Energetics

Energetics
CAPE Chemistry
Grade 12
Miss Bartley

Objectives- By the end of this
topic you should be able to:
 State that chemical reactions take place through energy
changes usually in the form of heat associated with the
making and breaking of bonds
 State that energy changes occur in chemical reactions
associated with the making and breaking of bonds
 Explain the differences between exothermic and
endothermic reactions using energy profile diagrams
 Explain the term ‘bond energy’

Objectives- By the end of this
topic you should be able to:
 Explain how bond energy data may be used to show
the relationship between the strength of covalent bonds
and reactivity of covalent molecules.
 Apply concepts associated with enthalpy changes.
 Explain the effect of ionic charge and radius on the
magnitude of lattice energy.
 State Hess law of constant heat summation
 Calculate enthalpy changes from appropriate
experimental data.

Thermochemistry
 What is thermochemistry?
 Thermochemistry is the study of energy changes
associated with chemical reactions.

Enthalpy
What is enthalpy?
 Enthalpy (H) is the total energy (Ek and Ep) of
system. It is in the molecules, the bonds, and
stored chemically.
 Enthalpy change (Δ H) is the measure of the
heat change in a system.
 System is the chemical reactants and products
and surrounding is the atmosphere, container,
any solvent that do not react etc.

Video on enthalpy.
 First 5 mins
 https://www.youtube.com/watch?v=SV7U4yA
XL5I

Endothermic vs exothermic
 What is the difference between the two?
 Endothermic reactions have more energy stored in
the new bonds (products) than were in the old
bonds (reactants) which takes in energy.
 Exothermic reactions have more energy in the old
bonds (reactants) than in the new bonds (products).

Enthalpy change
 What is the value of enthalpy change for
exothermic reactions?
Exothermic reaction enthalpy change is
negative due to Δ H = enthalpy products
– enthalpy reactants.
Endothermic reaction would be positive
because of the same equation.

Enthalpy change
 How do we show enthalpy change of a reaction?
 It is usually written with the equation.
 + represents endothermic, - represents exothermic
 It is given in the units of kJ/mol (or just kJ) because
it changes with the amount of reactants and limiting
reactants.
 This is at thermochemical standard conditions which
are 25 degrees C, 1 atm and solution concentrations
of 1 mol/dm3.

Enthalpy change
 What is the difference between heat and
temperature?
 Temperature is the measure of the kinetic energy of
the particles and does not depend on the amount
present.
 Heat is the measure of the total energy of a
substance that increases and decreases with amount
of substance.

Enthalpy change
 How do we find heat energy?
 Heat energy = mass x specific heat capacity x change
in temperature (m x c x ΔT).
 Specific heat capacity depends on the specific
material involved.

Calorimetry
 What is Calorimetry?
 This is a technique to measure the energy change in
a reaction.
 Total heat = (m c ΔT)liquid + (m c ΔT)solid

Video on calorimetry
 https://www.youtube.com/watch?v=JuWtBR-
rDQk

Questions
 Calculate the enthalpy change for the reaction
when 0.63g of zinc is added to 75cm3 of copper
sulfate solution in a calorimeter. The tem-
perature rose from 18C to 28.4C. Rel. atomic
mass of of zinc is 65.4

Enthalpy of Combustion by
Calorimetry

Calorimetry Question
 When 8.00 g of ammonium nitrate dissolves in 100.0
cm3 of water, the temperature dropped from 19.0 C to
14.5 C. What is the enthalpy of solution of ammonium
nitrate?

BOND ENTHALPIES
 Enthalpy change (H) can also be calculated
directly from bond enthalpies.
 Bond enthalpy for a diatomic molecule is
defined as the enthalpy change for the
following process:
X-Y(g) ↔ X(g) + Y(g)

BOND ENTHALPY
 The bond dissociation enthalpy is the energy
needed to break one mole of the bond to give
separated atoms - everything being in the gas
state.
 Important! The point about everything being in
the gas state is essential. You cannot use bond
enthalpies to do calculations directly from
substances starting in the liquid or solid state.

BOND ENTHALPY
 As an example of bond dissociation
enthalpy, to break up 1 mole of gaseous
hydrogen chloride molecules into separate
gaseous hydrogen and chlorine atoms
takes 432 kJ. The bond dissociation
enthalpy for the H-Cl bond is +432 kJ mol-
1.

