121 Types of Enthalpy Change 122 BornHaber Cycles 123 Enthalpy Changes Enthalpy of Solution 124 Mean Bond Enthalpy 125 Entropy 121 Enthalpy Change Ionic Compounds Learning Objectives ID: 526792
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Slide1
12 Thermodynamics
12.1 Types of Enthalpy Change
12.2 Born-Haber Cycles
12.3 Enthalpy Changes – Enthalpy of Solution
12.4 Mean Bond Enthalpy
12.5 EntropySlide2
12.1 Enthalpy Change – Ionic Compounds
Learning Objectives:
Describe what is meant by the term enthalpy change.
Describe the different types of enthalpy changes (formation, atomisation, ionisation energy, electron affinity, lattice formation, hydration, solution, bond enthalpy).
Calculate the enthalpy changes on forming ionic compounds.Slide3
Enthalpy Review
Enthalpy change
is the
heat change
at constant pressure.
Standard conditions: 100kPa, 298 K (starting temperature)
Remember that
heat
and
temperature
are not the same.
Heat is a type of
energy
and is measured in
joules
and heat changes lead to
temperature
changes, which is measure in
Kelvins
.Slide4
Types of Enthalpy Changes
Enthalpy of Formation
Enthalpy of Atomisation
First Ionisation Energy
/Second Ionisation Energy
First Electron Affinity/Second Electron Affinity
Lattice Enthalpy of Formation
Enthalpy of Lattice DissociationEnthalpy of HydrationEnthalpy of SolutionMean Bond Enthalpy
Write down the symbol and the definitionSlide5
Standard Enthalpy of Formation
∆
H
Ꝋ
f
Enthalpy change when
one mole of a compound is formed
from its constituent elements under standard conditions
all reactants and products in their standard states.
change
s
tandard conditions
formationSlide6
Standard Enthalpy of Atomisation
∆
H
Ꝋ
at
Enthalpy change when
one mole of gaseous atoms
is formed from the element
In it’s standard state
under standard conditionsSlide7
First Ionisation Energy
First IE
Enthalpy change when
one mole of gaseous atoms
is converted into
one mole of gaseous +1 ions
under standard conditionsSlide8
Second Ionisation Energy
Second IE
Enthalpy change when
one mole of gaseous +1 ions
is converted into
one mole of gaseous +2 ions
under standard conditionsSlide9
First Electron Affinity
First ∆
H
Ꝋ
ea
Enthalpy change when
one mole of gaseous atoms
is converted into one mole of gaseous -1 ions
under standard conditionsSlide10
Second Electron Affinity
Second ∆
H
Ꝋ
ea
Enthalpy change when
one mole of gaseous -1 ions
is converted into one mole of gaseous -2 ions
under standard conditionsSlide11
Lattice Formation Enthalpy
∆H
Ꝋ
L
Enthalpy change when
one mole of solid ionic compound
is formed
from it’s gaseous ionsunder standard conditions
(always negative, energy released)Slide12
Enthalpy of Lattice Dissociation
-∆H
Ꝋ
L
Enthalpy change when
one mole of solid ionic compound
d
issociates intoit’s gaseous ionsunder standard conditions
(always positive, energy is absorbed)Slide13
Standard Enthalpy of Hydration
∆
H
Ꝋ
hyd
Enthalpy change when
one mole of gaseous atoms
is surrounded by water molecules under standard conditionsSlide14
Standard Enthalpy of Solution
∆
H
Ꝋ
sol
Enthalpy change when
one mole of solute
completely dissolvesi
n sufficient solvent to form a solution in which the molecules are ions do not interact
under standard conditionsSlide15
Mean Bond Enthalpy
∆
H
Ꝋ
diss
Enthalpy change when
one mole of gaseous molecules
breaks a covalent bond forming two free radicals
a
veraged over a range of compounds
a
t standard conditionsSlide16
For each type…
W
rite
an equation to represent the chemical reaction being
described
Tell me if the process is likely to be positive or negative
Explain why.Slide17
Standard Enthalpy of Formation
∆
H
Ꝋ
f
H
2 (g)
+
O
2 (g)
H
2
O (l)
Usually going to be negative.Molecules usually form because the molecule is more stable (lower in energy) than the constituent elements.
Remember: Bond making releases energy.
Slide18
Standard Enthalpy of Atomisation
∆
H
Ꝋ
at
Br
2(l)
Br
(g)
Usually positive.
Molecules form because that is a more stable form, so gaseous atoms are less stable (higher in energy). Remember: Bond breaking require energy.
