Chapter 7

Energy, Rate, and Equilibrium

Denniston Topping Caret

5th Edition

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

7.1 Thermodynamics

• Thermodynamics - the study of energy, work, and heat– Applied to chemical change

• Calculate the quantity of heat obtained from combustions of one gallon of fuel oil

– Applied to physical change• Determine the energy released by boiling water

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

Basic Concepts – from Kinetic Molecular Theory

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

– Molecules and atoms frequently collide with each other

– Only some collisions, those with sufficient energy, will break bonds in molecules

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

The Chemical Reaction and Energy7.

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Change in Energy and Surroundings

• Absolute value for energy stored in a chemical system cannot be measured

• Can measure the change in energy during these chemical changes

• System - contains the process under study

• Surroundings - the rest of the universe

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Changes in the System• Energy can be lost from the system to the surroundings

• Energy may be gained by the system at the expense of the surroundings

– This energy change is usually in the form of heat

– This change can be measured

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

• This law is also called the Law of Conservation of Energy

• Where does the reaction energy come from that is released and where does the energy go when it is absorbed?7.

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sLaw of Conservation of Energy

Changes in Chemical Energy

• Consider the reaction converting AB and CD to AD and CB

• Each chemical bond is stored chemical energy

• If a reaction will occur: – Bonds must break– Breaking bonds requires energy7.

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

If the energy required to break the bonds is less than the energy released when the bonds are formed, there is a net release of energy

– This is called an exothermic reaction– Energy is a product in this reaction

These bonds must be broken in the

reaction, requiring energy

These bonds are formed, releasing

energy

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

Exothermic Reactions

If the energy required to break the bonds islarger than the energy released when thebonds are formed, there will need to be an external supply of energy

– This is called an endothermic reaction

These bonds must be broken in the

reaction, requiring energy

These bonds are formed, releasing

energy

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

Endothermic Reactions

Endothermic Reaction

Decomposition

22 kcal + 2NH3(g)

N2(g) + 3H2(g)

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Exothermic Reaction

Combustion

CH4(g) + 2O2(g)

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

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Enthalpy• Enthalpy - represents heat energy

• Change in Enthalpy (Ho) - energy difference between the products and reactants of a chemical reaction

• Energy released, exothermic reaction, enthalpy change is negative

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

• Energy absorbed, endothermic, enthalpy change is positive

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

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Spontaneous and Nonspontaneous Reactions

• Spontaneous reaction - occurs without any external energy input

• Most, but not all, exothermic reactions are spontaneous

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

• Another factor is needed, Entropy

S o is positive

So is negative

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Spontaneous and Nonspontaneous Reactions

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Spontaneous and Nonspontaneous Reactions

Are the following processes exothermic or endothermic?

– Fuel oil is burned in a furnace

– C6H12O6(s) 2C2H5OH(l) + 2CO2(g)

H = –16 kcal

– N2O5(g) + H2O(l) 2HNO3(l) + 18.3 kcal

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Entropy

• 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, the absence of a regular, repeating pattern

• Low entropy – well organized system such as a crystalline structure

• No such thing as negative entropy

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

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Entropy of Reactions

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

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

All of these processes have a positive So

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Processes Having Positive Entropy

Phase change

Melting

Vaporization

Dissolution

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Entropy and Reaction Spontaneity

• 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

Greatest Entropy

• Which substance has the greatest entropy?

– He(g) or Na(s)

– H2O(l) or H2O(g)

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

• Free energy predicts spontaneity of chemical reactions

– Negative Go…Always Spontaneous

– Positive Go…Never Spontaneous

Go = Ho - TSo

T in Kelvin

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Free Energy

Free Energy and Reaction Spontaneity

• Need to know both H and S to predict the sign of G, making a statement on reaction spontaneity

• Temperature also may determine direction of spontaneity H +, S - : G always +, regardless of T H -, S + : G always -, regardless of T H +, S + : G sign depends on T H -, S - : G sign depends on T7.

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7.2 Experimental Determination of Energy Change in Reactions

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

• Calorimeter - device which measures heat changes in calories

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

• Change in temperature of a solution, caused by a chemical reaction, can be used to calculate the gain or loss of heat energy for the reaction

– Exothermic reaction – heat released is absorbed

– Endothermic reaction – reactants absorb heat from the solution

• Specific heat (SH) - the number of calories of heat needed to raise the temperature of 1 g of a substance 1oC

sss SHTmQ ××=

Heat Energy in Reactions7.

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• Specific heat of the solution along with the total number of grams of solution and the temperature change, permits calculation of heat released or absorbed during the reaction

• SH 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)

Heat Energy in Reactions7.

