Thermochemistry
Chapter 6
6.1 The Nature of Energy & Types of Energy
6.2 Energy Changes in Chemical Reactions
6.3 Introductions to Thermodynamics
6.4 Enthalpy of Chemical Reactions
6.5 Calorimetry
6.6 Standard Enthalpy of Formation & Reaction
6.7Heat of Solution & Dilution
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6.1 The Nature of Energy and Types of Energy
Energy is the capacity to do work.
•
Radiant energy comes from the sun and is
earth’s primary energy source.
•
Thermal energy is the energy associated
with the random motion of atoms and
molecules.
•
Chemical energy is the energy stored
within the bonds of chemical substances.
•
Nuclear energy is the energy stored within
the collection of neutrons and protons in
the atom.
•
Potential energy is the energy available by
virtue of an object’s position.
6.1
Future Sources of Energy
Solar - The energy from the sun is inexhaustible, but we can only collect it poorly
at present. It is a good energy source to not replace our current sources, but to
supplement them and allow us to conserve our fossil fuels and use them more
wisely.
Geothermal - There are only a few areas around the world where there is
enough to be useful and then only for the local population.
Tidal - Again there are only a few places around the world where it can be used,
and for those local residents only.
Nuclear Fission - People are scared of nuclear energy, so as we know it, it will
be impossible to get people to accept it!
Nuclear Fusion - A form of nuclear energy that has tremendous potential, but at
the current time cannot be controlled safely. The internal confinement and laser
fusion methods both have tremendous potential and have the best chance for
being our new energy source for the future.
Cold Fusion - An expensive joke, and bad science!
6.2 Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies
that are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
6.2
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
SYSTEM
Surrounding
open
Exchange: mass & energy
closed
isolated
energy
nothing
6.2
Exothermic process is any process that gives off heat
– transfers thermal energy from the system to the
surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has
to be supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
6.2
Schematic of Exothermic and Endothermic Processes
7
6.3 Introduction to Thermodynamics
Thermodynamics is the branch of science which
studies the transformation of energy from one form to
another – interconversion of heats and other kinds of
energy.
Chemical thermodynamics (thermochemistry) looks
at the energy transformations which occur as a
result of chemical reactions.
Changes in the state of the
system (composition, energy,
temperature, pressure, volume)
6.3
Thermodynamics is the study of changes in the state of a
system.
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy , pressure, volume, temperature
DE = Efinal - Einitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
9
First Law of Thermodynamics
( Law of Conservation of Energy )
Energy can be converted from one form to another, but
cannot be created or destroyed.
“The Total Energy of the Universe is Constant”
One of the characteristics of any system is the total
amount of energy it contains. This characteristic is
called the total internal energy (symbol: E). E is a state
function.
The change in internal energy, ΔE is given by:
DE = Efinal − Einitial
Considering the change in a chemical reaction:
DE = Eproduct(s) − Ereactant(s)
The energy of a system can be changed by moving
energy in from the surroundings or by moving energy
out to the surroundings – using devices such as
furnaces and refrigerators.
DEsystem + DEsurroundings = 0
or
DEsystem = − DEsurroundings
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
Energy is only transferred and not changed in quantity
by such processes.
11
Another form of the first law for DE :
DE = q + w = q - PΔV
DE is the change in internal energy of a system
q is the heat exchange between the system and the
surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant
external pressure
6.3
Heat can flow into,
and out of, a system
to its surroundings.
Work can be done
by the system on
the surroundings,
and vice versa.
The internal energy of a system can increase or
decrease by any of these processes.
Thus energy as heat, work, total internal energy, or any
other form must be an algebraic quantity  a quantity
which has a positive sign (for increase) or a negative
sign (for decrease).
Increases or decreases in heat and work are more easily
visualized as heat or work flows between system and
surroundings and are looked at from the point of view of
the system:
•If q is negative, heat flows
OUT of the system (to the
surroundings).
•If q is positive, heat flows
INTO the system (from the
surroundings).