Slide19
First Ionisation Energy
First IE
Na
(g)
Na
+
(g)
+ e
-
Positive
Removing an electron takes energySlide20
Second Ionisation Energy
Second IE
Na
+
(g)
Na
2+
(
g)
+ e-
Very positiveRemoving electron from positive ion require a lot of energy.Slide21
First Electron Affinity
First ∆
H
Ꝋ
ea
O
(g)
+ e
-
O
-
(g)Usually NegativeEnergy is gained when electrons are added.Slide22
Second Electron Affinity
Second ∆
H
Ꝋ
ea
O
-
(g) + e
-
O2-
(g)Usually Positive
Because of repulsion, adding the second electron requires more energy than is gained.Slide23
Lattice Formation Enthalpy
∆H
Ꝋ
L
Na
+
(g) + Cl
-
(g)
NaCl
(s)Always negative
Bond making releases energy, more stable in lattice form.Slide24
Enthalpy of Lattice Dissociation
-∆H
Ꝋ
L
NaCl
(s)
Na
+
(g) + Cl
-
(g)Always positiveThis is opposite of lattice formation, breaking bonds requires energy.Slide25
Standard Enthalpy of Hydration
∆
H
Ꝋ
hyd
Na
+
(g)
+
aq
Na+
(aq
)Cl-
(g) +
aq Cl-
(
aq
)
Usually negative
Water molecules stabilise the charges of the ions.Slide26
Standard Enthalpy of Solution
∆
H
Ꝋ
sol
NaCl
(s) +
aq
Na
+
(aq)
+ Cl-
(aq)
Usually slightly positiveBreaking the bonds of the lattice requires energy, however, the water molecules stabilise the ions so overall only small amount of energy absorbed.Slide27
Mean Bond Enthalpy
∆
H
Ꝋ
diss
CH
4
(g)
C
(g)
+ 4H (g)
Always positiveBond breaking requires energy.Slide28
12.2 Born-Haber Cycles
Learning Objectives:
Describe Hess’ Law.
Use Born-Haber Cycles to calculate enthalpy changesSlide29
Hess’s Law of Thermodynamics
The enthalpy change for a reaction is the same, no matter what route is taken.
For example:
CH
4
(g)
+ O2 (g)
CO
2
(g) + H
2O
(g)C (s)
+ H2 (g) + O
2 (g)Slide30
Born-Haber Cycles
Born-Haber Cycles are just another method to solve for the unknown enthalpy change of a chemical reaction by using enthalpy changes that we DO know.
It uses a diagram to represent the enthalpy changes on a vertical scale. Increases in energy are UP ( ) arrows, decreases in energy are DOWN ( )arrows.Slide31
Molly started out with £0. Then she received £100 for her birthday. She went out to dinner, this costed £30. Then she bought some new shoes. At the end of the day to had spent all of her birthday money. How much did her new shoes cost?
With Birthday Money
After Dinner
Broke
∆
M
bd
= +£100
∆
M
din
= -£30
∆
M
shu
= ? = -£70Slide32
Formation of an Ionic Compound
Electrons are transferred to atoms to form ions.
Ions then attract and are arranged into an ionic lattice.
This is how ionic lattices are formed.Slide33
Enthalpy Change in Formation of Ionic Compounds
Na
(s)
+
Cl
2 (g)
NaCl
(s)
∆
HꝊ
f = ?
What are the steps for this complete reaction to occur?Atomisation of Na
Atomisation of Chlorine
Ionisation of Na
Electron affinity of Cl
Formation of lattice
Slide34
Enthalpy Change in Formation of Ionic Compounds
Na
(s)
+
Cl
2 (g)
NaCl
(s)
∆
H
Ꝋf = ?
What are the steps for this complete reaction to occur?
Atomisation of Na Na
(s)
Na
(g)
Atomisation
of Chlorine
Cl
2 (g)
Cl
(g)
Ionisation of Na Na
(g)
Na
+
(g)
Electron affinity of Cl Cl
(g)
Cl
-
(g)
Formation of lattice Na
+
(g)
+ Cl
-
(g)
NaCl
(s)
Slide35
Enthalpy Change in Formation of Ionic Compounds
Na
(s)
+
Cl
2 (g)
NaCl
(s)
∆
H
Ꝋf = ?
What are the steps for this complete reaction to occur?