CALCULATING BOND ENTHALPIES
 Breaking bonds: H = positive (endothermic)
 Forming bonds: H = negative (exothermic)
 Calculating H from bond enthalpies:
Hrxn = (bonds broken) + (bonds formed)

Table of bond enthalpies
Single Bonds Double Bonds Triple Bonds
H-H 436 C=C 612 C≡C 837
C-C 348 O=O 496 N≡N 944
C-H 412
O-H 463
N-H 388
N-N 163

Example: Calculate H for the
hydrogenation of ethane
There is more energy released than absorbed,
so the rxn is exothermic.
H = 2696 – 2820 = -124 kJ mol-1
H-H 436 C=C 612 C≡C 837
C-C 348 O=O 496 N≡N 944
C-H 412
O-H 463
N-H 388
N-N 163
C=C 612
4 C-H 4 x 412
H-H 436
2696 kJ
C-C 348
6 C-H 6 x 412 2820 kJ
Energy absorbed to break bonds: Energy released as bonds form:
HH (g)
(g) (g)
+

Example: Calculate H for the
combustion of hydrazine in oxygen.
This reaction has been used to power
spacecraft.
There is more energy released than absorbed,
so the rxn is exothermic.
H = 2211 – 2796 = -585 kJ mol-1
H-H 436 C=C 612 C≡C 837
C-C 348 O=O 496 N≡N 944
C-H 412
O-H 463
N-H 388
N-N 163
N-N 163
4 N-H 4 x 388
O=O 496
2211 kJ
N≡N 944
4 O-H 4 x 463 2796 kJ
Energy absorbed to break bonds: Energy released as bonds form:
O=O (g) (g)
+
(g) N≡N (g) + 2

WHAT HAPPENS WHEN LIQUIDS
ARE PRESENT
 Remember that you can only use bond enthalpies
directly if everything you are working with is in the gas
state.
 If you have one or more liquids present, you need an
extra energy term to work out the enthalpy change
when you convert from liquid to gas, or vice versa.
That term is the enthalpy change of vaporisation,
and is given the symbol ΔHvap or ΔHv.
 This is the enthalpy change when 1 mole of the liquid
converts to gas at its boiling point with a pressure of 1
bar (100 kPa).

ARE PRESENT
 For water, the enthalpy change of vaporisation is
+41 kJ mol-1. That means that it take 41 kJ to
change 1 mole of water into steam. If 1 mole of
steam condenses into water, the enthalpy change
would be -41 kJ. Changing from liquid to gas
needs heat; changing gas back to liquid releases
exactly the same amount of heat.

WHAT HAPPENS WHEN
LIQUIDS ARE PRESENT
 To see how this fits into bond enthalpy calculations, we
will estimate the enthalpy change of combustion of
methane - in other words, the enthalpy change for this
reaction:

ARE PRESENT
 Notice that the product is liquid water. You
cannot apply bond enthalpies to this. You must
first convert it into steam. To do this you have to
supply 41 kJ mol-1.
bond enthalpy (kJ mol-1)
C-H +413
O=O +498
C=O in carbon dioxide +805
O-H +464

ARE PRESENT

ARE PRESENT
 Before you go on, make sure that you can see why
every single number and arrow on this diagram is there.
 In particular, make sure that you can see why the first 4
appears in the expression "4(+464)". That is an easy
thing to get wrong.

ARE PRESENT
 That's the hard bit done - now the calculation:
 ΔH + 2(805) + 2(41) + 4(464) = 4(413) + 2(498)
 ΔH = 4(413) + 2(498) - 2(805) - 2(41) - 4(464)
 ΔH = -900 kJ mol-1
 The measured enthalpy change of combustion is -890
kJ mol-1, and so this answer agrees to within about 1%.
As bond enthalpy calculations go, that's a pretty good
estimate.

