Ch 5-Thermochem 2013

FOUNDATION CHEMISTRY I

CHM092

CHAPTER 5:

THERMOCHEMISTRY

PREPARED BY

SYED ABDUL ILLAH ALYAHYA

Topic CHAPTER 5

5.1 Introduction to Thermochemistry

5.2 System and Surroundings

5.3 Exothermic and Endothermic Reactions

5.4 Specific Heat Capacity

5.5 Enthalpy and Calorimetry

5.6 Enthalpy Changes for Chemical Reactions

5.7 Hess’s Law

5.8 Lattice Energy and Born-Harber Cycle

Topic

5.1 Introduction to

Thermochemistry

• All matter contains energy- whenever matter

undergoes a change, the quantity of energy that the

matter contains also changes

• Example – burning candle melts a piece of ice, both

chemical and physical changes

In chemical change- higher energy reactants (wax and O2 form

lower energy products (CO2 and H2O)

In physical change- heat is absorbed ice turns to water

• Thermodynamics is the study of energy and its

transformations

• Thermochemistry–the branch of thermodynamics

that deal with heat in chemical and physical changes

The Nature of Energy 5.1 Introduction to Thermochemistry

Manifestations of Energy


Some Forms of Energy

• Electrical – kinetic energy associated with the flow of electrical charge

• Heat or Thermal Energy – kinetic energy associated with molecular motion

• Light or Radiant Energy – kinetic energy associated with energy transitions in an atom

• Nuclear – potential energy in the nucleus of atoms

• Chemical – potential energy due to the structure of the atoms, the

attachment between atoms, the atoms’ positions relative to each other in the molecule, or the molecules, relative positions in the structure


• The amount of kinetic energy an object has is directly

proportional to its mass and velocity

KE = ½mv2

Units of Energy


Units of Energy

• Joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter

– 1 J = 1 N∙m = 1 kg∙m2/s2

• Calorie (cal) is the amount of energy needed to raise the temperature of one gram of water 1°C

– kcal = energy needed to raise 1000 g of water 1°C

– food Calories = kcals

Energy Conversion Factors

1 calorie (cal) = 4.184 joules (J) (exact)

1 Calorie (Cal) = 1000 cal = 1 kcal = 4184 J

1 kilowatt-hour (kWh) = 3.60 x 106 J


Conservation of Energy

• The Law of Conservation of

Energy states that energy

cannot be created nor destroyed

• When energy is transferred

between objects, or converted

from one form to another, the

total amount of energy present

at the beginning must be

present at the end


Temperature and Heat • The temperature of an object is a measure of its ability to

transfer energy as heat.

• When two objects at different temperature are brought into contact, energy will be transferred as heat from one at the higher temperature to the one at the lower temperature.

• Temperature determines the direction of thermal energy transfer

• The higher temperature of a given object, the greater the thermal energy of its atoms, ion, or molecules (energy associated with molecular motion)

• Heating and cooling are processes by which energy transferred as heat – from high temperature to lower temperature. – Heat is not a substance


Topic

5.2 System and

Surroundings

Definition

• We define the system as the material or process within which we are studying the energy changes within

• We define the surroundings as everything else with which the system can exchange energy with

• What we study is the exchange of energy between the system and the surroundings

Surroundings

System

Surroundings

System


Comparing the Amount of Energy in the

System and Surroundings During Transfer • Conservation of energy means that the amount of energy

gained or lost by the system has to be equal to the amount

of energy lost or gained by the surroundings


Energy Flow and

Conservation of Energy

• Conservation of energy requires that the sum of the energy

changes in the system and the surroundings must be zero

ΔEnergyuniverse = 0 = ΔEnergysystem + ΔEnergysurroundings

Δ Is the symbol that is used to mean change

• final amount – initial amount

DE = Efinal - Einitial


Energy Flow

• When energy flows out of a

system, it must all flow into the

surroundings

• When energy flows out of a

system, ΔEsystem is ─ve

• When energy flows into the

surroundings, ΔEsurroundings is

+ve

• Therefore:

─ ΔEsystem= ΔEsurroundings

Surroundings DE +

System DE ─


Energy Flow

• When energy flows into a

system, it must all come from

the surroundings

• When energy flows into a

system, ΔEsystem is +ve

• When energy flows out of the

surroundings,

ΔEsurroundings is ─ve

• Therefore:

ΔEsystem= ─ ΔEsurroundings

Surroundings

DE ─

System DE +


Open systems – both matter (mass) and energy can move into and out of an open system. The earth, for example, is an open system. Energy comes in from the sun, and is re-radiated back to space. Meteors strike the Earth, and cause parts of the Earth to be ejected into space. We send satellites into interstellar space. These are difficult systems to study scientifically because there are too many variables to track. Closed systems – matter can’t be transferred from a closed system, but energy can. We study these frequently, and a great example is the ubiquitous “coffee cup” calorimeter used in many general chemistry laboratories. These are great ways to see how a system (some chemical reaction) causes a change in temperature in the surroundings (the solvent, often water) that is contained in the cup. Isolated systems – neither matter or energy can be transferred from these systems to the surroundings

Types of System


Open systems – both matter (mass) and energy can move into

and out of an open system. The earth, for example, is an open

system. Energy comes in from the sun, and is re-radiated back

to space.

Closed systems – matter can’t be transferred from a closed

system, but energy can. We study these frequently, and a great

example is the ubiquitous “coffee cup” calorimeter used in many

general chemistry laboratories. These are great ways to see

how a system (some chemical reaction) causes a change in

temperature in the surroundings (the solvent, often water) that

is contained in the cup.

Isolated systems – neither matter or energy can be transferred

from these systems to the surroundings

Energy Flow in Chemical Reaction


Many chemical reactions are accompanied by the liberation

or absorption of heat.

Total heat absorbed or liberated in a chemical reaction is

called the heat of reaction

In chemical reaction, - systems are the reactants and

products. Surroundings is the vessel or container in which

the reaction take place and the air or other material in thermal

contact with the reaction system.