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• Q is the product – ms is the mass of solution in the calorimeter Ts is the change in temperature of the solution

from initial to final state– SHs is the specific heat of the solution

• Calculate with this equation

– Units are: calories = gram x ºC x calories/gram - ºC

Calculation of Heat Energy in Reactions

7.2

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sss SHTmQ ××=

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nsCalculating Energy Involved in

Calorimeter ReactionsIf 0.10 mol HCl is mixed with 0.10 mol 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.

Ts = 29.4oC - 25.0oC = 4.4oC

Q = ms x Ts x SHs

= 1.50 x 102 g solution x 4.4oC x 1.00 cal/goC

= 6.6 x 102 cal

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Calorimeter Reactions

Is the reaction endothermic or exothermic?

– 0.66 kcal of heat energy was released to the surroundings—the solution

– The reaction is exothermic

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

– 0.10 mol HCl used in the original reaction

– [6.6 x 102 cal / 0.10 mol HCl] x 10 = 6.6 kcal

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Measurement of Calories in FoodsNutritional Calorie

(large “C” Calorie) = – 1 kilocalorie (1kcal)

– 1000 calories• The fuel value of

food• Bomb calorimeter

is used to measure nutritional Calories

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nsCalculating the Fuel Value

of Foods1 g of glucose was burned in a bomb calorimeter. 1.25 x 103 g H2O was warmed from 24.5oC to 31.5oC. Calculate the fuel value of the glucose (in Kcal/g).

Ts = 31.5oC - 24.5oC = 6.1oCSurroundings of calorimeter is water with specific heat capacity = 1.00 cal/g H2O

oC

Fuel Value =

= 1.25 x 103 g H2O x 6.1oC x 1.00 cal/g H2OoC

= 7.6 x 103 cal

7.6 x 103 cal x 1 Calorie / 103 cal = 7.6 nutritional Calories

sss SHTmQ ××=

7.3 Kinetics

• Thermodynamics determines if a reaction will occur spontaneously, but tells us nothing about the amount of time the reaction will take

• Kinetics - the study of the rate (or speed) of chemical reactions– Also supplies an indication of the mechanism –

step-by-step description of how reactants become products

Kinetic Information

• Kinetic information represents changes over time, seen here:– Disappearance of reactant, A– Appearance of product, B

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7.3

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sAlternative Presentation of

Kinetic DataRather than the graph shown before, this figure demonstrates the change from purple reactant to green product over time from the molecular perspective

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sKinetic Data Assessed by

Color Change• Change in color over time can be used to monitor the progress of a chemical reaction• The rate of color change can aid in calculating the rate of the chemical reaction

7.3

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sThe Chemical Reaction

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

• Energy is required to break the bonds

– This energy comes from the collision of the molecules

– If sufficient energy available, bonds break and atoms recombine in a lower energy arrangement

– Effective collision is one that produces product molecules

– Leads to a chemical reaction

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

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etic

sActivation Energy and the

Activated Complex• Activation energy - the minimum amount of

energy required to initiate a chemical reaction

• Picture a chemical reaction in terms of changes in potential energy occurring during the reaction

– Activated complex - an extremely unstable, short-lived intermediate complex

– Formation of this activated complex requires energy (Ea) to overcome the energy barrier to start the reaction

• Reaction proceeds from reactants to products via the activated complex

• Activated complex - can’t be isolated from the reaction mixture

• Activation energy (Ea) is the difference between the energy of the reactants and that of the activated complex

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sActivation Energy and the

Activated Complex

•To be an Exothermic reaction requires a net release of energy (Ho)

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sActivation Energy in the Endothermic Reaction

• This figure diagrams an endothermic reaction

• Reaction takes place slowly due to the large activation energy required

• The energy of the products is greater than that of the reactants

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sFactors That Affect

Reaction Rate

1. Structure of the reacting species

2. Concentration of reactants

3. Temperature of reactants

4. Physical state of reactants

5. Presence of a catalyst

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sStructure of Reacting Species

• Oppositely charged species react more rapidly• Dissociated ions in solution whose bonds are already broken

have a very low activation energy

• Ions with the same charge do not react• Bond strength plays a role

• Covalent molecules bonds must be broken with the activation energy before new bonds can be formed

• Magnitude of the activation energy is related to bond strength

• Size and shape influence the rate• Large molecules may obstruct the reactive part of the

molecule• Only molecular collisions with correct orientation lead to

product formation

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sThe Concentration of Reactants

• Rate is related to the concentration of one or more of the reacting substances

• Rate will generally increase as concentration increases

– Higher concentration means more reactant molecules per unit volume

– More reactant molecules means more collisions per unit time

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sThe Temperature of Reactants

• Rate increases as the temperature increases

– Increased temperature relates directly to increased average kinetic energy

– Greater kinetic energy increases the speed of particles

– Faster particles increases likelihood of collision

• Higher kinetic energy means a higher percentage of these collisions will result in product formation