Limitations of Using
Average Bond Enthalpies
 Average bond enthalpies can only be used if all
reactants and products are gases.
◦ If the H2O product in the previous example were a liquid, then
even more heat would be evolved since the Hvap for H2O
would also need to be included in the calculation.
 Average bond enthalpies are obtained by
considering a number of similar compounds, but
in reality the energy of a particular bond will vary
slightly in different compounds (it will be
affected by neighboring atoms).
 Thus, H values obtained using average bond
enthalpies are not necessarily very accurate.

Bong energy and strength of
covalent bonds
 Enthalpy of reaction can determine the likelihood of a
reaction occurring. Exothermic reactions are more
likely to happen than endothermic ones.
Hence if the overall energy released when new bonds
form in the product is greater than the energy required to
break the bond the reaction will be more likely to occur.
 Bond energy is dependent on the molecular
environment of the bond. Thus strength of a C-C bond
varies slightly with the nature of the different atoms
attached to the two carbon atoms.

Lack of reactivity of nitrogen
 The chemistry of nitrogen is dominated by the ease
with which nitrogen atoms form double and triple
bonds. A neutral nitrogen atom contains five valence
electrons: 2s2 2p3. A nitrogen atom can therefore
achieve an octet of valence electrons by sharing three
pairs of electrons with another nitrogen atom.

Lack of reactivity of nitrogen
 Because the covalent radius of a nitrogen atom is
relatively small (only 0.070 nm), nitrogen atoms come
close enough together to form very strong bonds. The
bond-dissociation enthalpy for the nitrogen-nitrogen
triple bond is 946 kJ/mol, almost twice as large as that
for an O=O double bond.
 The strength of the nitrogen-nitrogen triple bond
makes the N2 molecule very unreactive.

Factors that affect bond energy
 1:-Greater the size of the atom, greater is the bond
length and less is the bond dissociation energy i.e. less is
the bond strength .
 2:-For the bond between two similar atoms, greater is
the multiplicity of the bond, greater is the bond
dissociation energy .
 3:- Greater the number of lone pairs of electrons
present on the bonded atoms, greater is the repulsion
between the atoms and hence less is the bond
dissociation energy.

Factors that affect bond energy
 4:- The bond energy increases as the hybrid orbitals
have greater amount of s orbital contribution. Thus,
bond energy decreases in the following order, sp 〉 sp²
〉sp³.
 5:-Greater the electronegativity difference, greater is the
bond polarity and hence greater will be the bond
strength.

QUESTIONS
1. Given the following bond enthalpies (bond energies) in kJmol-1
bond ΔHBE: C-H 412; C-C 347; O-H 464; O=O 498; C=O 805 (for CO2 only);
C-O 358
(i) Calculate the enthalpy of combustion of ethane assuming all the
species are gaseous.
(ii) Why in (i) do you not get the value of -1560 kJ mol-1?
2. Cl2(g) + C3H8(g) ====> C3H7Cl(g) + HCl(g)
(c) Given the following average bond energies in kJmol-1
bond ΔHBE: C-H 412 ; Cl-Cl 242 ; C-Cl 338 ; H-Cl 431
Calculate the enthalpy change for the reaction

Graded COURSE WORK
QUESTIONS cont’d
1. The student measured 50.0 cm3 of water into the
beaker and lit the burner. When the temperature of
the water had gone up by 12.8 °C, he found that
0.100 g of propan-1-ol had been burnt.
(i) Calculate the energy, in kJ, produced by burning
0.100 g of propan-1-ol. The specific heat capacity of
water is 4.18 J g–1 K –1 . energy = ................ kJ [2]

Graded COURSE WORK
QUESTIONS cont’d
(i) Calculate the number of moles of propan-1-ol in 0.100 g.
(ii) number of moles = ................ [2]
(iii) Calculate the enthalpy change of combustion, in kJ mol–1 , of propan-1-
ol. enthalpy change .................. kJ mol–1 [

Graded COURSE WORK
QUESTIONS cont’d
1. A student carries out an experiment to determine the enthalpy change of
combustion of glucose. In the experiment, 0.831 g of glucose is burned.
The energy released is used to heat 100 cm3 of water from 23.7 °C to 41.0
°C.
(i) Calculate the energy released, in kJ, during combustion of 0.831 g glucose.
The specific heat capacity of water = 4.18 J g–1 K –1 . Density of water =
1.00 g cm–3 . energy = ......................... kJ [2]
(ii) Calculate the amount, in moles, of glucose that is burned. amount =
......................... mol [2]
(iii) Calculate the enthalpy change of combustion of glucose. Give your
answer to three significant figures. ΔHC = ..................... kJ mol–1 [2] [Total
6 marks]

Definition of Enthalpy Changes
 Standard enthalpy change of reaction: Standard
enthalpy change per mole for the conversion of
reactants to products. This applies to any type of
reaction eg. Combustion etc
 Standard condition means pressure is 1 bar,
temperature is 25 °C, enthalpy is per mole, all
substances are in their standard states at the
stated temperature.

