Chapter 5 Thermochemistry. Topics Energy and energy changes Introduction to thermodynamics Enthalpy Calorimetry Hess’s Law Standard enthalpies

Chapter 5 Chapter 5 ThermochemistryThermochemistry

TopicsTopics

Energy and energy changesEnergy and energy changes Introduction to thermodynamicsIntroduction to thermodynamics EnthalpyEnthalpy CalorimetryCalorimetry Hess’s LawHess’s Law Standard enthalpies of formationStandard enthalpies of formation

Copyright McGraw-Hill 2009Copyright McGraw-Hill 200933

5.15.1 Energy and Energy ChangesEnergy and Energy Changes

Energy is involved in all types of Energy is involved in all types of physical and chemical changes physical and chemical changes

EnergyEnergy: the capacity to do work or : the capacity to do work or transfer heattransfer heat

All forms of energy are eitherAll forms of energy are either KineticKinetic Potential Potential

Kinetic energyKinetic energy - energy of motion - energy of motion Defining equationDefining equation

Where Where mm is mass and is mass and uu is velocity is velocity Thermal energyThermal energy - one form of kinetic - one form of kinetic

energy associated with random energy associated with random motionmotion• Changes in thermal energy are Changes in thermal energy are

monitored via changes in monitored via changes in temperature temperature

Potential energyPotential energy - energy of - energy of position position or compositionor composition Chemical energyChemical energy is stored within is stored within

structural units of chemical structural units of chemical substances. substances.

Electrostatic energyElectrostatic energy is energy is energy resulting from the interaction of resulting from the interaction of charged particles. charged particles. •Dependent on charges and Dependent on charges and distance between charges (distance between charges (QQ = = charge and charge and dd = distance = distance) )

•Defining equationDefining equation

•+ + EEelel: repulsive: repulsive

•− − EEelel: attractive: attractive

Law of conservation of energyLaw of conservation of energy Energy can be converted from Energy can be converted from

one form to another but not one form to another but not created or destroyed. created or destroyed. The total amount of energy in the The total amount of energy in the

universeuniverse is constant. is constant. Example Example

•A chemical reaction (potential) A chemical reaction (potential) gives off heat (thermal) gives off heat (thermal)

SystemSystem is the part of the universe of is the part of the universe of interest.interest.• ExampleExample

The reactants NaOH and HCl The reactants NaOH and HCl SurroundingsSurroundings are the are the rest of the rest of the

universeuniverse..• ExampleExample

When heat is given off from the When heat is given off from the reaction of NaOH and HCl, the reaction of NaOH and HCl, the energy is transferred from the energy is transferred from the system to the surroundings. system to the surroundings.

system: what is inside the container. system: what is inside the container. Reactants or products of a chemical Reactants or products of a chemical reactionreaction

surroundings: container ,room, etc.surroundings: container ,room, etc.

Energy changes in chemical reactionsEnergy changes in chemical reactions


The study of the transfer of heat The study of the transfer of heat (thermal energy) in chemical reactions(thermal energy) in chemical reactions. .

ExothermicExothermic - transfer of heat from the - transfer of heat from the system to the surroundings system to the surroundings

2H2H22((gg) + O) + O22((gg) ) 2H 2H22O(O(ll) ) + energy+ energy EndothermicEndothermic - the transfer of heat from - the transfer of heat from

the surroundings to the system the surroundings to the system

energy energy + 2HgO(+ 2HgO(ss) ) 2Hg( 2Hg(ll) + O) + O22((gg))

ThermochemistryThermochemistry

Transfer of EnergyTransfer of Energy

ExothermicExothermic processprocess energy is produced in reactionenergy is produced in reaction energy flows out of systemenergy flows out of system container becomes hot when touchedcontainer becomes hot when touched CHCH44(g) + 2O(g) + 2O22(g) CO(g) CO22(g) + 2H(g) + 2H22O(g) + heatO(g) + heat

Endothermic processEndothermic process energy is consumed by the reactionenergy is consumed by the reaction energy flows into the systemenergy flows into the system container becomes cold when touchedcontainer becomes cold when touched NN22(g) + O(g) + O22(g) + Energy 2NO(g) (g) + Energy 2NO(g)

http://college.hmco.com/chemistry/shared/media/zumdahl/visual/visual.html?name=CH09&vis=_v9a