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sThe Physical State of Reactants

Reactions occur when reactants can collide frequently with sufficient energy to react

•Solid state: atoms, ions, compounds are close together but restricted in motion

•Gaseous state: particles are free to move but often are far apart causing collisions to be relatively infrequent

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

•Reactions to be fastest in the liquid state and slowest in the solid state

• Liquid > Gas > Solid

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sThe Presence of a Catalyst

• Catalyst - a substance that increases the reaction rate– Undergoes no net change– Does not alter the final product of the reaction– Interacts with the reactants to create an alternative pathway

for product production

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sUse of a Solid Phase Catalyst

Haber Process is a synthesis of ammonia facilitated by a solid phase catalyst

– Diatomic gases bind to the surface– Bonds are weakened– Dissociation of diatomic gases and reformation of NH3

– Newly formed NH3 leaves the solid surface with the catalyst unchanged

N2+3H2 2NH3

• Consider a decomposition reaction with the following balanced chemical equation:

• When heated N2O5 decomposes to 2 products – NO2 and O2

• When holding all factors constant, except concentration, rate of reaction is proportional to the concentration

)(O)(NO4)(ON2 2252 ggg + →

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sMathematical Representation of

Reaction Rate

• Reaction rate is proportional to reactant concentration – – Concentration of N2O5 is denoted as [N2O5]

– Replace proportionality symbol with (=) and proportionality constant k

– k is called the rate constant

)(O)(NO4)(ON2 2252 ggg + →

]O[Nrate 52∝

]O[N rate 52k=

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sMathematical Representation of

Reaction Rate

• For a reaction Aproducts we write the equation:

rate = k[A]n

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

• The exponent n is the order of the reaction– If n=1, first order

– If n=2, second order

– n must be determined experimentally

– This exponent is not the same as the coefficient of the reactant in the balanced equation

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sRate Equation

• 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

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sRate Equation

• Write the form of the rate equation for the oxidation of ethanol (C2H5OH)

• The reaction has been experimentally determined to be first order in ethanol and third order in oxygen– Rate expression involves only the reactants

– Concentrations: [C2H5OH][O2]– Raise each to exponent corresponding to its order

rate = k [C2H5OH][O2]3

– Remember that 1 as an exponent is understood and NOT written

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

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sWriting Rate Equations

7.4 Equilibrium

Rate and Reversibility of Reactions• Equilibrium reactions - chemical reactions

that do not go to completion– Completion – all reactants have been converted

to products– Equilibrium reactions are also called incomplete

reactions– Seen with both physical and chemical processes

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

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Physical Equilibrium• Physical equilibria are reversible reactions

– Dissolved oxygen in lake water– Stalactite and stalagmite formation– Sugar dissolved in water

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

• Dynamic equilibrium - the rate of the forward process in a reversible reaction is exactly balanced by the rate of the reverse process

1. If you add 2-3 g of sugar into 100 mL water• All will dissolve with stirring in a short time• No residual solid sugar, sugar dissolved completely

Sugar(s) Sugar(aq)

2. If add 100 g of sugar in 100 mL of water

• Not all of it will dissolve even with much stirring

• Over time, you observe no further change in the amount of dissolved sugar

• Appears nothing is happening – Incorrect!

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Sugar in Water

Appears nothing is happening is incorrect!– Individual sugar molecules are constantly going into and

out of solution

– Both happen at the same rate

– Over time the amount of sugar dissolved in the measured volume of water does not change

• An equilibrium situation has been established

• Some molecules dissolve and others return to the solid state – the rate of each process is equal

Sugar(s) Sugar(aq)

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Sugar in Water

sugar(s) sugar(aq)

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Dynamic Equilibrium

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

• Continuous change is taking place without observable change in the amount of sugar in either the solid or the dissolved form

)][sugar(

)][sugar(Keq s

aq

k

k

r

f ==

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• ratef = forward rate rater = reverse rate

• at equilibrium: ratef = rater

• ratef = kf[sugar(s)]

• rater = kr[sugar(aq)]

• kf[sugar(s)]=kr[sugar(aq)]• Equilibrium constant (Keq) - ratio of the two rate

constants

Equilibrium Constant

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Chemical Equilibrium

The Reaction of N2 and H2

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

• Mix components at elevated temperature

• Some molecules will collide with sufficient energy to break N-N and H-H bonds

• Rearrangement of the atoms will produce the product NH3

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Chemical Equilibrium

• Initially the forward reaction is rapid– Reactant concentrations are

high– Product concentration

negligible

• Forward reaction rate decreases with time– Concentrations of reactants

are decreasing– Product concentration

increasing

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

Equilibrium occurs when the rate of reactant depletion is equal to the rate of product depletion Rates of forward and reverse reactions are Equal