 Standard enthalpy change of formation: Standard
enthalpy change when one mole of a compound is
formed from its elements.
 Value for enthalpy change of an element in its normal
state at room temperature is zero.
 Eg. 2Al(s) + 3/2 O2(g) Al2O3(s) dH=-1676kJ/mol
substances are in their standard states at the stated
temperature.

 Standard enthalpy change of combustion:
Standard enthalpy change when one mole of a
substance is combusted in oxygen.
 Eg. CH4(g) + 2 O2(g) CO2(g) + 2H2O(l)
dH=-890.7kJ/mol
stated temperature.

 Standard enthalpy change of fusion: Standard
enthalpy change when one mole of a solid
substance is melted at its melting point.
 Eg. Al(s)  Al(l) + 2H2O(l) dH= +10.7kJ/mol
stated temperature.

 Standard enthalpy change of vaporisation:
Standard enthalpy change when one mole of a
liquid substance is vaporised at its boiling point.
 Eg. H2O(l)  H2O (g) dH=+40.7 kJ/mol
stated temperature.

 Standard enthalpy change of neutralization:
Standard enthalpy change when an acid and a
base reacts to form one mole of water.
 NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
 dH= -53kJ/mol
stated temperature.

 Standard enthalpy change of hydration: Standard
enthalpy change when a hydrated ion is
produced from an ion in gas phase.
 Na+(g) + H2O (l) Na+ (aq) dH= -406kJ/mol
stated temperature.

 Standard enthalpy change of solution: Standard
enthalpy change when a solid is dissolved in a
large excess of water.
 NaOH(s) + H2O  NaOH(aq)
dH=-42.7kJ/mol
stated temperature.

 Standard enthalpy change of atomisation:
Standard enthalpy change when separate gaseous
atoms are produced from an element in its
standard state.
 Na(s)  Na(g) dH= +108kJ/mol
stated temperature.

 First ionization energy/ enthalpy: Standard
enthalpy change when one mole of electron is
removed from a single gaseous atom.
 Na(g)  Na+(g) + e dH= +498kJ/mol
stated temperature.

 Electron affinity/ electron gain enthalpy:
Standard enthalpy change when one mole of
electron is added to a single gaseous atom.
 Cl(g) + e Cl-(g) dH= -351kJ/mol
stated temperature.

 Lattice formation enthalpy: Standard enthalpy
change when one mole of an ionic substance is
formed from its gaseous ions.
 Na+ (g)+ Cl- (g)  NaCl(s) dH= -787kJ/mol
stated temperature.

HESS’ LAW OF CONSTANT
HEAT SUMMATION
The energy change in a reaction is
independent of the route taken from
reactants to products provided the initial and
final conditions are the same.

HESS’ LAW
The value of H for a reaction is the same
whether it occurs in one step or in a series of
steps.
H for the overall equation is the sum of the
H’s for the individual equations.
Hrxn = H1 + H2 + …

HESS’ LAW
 This law is simply a statement of the law of
conservation of energy:
 Energy may be exchanged between the
materials and the surroundings but the total
energy of the materials and the
surroundings remain constant
 It can be used to determine enthalpy
changes which cannot be measured
directly by experiment.

Hess’s Law
 What does Hess’s Law say?
 It states that if you can add 2 or more different
equations to produce the final equation, you can add
the individual enthalpy changes to find the total.
 This means that if you break down the steps
involved, the overall change will be the same.

HESS’ LAW
 Another way of thinking about Hess’ Law is
graphically in terms of “energy cycles.”
 The enthalpy change for a reaction depends only
on the difference between the enthalpy of the
products and the enthalpy of the reactants. It is
independent of the reaction pathway.