Atomisation of Na Na
(s)
Na
(g)
∆
H
Ꝋ
at
=
+108 kJ/
mol
Atomisation
of Chlorine
Cl
2 (g)
Cl
(g)
∆
H
Ꝋ
at
=
+122 kJ/
mol
Ionisation of Na Na
(g)
Na
+
(g)
first IE = +496 kJ/
mol
Electron affinity of Cl Cl
(g)
Cl
-
(g)
first EA = -349 kJ/
mol
Formation of lattice Na
+
(g)
+ Cl
-
(g)
NaCl
(s)
∆
H
Ꝋ
L
=
-788 kJ/
mol
Slide36
Born-Haber Cycle Formation of
NaCl
Na
(s)
+
Cl
2
(g)
Na
(g)
+
Cl
2
(g)
Na
(g)
+
Cl
(g)
Na
+
(g)
+
Cl
(g)
Na
+
(g)
+
Cl
-
(g)
NaCl
(s)
∆
H
Ꝋ
at
= +108 kJ/
mol
∆
H
Ꝋ
at
=
+122
kJ/
mol
First IE
=
+496
kJ/
mol
First EA
=
-349
kJ/
mol
∆
H
Ꝋ
L
=
-788
kJ/
mol
∆
H
Ꝋ
f
= ?Slide37
Born-Haber Cycle Formation of
NaCl
Na
(s)
+
Cl
2
(g)
Na
(g)
+
Cl
2
(g)
Na
(g)
+
Cl
(g)
Na
+
(g)
+
Cl
(g)
Na
+
(g)
+
Cl
-
(g)
NaCl
(s)
∆
H
Ꝋ
at
= +108 kJ/
mol
∆
H
Ꝋ
at
=
+122
kJ/
mol
First IE
=
+496
kJ/
mol
First EA
=
-349
kJ/
mol
∆
H
Ꝋ
L
=
-788
kJ/
mol
∆
H
Ꝋ
f
=
-411 kJ/
molSlide38
Example: Lattice Formation Enthalpy of MgCl
2
Write out the overall equation for the formation of magnesium chloride.
Write equations for all of the steps in the formation of magnesium chloride.
HINT: there are six stepsSlide39
Draw a Born-Haber Diagram for MgCl
2
HINT: there is a “trick” step, can you catch it? Remember your definitions
∆
H
Ꝋ
at
Mg = +148 kJ/mol
∆
H
Ꝋ
at Cl = +122 kJ/mol
First IE Mg= +738 kJ/molSecond IE Mg = +1451 kJ/molFirst EA Cl = -349 kJ/
mol∆HꝊf
MgCl2 = -641 kJ/mol
Use your Born-Haber Diagram to Calculate the Lattice Formation EnthalpySlide40
Mg
(s)
+
Cl
2
(g)
Mg
(g)
+
Cl
2
(g)
Mg
(g)
+ 2
Cl
(g)
Mg
2+
(g)
+ 2
Cl
(g)
Mg
2+
(g)
+ 2
Cl
-
(g)
MgCl
2
(s)
∆
H
Ꝋ
at
= +
148
kJ/
mol
2 x
∆
H
Ꝋ
at
=
+122 kJ/
mol
x 2
= +244 kJ/
mol
First IE
=
+738
kJ/
mol
2 x
First EA
=
-349 kJ/
mol
x 2 = -698 kJ/
mol
∆
H
Ꝋ
L
=
-2524
kJ/
mol
∆
H
Ꝋ
f
= ?
Second IE
=
+1451
kJ/
mol
Mg
+
(g)
+ 2
Cl
(g)Slide41
Mg
(s)
+
Cl
2
(g)
Mg
(g)
+
Cl
2
(g)
Mg
(g)
+ 2
Cl
(g)
Mg
2+
(g)
+ 2
Cl
(g)
Mg
2+
(g)
+ 2
Cl
-
(g)
MgCl
2
(s)
∆
H
Ꝋ
at
= +
148
kJ/
mol
2 x
∆
H
Ꝋ
at
=
+122 kJ/
mol
x 2
= +244 kJ/
mol
First IE
=
+738
kJ/
mol
2 x
First EA
=
-349 kJ/
mol
x 2 = -698 kJ/
mol
∆
H
Ꝋ
L
=
-2524
kJ/
mol
∆
H
Ꝋ
f
=
-641 kJ/
mol
Second IE
=
+1451
kJ/
mol
Mg
+
(g)
+ 2
Cl
(g)Slide42
12.3 More Enthalpy Changes
Learning Objectives:
Calculate enthalpy change of solution.
Describe how lattice enthalpy calculations support models for ionic bonding.
Explain how ions can become polarised.Slide43
Enthalpy of Solution
Ionic solids can dissolve in polar solvents.
This is called hydration if the solvent is water.
Hydration is when the water molecules surround ions.
What are the steps for process of forming a solution?
Breaking the ionic lattice (enthalpy of lattice dissociation).
Hydrating the positive ions (enthalpy of hydration).