Internal Energy and Energy Exchange


The internal energy (U) in a chemical system is the sum of

the potential and kinetics energies of the atoms, molecules or

ions in the system.

Change in internal energy (∆U) is a measure the energy

transferred as heat and work to or from the system.

DU = q + w

The sum of the energy transferred as heat between the

system and surrounding (q)

The energy transferred as work between the system and its

surroundings (w)

U is represent the internal energy, but some textbook use E, doesn’t matter

Energy Diagrams Energy diagrams are a “graphical” way of showing the

direction of energy flow during a chemical process

• If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be +ve

• If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be ─ve


energy released DErxn = ─

energy absorbed DErxn = +

Energy Flow in a Chemical Reaction

• The total amount of internal energy in

1mol of C(s) and 1 mole of O2(g) is

greater than the internal energy in 1 mole

of CO2(g)

– at the same temperature and pressure

• In the reaction C(s) + O2(g) → CO2(g), there

will be a net release of energy into the

surroundings

– −ΔEreaction = ΔEsurroundings

• In the reaction CO2(g) → C(s) + O2(g), there

will be an absorption of energy from the

surroundings into the reaction

– ΔEreaction = − ΔEsurroundings In

tern

al E

ner

gy

CO2(g)

C(s), O2(g)

Surroundings

System C + O2 → CO2

System CO2 →C + O2


Energy transferred as heat only (q) and does no work (w = 0)

ΔE = q + 0 = q

The two cases where energy is transferred as heat only.

1. Heat flowing out from a system.

2. Heat flowing into a system

The system releases heat. q = -ve The system absorbs heat. q = +ve

Energy Exchange in Chemical Reactions


Energy transferred as work only (w) and q = 0, ΔE = 0 + w = w

1. Work done by a system = work done on the surroundings

The system does work on the surroundings. w = -ve

H2(g) + Zn2+(aq) + 2Cl-(aq)

Zn(s) + 2H+(aq) + 2Cl-(aq)



2. Work done on a system = work done by a surroundings

The system has work done on it by the surroundings. w = +ve



The values of q and w can have either a positive or negative



q w + = ∆E

+

+

-

-

-

-

+

+

+

-

depends on sizes of q and w

depends on sizes of q and w



Example 1 Determining the Change in Internal

Energy of a System

PROBLEM: When gasoline burns in a car engine, the heat released

causes the products CO2 and H2O to expand, which

pushes the pistons outward. Excess heat is removed by

the car’s radiator. If the expanding gases do 451 J of

work on the pistons and the system releases 325 J to the

surroundings as heat, calculate the change in energy

(ΔE) in J, kJ, and kcal.

Define the system and surroundings and assign signs to q

and w correctly. Then ΔE = q + w. The answer can then be

converted from J to kJ and to kcal.

PLAN:

SOLUTION:

Heat is given out by the system, so q = - 325 J

The gases expand to push the pistons, so the system does work on

the surroundings and w = - 451 J

ΔE = q + w = -325 J + (-451 J) = -776 J

-776 J x 103J

1 kJ = -0.776 kJ -0.776 kJ x

4.184 kJ

1 kcal = -0.185 kcal

Heat is given out by a chemical reaction, so it makes sense to define

the system as the reactants and products involved. The pistons, the

radiator and the rest of the car then comprise the surroundings.

Topic

5.3 Endothermic and Exothermic

Reactions

Endothermic & Exothermic Reactions

• When ΔE is ─, heat is being released by the system

• Reactions that release heat are called exothermic reactions

• When ΔE is +, heat is being absorbed by the system

• Reactions that absorb heat are called endothermic reactions

• Chemical heat packs contain iron filings that are oxidized in an

exothermic reaction ─ your hands get warm because the released heat

of the reaction is transferred to your hands

• Chemical cold packs contain NH4NO3 that dissolves in water in an

endothermic process ─ your hands get cold because the pack is

absorbing your heat

• For an exothermic reaction, the surrounding’s temperature rises due to release of thermal energy by the reaction

• This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat

• During the course of a reaction,old bonds are broken and new bonds are made

• The products of the reaction have less chemical potential energy than the reactants

• The difference in energy is released as heat

Molecular View of Exothermic Reactions

Molecular View of Endothermic Reactions

• In an endothermic reaction, the surrounding’s temperature

drops due to absorption of some of its thermal energy by

the reaction

• During the course of a reaction, old bonds are broken and

new bonds are made

• The products of the reaction have more chemical potential

energy than the reactants

• To acquire this extra energy, some of the thermal energy

of the surroundings is converted into chemical potential

energy stored in the products

Topic


Heat Exchange

• Heat is the exchange of thermal energy between the

system and surroundings

• Heat exchange occurs when system and surroundings

have a difference in temperature

• Temperature is the measure of the amount of thermal

energy within a sample of matter

• Heat flows from matter with high temperature to matter with

low temperature until both objects reach the same

temperature - thermal equilibrium is achieved

• After thermal equilibrium is attained, the object whose

temperature increased has gained thermal energy, and the

object whose temperature decreased has lost thermal

energy


Quantity of Heat Energy Absorbed:

Heat Capacity

• When a system absorbs heat, its temperature increases

• The increase in temperature is directly proportional to the amount of heat absorbed

• The proportionality constant is called the heat capacity, C

– units of C are J/°C or J/K

q = C x ΔT

• The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat


Factors Affecting Heat Capacity

• The heat capacity of an object depends on its amount of

matter

– usually measured by its mass

– 200 g of water requires twice as much heat to raise its

temperature by 1°C as does 100 g of water

• The heat capacity of an object depends on the type of

material

– 1000 J of heat energy will raise the temperature of 100

g of sand 12 °C, but only raise the temperature of 100 g

of water by 2.4 °C


Specific Heat Capacity

• Measure of a substance’s intrinsic

ability to absorb heat

• The specific heat capacity is the

amount of heat energy required to raise

the temperature of one gram of a

substance 1°C

– Symbol is Cs

– units are J/(g∙°C)