•If w is negative, work is done BY the system (on the
surroundings).
•If w is positive, work is done ON the system (by the
surroundings).
EXAMPLE: A sample of nitrogen gas expands in volume from
1.6 L to 5.4 L at constant temperature. What is the work done in
joules if the gas expands
(a) against a vacuum ?
(b) against a constant pressure of 3.7 atm?
w = -P DV
(a)
DV = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = -14.1 L•atm
w = -14.1 L•atm x
101.3 J = -1430 J
1L•atm
16
Question #1:
When gasoline burns in a car engine, the heat release causes the
products CO2 and H2O to expand, which pushes the pistons outward.
Excess heat is removed by the car’s cooling system. If the expanding
gases do 451 J of work on the pistons and the system loses 325 J to
the surroundings as heat, calculate the change in energy, ΔE in kJ.
Answer:
Question #2:
In the internal combustion engine, the heat produced by the
combustion of the fuel causes the carbon dioxide and water that
is produced during the combustion to expand, pushing the
pistons. Excess heat is removed by the cooling system. Determine
the change in energy (DE) in J, kJ, and kcal if the expanding gases
do 515 J of work on the pistons, and and the system loses 407 J of
heat to the cooling system.
Answer:
Question #3 :
a) A system receives 425 J of heat and delivers 425 J
of work to its surroundings. What is the change in
internal energy of the system (in Joule)?
b) What is the change in internal energy (in Joule) of
a system that releases 675 J of thermal energy to
its surroundings and has 525 calories of work
done on it?
Answer:
ENTHALPY
In thermodynamics, the quantity enthalpy,
symbolized by H, also called heat content,
is the sum of the internal energy of a
system plus the energy associated with
work done by the system on the
atmosphere which is the product of the
pressure times the volume.
qreaction (at constant pressure) = ΔH
Enthalpy (H) is used to quantify the heat flow into or out
of a system in a process that occurs at constant
pressure.
DH = H (products) – H (reactants)
Enthalpy and the First Law of Thermodynamics
DE = q + w
At constant pressure: q = DH and w = -PDV
DE = DH - PDV
DH = DE + PDV
21
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
22
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
DH = 6.01 kJ/mol
23
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) DH = -890.4 kJ/mol
24
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ/mol
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ/mol
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
25
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ/mol
H2O (l)
H2O (g)
DH = 44.0 kJ/mol
EXAMPLE: How much heat is evolved when 266 g of
white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
DH = -3013 kJ/mol
3013 kJ
= 6470 kJ
1 mol P4
26
Fig. 6.13
Question #4 :
Consider the following thermochemical equation:
H2 (g) + Cl2 (g) → 2 HCl (g)
ΔH = – 185 kJ
Calculate ΔH when
(a) 1.00 mol of HCl is formed
(b) 2.50 L of HCl (g) at 50.0 oC and 725 mmHg is formed
Question #5 :
Consider the decomposition of methanol into CO and H2:
CH3OH (ℓ) → CO (g) + 2 H2 (g)
ΔH = 90.7 kJ
a) How much heat transfer into the system is required to decompose
100.0 kg of methanol?
b) If 400.0 mol CO (g) is produced, what quantity of energy must be
transferred?
c) Suppose that the reverse reaction were to occur, in which CO
and H2 combine to form methanol. If 43.0 g CO (g) reacts with an
excess of H2, what is the heat transfer?
Moore et. al. 65
Question #6 :
Elemental chromium is produced by the Goldschmidt reaction of
elemental aluminum with a metal oxide such as chromium. What
mass of chromium metal and how much heat would be produced by
the reaction of 41.0 g of Al, and 255 g of chromium(III) oxide?
Cr2O3 (s) + 2 Al(s)  Al2O3 (s) + 2 Cr(ℓ)
Answer:
DH = –536 kJ
A Comparison of DH and DE
2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol
2Na (s) + 2H2O (l)
DE = DH - PDV
At 25 0C, 1 mole H2 = 24.5 L at 1 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
DE & DH are
approximately the same.
