what are the factors that affect rate?

• structure

• concentration

• temperature

• physical state

• catalyst

Hold those factors in yellow constant and observe how concentration affects rate...

8.3

Kin

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sMathematical Representation of Reaction Rate 7

• Consider the following decomposition:

)(O)(NO4)(ON2 2252 ggg

• When holding all factors except concentration constant, find:

]O[Nrate 52 or ]Ok[N rate 52

k is called the rate constant

8.3

Kin

etic

s• For a reaction Aproducts the

equation will be:

rate = k[A]n

• this is called the rate equation (or rate law)

• n is the order of the reaction• n=1, first order

• n=2, second order

• etc.

• n must be determined experimentally

8.3

Kin

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s• For the equation A + B products

the rate equation is:

rate = k[A]n[B]m

• What would be the general form of the rate equation for the reaction:

CH4+2O2CO2+2H2O

• Rate = k[CH4]n[O2]m

• Knowing the rate equation and the order helps industrial chemists determine the optimum conditions for preparing a product.

8.4 Equilibrium

Rate and Reversibility of Reactions

• Equilibrium reactions - chemical reactions that do not go to completion.– Incomplete reactions

• After no further obvious change, measurable quantities of reactants and products remain.

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8.4

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Physical Equilibrium

• Reversible reaction - a process that can occur in both directions– Use the double arrow symbol

• Dynamic equilibrium - the rate of the forward process is exactly balanced by the rate of the reverse process

• Example: Sugar in Water– If put 2-3 g of sugar in 100 mL water, all will

dissolve.– sugar (s) sugar (aq)

8.4

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• If dissolving 100 g in 100 mL of water, not all of it will dissolve.– Over time, you observe no further change in

the amount of dissolved sugar.– Individual sugar molecules are constantly

going into and out of solution.– Happens at the same rate.

• The double arrow serves as– an indicator of a reversible process– an indicator of an equilibrium process, and– a reminder of the dynamic nature of the

process.

sugar(s) sugar(aq)

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• ratef = forward rate

• rater = reverse rate

• at equilibrium: ratef=rater

• We learned from Kinetics that

• ratef=kf[sugar(s)]

• rater=kr[sugar(aq)]

• so

• kf[sugar(s)]=kr[sugar(aq)]

sugar(s) sugar(aq)

8.4

Eq

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)][sugar(

)][sugar(

s

aq

k

k

r

f

kf[sugar(s)]=kr[sugar(aq)]

• Equilibrium constant (Keq)- ratio of the two rate constants

)][sugar(

)][sugar(Keq s

aq

k

k

r

f

8.4

Eq

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Chemical Equilibrium

• Example: The Reaction of N2 and H2

N2(g) + 3H2(g) qe 2NH3(g)

• Forward and reverse reaction rates are equal.

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N2(g) + 3H2(g) qe 2NH3(g)

subdivides into:

• forward rxn: N2(g) + 3H2(g) 2NH3(g)

• reverse rxn: 2NH3(g) N2(g) + 3H2(g)

and

• ratef = kf[N2]n[H2]m

• rater = kr[NH3]p

• ratef=rater

m2

n2

p3

eq ]H[][N

][NHK

r

f

k

k

8.4

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2n

2

p3

eq ]H[][N

][NHK

r

f

k

k

• The exponents in the rate expression are numerically equal to the coefficients.

• So,

• Keq is a constant at constant temperature.

322

23

eq ]H][[N

][NHK

8.4

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The Generalized Equilibrium-Constant Expression for a Chemical Reaction

ba

dc

B][[A]

D][C][Keq

9aA + bB cC + dD

8.4

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Writing Equilibrium-Constant Expressions

• Each chemical reaction has a unique equilibrium constant value at a specified temperature.

• The brackets represent molar concentration.

• All equilibrium constants are shown as unitless.

• Only the concentration of gases and substances in solution are shown.

• concentration for pure liquids and solids are not shown.

8.4

Eq

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Write equilibrium expressions for the following equilibria:

1. 2HI(aq) H2(g) + I2(g)

2. BaCl2(s) Ba2+(aq) + 2Cl-(aq)

3. CaCO3(s) CaCO(s) + CO2(g)

8.4

Eq

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Interpreting Equilibrium Constants

• The numerical value of the equilibrium constant tells us the extent to which reactants have converted to products.

1. Keq greater than 1 x 102.

• Large value of Keq: numerator > denominator.

• At equilibrium mostly product present.

2.Keq less than 1 x 10-2.

• Which is larger, numerator or denominator?

• Ans: Denominator.

• At equilibrium which is present in the larger quantity? Reactants or Products?

• Ans: Reactants.

3.Keq between 1 x 10-2 and 1 x 102

• Equilibrium mixture contains significant concentration of both reactants and products.

8.4

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8.4

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Calculating Equilibrium Constants

• A reversible reaction is allowed to proceed until the system reaches equilibrium

• Amount of reactants and products no longer changes.

2NO2(g) N2O4(g)

• Reaction mixture is analyzed to determine the molar concentrations of each of the product and reactants.

• Concentrations are substituted into the expression.

• What would the equilibrium expression be for the reaction:

22

42eq ]NO[

]O[NK

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2NO2(g) N2O4(g)

8.4

Eq

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22

42eq ]NO[

]O[NK

• At equilibrium, [NO2]=0.0547 M and [N2O4]=0.643 M.

• What is the value for Keq?

• 215

• Would the value for Keq be the same if the equation were written:

N2O4(g) 2NO2(g)?

NO

108.

4 E

qu

ilib

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mLeChateleir’s Principle

• LeChateleir’s Principle - if a stress is placed on a system at equilibrium, the system will respond by altering the equilibrium composition in such a way as to minimize the stress.

108.

4 E

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mLeChateleir’s Principle

We will examine the following “stresses.”

1. Effect of Concentration

2. Effect of Heat

3. Effect of Pressure

4. Effect of Catalyst

8.4

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1. Effect of Concentration

• Adding or removing reactants and products at a fixed volume.

• Addition of N2 or H2. To minimize the stress, which way will the reaction shift?

• To the right. Forming more products.

• If NH3 is put in the reaction vessel?

• Equilibrium shifts to the left, forming more reactants.

N2(g) + 3H2(g) 2NH3(g)

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2. Effect of Heat

• Addition of heat is similar to increasing the amount of product.

• If heat is added by raising the temperature, which way will the equilibrium shift?

• To the left.

• Exothermic reactions: treat heat as a product

N2(g) + 3H2(g) 2NH3(g) + 22 kcal

8.4

Eq

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• Which way will this reaction shift if heat is added by increasing the temperature?

• To the right.

• Endothermic Reaction - treat heat as a reactant.

39 kcal + 2N2(g) + O2(g) 2NH3(g)

8.4

Eq

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3. Effect of Pressure

• Pressure affects the equilibrium only if one or more substances in the reaction are gases

• Relative number of gas moles on reactant and product side must differ.

• When pressure goes up…shift to side with less moles of gas

• When pressure goes downs…shifts to side with more moles of gas

8.4

Eq

uil

ibri

um • If increase pressure, which way will

the equilibrium shift?

• Right. Producing more NH3

N2(g) + 3H2(g) 2NH3(g)

• If increase the pressure, which way will the equilibrium shift?

• No change!

2HI(g) H2(g) + I2(g)

8.4

Eq

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4. Effect of a Catalyst

• A catalyst has no effect on the equilibrium composition.

• It increases the rate of both the forward and reverse reaction to the same extent.

The EndChapter 8