26771935 matriculation chemistry thermochemistry
TRANSCRIPT
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9.0 THERMOCHEMISTRY
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Concept of Enthalpy
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Important Terms
Heat is energy transferred between two bodies ofdifferent temperatures
System is any specific part of the universe
Surroundings is everything that lies outside the
system
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Open system is a system that can exchangemass and energy with its surroundings
Closed system is a system that allows theexchange of energy with its surroundings
Isolated system is a system that does not allowthe exchange of either mass or energy with itssurroundings
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Energy is the ability to do work
SI unit of energy is kg m2
s-2
orJoule (J) Non SI unit of energy is calorie (Cal)
1 Cal = 4.184 J
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Thermochemistry
A study of heat change in chemical reactions.
Two types of chemical reactions:
Exothermic Endothermic
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Exothermic reactions
Enthalpy of products < Enthalpy of reactants, H is
negative.
Energy is released from the system to the
surroundings.
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Consider the following reaction:
A (g) + B (g) C (g) H = ve
(reactants) (product)
reactantsenthalpy
reaction pathway
= -veH
products
Energy profile diagram for exothermic reaction
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Enthalpy of products > enthalpy of reactants, H ispositive
Energy is absorbed by the system from the surrounding
Endothermic Reactions
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Consider the following reaction
A (g) + B (g) C(g) H = + ve
(reactants) (product)
Energy profile of diagram endothermic reactions
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Enthalpy, H
The heat content of a system or total energy in thesystem
Enthalpy, H of a system cannot be measured whenthere is a change in the system.
Example: system undergoes combustion or ionisation.
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Enthalpy of Reaction, H andStandard Condition
Enthalpy of reaction:
The enthalpy change associated with a chemicalreaction.
Standard enthalpy, H
The enthalpy change for a particular reaction thatoccurs at 298K and 1 atm(standard state)
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Thermochemical Equation The thermochemical equation shows the enthalpy changes.
Example : H2O(s) H2O(l) H = +6.01 kJ
1 mole ofH2O(l) is formed from 1 mole ofH2O(s) at 0C,H = +6.01 kJ
However, when 1 mole ofH2O(s)is formed from 1 mole ofH2O(l), the magnitude ofH remains the same with theoppositesignof it.
H2O(l) H2O(s) H = 6.01 kJ
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Types of Enthalpies
There are many kind of enthalpies such as:
Enthalpy of formation
Enthalpy of combustion
Enthalpy of atomisation
Enthalpy neutralisation
Enthalpy hydration
Enthalpy solution
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Enthalpy of Formation, Hf
The change of heat when 1 mole of a compound is
formed from its elements at their standard states.
H2(g) + O2(g) H2O (l) Hf= 286 kJ mol1
The standard enthalpy of formation of any element in
its most stable state form is ZERO.
H (O2 ) = 0 H (Cl2) = 0
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Enthalpy of Combustion, Hc
The heat released when 1 mole of substance is
burned completely in excess oxygen.
C(s) + O2(g) CO2(g) Hc = 393 kJ mol1
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Enthalpy of Atomisation, Ha
The heat change when 1 mole ofgaseous atoms is formedfrom its element
Ha is always positive because it involves only breaking ofbonds
e.g:
Na(s)
Na(g)
Ha = +109 kJ mol-1
Cl2(g) Cl(g) Ha = +123 kJ mol-1
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Enthalpy of Neutralization, Hn
The heat change when 1 mole of water, H2O is formed from the
neutralization of acid and base .
HCl(aq)+ NaOH(aq) NaCl(aq) +H2O(aq) Hn = 58 kJ mol1
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Enthalpy of Hydration, Hhyd
The heat change when 1 mole ofgaseous ionsis
hydratedin water.
e.g:Na+(g) Na
+(aq) Hhyd = 406 kJ mol
-1
Cl-(g) Cl-(aq) Hhyd = 363 kJ mol
-1
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Enthalpy of Solution, Hsoln
The heat change when 1 mole of a substance is
dissolves in water.
e.g:KCl(s) K
+(aq) + Cl
(aq) Hsoln = +690 kJ mol
-1
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Enthalpy of Sublimation, Hsubl
The heat change when one mole of a substance
sublimes (solid into gas).
I2 (s) I2(g)
sublH
Hsubl= +106 kJ mol1
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Calorimetry
A method used in the laboratory to measure the
heat change of a reaction.