• The molar heat capacity is the amount

of heat energy required to raise the

temperature of one mole of a substance

1°C ( Molar heat capacity for water is

75.4 J/mol.oC)


Specific Heat of Water

• The rather high specific heat of water allows water to

absorb a lot of heat energy without a large increase in its

temperature

• The large amount of water absorbing heat from the air

keeps beaches cool in the summer – without water, the Earth’s temperature would be about the same as

the moon’s temperature on the side that is facing the sun (average

107 °C or 225 °F)

• Water is commonly used as a coolant because it can

absorb a lot of heat and remove it from the important

mechanical parts to keep them from overheating – or even melting

– it can also be used to transfer the heat to something else

because it is a fluid


Quantifying Heat Energy

• The heat capacity of an object is proportional to its mass

and the specific heat of the material

• So we can calculate the quantity of heat absorbed by an

object if we know the mass, the specific heat, and the

temperature change of the object

Heat = (mass) x (specific heat) x (temp. change)

q = (m) x (Cs) x (ΔT)


Example 3: How much heat is absorbed by a copper

penny with mass 3.10 g whose temperature rises from

−8.0 °C to 37.0 °C?

q = m ∙ Cs ∙ DT

Cs = 0.385 J/g•ºC (Table 6.4)

the unit is correct, the sign is reasonable as

the penny must absorb heat to make its

temperature rise

T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g

q, J

Check: • Check

Solution: • Follow the

conceptual

plan to

solve the

problem

Conceptual

Plan:

Relationships:

• Strategize

Given:

Find:

• Sort

Information

Cs m, DT q

Exercise 2 – Calculate the amount of heat released

when 7.40 g of water cools from 49° to 29 °C

(water’s specific heat is 4.18 J/gºC)

Exercise 2 – Calculate the amount of heat released

when 7.40 g of water cools from 49° to 29 °C

q = m ∙ Cs ∙ DT

Cs = 4.18 J/gC (Table 6.4)

T1= 49 °C, T2 = 29 °C, m = 7.40 g

q, J

Check: • Check

Solution: • Follow the

concept plan

to solve the

problem

Conceptual

Plan:

Relationships:

• Strategize

Given:

Find:

• Sort

Information

Cs m, DT q

the unit is correct, the sign is reasonable as

the water must release heat to make its

temperature fall

Thermal Energy Transfer

• When two objects at different temperatures are placed in

contact, heat flows from the material at the higher

temperature to the material at the lower temperature

• Heat flows until both materials reach the same final

temperature

• The amount of heat energy lost by the hot material equals

the amount of heat gained by the cold material

qhot = −qcold

mhotCs,hotDThot = −(mcoldCs,coldDTcold)


1. Heat lost by the metal > heat gained by water

2. Heat gained by water > heat lost by the metal

3. Heat lost by the metal > heat lost by the water

4. Heat lost by the metal = heat gained by water

5. More information is required

1. Heat lost by the metal > heat gained by water

2. Heat gained by water > heat lost by the metal

3. Heat lost by the metal > heat lost by the water

4. Heat lost by the metal = heat gained by water

5. More information is required

A piece of metal at 85 °C is added to water at 25 °C, the final

temperature of the both metal and water is 30 °C. Which of

the following is true?

q = m ∙ Cs ∙ DT

Cs, Al = 0.903 J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4)

Example 4: A 32.5-g cube of aluminum initially at 45.8 °C is

submerged into 105.3 g of water at 15.4 °C. What is the final

temperature of both substances at thermal equilibrium?

the unit is correct, the number is reasonable as the final

temperature should be between the two initial temperatures

mAl = 32.5 g, Tal = 45.8 °C, mH20 = 105.3 g, TH2O = 15.4 °C

Tfinal, °C

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

Cs, Al mAl, Cs, H2O mH2O DTAl = kDTH2O qAl = −qH2O Tfinal

Exercise 3 – A hot piece of metal

weighing 350.0 g is heated to 100.0 °C.

It is then placed into a coffee cup

calorimeter containing 160.0 g of water

at 22.4 °C. The water warms and the

copper cools until the final temperature

is 35.2 °C. Calculate the specific heat of

the metal and identify the metal.

Exercise 3 – Calculate the specific heat and

identify the metal from the data

q = m x Cs x DT; qmetal = −qH2O

the units are correct, the number indicates

the metal is copper

metal: 350.0 g, T1 = 100.0 °C, T2 = 35.2 °C

H2O: 160.0 g, T1 = 22.4 °C, T2 = 35.2°C, Cs = 4.18 J/g °C

Cs , metal, J/gºC

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

m, Cs, DT q

Additional Problems


1. How much energy must be transferred to raise the temperature of a

cup of coffee (250mL) from 20.5oC to 95.6oC. (Assume that water

and coffee have the same density 1.00g/mL and specific heat

capacity 4.184 J/g.oC

2. A iron block whose temperature is 78.8oC is placed in a calorimeter

containing 244g of water at 18.8oC. After thermal equilibrium is

attained at 22oC. What is the mass of iron block? (Assume no energy

is lost to calorimeter and its surrounding)

Topic


Pressure –Volume Work

• PV work is work caused by a volume change against an

external pressure

• When gases expand, ΔV is +ve, but the system is doing

work on the surroundings, so wgas is ─ve

• As long as the external pressure is kept constant

Workgas = External Pressure x Change in Volumegas

w = ─PΔV

– to convert the units to joules use 101.3 J = 1 atm∙L


Enthalpy

• The enthalpy, H, of a system is the sum of the internal

energy of the system and the product of pressure and

volume

– H is a state function

H = E + PV

• The enthalpy change, ΔH, of a reaction is the heat

evolved in a reaction at constant pressure

ΔHreaction = qreaction at constant pressure

• Usually ΔH and ΔE are similar in value, the difference is

largest for reactions that produce or use large quantities of

gas


The Difference between ΔH and ΔE

• Lighters are usually fueled by butane (C4H10). When 1 mole

of butane burns at constant pressure, it produces 2658 kJ of

heat and does 3 kJ of work. What are the values of ΔH and

ΔE for the combustion of one mole butane?

