Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

Jeffrey MackJeffrey MackCalifornia State University,California State University,

SacramentoSacramento

Chapter 5Chapter 5

Principles of Chemical Principles of Chemical Reactivity: Reactivity: Energy and Chemical Energy and Chemical ReactionsReactions

Questions that need to be addressed:

•How do we measure and calculate the energy changes that are associated with physical changes and chemical reactions?

•What is the relationship between energy changes, heat, and work?

•How can we determine whether a chemical reaction is product-favored or reactant-favored at equilibrium?

•How can we determine whether a chemical reaction or physical change will occur spontaneously, that is, without outside intervention?

Energy & ChemistryEnergy & Chemistry

1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C.

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food “calorie”)

SI units for energy: joule (J)

1 cal = exactly 4.184 J

James Joule1818-1889

Units of EnergyUnits of Energy

Some Basic PrinciplesSome Basic Principles

•Energy as the capacity to do work or transfer heat (q).•Heat (q) is NOT temperature, temperature is a measure of kinetic energy.•Energy is divided into two basic categories:

– Kinetic energy (the energy associated with motion)

– Potential energy (energy that results from an object’s position).

•The law of conservation of energy requires that energy can neither be created nor destroyed. •However, energy can be converted from one type into another.


Energy can be divided into two forms:

Kinetic: Energy of motion:

Thermal

Mechanical

Electrical

Potential: Stored energy:

Gravitational

Electrostatic

Chemical & Nuclear


Kinetic energyKinetic energy — energy of motion— energy of motion

• Translation

Potential & Kinetic EnergyPotential & Kinetic Energy

Potential energyPotential energy — energy a motionless body — energy a motionless body has by virtue of its composition and position.has by virtue of its composition and position.

Potential & Kinetic Energy Potential & Kinetic Energy

• Burning peanuts supply sufficient energy to boil a cup of water.

• Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)


• Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower temperature.

• Transfer of energy as heat continues until both objects are at the same temperature and thermal equilibrium is achieved.

• At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a temperature decrease has lost thermal energy.

Thermal EquilibriumThermal Equilibrium

• SYSTEMSYSTEM– The object under study

• SURROUNDINGSSURROUNDINGS– Everything outside the

system

• Energy flows between the two

System & SurroundingsSystem & Surroundings

• When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMICEXOTHERMIC..

• In the case of thermal energy, the temperature of the system decreases. (qsystem < 0)

Tsystem = (Tfinal – Tinitial) < 0

Directionality of Energy TransferDirectionality of Energy Transfer

• When energy enters the system and from the surroundings, the process is said to be ENDOTHERMICENDOTHERMIC..

• In the case of thermal energy, the temperature of the system increases. (qsystem > 0)

Tsystem > 0

Directionality of Energy TransferDirectionality of Energy Transfer

• Heat flows between the system and surroundings.

• U is defined as the internal energy of the system.

• q = the heat absorbed or lost

Heat & Changes in Internal EnergyHeat & Changes in Internal Energy

If heat enters the system:

U > 0 therefore qq is positive (+)

If heat leaves the system:

U < 0 therefore qq is negative (–)

The sign of q is a “conventionconvention”, it designates the direction of heat flow between the system and surroundings.


Surroundings

System

heat in

qsystem > 0(+)

Usystem > 0

heat out

qsystem < 0(–)

Usystem < 0increases decreases

q = the heat absorbed or lost by the system.

UUsystemsystem

UUsy

stem

syst

em


The amount of heat (q) transfer related to an object and temperature is given by:

When heat is absorbed or lost by a body, the temperature must change as long as the phasephase (s, g or l) remains constant.

q = heat lost or gained (J)

C = Heat Capacity of an object

J

C or K

T = Tfinal Tinitial is the temperature change (°C or K)

Heat CapacityHeat Capacity

The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by:

When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant.

q = heat lost or gained (J) m = mass of substance (g)

C = the Specific Heat Capacity of a compound J

g C or K

T = Tfinal Tinitial is the temperature change (°C or K)

Heat & Specific Heat CapacityHeat & Specific Heat Capacity

Recall that T = Tf –Tin

50.0 °C = 50.0 °C + 273.2 = 323.2 K

25.0 °C = (25.0 °C + 273.2) = 298.2 K

= 25.0 °C = 25.0 K

The are the same!