HESS’ LAW
The enthalpy change going from A to B is the same
whether the reaction proceeds directly to A or
whether it goes via an intermediate.
A
B
C
H1
H2
H3
H1 = H2 + H3

HESS’ LAW
For example, the enthalpy of combustion of both carbon and
carbon monoxide to form carbon dioxide can easily be measured
directly, but the combustion of carbon to carbon monoxide
cannot. This can be represented by an energy cycle.
C(s) + ½O2(g) CO(g)
CO2(g)
½O2(g)
-393
kJ mol-1
Hx
-283 kJ mol-1
-393 = Hx + (-283)
O2(g)
Hx = -393 + 283 = -110 kJ mol-1

Example: Calculate the standard enthalpy change when one mole of
methane is formed from its elements in their standard states. The
standard enthalpies of combustion ( ) of carbon, hydrogen and
methane are -393, -286 and -890 kJ mol-1 respectively.
 Step 1: Write the equation for enthalpy change with the
unknown H value. Call this value Hx.

 c
H
C(s) + 2H2(g) CH4(g)
Hx

 Step 2: Construct an energy cycle showing the different
routes to the products (in this case the products of
combustion).

 c
H
C(s) + 2H2(g) CH4(g)
Hx
CO2(g) + 2H2O(g)
2O2(g)
O2(g) O2(g)

 Step 3: Use Hess’ Law to equate the energy changes for the
two different routes.

 c
H
C(s) + 2H2(g) CH4(g)
Hx
CO2(g) + 2H2O(g)
2O2(g)
O2(g) O2(g)
)
CH
(
H
H
)
H
(
H
2
)
C
(
H 4
c
x
2
c
c











direct route route via methane

 Step 4: Rearrange the equation and substitute values to solve
for the unknown H value.

 c
H
)
CH
(
H
H
)
H
(
H
2
)
C
(
H 4
c
x
2
c
c











)
CH
(
H
)
H
(
H
2
)
C
(
H
H 4
c
2
c
c
x











(-890)
-
2(-286)
393
Hx 


 
-1
x mol
kJ
75
H 

 

Lattice Energy and Born-Haber
Cycles
 Lattice energies are not easily empirically determ
ined as it’s difficult to isolate gaseous ions: rath
er, lattice energies are calculated using Born‐
Haber cycle methodology, which is an applic
ation of Hess’ Law.

Lattice Enthalpy
Lattice enthalpy depends on
 The charges on the ions. The greater the charges
on the ions, the greater the attraction between
them and greater lattice energy. The magnitude
of the attraction is proportional to the product
of the charges.
 The distance between the ions. The smaller the
distance between the ions, the greater the
attraction.
 The detailed crystal structure of the compound

Lattice Enthalpy
 The exothermic lattice formation enthalpy must
be large enough to compensate for all the
endothermic processes in the Born Haber if the
compound is to be stable and not form back its
elements easily on heating.
 Eg. NaO does not exist because the electron
gain enthalpy is too large and endothermic and
the lattice enthalpy of the hypothetical NaO is
not large enough to compensate for it.

Use of Born-Haber Cycles
 Empirical value of ΔHo
lat is found using Born-
Haber cycle.
 Theoretical value of ΔHo
lat can be found by
summing the electrostatic attractive and
repulsive forces between the ions in the crystal
lattice.

Trends
ΔHo
lat Change from NaCl
MgO 3889 Increased ionic charge
NaCl 771 ------
KBr 670 Larger ions

Compound Empirical value Theoretical value
NaCl 771 766
KBr 670 667
KI 632 631
AgCl 905 770

Agreement
 Usually there is good agreement between
empirical and theoretical values
 If there isn’t good agreement
 Implying that the description of the compound as
ionic is inappropriate
 There could be a significant degree of covalent
character in the bonding (EN difference less than
1.7)
 Presence of covalent character leads to an increase in
ΔHo
lat

Born-Haber Cycle
 A series of hypothetical steps and their enthalpy
changes needed to convert elements to an ionic
compound and devised to calculate the lattice
energy.
 Using Hess’s law as a means to calculate the
formation of ionic compounds

Overview of Born-Haber Cycle
Steps
1. Elements (standard state) converted into
gaseous atoms
2. Losing or gaining electrons to form cations and
anions
3. Combining gaseous anions and cations to form
a solid ionic compound
Hence to calculate enthalpy of formation using
Born Haber cycle, we need atomisation enthalpy,
ionization enthalpy, electron affinity and lattice
enthalpy.