Answer

ΔH represents only the heat exchanged; therefore ΔH = -

2658 kJ

But ΔE represents the heat and work exchanged ; therefore

ΔE = -2661 kJ

Notice that both are negative because heat and work are

flowing out of the system and into surroundings- same

magnitude


Enthalpy of Reaction

• The enthalpy change in a chemical reaction is an extensive property – the more reactants you use, the larger the enthalpy change

• By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = −2044 kJ


Relationships Involving ΔHrxn

• When reaction is multiplied by a factor, ΔHrxn is

multiplied by that factor

– because ΔHrxn is extensive

C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = 2(−393.5 kJ) = −787.0 kJ

• If a reaction is reversed, then the sign of ΔH is

changed CO2(g) → C(s) + O2(g) ΔH = +393.5 kJ


Exchanging Energy Between

System and Surroundings

• Exchange of heat energy

q = mass x specific heat x ΔTemperature

• Exchange of work

w = −Pressure x ΔVolume

ΔE = m x Cs x ΔT + (-PΔV)


Measuring ΔE,

Calorimetry at Constant Volume

• Because ΔE = q + w, we can determine ΔE by measuring q and w

• In practice, it is easiest to do a process in such a way that there is no change in volume, so w = 0 – at constant volume, ΔEsystem = qsystem

• In practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we measure the temperature change in the surroundings – use insulated, controlled surroundings

– qsystem = −qsurroundings

• The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water

qsurroundings = qcalorimeter = ─qsystem

Bomb Calorimeter

• Used to measure ΔE

because it is a constant

volume system

• The heat capacity of the

calorimeter is the amount of

heat absorbed by the

calorimeter for each degree

rise in temperature and is

called the calorimeter

constant Ccal, kJ/ºC

qcal = Ccal x ΔT qcal = -qrxn

Measuring ΔH

Calorimetry at Constant Pressure

• Reactions done in aqueous solution are

at constant pressure

– open to the atmosphere

• The calorimeter is often nested foam

cups containing the solution

qreaction = ─ qsolution

= ─(masssolution x Cs, solution x ΔT)

ΔHreaction = qconstant pressure = qreaction

– to get ΔHreaction per mol, divide by the

number of moles

Example 7: What is ΔHrxn/mol Mg for the reaction

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in

100.0 mL of solution and changes the temperature from 25.6 °C to 32.8

°C?

the sign is correct and the value is reasonable because the reaction is exothermic

0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C, assume d = 1.00 g/mL, Cs = 4.18 J/g∙°C H, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

qrxn, mol DH Cs, DT, m qsol’n qrxn

qsol’n = m x Cs, sol’n x DT = −qrxn, MM Mg= 24.31 g/mol

6.499 x 10−3 mol Mg, 1.00 x 102 g, DT = 7.2 °C,,Cs = 4.18 J/g∙°C

DH, kJ/mol

Exercise 6 – What will be the final temperature of the solution

in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M

HCl(aq) is added to a 50.00 mL sample of 0.250 NaOH(aq).

The initial temperature is 19.50 °C and the DHrxn is −57.2

kJ/mol NaOH.

(assume the density of the solution is 1.00 g/mL and the

specific heat of the solution is 4.18 J/g∙°C)

qsol’n = m x Cs, sol’n x DT = −qrxn, 0.250 mol NaOH = 1 L

Exercise 6 – What will be the final temperature of the solution in a coffee cup

calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added to a 50.00 mL

sample of 0.250 NaOH(aq). The initial temperature is 19.50 °C.

the sign is correct and the value is reasonable, because the reaction is

exothermic we expect the final temperature to be higher than the initial

50.00 mL of 0.250 M HCl, 50.00 mL of 0.250 M NaOH T1 = 19.50 °C,

d = 1.00 g/mL, Cs = 4.18 J/g∙°C, D H = −57.2 kJ/mol

T2, °C

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

s

nsol'

Cm T

D

qHmol rxn Dq rxnnsol' qq

mol rxnq

H D

1.25 x 10−2 mol NaOH, 1.00 x 102 g, T1 = 19.50 °C,

DH =−57.2 kJ/mol, Cs = 4.18 J/g∙°C

T2, °C

qrxn qsol’n DH DT Cs, qsol’n, m T2

Example 8 Reaction involving limiting reactant

You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at

25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After

stirring, the final temperature is 27.210C. Calculate qsoln (in J) and ΔHrxn

(in kJ/mol). (Assume the total volume is the sum of the individual

volumes and that the final solution has the same density and specfic heat

capacity as water: d = 1.00 g/mL and s = 4.18 J/g*K). The heat capacity

of the calorimeter can be ignored.

Solution

We need to determine the limiting reactant from the net ionic equation.

The moles of NaOH and HCl as well as the total volume can be

calculated. From the volume, we use density to find the mass of the

water formed. At this point qsoln can be calculated using the mass, Cs,

and DT. The heat divided by the moles of water will give us the heat per

mole of water formed.

For NaOH 0.500 M x 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M x 0.0250 L = 0.0125 mol H+

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)

total volume after mixing = 75 mL

75 mL x 1.00 g/mL = 75.0 g of water

qreaction = -(qsolution) = -(mass x specific heat x ΔT)

= -[75.0 g x 4.18 J/g*0C x (27.21-25.00)0C]

= -693 J

(-693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed

HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn.