T T

Does it matter if we calculate a temperature change in Kelvin or degrees C?

Tf =

–Tin =

let Tin = 25.0 °C and Tf = 50.0 °C

Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 CThe specific heat capacity of the metal is:

J2.72

g C

Data Information: Mass, initial temp, heat capacity of metal, heat (q) absorbed.

Solve q = mCT for Tfinal, plug in data.

Final temperature of the metal is determined.

Find Tfinal of the metal after it absorbs the energy

Strategy Map:Strategy Map:

Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 CThe specific heat capacity of the metal is:

J2.72

g C

rearranging:

Tf =+255 cal

4.184 J

1 cal

25.0 gJ

2.72 g C

+ 17.0 °C = 32.7 °C

Tf > Tin as

expected

Example Problem:Example Problem:• 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially

at 21.0 °C.

• At thermal equilibrium, the water and iron are both at 23.1 °C

• What is the specific heat capacity of the metal?

Solution:Solution:

Heat is transferred from the hot metal to the colder water.

Energy is conserved so:


at 21.0 °C.



Solution:Solution:


at 21.0 °C.



Solution:Solution:


at 21.0 °C.



Solution:Solution:


at 21.0 °C.



Fe

J 1 K225 g 4.184 (23.1 C 21.0 C)

Jg K 1 CC 0.469

1 K g K55.0 g (23.1 C 99.8 C) 1 C

• When matter absorbs heat, its temperature will rise until it undergoes a Phase ChangePhase Change.

• The matter will continue to absorb energy, however during the phase change its temperature remains constant: Phase changes are “IsothermalIsothermal” processes.

Energy & Changes of StateEnergy & Changes of State

• Some changes of state (phase changes) are endothermic:

• When you perspire, water on your skin evaporates.

• This requires energy.• Heat from your body is absorbed by

the water as it goes from the liquid state to the vapor state, as a result you cool down.

+ energy

Energy Transfer & Changes of StateEnergy Transfer & Changes of State

• Some changes of state (phase changes) are exothermic:

• When it is muggy outside, water condenses on your skin.

• This releases energy.

• Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state.

• As a result you feel hot.

+ energy

Energy Transfer & Changes of StateEnergy Transfer & Changes of State

Heating/Cooling Curve for WaterHeating/Cooling Curve for Water

Note the isotherms at each phase

transition.

• The energy associated with a change of state is given by the Enthalpy or Heat of the phase change.

• Since there is no temperature change associated with the process, the units are most often in J/g or J/mol.

Sublimation: subH > 0 (endothermic)

Vaporization: vapH > 0 (endothermic)

Melting or Fusion: fusH > 0 (endothermic)

Deposition: depH < 0 (exothermic)

Condensation: conH < 0 (exothermic)

Freezing: freH < 0 (exothermic)

Where H refers to the “Heat” of a phase change

Heat & Changes of StateHeat & Changes of State

q(heating or cooling) = m C T

q(phase change) = (phase change)H n

Heating & Cooling:

Heat absorbed or lost in a Phase change:

(n = moles or grams) (n = moles or grams)

Energy Change CalculationsEnergy Change Calculations

Problem:Problem:

What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C?