Apparatus used is known as the calorimeter Examples of calorimeter
Simple calorimeter
Bomb calorimeter
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The outer Styrofoam cup
insulate the reaction
mixture from the
surroundings (it is
assumed that no heat is
lost to the surroundings)
Heat release by the
reaction is absorbed by
solution and the
calorimeter
Simple calorimeter
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Important Terms in Calorimeter
Specific heat capacity, c
Specific heat capacity, cof a substance is the amount of
heatrequired to raise the temperature ofone gramof the
substance by one degree Celsius(Jg 1 C 1). Heat capacity, C
Heat capacity,Cis the amount of heat required to raise
the temperature ofagiven quantity of the substance by
one degree Celsius (J C 1)
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q = mcT
Heat released by
substance=
Heat absorbed
by calorimeter
q = heat released by substance
m= mass of substance
C= specific heat capacityT = temperature change
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Basic Principle in Calorimeter
Heat released
by a reaction=
Heat absorbed
by surroundings
Surroundings may refer tothe:i. Calorimeter itself or;
ii. The water and calorimeter qreaction= mcT or CT
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Example 1
In an experiment, 0.100 g of H2 and excess of O2 were compressed
into a 1.00 L bomb and placed into a calorimeter with heat capacity
of9.08 x 104 J0C 1. The initial temperature of the calorimeter was
25.0000C and finally it increased to 25.155 0C. Calculate the
amount of heat released in the reaction to form H2O, expressed in kJ
per mole.
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Solution
Heat released = Heat absorbed by the
calorimeter
q = CT
= (9.08 X 104 J0C-1) X (0.1550C)
= 1.41 X 104 J
= 14.1 kJ
H2(g) + O2(g) H2O(c)
mole of H2 = 0.1002.016
= 0.0496 mol
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moles of H2O = mole of H2
0.0496 mol of H2O released 14.1 kJ energy
1 mol H2O released = kJ
= 284 kJ
Heat of reaction, H = - 284 kJ mol 1
0496.0
1.14
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Example 2
1. Calculate the amount ofheat released in a reaction inan aluminum calorimeter with a mass of3087.0g andcontains 1700.0 mL of water. The initial temperature ofthe calorimeter is 25.0C and it increased to 27.8C.
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Given:
Specific heat capacity of aluminum = 0.553Jg-1C-1
Specific heat capacity of water = 4.18 Jg-1C-1
Water density = 1.0 g mL-1T = (27.8 -25.0 )C = 2.8C
Solution
q = mwcwT + mcccT
= (1700.0 g)(4.18 Jg-1C-1)(2.8 C) +
(3087.0 g)(0.553 Jg-1 C-1)(2.8C)
= 24676.71 J
= 24.7 kJ
Heat absorbed bywater
Heat absorbed byaluminium
calorimeter
Heat released = +
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HESSS LAW
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Hess Law
Hesss Law states that when reactants are converted to
products, the change in enthalpy is the same whether the
reaction takes place in one step or in the series of steps.
The enthalpy change depends only on the nature of thereactants and products and is independent of the route
taken.
1HA B
C
3H
2H
321HHH
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i. List all the thermochemical equations involved
Algebraic Method
1-
-1560kJmolH)(2
3)(2
2)(22
7)(62
.
1-
-286kJmolH)(2)(22
1)(2
.
1-393kJmol-H
)(2)(2)(.
gOH
gCO
gO
gHCiii
gOH
gO
gHii
gCO
gO
SCi
Step 1
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i. List all the thermochemical equations involved
Algebraic Method
1-
-1560kJmolH)(2
3)(2
2)(22
7)(62
.
1-
-286kJmolH)(2)(22
1)(2
.
1-393kJmol-H
)(2)(2)(.
gOH
gCO
gO
gHCiii
gOH
gO
gHii
gCO
gO
SCi
Step 1
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ii. Write the enthalpy of formation reaction for C2H6
)(62
?
)(23
)( gHC
fH
gH
sC
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iii. Add the given reactions so that the result is the desiredreaction.
-84kJ
3H
2H
1H
fH
84kJ-)(62
?
)(23
)(2
_______________________________________________________________
1560kJ3H)(227)(62)(232(g)2CO(iii)
-286kJ32
H)(2
3)(22
3)(2
33)(
kJ-39321
H)(2
2)(2
2)(
22)(
gHC
fH
gH
sC
gOgHCgOHreverse
gOH
gO
gHii
gCO
gO
SCi
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Energy Cycle Method
Draw the energy cycle and apply Hesss Law to calculate theunknown value.
2C(s) + 3H2(g) C2H6(g)
2CO2(g) + 3H2O(g)
2O2(g) 32O2(g)HH
H
H72O2(g)
OO
O
Of
12
3=2(-393)
=3(-286)
=-(-1560)
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1-kJmol-84
1560858--786
3
H+)2
H3(+)1
H2(=f
H
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Example 1
The thermochemical equation of combustion of carbonmonoxide is shown as below.