Step 1: Atomisation
 The standard enthalpy change of atomisation is the
ΔH required to produce one mole of gaseous
atoms
 Na(s)  Na(g) ΔHo
at = +109 kJmol-1
 Cl2(g)  Cl(g) ΔHo
at = +121 kJmol-1

Step 2: Formation of gaseous ions
 Electron Affinity
Enthalpy change when one mole of
gaseous atoms or anions gains electrons
to form a mole of negatively charged
gaseous ions.
 Cl(g) + e-  Cl-(g) ΔHo = -364 kJmol-1
For most atoms = exothermic, but
gaining a 2nd electron is endothermic due
to the repulsion between the anion and
the electron

Becoming cations
 Ionisation energy
 Enthalpy change for one mole of a gaseous element
or cation to lose electrons to form a mole of
positively charged gaseous ions
 Na(g)  Na+(g) + e- IE1= +494 kJmol-1

Lattice Enthalpy
 Energy required to form one mole of a solid
compound from gaseous ions.
 Na+(g) + Cl-(g) NaCl (s)
ΔHo
lat = -771kJmol-1
 It is highly endothermic
 We cannot directly calculate ΔHo
lat , but values
are obtained indirectly through Hess’s law for the
formation of the ionic compound

Born-Haber Cycle
 This is a Born‐Haber Cycle (Born
Haber ) for the reaction between Sodium and
chlorine.
 The enthalpy of formation (indicated by blue
arrow) = sum of all the enthalpies indicated by
black arrows which all go in the opposite
direction.

Calculations
 Calculate the lattice energy of NaCl(s) using the
following: (kJmol-1)
 Enthalpy of formation of NaCl = - 411
 Enthalpy of atomisation of Na = +109
 Enthalpy of atomisation of Cl = +121
 Electron affinity of Cl = - 364
 Ionisation energy of Na = + 494
 Enthalpy of atomisation + electron affinity +
ionisation = enthalpy of formation + lattice
energy

Past Paper Question May/Jun
2008 Paper 2

Graded Homework
 Question 1
a) Draw an Energy cycle to show the relationship
between lattice enthalpy, hydration enthalpy
and enthalpy of solution for sodium chloride.
[6 marks]
b) Calculate the molar enthalpy of solution if
lattice enthalpy= +788kJ/mol and the sum of
hydration enthalpy of the chloride and sodium
ions = -784kJ/mol. [2 marks]

Graded Homework
Question 2
a) Draw a Born-Haber cycle for the formation of Calcium Oxide. [12 marks]
Using the following data answer the questions that follow.
dH (atomisation) Ca= +193 kJ/mol
dH (atomisation) ½ O2 = +248 kJ/mol
dH (1st ionization energy) Ca= +590 kJ/mol
dH (2nd ionization energy) Ca = +1150 kJ/mol
dH (lattice) CaO= -3513 kJ/mol
dH (formation) CaO= -635 kJ/mol
b) Given that the second electron affinity of oxygen in +844kJ/mol, what is the
first electron affinity? [3 marks]
c) Would the first ionization enthalpy for calcium be larger or smaller than it is
for magnesium? Explain your answer. [2 marks]

Are you able to:
 State that chemical reactions take place through energy
changes usually in the form of heat associated with the
making and breaking of bonds
 State that energy changes occur in chemical reactions
associated with the making and breaking of bonds
 Explain the differences between exothermic and
endothermic reactions using energy profile diagrams
 Explain the term ‘bond energy’

Are you able to:
 Explain how bond energy data may be used to show
the relationship between the strength of covalent bonds
and reactivity of covalent molecules.
 Apply concepts associated with enthalpy changes.
 Explain the effect of ionic charge and radius on the
magnitude of lattice energy.
 State Hess law of constant heat summation
 Calculate enthalpy changes from appropriate
experimental data.

Energetics

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