+333 J/g +2260 J/g

melt the ice form liquid water at 0 °C heat the water to 100 °C boil water

fusH

Cwater

VapH

Equations:

qphase change = mH

qheat = mCT

Problem:Problem:


Ice H2O(s) (0 °C)

fusHwater H2O(l) (0 °C)

water H2O(l)

(100 °C)

Cwater vapHsteam H2O(g)

(100 °C)

Problem:Problem:


qtotal =

melt ice heat water boil water

Problem:Problem:


Ice H2O(s) (0 °C)


water H2O(l)

(100 °C)


(100 °C)

total

6

Jq 500. g 333

g

J 500. g 4.18 (100.0 C 0.0 C)

g C

J 500. g 2260 1.51 10 J

g

total

6

Jq 500. g 333

g

J 500. g 4.18 (100.0 C 0.0 C)

g C

J 500. g 2260 1.51 10 J

g

Problem:Problem:


Ice H2O(s) (0 °C)


water H2O(l)

(100 °C)


(100 °C)

The total internal energy of an isolated system is constant.

The “systemsystem” is that which we are interested in.

The “surroundingssurroundings” are everything in contact with the system.

Together:System + Surroundings = Universe

First Law of ThermodynamicsFirst Law of Thermodynamics

∆Usystem = Ufinal – Uinitial

Any energy lost by the system is transferred to the surroundings and vice versa.

Any change in energy is related to the final and initial states of the system.

The same holds for the surrounding!

EnergyEnergy

EnergyEnergy

First Law of ThermodynamicsFirst Law of Thermodynamics

Energy cannot be created or destroyed.

Energy of the universes (system + surroundings) is constant.

Any energy transferred from a system must be transferred to the surroundings (and vice versa).

From the first law of thermodynamics: When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system:

U q w change in

system energy

=heat

lost or gainedby the system

+work done by or on the system

Relating Relating U to Heat and WorkU to Heat and Work

energy transfer outenergy transfer out(exothermic), -q(exothermic), -qenergy transfer outenergy transfer out(exothermic), -q(exothermic), -q

energy transfer inenergy transfer in(endothermic), +q(endothermic), +qenergy transfer inenergy transfer in(endothermic), +q(endothermic), +q

SYSTEM

∆U = q + w

w transfer inw transfer in(+w)(+w)w transfer inw transfer in(+w)(+w)

w transfer outw transfer out(-w)(-w)w transfer outw transfer out(-w)(-w)

U initial

U Final

ener

gy

q in

work in

Uf > Ui

Usystem > 0 (+)

q out

work out

Uf < Ui

Usystem < 0 (–)

U Final

U initial

Energy in Energy out

Heat

Work

Usystem = 0

work and heat can balance!

Energy is the capacity to do work.Work equals a force applied through a distance.When you do work, you expend energy.

When you push down on a bike pump, you do work.

change in volume V = x3

32

FP V = x = F x

x

P × V = work

Therefore work is equal to a change of volume at constant pressure.

Enthalpy, ““HH”” is the heat transferred between the system and surroundings under conditions of constant pressure.

H = qp the subscript the subscript ““pp”” indicates constant indicates constant pressurepressure

if no if no ““PVPV”” work is work is done by the done by the system, system, V = 0V = 0

0

H = Up

the change in the change in enthalpy is the enthalpy is the change in internal change in internal energy at constant energy at constant pressurepressure

EnthalpyEnthalpy

For a “system” the overall change in EnthalpyEnthalpy is path independent.

No matter which path is taken (AB) the results are the same:

Final – Initial

Since individual Enthalpies cannot be directly measured, we only deal with enthalpy changes

(H = Hf – Hi)

A

B

Enthalpy is a Enthalpy is a “State“State Function”Function”

• Since Enthalpies are state functions, one must specify the conditions at which they are measured.

• H(T,P): Enthalpy is a function of temperaturetemperature and pressurepressure.

• “H°” indicates that the Enthalpy is taken at Standard StateStandard State conditions.

• Standard State ConditionsStandard State Conditions are defined as:

• 1 atm = 760 mm Hg or 760 torr & 298.15 K or 25 °C

Enthalpy ConditionsEnthalpy Conditions

Reactants Products

Ene

rgy

H (Products)

H (Reactants) H (Products)

H (Reactants)

rH > 0 rH < 0

EndothermicEndothermic ExothermicExothermic

H = Hfinal Hinitial

Enthalpy of reaction = rH = Hproducts Hreactants

Enthalpies & Chemical Reactions: Enthalpies & Chemical Reactions: rrHH

Just like a regular chemical equation, with an energy term.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ+ 802 kJ

What does this imply?What does this imply? CONVERSION FACTORS!!!CONVERSION FACTORS!!!