Comparison of Endothermic and Exothermic Comparison of Endothermic and Exothermic ProcessesProcesses

JouleJoule (J) is the SI unit for energy. (J) is the SI unit for energy. •The amount of energy The amount of energy possessed by a 2 kg mass possessed by a 2 kg mass moving at a speed of 1 m/s moving at a speed of 1 m/s

Units of EnergyUnits of Energy


CalorieCalorie (cal) (cal) - commonly - commonly used on food labels used on food labels

1 cal 4.184 J 1 cal 4.184 J 1000 cal = 1 Cal = 1 kcal1000 cal = 1 Cal = 1 kcal Food calories (Cal) are really Food calories (Cal) are really

1000 calories (cal). 1000 calories (cal).

Calculate the kinetic energy of a neon atom Calculate the kinetic energy of a neon atom

moving at a speed of 98 m/smoving at a speed of 98 m/s. .

226k kg)(98m/s)10(3.352

2

1 E

J101.6 22k

E

kg103.352g 1000

kg 1amu

g101.661amu 20.18 26

24

Example

5.25.2 Introduction to ThermodynamicsIntroduction to Thermodynamics

Types of systems: Types of systems: openopen (exchange of mass and (exchange of mass and

energy)energy) closedclosed (exchange of energy) (exchange of energy) isolatedisolated (no exchange) (no exchange)

State functions depend only on initial and State functions depend only on initial and final states of the system and not on how final states of the system and not on how the change was carried out. the change was carried out. Energy (Energy (EE)) Pressure (Pressure (PP)) Volume (Volume (VV)) Temperature (Temperature (TT) )

StateState functionfunction

State FunctionState Function

energy changeenergy change is is independentindependent of of pathway thus it is a pathway thus it is a state functionstate function

(state property)(state property) State function: a property of a system State function: a property of a system

that depends only on its present state that depends only on its present state (it does not depend on system(it does not depend on system’’s past or s past or future, i.e, it does not depend on how future, i.e, it does not depend on how the system arrived at the present the system arrived at the present state)state)

work and heatwork and heat dependdepend on pathway so on pathway so they they are not state functions

First Law of ThermodynamicsFirst Law of Thermodynamics

Energy can be converted from one form to Energy can be converted from one form to another but cannot be created or destroyed. another but cannot be created or destroyed. Based on the law of conservation of energyBased on the law of conservation of energy

Internal energyInternal energy ( (UU))

It is the sum of potential and kinetic energy of It is the sum of potential and kinetic energy of all particles in the systemall particles in the system

U = PU = PEE + K + KEE

Kinetic energyKinetic energy - molecular motion - molecular motion Potential energy Potential energy - attractive/repulsive - attractive/repulsive

interactionsinteractions

The The change in internal energy of a change in internal energy of a system system between final (f) and initial (i) between final (f) and initial (i) states is defined as: states is defined as:

UU = = UUff UUii

For a chemical systemFor a chemical system CannotCannot calculate the calculate the total internal total internal

eenergy with any certaintynergy with any certainty CanCan calculate the calculate the changechange in energy in energy of of

the system experimentallythe system experimentally

UU = = UU(products)(products) UU(reactants)(reactants)

Consider:Consider:

S(S(ss)) + O + O22((gg)) SO SO22((gg))

This reaction releases heat, This reaction releases heat, therefore therefore UU is negative. is negative.

UUsystemsystem + + UUsurroundings surroundings = 0 = 0

When a system releases heat, some When a system releases heat, some of the chemical energy is released as of the chemical energy is released as thermal energy thermal energy to the surroundings to the surroundings but this but this does not change the total does not change the total energy of the universe. energy of the universe.