Chemical Equilibrium

Basic equation divides into 2 parts:

• forward rxn: N2(g) + 3H2(g) 2NH3(g)

• reverse rxn: 2NH3(g) N2(g) + 3H2(g)

• ratef = kf[N2]n[H2]m

• rater = kr[NH3]p

• ratef = rater

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

m2

n2

p3

eq ]H[][N

][NHK ==

r

f

kk

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m2

n2

p3

eq ]H[][N

][NHK ==

r

f

kk

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

• Keq is a constant at constant temperature

322

23

eq ]H][[N

][NHK =

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Chemical Equilibrium

ba

dc

B][[A]

D][C][Keq =

aA + bB cC + dD

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The Generalized Equilibrium-Constant Expression for a

Chemical Reaction

• A and B are reactants

• C and D are products

• a, b, c, and d are the coefficients of the balanced equation

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Writing Equilibrium-Constant Expressions

• Equilibrium constant expressions can only be written after a correct, balanced chemical equation

• 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

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Writing an Equilibrium-Constant Expression

Write an equilibrium-constant expression for the reversible reaction:

H2(g) + F2(g) 2HF(g)•No solids or liquids are present

– All reactants and products appear in the expression– Numerator term is the product term [HF]2

– Denominator term is the reactants [H2] and [F2] – Each term contains an exponent identical to the

corresponding coefficient in the balanced equation

Keq = [HF]2

[H2][F2]

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Writing an Equilibrium-Constant Expression

Write an equilibrium-constant expression for the reversible reaction:

MnO2(s) + 4H+(aq) + 2Cl-(aq) Mn2+(aq) + Cl2(g) + H2O(l)

• MnO2 is a solid

• H2O(l) is a product, but negligible compared to solvent water

– Numerator term is the product terms [Mn2+] and [Cl2]

– Denominator term is the reactants [H+]4 and [Cl-]2

– Each term contains an exponent identical to the corresponding coefficient in the balanced equation

Keq = [Mn2+] [Cl2] [H+]4 [Cl-]2

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Interpreting Equilibrium Constants

• Reversible arrow in chemical equation indicates equilibrium exists

• 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 indicates numerator (product term) >>> denominator (reactant term)

• At equilibrium mostly product present

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Interpreting Equilibrium Constants

2. Keq less than 1 x 10-2

• Small value of Keq indicates numerator (product term) <<< denominator (reactant term)

• At equilibrium mostly reactant present

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

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

2NO2(g) N2O4(g)

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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

• Analyze reaction mixture to determine molar concentrations of each product and reactant

HI placed in a sealed container and comes to equilibrium; equilibrium reaction is:

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

•Equilibrium concentrations:– [HI] = 0.54 M– [H2] = 1.72 M– [I2] = 1.72 M

•Substitute concentrations:7.4

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Calculating an Equilibrium Constant

Keq = [H2] [I2] [HI]2

Keq= [1.72] [1.72] = 2.96

[0.54]2 0.29 = 10.1 or 1.0 x 101

2 significant figures

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Le Chateleir’s Principle

• Le Chateleir’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

• If reactants and products are present in a fixed volume and more NH3 is added into the container, the system will be stressed

– Stressed = the equilibrium will be disturbed

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

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Le Chateleir’s Principle

• Adding NH3 to the system causes stress– To relieve stress, remove as much of added

material as possible by converting it to reactants

• Adding N2 or H2 to the system causes stress also– To relieve stress, remove as much of added

material as possible by converting it to product N2(g) + 3H2(g) 2NH3(g)

Equilibrium shiftedProduct introduced:Reactant introduced:

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

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Effect of Concentration

• Adding or removing either reactants or products at a fixed volume is saying that the concentration is changed

• Removing material decreases concentration

• System will react to this stress to return concentrations to the appropriate ratio

A B C

A: Reaction at equilibriumB: Shift to reactant with more red colorC: Shift to product with loss of red color

• Exothermic reactions: treat heat as a product

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

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Effect of Heat

• Addition of heat is treated as increasing the amount of product

• More product shifts equilibrium to the left

– Increases amount of reactants

– Decreases amount of product Heat favors the blue species

while cold favors the pink

• Endothermic Reaction – treat heat as a reactant

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

• This reaction shift will shift to the right if heat is added by increasing the temperature

7.4

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Effect of Heat

7.4

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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

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

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

– Increased pressure favors decreased volume with more product (2 moles) formed and less reactant (4 moles)

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

• If pressure increases in this reaction, which way will the equilibrium shift?

– No shift in equilibrium as both reactant and product have 2 moles of gas

7.4

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Effect of Pressure

7.4

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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

• While equilibrium composition and concentration do not change, equilibrium is reached in a shorter time