From the equation:

energy out…Exothermic

Energy is a product just like CO2 or H2O!

Thermochemical EquationsThermochemical Equations

Problem:Problem: How many kJ of energy are released when 128.5 g of methane, CH4(g) is combusted?

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

6.43103 kJ

g mols J

molar molar massmass

ReactionReactionenthalpyenthalpy

rHo = 802 kJ

When the reaction is reversed , the sign of H reverses:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) rHo = +802 kJ

EndothermicEndothermic


ExothermicExothermic


[ ][ ]

H scales with the reaction:


rHo = 401 kJ

Yes you can write the reaction with fractions, so long as you are writing it on a mole basis…


Change in enthalpy depends on state:

H2O(g) H2O(l)

This means that waterThis means that water’’s liquid state lies 44 s liquid state lies 44 kJ/mol lower than the gas statekJ/mol lower than the gas state

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rH° = 802 kJ

rH° = 890 kJ

H = 44 kJ/mol

= 88kJ

From your text:

[ ] 2 = 88kJThe difference in the rxn His due to the change in state!!


CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

CO2(g) + 2H2O(l)

rHo = 802 kJ

rHo = 890 kJ

2 44 kJ

The difference is the The difference is the 88 88 kJ released when 2 mols kJ released when 2 mols of water go from gas to of water go from gas to liquid.liquid.

Comparing the reaction with water as a gas or liquid:

Ent

halp

y (H

)

productsproducts

productsproducts

reactantsreactants

either pathway either pathway gives the same gives the same results!results!

• A constant pressure calorimeter can be used to measure the amount of energy transferred as heat under constant pressure conditions, that is, the enthalpy change for a chemical reaction.

Constant Pressure Calorimetry, Constant Pressure Calorimetry, Measuring Measuring HH

• The constant pressure calorimeter used in general chemistry laboratories is often a “coffee-cup calorimeter.” This inexpensive device consists of two nested Styrofoam coffee cups with a loose-fitting lid and a temperature-measuring device such as a thermometer or thermocouple.


• Because the “coffee-cup calorimeter.” is an isolated system, “What happens in the coffee cup, stays in the coffee cup!”

• No mass loss to the surroundings.

• No heat loss to the surroundings.


Problem:Problem:• 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-

cup calorimeter. • This resulted in a decrease in temperature from 18.6 °C to 16.2 °C. • Calculate the enthalpy change for dissolving NH4NO3(s) in water in

kJ/mol. • Assume the solution has a specific heat capacity of 4.18 J/g ? K.




Data Information: Mass of reactant, C, water & temperature change.

Cals. Moles of NH4NO3Step 1:

Eq. gives mole ratios (stoichiometry)

Determine qsolution = mCT

Step 2:

qsolution

qsolution + qrxn = 0Step 3:

qrxn = q(NH4NO3)

rH =qrxn/mol NH4NO3 Step 4:

Enthalpy per mole of reactant




4 34 3 4 3

solution solution

solution

3solution

1 mol NH NO5.44 g NH NO 0.0680 moles NH NO

80.04 g

q m C T

Jq 154.4 g 4.18 (16.2 C 18.6 C)

g C

q 1.55 10 J

4 34 3 4 3

solution solution

solution

3solution

1 mol NH NO5.44 g NH NO 0.0680 moles NH NO

80.04 g

q m C T

Jq 154.4 g 4.18 (16.2 C 18.6 C)

g C

q 1.55 10 J




The sign is positive indicating an endothermic process.

Under conditions of constant volume, any heat transferred is equal to a change of internal energy rU.

qV = rU

Constant Volume Calorimetry, Constant Volume Calorimetry, Measuring Measuring UU

Heats of combustion (rUo

combustion ) are measured using a device called a Bomb Bomb CalorimeterCalorimeter.