C(s) + O2(g) CO(g) = ?
given :
C(s) + O2(g) CO2(g) H= -394 kJ mol-1
CO(s)
+ O2(g)
CO2(g)
H= -283 kJ mol-1
Calculate the enthalpy change of the combustion of carbon tocarbon monoxide.
H
H
H
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Example 2
Calculate the standard enthalpy of formation of methane if the
enthalpy of combustion of carbon, hydrogen and methane are
as follows:H [C(s)] = -393 kJ mol
-1
H [H2(s)] = -293 kJ mol-1
H [CH4(s)] = -753 kJ mol-1
cH
cH
cH
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Example 3
Standard enthalpy of formation of ammonia, hydrogenchloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJmol-1, 314.4 kJ mol-1 respectively. Write the thermochemicalequation for the formation of each substance and calculatethe enthalpy change for the following reaction.
NH3(g) + HCl (g) NH4Cl(s)
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Exercise
1.Calculate the enthalpy of formation of benzene
if :
H (CO2(g) ) = -393.3 kJ mol-1
H (H2O(l) ) = -285.5 kJ mol-1
H (C6H6(l) ) = -3265.3 kJ mol-1
fH
fH
fH
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Born-Haber
Cycle
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Lattice Energy, Hlattice
is the energy required to completely separate one mole of a
solid (ionic compound) into gaseous ions
e.g:
NaCl(s) Na+(g) + Cl-(g) Hlattice = +771 kJ mol-1(lattice dissociation)
Na+(g) + Cl-(g) NaCl(s) Hlattice = -771 kJ mol-1
(lattice formation)
The magnitude of lattice energy increases as the ionic charges increase
the ionic radii decrease
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There is a strong attraction between small ions and
highly charged ions so the
H is more negative.
H for MgO is more negative than H for Na2O
because Mg2+ is smaller in size and has bigger charge
than Na+, therefore
Hlattice (MgO) > Hlattice (Na2O)
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Na+ and Cl- ions in the solid crystal are separated fromeach other and converted to the gaseous state (Hlattice)
The electrostatic forces between gaseous ions and polarwater molecules cause the ions to be surrounded by watermolecules (Hhydr)
Hsoln = Hlattice + Hhdyr
Hydration Process of Ionic Solid
Na+ and Cl- ion in
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Heat of Solution
Na+ and Cl- ion in
the gaseous state
Na+ and Cl- ion inthe solid state
Hydrated Na+ and Cl- ion
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Born-Haber Cycle
The process of ionic bond formation occurs in a few stages. Ateach stage the enthalpy changes are considered.
The Born Haber cycle is often used to calculate the lattice
energy of an ionic compound. In the Born-Haber cycle energy diagram, by convention,
positive values are denoted as going upwards, negative valuesas going downwards.
Consider the enthalpy changes in the formation of sodium
chloride.
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Example :
Given;
i. Enthalpy of formation NaCl = -411 kJmol-1
ii. Enthalpy of sublimation of Na = +108 kJmol-1
iii. First ionization energy of Na = +500 kJmol-1
iv. Enthalpy of atomization of Cl = +122 kJmol-1
v. Electron affinity of Cl = -364 kJmol-1
vi. Lattice energy of NaCl = ?
(s)2(g)21
(s) NaClClNa
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Example: A Born-Haber cycle for NaCl
Na(s) + Cl2(g)
Na(g) + Cl2(g)
NaCl(s)
energy
E=0
Na(g) + Cl(g)
Na+(g) + e + Cl(g)
Na+(g) + Cl- (g)
HaNa
HaCl
IonisationEnergy of Na
Electron Affinity of Cl
Lattice energy
Hf NaCl
From Hesss Law:
HfNaCl = HaNa + HaCl +IENa +
EACl + Lattice Energy-ve
+ve
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Calculation:
kJ777H
kJ364kJ122kJ500kJ108kJ411H
EAHIEHHH
HEAHIEHH
lattice
lattice
)Cl(aS
0
flattice
lattice)Cl(aS
0
f
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Exercise:
Construct a Born-Haber cycle to explain why ionic compoundNaCl2 cannot form under standard conditions. Use the databelow:
i. Enthalpy of sublimation of sodium = +108 kJmol-1
ii.
First ionization energy of sodium = +500 kJmol
-1
iii. Second ionization energy of sodium = +4562 kJmol-1
iv. Enthalpy of atomization of chlorine = +121kJmol-1
v. Electron affinity of chlorine = -364 kJmol-1
vi. Lattice energy of NaCl2 = -2489 kJmol-1