Hydrating the negative ions (enthalpy of hydration).Slide44
Example:
NaClSlide45
Ionic Bonding Models
For most ionic compounds the
theoretical values
calculated from Born-Haber cycles
agrees
with
experimental values
.This proves that the model for ionic bonding (
lattice
) is correct.
However, some ionic compounds have theoretical and experimental values that
DO NOT agree.Another model needed to be found to explain these discrepancies.Slide46
Polarisation
ZnSe
experimental lattice formation enthalpy = -3611 kJ/
mol
t
heoretical lattice formation enthalpy = -3305 kJ/
molWHY?
Zn
2+
is very small and has a high + charge
Se2- is very large and has a high – chargeZn2+ moves closely to electron density of Se
2- and attracts the e-Since Se2-
is large, the e- are far from the nucleus and easily pulled awayThis distorts the electron cloud surrounding Se2-Slide47
Polarisation
The
distortion
causes their to be some electron density shared between the two ions (slightly covalent nature).
The Se
2-
ion is said to be polarised
.This causes the enthalpy change to be greater than expected.Slide48
When does polarisation happen?
Cation = small size, high charge
Anion = large size, high chargeSlide49
12.4 Mean Bond Enthalpy
Learning Objectives:
Explain the term mean bond enthalpy.
Calculate enthalpy changes using mean bond enthalpy.
Explain why this method is not as accurate.Slide50
Mean Bond Enthalpy
The average bond enthalpy term is the average amount of energy needed to break a specific
covalent bond
, measured over a wide variety of different molecules.
A measure of strength of a covalent bond.
In comparison, lattice enthalpy is a measure of the strength of an ionic bond.Slide51
Predicting reactions
Mean bond enthalpies can be used to predict how molecules may react.
We can predict which bonds may be more likely to break.
Which bond is most likely to break?
C-H 413 kJ/
mol
C-C 437 kJ/
mol
C-Br 290 kJ/
molSlide52
Predicting reactions
Mean bond enthalpies can also be used to compare
reactivities
of different molecules.
Which
haloalkane
is more reactive?
C-F 467 kJ/mol
C-Cl 346 kJ/
mol
C-Br 290 kJ/
molC-I 228 kJ/molSlide53
Calculating Approximate Enthalpy Changes
Hess’s Law can be applied.
One possible route to products would be to break all bonds in the reactants and then form all of the bonds for the products.
The enthalpies for these two processes can then be summed up to find the total enthalpy change.
Remember:
bond breaking requires energy (+ value)
bond formation releases energy (- value)Slide54Slide55Slide56
12.5 Why do chemical reactions take place?
Learning Objectives:
Explain the concept of entropy.
Calculate using enthalpy and entropy whether a reaction will spontaneously occur.
Analyse the effects of temperature on feasibility of a reaction.Slide57
Is a reaction feasible or spontaneous?
Reactions that will take place on their own are called spontaneous.
If it is possible for a reaction to take place on their own, the reaction is feasible.
What determines if a reaction is feasible?
If ΔH (enthalpy) is negative, the reaction is exothermic
If ΔS (entropy) is positive, the reaction increases in randomnessSlide58
Entropy
Entropy is a mathematical measure of the randomness of a system.
Change in entropy is represented as ΔS.
The universe prefers randomness (higher entropy) and is always moving towards disorder.
Values for entropy of different substances are determined mathematically, you will not be expected to calculate these, only how to use them. (see pg. 179)Slide59
Calculating Entropy Changes
Calculate the difference in entropy from reactants to products to find the ΔS of a reaction.
ΔS =
S
products
–
S
reactants
If ΔS is positive, entropy is increasing, disorder is increasing. The products are more disordered than the reactants.
If ΔS is
negative,
entropy is decreasing
, disorder is decreasing. The products are less disordered than the reactants.Slide60
Gibbs Free Energy
ΔG represents the Gibbs free energy and combines both enthalpy and entropy.
It is used to determine whether or not a reaction is feasible.
ΔG = ΔH – TΔS
If ΔG is negative (-) the reaction is feasible.
If ΔG is positive (+) the reaction is NOT feasible.Slide61
What happens if ΔG = 0?
There will be a temperature where
ΔG =
0.
This is the temperature at which the reaction is just feasible.
In a closed system an equilibrium between products and reactants occur.
ΔG =
0 can also be used to calculate ΔS.Cases where both forms are equally likely (ie melting point), ΔG =
0.Slide62
Thermodynamics does not predict the rate of a reaction
Thermodynamics = Kinetics
Thermodynamics only predicts whether a reaction is feasible.
It DOES NOT predict how quickly the reaction may take place.
Kinetics is the branch of chemistry dealing with rate of reaction.