A combustible sample is reacted with excess O2

The heat capacity of the bomb is constant.

The heat of reaction is found by:

Constant Volume Calorimetry, Constant Volume Calorimetry, Measuring Measuring UU

Octane, the primary component of gasoline combusts by the reaction:

C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9H2O(l)

A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb.

The temperature of the water rises to 33.20 °C from 25.00 °C when the octane is reacted.

If the heat capacity of the bomb is 837 J/°C, calculate the heat of reaction per mole of octane.

–qrxn = qwater + qbomb

Since the temperature of the water rose, the reaction must have been exothermic:Therefore one can write:

Calculating Heat in an Exothermic Calculating Heat in an Exothermic ReactionReaction

–qrxn = mwaterCwaterTwater + qbombTwater

4.184J

g C 33.20 25.00 C

J837

C

33.20 25.00 C qqwaterwater

qqbombbomb

qRXN = – 4803 J or – 48.0 kJ

Heat transferred per mole qV:

Calculating Heat in an Exothermic Calculating Heat in an Exothermic ReactionReaction

The overall enthalpy change for a reaction is equal to the sum The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.of the enthalpy changes for the individual steps in the reaction.

Reactants ProductsrH = ??

unknown!

Intermediate Reaction

rH1

know

n

r H

2

known rH1 + rH2 = rH

The sum of the H’s in one direction must equal the sum in the other direction.

What if the enthalpy What if the enthalpy changes through changes through another path are know?another path are know?

Why?Why?

Because enthalpy is a state function… Path independentPath independent!

so we can write…so we can write…

Hess’s LawHess’s Law

Forming CO2 can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed.

Hess’s law & Energy Level Hess’s law & Energy Level DiagramsDiagrams

Notice that the path from reactants to products in the desired reaction goes through an ““intermediate intermediate compoundcompound”” in the given reactions.

This means that the path for hydrogen and nitrogen to produce ammonia goes through hydrazine (N2H4).

Therefore, the path to the enthalpy of the reaction must be a sum of the two given reactions!

Hess’s Law Problem:Hess’s Law Problem:Example: Determine the rH for the reaction:

3H2(g) + N2(g) 2NH3(g) rrHHoo = ???= ???

Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆rH°1 = +95.4 kJ

(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ

adding equations (1) & (2) yields:

2 H2(g) + N2(g)+ N2H4(g) + H2(g) N2H4(g) + 2NH3(g)

Hess’s Law Problem:Hess’s Law Problem:

Example: Determine the rH for the reaction:

3H2(g) + N2(g) 2NH3(g) rrHHoo = ???= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ



3H2(g) + N2(g) 2NH3(g) rrHHoo = ???= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ



Look what happens…

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



3H2(g) + N2(g) 2NH3(g) rrHHoo = ???= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆rH°2 = –187.6 kJ



Look what happens…

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

rH = ∆rH°1 + ∆rH°2 = +95.4 kJ + (–187.6 kJ)= –92.2 kJ

therefore…therefore…

• Reaction (1) is endothermic by 95.4 kJ• Reaction (2) is exothermic by 187.6 kJ• Since the exothermicityexothermicity has a greater magnitude

than the endothermicityendothermicity, the overall process is exothermic (rH < 0).

intermediatesH2(g) + N2H4(g)

reactants3H2(g) + N2(g)

products2N3 (g)

(Step 1)Ho = 95.4kJ

(Step 2)Ho = 187.6kJ

(Overall)Ho = 92.2kJ

When 1 mole of compound is formed from its elements, the enthalpy change for the reaction is called the enthalpy of enthalpy of formationformation, fHo (kJ/mol).

These enthalpies are always reported at Standard Standard conditionsconditions:

1 atm and 25 °C (298 K).The standard enthalpies of formation of the most stable form of any element is zero:

fH (element) = 0

fH° O2(g) = 0 fH° O(g) ≠ 0elemental formelemental form NOT the elemental formNOT the elemental form

Standard Enthalpies of FormationStandard Enthalpies of Formation

H2(g) + ½ O2(g) H2O(gg) ∆rH˚ = -242 kJ

H2(g) + ½ O2(g) H2O(liqliq) ∆rH˚ = -286 kJ

Same reaction, different phases, different enthalpies.