UUsystemsystem = = UUsurroundingssurroundings

When a system undergoes a change in When a system undergoes a change in energy, the surroundings must energy, the surroundings must undergo a change in energy undergo a change in energy equal in equal in magnitude and opposite in sign. magnitude and opposite in sign.

where where qq is heat, is heat, ww is work is work

Work and heatWork and heat UU can be changed (- or + can be changed (- or + ∆U)∆U) by a by a

flow of work, heat, or bothflow of work, heat, or both

UUsyssys = q + w = q + w

Sign Conventions of q and w

Calculate the overall change in Calculate the overall change in internalinternal

energy for a system that absorbs 125 energy for a system that absorbs 125 J ofJ of

heat and does 141 J of work on theheat and does 141 J of work on the

surroundingssurroundings . .

qq is + (heat absorbed) is + (heat absorbed)

ww is is (work done) (work done)

UUsyssys = = qq + + ww = (+125 J) + ( = (+125 J) + (141J) 141J) = = 16 J16 J

WorkWork

common types of common types of workwork ExpansionExpansion-- work work

done by the systemdone by the system CompressionCompression-- work work

done on the systemdone on the system

A

FP

hAPhFw

VPw

expansionexpansion+∆V+∆V-w-w

compressiocompressionn

-∆V-∆V+w+w

PP is external is external pressurepressure

Work is force applied over distance

Pressure-volume work, w, done by a Pressure-volume work, w, done by a system issystem is

w = w = PPVV UUsyssys = q + w = q + w UUsyssys = q = q P P V V

Constant volume, Constant volume, V = 0 V = 0 qqvv = = UUsyssys

qqv v means a constant-volume processmeans a constant-volume process

5.35.3 Reactions carried out at constant Reactions carried out at constant volumevolume

or at constant pressureor at constant pressure

UU = = qq + + ww

UU = = qqpp PPVV

qqpp = = UU + P + PVV

Reactions carried out at constant Reactions carried out at constant pressurepressure

Enthalpy and enthalpy changesEnthalpy and enthalpy changes

Enthalpy is a property of a systemEnthalpy is a property of a system definition: H definition: H = = U + PV U + PV since since U, P and V are all state functions, then H is U, P and V are all state functions, then H is

also a state functionalso a state function

The term The term enthalpyenthalpy is composed of the prefix is composed of the prefix en-en-, meaning ", meaning "to put intoto put into" and the " and the Greek word word --thalpeinthalpein, meaning , meaning "to heat","to heat", that is that is “to put heat into”

http://en.wikipedia.org/wiki/Greek_language

Enthalpy changeEnthalpy change

H = U + PV

The change in enthalpy

H = U + (PV)

When pressure is held constant

H = U + PV

Since

qp = U + PV

qp = H

At constant pressureAt constant pressure

qqpp = = HH Enthalpy of reactionEnthalpy of reaction

HH is + for endothermic changes. is + for endothermic changes. HH is − for exothermic changes. is − for exothermic changes.


Equations that represent both mass and Equations that represent both mass and enthalpy changesenthalpy changes

Endothermic reactionEndothermic reactionHH22O(O(ss)) H H22O(O(ll)) HH = + 6.01 kJ/mol = + 6.01 kJ/mol

• Exothermic reaction Exothermic reaction

Thermochemical EquationsThermochemical Equations


Comparison of Endothermic and Exothermic Changes

Always specify state of reactants Always specify state of reactants and products. and products.

When multiplying an equation by When multiplying an equation by a factor (a factor (nn), multiply the ), multiply the HH value by same factor. value by same factor.

Reversing an equation changes Reversing an equation changes the sign but not the magnitude the sign but not the magnitude of the of the HH. .

Thermochemical equation guidelines


Given the following equationGiven the following equation , ,

CC66HH1212OO6(6(ss) ) + 6O+ 6O2(2(gg) ) → 6CO→ 6CO2(2(gg) ) + 6H+ 6H22OO((ll))

ΔΔHH = = 2803 kJ/mol 2803 kJ/mol

calculate the energy released when calculate the energy released when 45.00 g of glucose is burned in 45.00 g of glucose is burned in oxygenoxygen..

kJ 700.0OHC mol 1

kJ 2803

OHC g 180.2

OHC mol 1OHC g 45.00

61266126

61266126

5.45.4 CalorimetryCalorimetry

The science of measuring The science of measuring

heat changesheat changes

calorimetercalorimeter-- device used to device used to experimentally find the heat associated experimentally find the heat associated with a chemical reactionwith a chemical reaction

substances respond differently when substances respond differently when heated ( heated ( to raise T for two substances to raise T for two substances by 1 degree, they require different by 1 degree, they require different amount of heat) amount of heat)

Specific heat Specific heat ((ss)) Specific heat Specific heat ((ss) ) - the amount of - the amount of

heat required to raise the temp of 1 heat required to raise the temp of 1 g of a substance by 1g of a substance by 1C. C. Units: J/g Units: J/g CC Relation to amount of heat (Relation to amount of heat (qq))

where qwhere q is heat, is heat, mm is mass, is mass, ss is is specific heatspecific heat

and and TT = change in temp = change in temp

((TT = = TTfinalfinal TTinitialinitial))



Heat capacityHeat capacity ( (CC) ) - the amount of - the amount of heat required to raise the temp of heat required to raise the temp of an object by 1an object by 1C. C. Units: J/Units: J/CC Relation to amount of heat (Relation to amount of heat (qq))

where qwhere q is heat, is heat, CC is heat capacity is heat capacity

and and TT = change in temp = change in temp

((TT = = TTfinalfinal TTinitialinitial))

Calculate the amount of energy Calculate the amount of energy requiredrequired

to heat 95.0 grams of water from 22.5to heat 95.0 grams of water from 22.5CCto 95.5to 95.5C. C.

TT = = TTfinalfinal – – TTinitialinitial = 95.5 = 95.5 ooC − 22.5C − 22.5ooC C

TT = 73.0 = 73.0 ooCC

qq = (95.0 g) (4.184 J/g = (95.0 g) (4.184 J/gC) (73.0C) (73.0C)C)

qq = 2.90 x 10 = 2.90 x 1044 J or 29.0 kJ J or 29.0 kJ

Constant-Pressure CalorimetryConstant-Pressure Calorimetry uses simplest calorimeter (like coffee-cup uses simplest calorimeter (like coffee-cup

calorimeter) since it is open to aircalorimeter) since it is open to air used to find heat exchangeused to find heat exchange

between the system andbetween the system and

surroundings of a chemicalsurroundings of a chemical

reactionreaction

i.e., changes in enthalpyi.e., changes in enthalpy

for reactions occurring for reactions occurring

in a solutionin a solution

since since qqPP = ∆H = ∆H

System: Reactants & ProductsSystem: Reactants & Products

Surroundings: the water Surroundings: the water

System: System: reactants and productsreactants and products

SurroundingsSurroundings: water : water Assuming that the calorimeter Assuming that the calorimeter does not leak or absorb heat does not leak or absorb heat

coffee cup calorimeter

Constant-pressure calorimetryConstant-pressure calorimetryFor an exothermic reaction: For an exothermic reaction:



A metal pellet with a mass of 85.00 grams at an A metal pellet with a mass of 85.00 grams at an originaloriginal

temperature of 92.5temperature of 92.5C is dropped into a calorimeter\C is dropped into a calorimeter\

with 150.00grams of water at an original with 150.00grams of water at an original temperature oftemperature of

23.123.1C. The final temperature of the water and the C. The final temperature of the water and the pelletpellet

Is 26.8Is 26.8C. Calculate the heat capacity and the C. Calculate the heat capacity and the specificspecific

heat for the metal. heat for the metal.

Example

qqwaterwater = = msmsTT

= (150.00 g) (4.184 J/g= (150.00 g) (4.184 J/gC) (3.7C) (3.7C)C)

= 2300 J (water gained energy) = 2300 J (water gained energy)

= -2300 J (pellet released = -2300 J (pellet released energy) energy)

Heat capacity of pellet: Heat capacity of pellet: qq = = CCTT

CC = = qq//TT

= = 2300 J/2300 J/65.765.7C = 35 J/C = 35 J/CC Specific heat of pellet: J/gSpecific heat of pellet: J/gooCC

s = 35 J/oC85.00 g

= 0.41J/goC

bomb calorimeter

Constant-volume calorimetry

Isolated system

Here, the ∆V = 0 so -P∆V = w = 0 Here, the ∆V = 0 so -P∆V = w = 0

∆∆U = q + w = qU = q + w = qVV for constant for constant volumevolume

Typical procedure used in a Typical procedure used in a bomb calorimeterbomb calorimeter

• Known amount of sample placed in Known amount of sample placed in steel container and then filled with steel container and then filled with oxygen gasoxygen gas

• Steel chamber submerged in known Steel chamber submerged in known amount of wateramount of water

• Sample ignited electrically Sample ignited electrically

• Temperature increase of water is Temperature increase of water is determined determined


A snack chip with a mass of 2.36 g A snack chip with a mass of 2.36 g was burned in a bomb calorimeter. was burned in a bomb calorimeter. The heat capacity of the The heat capacity of the calorimeter 38.57 kJ/calorimeter 38.57 kJ/C. During the C. During the combustion the water temp rose by combustion the water temp rose by 2.702.70C. Calculate the energy in kJ/g C. Calculate the energy in kJ/g for the chipfor the chip


qqrxn rxn = − = − CCcalcal TT

(− = (− = 38.5738.57 kJ/kJ/CC( )( )2.702.70CC ) )

− = − = 104104 kJkJ

Energy content is a positive Energy content is a positive quantityquantity..

= = 104104 kJ/2.36 gkJ/2.36 g

= = 44.144.1 kJ/gkJ/g

Food Calories: 10.5 Cal/gFood Calories: 10.5 Cal/g

Example 3Example 3

Compare the energy released in the Compare the energy released in the combustion of Hcombustion of H22 and CH and CH44 carried out in carried out in a bomb calorimeter with a heat capacity a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of of methane produced a T change of 7.3°C while the combustion of 1.15 g of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of hydrogen produced a T change of 14.3°C. Find the energy of combustion 14.3°C. Find the energy of combustion per gram for each. per gram for each.

Example 3Example 3 methane: CHmethane: CH44

hydrogen: Hhydrogen: H22

The energy released by HThe energy released by H22 is about 2.5 is about 2.5 times the energy released by CHtimes the energy released by CH44

gkJgkJH

kJCC

kJH

TCH rcalorimete

/555.1/83

83)3.7()3.11(

gkJgkJH

kJCC

kJH

TCH rcalorimete

/14115.1/162

162)3.14()3.11(

5.55.5 Hess’s LawHess’s Law

Hess’s LawHess’s Law: The change in enthalpy : The change in enthalpy that occurs when reactants are that occurs when reactants are converted to products is the same converted to products is the same whether the reaction occurs in one whether the reaction occurs in one step or a series of stepsstep or a series of steps . .

It is sed for calculating enthalpy for a It is sed for calculating enthalpy for a reaction that cannot be determined reaction that cannot be determined

directlydirectly . .

Example 1Example 1

NN22(g) + 2O(g) + 2O22(g) (g) 2NO 2NO22(g)(g) ∆H = 68 kJ∆H = 68 kJ

OROR

NN22(g) + O(g) + O22(g) (g) 2NO(g) 2NO(g) ∆H = 180 kJ∆H = 180 kJ

2NO(g) + O2NO(g) + O22(g) (g) 2NO 2NO22(g)(g)∆H = -112 kJ∆H = -112 kJ

NN22(g) + 2O(g) + 2O22(g) (g) 2NO 2NO22(g)(g) ∆H = 68 kJ∆H = 68 kJ

RulesRules

1.1. If a reaction is If a reaction is reversedreversed, the sign , the sign of ∆H must be reversed as well.of ∆H must be reversed as well.

because the sign tells us the because the sign tells us the direction of heat flow at constant Pdirection of heat flow at constant P

2.2. The magnitude of The magnitude of ∆H is directly∆H is directly proportional to proportional to quantitiesquantities of of reactants and products in reactants and products in reaction. reaction.

If coefficients are multiplied by an If coefficients are multiplied by an integer, the integer, the ∆H must be multiplied ∆H must be multiplied in the same wayin the same way..

because ∆H is an extensive propertybecause ∆H is an extensive property


Given the following equationsGiven the following equations::

HH33BOBO33((aqaq) ) HBO HBO22((aqaq) + H) + H22O(O(ll)) HHrxnrxn = = 0.02 0.02 kJkJ

HH22BB44OO77((aqaq) + H) + H22O(O(ll) ) 4 HBO 4 HBO22((aqaq)) HHrxn rxn = = 11.3 kJ11.3 kJ

HH22BB44OO77((aqaq) ) 2 B 2 B22OO33((ss) + H) + H22O(O(ll)) HHrxnrxn = 17.5 kJ = 17.5 kJ

Find the Find the HH for this overall reaction for this overall reaction..

2H2H33BOBO33((aqaq) ) B B22OO33((ss) + 3H) + 3H22O(O(l)l)

2H2H33BOBO3(3(aqaq)) 2HBO 2HBO2(2(aqaq)) + 2H + 2H22OO((ll)) x 2x 2

HHrxnrxn = 2(−0.02 kJ) = −0.04 kJ = 2(−0.02 kJ) = −0.04 kJ

2HBO2HBO2(2(aqaq)) 1/2H1/2H22BB44OO7(7(aqaq)) + 1/2H + 1/2H22OO((ll)) reverse, reverse, ÷2÷2

HHrxn rxn = +11.3 kJ/2 = 5.65 kJ = +11.3 kJ/2 = 5.65 kJ

1/2H1/2H22BB44OO7(7(aqaq)) B B22OO3(3(ss))+ 1/2H+ 1/2H22OO((ll) ) ÷÷ 22

HHrxnrxn = 17.5 kJ/2 = 8.75 kJ = 17.5 kJ/2 = 8.75 kJ

2H2H33BOBO33((aqaq) ) B B22OO33((ss) + 3H) + 3H22O(O(l)l)

HHrxnrxn = 14.36 kJ = 14.36 kJ


5.65.6 Standard Enthalpies of Standard Enthalpies of FormationFormation

Symbol: Symbol: HHff

The enthalpy change that results The enthalpy change that results when 1 mole of a compound is when 1 mole of a compound is formed from its elements in their formed from its elements in their standard states. standard states. HHff

for an element in its standard for an element in its standard state is defined as zero.state is defined as zero.

Standard stateStandard state: 1 atm, 25: 1 atm, 25C C Values found in reference tablesValues found in reference tables Used to calculate the Used to calculate the HH

rxnrxn

Standard StatesStandard States For a For a compound::

for gas: P = 1 atmfor gas: P = 1 atm For pure substances, it is a pure For pure substances, it is a pure

liquid or pure solid stateliquid or pure solid state in solution: concentration is 1 Min solution: concentration is 1 M

For an For an element:: form that exists in at 1 atm and form that exists in at 1 atm and

25°C25°C

O: OO: O22(g)(g)K: K(s)K: K(s) Br: Br: BrBr22(l)(l)


HHrxnrxn

= = nnHHff (products) (products)

mmHHff(reactants)(reactants)

where where denotes summation denotes summation

nn and and m m are the coefficients in theare the coefficients in the

balanced equation balanced equation

Defining equation for enthalpy of reaction:

Calculate the Calculate the HHrxnrxn

for the following reactionfor the following reaction

from the table of standard values. from the table of standard values.

CHCH44((gg)) + 2O + 2O22((gg)) CO CO22((gg)) + 2H+ 2H22O(O(ll))

HHrxnrxn == nnHHff

(products) (products) - - mmHHff

(reactants)(reactants)

= [1(= [1(393.5) + 2(393.5) + 2(285.8)] 285.8)] [1( [1(74.8) + 74.8) + 2(0)]2(0)]

= = 890.3 kJ/mol (exothermic) 890.3 kJ/mol (exothermic)

Example:


Key PointsKey Points

First law of thermodynamics First law of thermodynamics Enthalpy (heat of formation; heat of Enthalpy (heat of formation; heat of

reaction) reaction) State function State function Calorimetry Calorimetry Specific heat Specific heat Hess’s law Hess’s law Calculations involving enthalpy, Calculations involving enthalpy,

specific heat, energy specific heat, energy

Documents

Chapter 5 Thermochemistry. Topics Energy and energy changes Introduction to thermodynamics Enthalpy Calorimetry Hess’s Law Standard enthalpies