Topic

5.6 Standard Enthalpy and

Enthalpy Changes of Chemical

Reactions

Standard Conditions • The standard state is the state of a material at a defined

set of conditions – pure gas at exactly 1 atm pressure

– pure solid or liquid in its most stable form at exactly 1 atm pressure

and temperature of interest usually 25 °C

– substance in a solution with concentration 1 M

• The standard enthalpy change, ΔH°, is the enthalpy

change when all reactants and products are in their

standard states

• The standard enthalpy of formation, ΔHf°, is the enthalpy

change for the reaction forming 1 mole of a pure compound

from its constituent elements – the elements must be in their standard states

– the ΔHf° for a pure element in its standard state = 0 kJ/mol • by definition

The Standard Enthalpy of Formation

The standard enthalpy of formation of a substance is the

enthalpy change that results when 1 mol of the substance in

its standard state is formed from its elements in their

standard states. The reference forms of the elements are

their most stable forms at 25 oC and 1 atm pressure

Example :

Ca (s) + C (s) + 3/2 O2 (g) → CaCO3 (s) = -1207 kJ mol-1

The standard heat of formation of CaCO3 (s) is –1207 kJ mol-1

The Standard heat of formation, ∆Hf , of a pure element in its most stable

form is zero, 0. 0

Examples :

for elements: Fe (s) , Na(s) , O2 (g), Cl (g) and others = 0 kJmol-1.

Standard Enthalpies of Formation

Formation Reactions

• Reactions of elements in their standard state to

form 1 mole of a pure compound

– if you are not sure what the standard state of an element

is, find the form in Appendix IIB that has a ΔHf° = 0

– because the definition requires 1 mole of compound be

made, the coefficients of the reactants may be fractions

Writing Formation Reactions

Write the Formation Reaction for CO(g)

• The formation reaction is the reaction between the elements in the compound, which are C and O

C + O → CO(g)

• The elements must be in their standard state – there are several forms of solid C, but the one with

DHf° = 0 is graphite

– oxygen’s standard state is the diatomic gas

C(s, graphite) + O2(g) → CO(g)

• The equation must be balanced, but the coefficient of the product compound must be 1 – use whatever coefficient in front of the reactants is

necessary to make the atoms on both sides equal without changing the product coefficient

C(s, graphite) + ½ O2(g) → CO(g)

Example 9 Write the formation reactions for CO2(g) and Al2(SO4)3 (S)

C(s, graphite) + O2(g) CO2(g)

2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g) Al2(SO4)3(s)

Exercise 7

Write balanced thermochemical equations to represent the

following standard enthalpies of formation.

NaHCO3 (s) = - 948 kJ mol-1

Fe2O3 (s) = - 824 kJ mol-1

Al2O3 (s) = - 1590 kJ mol-1

C2H5OH (l) = - 277 kJ mol-1

NH3 (g) = - 48.5 kJ mol-1

H2O (l) = - 286 kJ mol-1

ofDofDofD

ofDofD

ofD

Example 9: Write the formation reactions for the

following:

CO2(g)

Al2(SO4)3(s)

C(s, graphite) + O2(g) CO2(g)

2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g) Al2(SO4)3(s)

Standard Enthalpy of Combustion

The standard enthalpy (heat) of combustion of a substance is

the amount of heat released when 1 mole of the substance is

burnt completely in oxygen at standard state conditions.

Example 10

H2 (g) + ½ O2 (g) → H2O (l) = -286 kJ

The standard heat of combustion of hydrogen, H2 (g) is –286 kJ mol-1.

ocD

Example 10: How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?

1 kg = 1000 g, 1 mol C3H8 = −2044 kJ, Molar Mass = 44.09 g/mol

the sign is correct and the number is reasonable

because the amount of C3H8 is much more than 1 mole

13.2 kg C3H8,

q, kJ/mol

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

mol kJ kg g

Exercise 9

How much heat is evolved when a 0.483 g diamond is burned?

(ΔHcombustion = −395.4 kJ/mol C)

{-15.9 kJ)

Exercise 8

Write balanced thermochemical equations which represent the standard

enthalpies of combustion for the following substances.

∆Hc C4H10 (g) = − 2875 kJ mol-1

∆Hc Zn (s) = − 350.6 kJ mol-1

∆Hc Al (s) = − 795 kJ mol-1

∆Hc C12H22O11 (s) = − 5461 kJ mol-1

Exercise 10

The amount of heat liberated by the complete combustion of 0.3 mol of

C2H6 (g) is sufficient to boil a certain quantity of water with an initial

temperature of 30 oC in a calorimeter. The enthalpy of combustion of C2H6

(g) is –1560 kJ mol-1. Calculate the volume of water in the calorimeter

which was boiled by the heat released in the experiment. (The specific

heat capacity of water is 4.2 J g-1 k-1 ; The density of water is1 g cm-3 ). { 1.6 dm3}

Standard Enthalpy of Neutralization (∆Ho)

The standard enthalpy of neutralization is the enthalpy change when

one mole of H+ (aq) ions from an acid reacts with one mole of OH-(aq)

ions from an alkali to form one mole of water molecules, H2O (l) under

standard conditions.

H+(aq) + OH-(aq) → H2O(l) ∆H=std enthalpy of neutralization

Examples:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ∆Ho = −57.3 kJ mol−1

½Ca(OH)2(aq) + HNO3(aq) → ½CaNO3(aq) + H2O(l) ∆Ho = −57.3 kJ mol−1

The above are neutralization reactions of strong acids and strong alkalis.

Strong acids and strong alkalis dissociate almost completely in aqueous solutions

to give H+(aq) and OH-(aq), standard enthalpy of neutralization of any strong

acid and strong alkali is almost constant at about -57.3 kJ mol-1

That is , H+(aq) + OH-(aq) → H2O(l) ∆H = −57.3 kJ mol−1

If a weak acid or a weak base is used, the standard enthalpy change of

neutralization is always less than -57.3 kJ mol-1. This is because the weak acid

does not dissociate completely. Therefore, when the acid reacts, some of the heat

energy liberated during the neutralization reaction is absorbed to dissociate the

acid. For example,

KOH(aq) + HCN(aq) → KCN(aq) + H2O(l) ∆Ho = −11.7 kJ mol−1

NaOH(aq) + CH3COOH(aq) → CH3COONa(aq) + H2O(l) ∆Ho = −55.6 kJ mol−1

NH4OH(aq) + CH3COOH(aq) → CH3COONH4(aq) + H2O(l) ∆Ho = −50.4 kJ mol−1

Standard enthalpy of neutralization of HF (weak acid) and a strong base is

higher than -57.3 kJ, i.e about -69 kJ because HF dissociates with liberation of

energy (exothermic). Similarly with H2SO4, it releases a lot of heat with dilution

when the aqueous alkali is added to it.

NaOH(aq) + HF(aq) → NaF(aq) + H2O(l) ∆Ho = −69.0 kJ mol−1

KOH(aq) + H2SO4(aq) → ½K2SO4(aq) + H2O(l) ∆Ho = −66.0 kJ mol−1

Calculating Standard Enthalpy Change

for a Chemical Reaction

• Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products

• The ΔH° for the reaction is then the sum of the ΔHf° for the component reactions

ΔH°reaction= Σ n ΔHf°(products)−Σ n ΔHf°(reactants)

– Σ means sum

– n is the coefficient of the reaction

CH4(g) → C(s, graphite) + 2 H2(g) D H° = + 74.6 kJ

2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ

CH4(g)+ 2 O2(g)→ CO2(g) + H2O(g)

C(s, graphite) + 2 H2(g) → CH4(g) DHf°= − 74.6 kJ/mol CH4

C(s, graphite) + O2(g) → CO2(g) DHf°= −393.5 kJ/mol CO2

H2(g) + ½ O2(g) → H2O(g) DHf°= −241.8 kJ/mol H2O

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) DH° = −802.5 kJ

DH° = [((−393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))]

= −802.5 kJ

DH° = [(DHf° CO2(g) + 2∙DHf°H2O(g))− (DHf° CH4(g) + 2∙DHf°O2(g))]

Example 11: Calculate the enthalpy change in the

reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

1. Write formation reactions for each compound

and determine the DHf° for each

2 C(s, gr) + H2(g) C2H2(g) DHf° = +227.4 kJ/mol

C(s, gr) + O2(g) CO2(g) DHf° = −393.5 kJ/mol

H2(g) + ½ O2(g) H2O(l) DHf° = −285.8 kJ/mol

2 C2H2(g) 4 C(s) + 2 H2(g) DH° = 2(−227.4) kJ

4 C(s) + 4 O2(g) 4CO2(g) DH° = 4(−393.5) kJ

2 H2(g) + O2(g) 2 H2O(l) DH° = 2(−285.8) kJ

2. Arrange equations so they add up to desired reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) DH = −2600.4 kJ


reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

DH°reaction = S n DHf°(products) − S n DHf°(reactants)

DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]

DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))]

DHrxn = −2600.4 kJ


reaction

2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l)

Exercise 11 – Calculate the DH° for decomposing

10.0 g of limestone, CaCO3, under standard

conditions.

CaCO3(s) → CaO(s) + O2(g)

Copyright © 2011 Pearson Education, Inc.

Exercise 11 – Calculate the DH° for decomposing 10.0 g of

limestone, CaCO3(s) → CaO(s) + O2(g)

MMlimestone = 100.09 g/mol,

the units and sign are correct

10.0 g CaCO3

DH°, kJ

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

DHf°’s DH°rxn per mol CaCO3

g CaCO3 mol CaCO3 kJ from

above

CaCO3(s) → CaO(s) + O2(g)

DH°rxn = +572.7 kJ

83

Exercise 12

Given the chemical equation below :

SO3 (g) + H2O (l) → H2SO4 (aq)

56.5 kJ of heat is released, when a certain quantity

of SO3 (g) gas is dissolved in water to give 24.5 g of H2SO4 in

the solution.

(a)Calculate the heat of reaction, ∆H, for the formation of

H2SO4 represented by the above thermochemical equation.

(b)Draw the enthalpy diagram for the above reaction.

{Ans: -226kJ}

Stoichiometric Calculations involve DH

Exercise 13 – How much heat is evolved when a

0.483 g diamond is burned?

1 mol C = −395.4 kJ, Molar Mass = 12.01 g/mol

the sign is correct and the number is reasonable

because the amount of diamond is less than 1 mole

0.483 g C, DH = −395.4 kJ/mol C

q, kJ/mol

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

mol kJ g

Exercise 14

(a) 2NO (g) + O2 (g) → 2NO2 (g)

57 kJ of heat energy is released for each mole of NO2 molecule formed

in the above reaction. Calculate the enthalpy change, ∆H, for the above

reaction. (-114 kJ)

(b)

Using the answer from (a) above, determine the enthalpy change, ∆H,

for the reactions represented by the following equations :

i) NO (g) + ½ O2 (g) → NO2 (g) (-57 kJ)

ii) NO2 (g) → NO (g) + ½ O2 (g) (+57 kJ)

iii) 2/3 NO (g) + 1/3 O2 (g) → 2/3 NO2 (g) {-38 kJ}

Topic

5.7 Hess’s Law

• DH is well known for many reactions, and it is inconvenient

to measure DH for every reaction in which we are

interested.

• Since H is a state function, the DH in going from some

initial state to some final state is independent of the

pathway.

• This means that in going from a particular set of reactants

to a particular set of products, the DH is the same whether

the reaction takes place in one step or in a series of steps.

• We can estimate DH using published values and the

properties of enthalpy- Thus, it is not necessary to make

calorimetric measurements for all reactions- This principle

is known as Hess’s Law

5.7 Hess’s Law

5.7 Hess’s Law

Consider the oxidation of carbon to form carbon monoxide.

C (graphite ) + ½ O2 (g) CO (g) ∆H = ?

Problems

-Even if a deficiency of oxygen is used, the primary product

of the reaction of carbon and oxygen is CO2.

-Even if a CO is formed, it reacts with O2 to form CO2. Thus

the reaction cannot be carried out in a way that allows CO to

be sole product- impossible to measure ∆H for this reaction

by calorimetry.

Solution

- ∆H for the forming CO (g) from C (graphite) and O2 (g) can be

determined indirectly, as long as ∆H for other reactions is

already measured.

5.7 Hess’s Law Because DH is a state function, the total enthalpy change

depends only on the initial state of the reactants and the

final state of the products.

C (graphite ) + ½ O2 (g) CO (g) ∆Horxn = ?

Eq 1 : CO (g) + ½ O2 (g) CO2 (g) ∆Ho1 = -283.0 kJ/mol

Eq 2 : C (graphite) + O2 (g) CO2 (g) ∆Ho2 = -393.5 kJ/mol

Thus , ∆Horxn = ∆Ho

1 + ∆Ho2

Eq 1 in reversed form so ∆Ho1 is +ve

CO2 (g) CO (g) + ½ O2 (g) ∆Ho1 = +283.0 kJ/mol

+

C (graphite) + O2 (g) CO2 (g) ∆Ho2 = -393.5 kJ/mol

C (graphite ) + ½ O2 (g) CO (g) ∆Horxn = -110.5 kJ/mol

Hess’s law states that “if a reaction is carried out in a

series of steps, DH for the overall reaction will be equal to

the sum of the enthalpy changes for the individual steps.”

Definition

5.7 Hess’s Law

[3 NO2(g) 3 NO(g) + 1.5 O2(g)] DH° = (+174 kJ)

[1 N2(g) + 2.5 O2(g) + 1 H2O(l) 2 HNO3(aq)] DH° = (−128 kJ)

[2 NO(g) N2(g) + O2(g)] DH° = (−183 kJ)

3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = − 137 kJ

Example 14: Hess’s Law

Given the following information:

2 NO(g) + O2(g) 2 NO2(g) DH° = −116 kJ

2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq) DH° = −256 kJ

N2(g) + O2(g) 2 NO(g) DH° = +183 kJ

Calculate the DH° for the reaction below:

3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) DH° = ?

[2 NO2(g) 2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+116 kJ)

[2 N2(g) + 5 O2(g) + 2 H2O(l) 4 HNO3(aq)] x 0.5 DH° = 0.5(−256 kJ)

[2 NO(g) N2(g) + O2(g)] DH° = −183 kJ

Exercise 15 – Hess’s Law


Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ

2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJ


Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ

Exercise 15 – Hess’s Law


Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ



Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ

CuCl2(s) Cu(s) + Cl2(g) DH° = +206 kJ


Cu(s) + CuCl2(s) 2 CuCl(s) DH° = +170. kJ

Extra Practice – Hess’s Law

Use Hess’s Law to calculate the enthalpy change for

the formation of CS2 (l) from C (s) and S (s) from the

following enthalpy values

C (s) + O2 (g) CO2 (g) ΔHof = -393.5 kJ/mol

S (s) + O2 (g) SO2 (g) ΔHof = -295.8 kJ/mol

CS2 (s) +3O2 (g) 2SO2 (g) +CO2 (g) ΔHof = -1103.9

kJ/mol

Topic

5.8 Lattice Energy and

The Born-Haber Cycle

Energetics of Ionic Bond Formation

• The ionization energy of the metal is endothermic

– Na(s) → Na+(g) + 1 e ─ ΔH° = +496 kJ/mol

• The electron affinity of the nonmetal is exothermic

– ½Cl2(g) + 1 e ─ → Cl─(g) ΔH° = −244 kJ/mol

• Generally, the ionization energy of the metal is larger than

the electron affinity of the nonmetal, therefore the formation

of the ionic compound should be endothermic

• But the heat of formation of most ionic compounds is

exothermic and generally large. Why?

– Na(s) + ½Cl2(g) → NaCl(s) ΔH°f = −411 kJ/mol

5.8 Lattice Energy & The Born-Haber Cycle

The Crystal Lattice

The extra energy that is released comes from the formation

of a structure in which every cation is surrounded by anions,

and vice versa

This structure is called a crystal lattice

The crystal lattice is held together by the electrostatic

attraction of the cations for all the surrounding anions

The crystal lattice maximizes the attractions between

cations and anions, leading to the most stable arrangement


The Crystal Lattice



Lattice Energy

• The extra stability that accompanies the formation of the

crystal lattice is measured as the lattice energy

• The lattice energy is the energy released when the solid

crystal forms from separate ions in the gas state

– always exothermic

– hard to measure directly, but can be calculated from

knowledge of other processes

• Lattice energy depends directly on size of charges and

inversely on distance between ions



The Standard Lattice Enthalpy

The standard lattice enthalpy of an ionic compound is the

heat released when 1 mole of the ionic crystal is formed from

its gaseous ions. The value is negative because the

process is exothermic.

Example

Na+(g) + Cl-(g) → NaCl (s) ∆H = -770 kJ



The Born–Haber Cycle

• The Born–Haber Cycle is a hypothetical series of

reactions that represents the formation of an ionic

compound from its constituent elements

• The reactions are chosen so that the change in enthalpy of

each reaction is known except for the last one, which is the

lattice energy



Born–Haber Cycle

• Use Hess’s Law to add up enthalpy changes of other

reactions to determine the lattice energy

ΔH°f(salt) = ΔH°f(metal atoms, g) + ΔH°f(nonmetal atoms, g) +

ΔH°f(cations, g) + ΔH°f(anions, g) + ΔH°(crystal lattice)

– ΔH°(crystal lattice) = Lattice Energy

– for metal atom(g) cation(g), ΔH°f = 1st ionization energy

• don’t forget to add together all the ionization energies to

get to the desired cation

– M2+ = 1st IE + 2nd IE

– for nonmetal atoms (g) anions (g), ΔH°f = electron affinity



The Standard Enthalpy of Ionization

The standard enthalpy of ionization is the enthalpy

required to remove 1 mole of electrons from 1 mole of

gaseous metallic atom to form a positive ion. The process is

endothermic because energy is absorbed to release an

electron from an atom.

The amount of energy required to remove the first valence

electron from a gaseous atom is called the first ionization

energy while that required to remove the second electron

from the gaseous positive ion is called the second ionization

energy. Example

M(g) - e → M+(g) ∆H = First ionization energy

M+(g) - e → M2+(g) ∆H = Second ionization energy



Electron Affinity

The electron affinity is the enthalpy liberated / released when

1 mole of gaseous atoms accept 1 mole of electrons to

become negative ions. It has a negative value, showing that

this reaction is exothermic.

Example

A(g) + e → A- (g) ∆H = First electron affinity

A-(g)+ e → A2-(g) ∆H =Second electron affinity



The Standard Bond Formation Enthalpy

The standard bond formation enthalpy is the energy that is

released when 1 covalent bond is formed between 2 free

gaseous atoms. This proses is exothermic.

Example

2H (g) → H2 (g) ∆H = - 432 kJ mol-1

The standard bond formation enthalpy of the H-H bond in H2

molecule is – 432 kJ mol-1.



Standard Bond Enthalpy (Bond Energy)

The standard bond dissociation enthalpy is the energy

that must be absorbed to separate the two atoms in a

covalent bond. This process is endothermic

Example

H2 (g) → 2H (g) ∆H = +432 kJ mol-1

The standard bond dissociation enthalpy of the H-H bond in

H2 molecule is +432 kJ mol-1



Standard Enthalpy of Atomization

The standard enthalpy of atomization is the heat absorbed

when 1 mole of free gaseous atom is formed from its element

in the standard state. The process is endothermic as heat is

absorbed.

Example

½ H2 (g) → H (g) ∆H = +216 kJ

Na(s) → Na(g) ∆H = +109 kJ




Born–Haber Cycle for NaCl

DH°f(NaCl, s) = DH°f(Na atoms,g) + DH°f(Cl atoms,g) + DH°f(Na+,g) + DH°f(Cl−,g) + DH°(NaCl lattice)

DH°f(NaCl, s) = DH°f(Na atoms,g) + DH°f(Cl–Cl bond energy) + Na 1st Ionization Energy + Cl Electron Affinity + NaCl Lattice Energy

Na(s) → Na(g) DHf(Na,g) ½ Cl2(g) → Cl(g) DHf(Cl,g) Na(g) → Na+(g) DHf(Na+,g) Cl (g) → Cl−(g) DHf(Cl −,g) Na+ (g) + Cl−(g) → NaCl(s) DH (NaCl lattice) Na(s) + ½ Cl2(g) → NaCl(s) DHf (NaCl, s)

Na(s) → Na(g) DHf(Na,g) ½ Cl2(g) → Cl(g) DHf(Cl,g) Na(g) → Na+(g) DHf(Na+,g) Cl (g) → Cl−(g) DHf(Cl −,g) Na+ (g) + Cl−(g) → NaCl(s) DH (NaCl lattice)

Na(s) → Na(g) +108 kJ ½ Cl2(g) → Cl(g) +½(244 kJ) Na(g) → Na+(g) +496 kJ Cl (g) → Cl−(g) −349 kJ Na+ (g) + Cl−(g) → NaCl(s) DH (NaCl lattice) Na(s) + ½ Cl2(g) → NaCl(s) −411 kJ

NaCl Lattice Energy = DH°f(NaCl, s) − [DH°f(Na atoms,g) + DH°f(Cl–Cl bond energy) + Na 1st Ionization Energy + Cl Electron Affinity ]

NaCl Lattice Energy = (−411 kJ) − [(+108 kJ) + (+122 kJ) + (+496 kJ) + (−349 kJ) ] = −788 kJ

Exercise 16 – Given the information below,

determine the lattice energy of MgCl2

Mg(s) Mg(g) DH1°f = +147.1 kJ/mol ½ Cl2(g) Cl(g) DH2°f = +122 kJ/mol Mg(g) Mg+(g) DH3°f = +738 kJ/mol Mg+(g) Mg2+(g) DH4°f = +1450 kJ/mol Cl(g) Cl−(g) DH5°f = −349 kJ/mol Mg(s) + Cl2(g) MgCl2(s) DH6°f = −641 kJ/mol

Exercise 16 – Given the information below,

determine the lattice energy of MgCl2

Mg(s) Mg(g) DH1°f = +147.1 kJ/mol 2{½ Cl2(g) Cl(g)} 2DH2°f = 2(+122 kJ/mol) Mg(g) Mg+(g) DH3°f = +738 kJ/mol Mg+(g) Mg2+(g) DH4°f = +1450 kJ/mol 2{Cl(g) Cl−(g)} 2DH5°f = 2(−349 kJ/mol) Mg2+(g) + 2 Cl−(g) MgCl2(s) DH°lattice energy = ? kJ/mol Mg(s) + Cl2(g) MgCl2(s) DH6°f = −641 kJ/mol


Exercise17

Given the following data :

The Standard heat of formation of BaCl2(s) ∆H1 = -860 kJ mol-1

The Standard heat of atomization of barium, Ba(g) ∆H2 = +175 kJ mol-1

The first ionization energy of Ba(s) ∆H3 = +500 kJ mol-1

The second ionization energy of Ba(s) ∆H4 = +1000 kJ mol-1

The Standard enthalpy of atomization of Cl2(g) ∆H5 = +121 kJ mol-1

The first electron affinity of Cl2 (g) ∆H6 = -364 kJ mol-1

(a) Write the thermochemical equations to represent all the above

reactions

(b) Construct a Born-Haber cycle for the formation of solid barium

chloride BaCl2 (s) and calculate the standard lattice enthalpy of

BaCl2 (s) .

{ -2049 kJ }


Exercise 18

The thermochemical equations given below represent all the reactions involved in a

Born-Haber cycle for the formation of solid sodium fluoride, NaF(s) , from its

elements in their standard states.

I) Na (s) + ½ F2 (g) → NaF(s) ∆H = x kJ

II) Na (s) → Na+(g) ∆H = +610 kJ

III) ½ F2 (g) → F-(g) ∆H = -271 kJ

IV) Na+(g) + F-(g) → NaF(s) ∆H = -915 KJ

(a) Use the above equations, construct a Born-Haber cycle for the

formation of sodium fluoride.

(b)(i) Name the enthalpy change for the reactions represented by

equations I, and IV.

(ii) calculate the value of x.

(c) Equations (II) and (III) are formed by the combination of 2

consecutive steps of reactions . Write equations for the 2

consecutive steps in each of the equations (II) and (III) . Name the

enthalpy change for each reaction step in the two equations.

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Ch 5-Thermochem 2013