Since Enthalpy is a state function, enthalpy values depend on the reaction conditions in terms of the phases of reactants and products.

Enthalpy ValuesEnthalpy Values

The formation of water is given by the reaction:

H2(g) + ½ O2(g) H2O(l)

Each element and the compound are represented by the physical state they take on at 25.0 °C and 1 atm pressure. (Standard State conditions)

A chemical reaction that describes the formation of one mole of a compound from its elements at standard state conditionsstandard state conditions is known as a ““formation formation reactionreaction””.

Formation ReactionsFormation Reactions

QuestionQuestion: : What is the formation reaction for potassium permanganate?

KMnO4

• salts are solids at standard state conditions.• metals are solids at standard state conditions.• oxygen is a gas at standard state conditions.• balance for one mole of the product

(s)K + Mn + O2(s)(s) (g)22

standard state conditionsstandard state conditions = 25 = 25ooC and 1 atmC and 1 atm

compoundelements

• All components of a reaction can be related back to their original elements.

• Each compound has an enthalpy of formation associated with it.

• Reactants require energy to return to component elements.

• Products release energy when formed form component elements.

• Since enthalpy is a state function, the sum of the above must relate somehow to the overall enthalpy of a reaction.

Enthalpy Changes for a Reaction:Enthalpy Changes for a Reaction:Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

C3H8(g) + 5O2(g) 3CO2(g) + 4 H2O(l)

• In order to make CO2(g) and H2O(l) one must break the propane up into its elements.

• This takes energy.• The elements carbon, hydrogen and oxygen then

combine to make the new compounds, CO2 and H2O.

• This process releases energy.

3C(s)+ 8H2(g)+ 5O2(g)

Consider the Combustion of Consider the Combustion of PropanePropane

The sum of the fH° for the productsproducts multiplied by the respective coefficients, subtracted by the sum of the fH° for the reactantsreactants multiplied by the respective coefficients, yields the fH° for the reaction.

n and m are the stoichiometric balancing coefficients.

In other words:In other words: Energy gained– Energy spent = Net Energy

Enthalpy Changes for a Reaction:Enthalpy Changes for a Reaction:Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

fH° [CaO(s)] = – 635.5 kJ/mol

CaO(s) + CO2(g) CaCO3(s)

fH° [CO2(g)] = – 393.5 kJ/mol

fH° [CaCO3(s)] = –1207 kJ/mol

ProblemProblem: : Calculate the rH° for CaCO3(s) given the following fH°

fH° [CaO(s)] = – 635.5 kJ/mol


fH° [CO2(g)] = – 393.5 kJ/mol




fH° [CaO(s)] = – 635.5 kJ/mol


fH° [CO2(g)] = – 393.5 kJ/mol


rH° = –1207 kJ/mol

1mol CaCO3

– (–635.5 kJ/mol

1mol CaO

–393.5 kJ/mol)

1mol CO2

rH° = – 178 kJ/mol

ProblemProblem: : The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.

Calculate the fH° for naphthalene given the following enthalpies of formation:

fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)

Step 1: Write the balanced equation for the reaction…



fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– {fH° [C8H10(l)] + 12 fH° [O2(g)]}

combustionH° =

Next recall that:Next recall that:

C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)

From the problem, all quantities are know but f 10 8H C H



fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

fH° [C10H8(l)] = 10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– {combH° + 12 fH° [O2(g)]}

elements = 0elements = 0



fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

fH° [C10H8(l)] = 10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– {combH° + 12 fH° [O2(g)]}

elements = 0elements = 0H° f [C10H8(l)] =

10 (–393.5 kJ/mol+ 4 (–285.7 kJ/mol)– (– 5156 kJ/mol)

= + 79 kJ/mol

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Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions