ch 5-thermochem 2013
DESCRIPTION
matriculation thermochemistryTRANSCRIPT
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
5.1 Introduction to Thermochemistry
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
5.1 Introduction to Thermochemistry
• The amount of kinetic energy an object has is directly
proportional to its mass and velocity
KE = ½mv2
Units of Energy
5.1 Introduction to Thermochemistry
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
5.1 Introduction to Thermochemistry
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
5.1 Introduction to Thermochemistry
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
5.1 Introduction to Thermochemistry
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
5.2 System and Surroundings
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
5.2 System and 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
5.2 System and Surroundings
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 ─
5.2 System and Surroundings
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 +
5.2 System and Surroundings
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
5.2 System and Surroundings
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
5.2 System and Surroundings
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
5.2 System and Surroundings
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
5.2 System and Surroundings
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
5.2 System and Surroundings
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
5.2 System and Surroundings
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)
5.2 System and Surroundings
Energy Exchange in Chemical Reactions
2. Work done on a system = work done by a surroundings
The system has work done on it by the surroundings. w = +ve
5.2 System and Surroundings
Energy Exchange in Chemical Reactions
The values of q and w can have either a positive or negative
5.2 System and Surroundings
Energy Exchange in Chemical Reactions
q w + = ∆E
+
+
-
-
-
-
+
+
+
-
depends on sizes of q and w
depends on sizes of q and w
5.2 System and Surroundings
Energy Exchange in Chemical Reactions
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
5.4 Specific Heat Capacity
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
5.4 Specific Heat Capacity
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
5.4 Specific Heat Capacity
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
5.4 Specific Heat Capacity
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)
5.4 Specific Heat Capacity
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
5.4 Specific Heat Capacity
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)
5.4 Specific Heat Capacity
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)
5.4 Specific Heat Capacity
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
5.4 Specific Heat Capacity
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
5.5 Enthalpy and Calorimetry
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
5.5 Enthalpy and Calorimetry
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
5.5 Enthalpy and Calorimetry
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
5.5 Enthalpy and Calorimetry
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
5.5 Enthalpy and Calorimetry
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
5.5 Enthalpy and Calorimetry
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)
5.5 Enthalpy and Calorimetry
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
Example 11: Calculate the enthalpy change in the
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
Example 11: Calculate the enthalpy change in the
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
Given the following information:
Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ
Exercise 15 – Hess’s Law
Given the following information:
Cu(s) + Cl2(g) CuCl2(s) DH° = −206 kJ
2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 kJ
Calculate the DH° for the reaction below:
Cu(s) + CuCl2(s) 2 CuCl(s) DH° = ? kJ
CuCl2(s) Cu(s) + Cl2(g) DH° = +206 kJ
2 Cu(s) + Cl2(g) 2 CuCl(s) DH° = − 36 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
5.8 Lattice Energy & The Born-Haber Cycle
The Crystal Lattice
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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.
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
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
5.8 Lattice Energy & The Born-Haber Cycle
Copyright © 2011 Pearson Education, Inc.
Copyright © 2011 Pearson Education, Inc.
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
Copyright © 2011 Pearson Education, Inc.
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 }
Copyright © 2011 Pearson Education, Inc.
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.