DH is smaller because
some of the internal
energy released is used to
do gas expansion work,
so less heat is evolved.
DE = DH – D(PV)
DE = DH – D(nRT)
DE = DH – RTΔn
Δn = mol product (gases) – mol reactant (gases)
6.4
CALORIMETRY
Calorimetry is the science of measuring the heat of
chemical reactions or physical changes.
Calorimetry involves the use of a calorimeter.
To find the enthalpy
change a reaction, the
initial and final (after the
reaction has finished)
temperatures are noted.
q = msΔt
A simple calorimeter
may just consist of a
thermometer attached to
an insulated container.
Δt = tfinal – tinitial
m = mass of substance (g)
s = specific heat capacity
(J/g.oC)
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C=mxs
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
33
EXAMPLE: How much heat is given off when an 869 g
iron bar cools from 940C to 50C?
Specific heat of Fe, s = 0.444 J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
Question #7 :
Calculate the quantity of heat required to heat
a gold ring with a mass of 21.63 g from
22.68oC to 255.05oC. The specific heat
capacity (C) of gold is 0.129 J/g.oC.
Answer:
6.5
Question #8 :
A home solar energy storage unit uses
400L of water for storing thermal energy.
On a sunny day, the initial temperature
of the water is 22.0oC. During the course
of the day, the temperature of the water
rises to 38.0oC as it circulates through
the water wall. How much energy has
been stored in the water?
J/g.oC.
Density of water at 22.0oC is 0.998 g/L,
specific heat of water is 4.184 J/g.oC.
Answer:
Constant-Volume Calorimetry (Bomb Calorimeter)
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = msDt
qbomb = CbombDt
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!
Common usage : to
measure heat release in a
combustion reaction.
Cbomb is usually determined
by burning an amount of
high purity crystalline
benzoic acid.
6.5
Question #9 :
A 3.30 g sample of glucose, C6H12O6 is placed in a bomb calorimeter,
ignited and burned to form CO2 and H2O. The temperature of the
water changed from 22.4 oC to 34.1 oC. If the calorimeter contained
850.0 g of water and had a heat capacity of 847 J/oC, what is the heat
of combustion of glucose?
Answer:
Constant-Pressure Calorimetry
(Coffee cup Calorimeter)
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = msDt
qcal = CcalDt
Reaction at Constant P: DH = qrxn
No heat enters or leaves!
Common usage:
•To determine specific heat capacity of a solid
that does not react or dissolve in water;
•To determine enthalpy of aqueous solution
reaction;
•To determine specific heat capacity of salt
sample.
6.5
Question #10 :
A 9.783 g sample of an unknown solid was heated to 95.70oC and
carefully added to a coffee-cup calorimeter containing 45.00 g of
water. The water temperature increased from 13.48oC to 24.26oC.
What is the heat capacity of the solid, assuming that no heat is lost to
the surroundings.
Answer:
Question #11 :
When 0.800 g of Mg is added to 250.0 mL of 0.40 M HCl
in a coffee cup calorimeter, the temperature of the
solution increases from 23.4 oC to 37.9 oC. Determine
the enthalpy change for the reaction.
Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
Answer:
ΔH = ??
Special ΔH’s of Reactions
When one mole of a substance combines with oxygen in a
combustion reaction, the heat of reaction is the heat of combustion
(ΔHcomb):
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(g)
ΔH = ΔHcomb
When one mole of a substance is produced from it’s elements, the
heat of reaction is the heat of formation (ΔHf ) :
Ca(s) + Cl2 (g) → CaCl2 (s)
ΔH = ΔHf
When one mole of a substance melts, the enthalpy change
is the heat of fusion (ΔHfus) :
ΔH = ΔHfus
HO → HO
2
(s)
2
(ℓ)
When one mole of a substance vaporizes, the enthalpy
change is the heat of vaporization ( ΔHvap) :
H2O(L) → H2O(g)
ΔH = ΔHvap
6.6: Standard Enthalpy of Formation
(ΔHof) and Reaction (ΔHorxn)
There is no way to measure the absolute value of the
enthalpy of a substance. Therefore, an arbitrary scale
with the standard enthalpy of formation (ΔHof) as a
reference point for all enthalpy expressions is
established.
Standard enthalpy of formation (DHf0) is the heat
change that results when one mole of a compound is
formed from its elements at a pressure of 1 atm.
The standard enthalpy of reaction (DHrxn0 ) is the
enthalpy of a reaction carried out at 1 atm.
6.6
Calculations of Standard Enthalpy of
Formation (ΔHof) and Reaction (ΔHorxn)
Direct Method
The standard enthalpy
change, ΔHo, for a
given thermochemical
equation is equal to the
sum of the standard
enthalpies of formation
of the product minus
the sum of the standard
enthalpies of formation
of the reactants.
Indirect Method
Hess’s Law
When reactants are
converted to products,
the change in enthalpy
is the same whether the
reaction takes place in
one step or in series of
steps.
Direct Method:
Calculation is done by substracting the sum of the
standard enthalpies of formation of the reactants from
the sum of the standard enthalpies of formation of the
products, as shown in the equation below.
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Consider a general equation:
aA + bB
cC + dD
0
DHrxn
= [cDH0f (C) + dDH0f (D) ] – [aDH0f (A) + bDH0f (B) ]
The standard enthalpy of formation of any element in
its most stable form is zero.
0 (C, graphite) = 0
0
DH
DH (O2) = 0
f
f
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
6.6
EXAMPLE: Benzene (C6H6) burns in air to produce carbon dioxide
and liquid water. How much heat is released per mole of benzene
combusted? The standard enthalpy of formation of benzene is
49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
DHrxn
= SnDH0f (products)- SmDH0f (reactants)
0
DHrxn
= [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
0
DHrxn
= [12 x (–393.5) + (6 x (–187.6))] – [2 x 49.04] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
6.6
Question #12 :
Methanol (CH3OH) is used as a fuel in high-performance
engines in race cars. Using the data on enthalpies of
formation, compare the standard enthalpy of combustion
per gram of methanol with that per gram of gasoline
(assume gasoline is liquid octane, C8H18). Given:
Compound
CH3OH(ℓ)
DHf0 (kJ/mol)
– 239.2
C8H18 (ℓ)
CO2 (g)
– 259.2
– 393.5
H2O(ℓ)
– 285.8
Question #13 :
Long-chain fatty acids such as palmitic acid, CH3(CH2)14COOH, are
one of the two major sources of energy in our diet. Calculate the
DHrxn0 for the combustion of palmitic acid and glucose, C6H12O6.
Based on the energy released upon combustion per gram, which is
the better fuel: glucose or palmitic acid? Given:
Averill pg 221
Compound
DHf0 (kJ/mol)
CH3OH(ℓ)
– 239.2
C8H18 (ℓ)
– 259.2
CO2 (g)
– 393.5
H2O(ℓ)
– 285.8
H2O(g)
– 241.8
48
Hess’s Law: When reactants are converted to products,
the change in enthalpy is the same whether the reaction
takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only
where you start and end.)
The enthalpy change of an overall process is the sum of
the enthalpy changes of its individual steps.
The basic procedure is:
1. Write out the thermochemical equations for the step reactions.
2. Write the balanced chemical equation for the target reaction.
3. If necessary, reverse step reactions so products/reactants
match the target reaction.
4. Scale step reactions so products/reactants that don't appear in
the target reaction will cancel out (use multiplication).
5. Add the step reactions.
6. Scale the resulting reaction so it matches the target reaction.
6.6
Example :
ΔH1
A
D
ΔH2
ΔH4
B
ΔH for A to B
(direct route)
ΔH3
C
= ΔH for A to B to C to D
(alternative route)
By Hess's Law: ΔH1 = ΔH2 + ΔH3 + ΔH4
C (graphite) + 1/2O2 (g)
CO (g) + 1/2O2 (g)
C (graphite) + O2 (g)
CO (g)
CO2 (g)
CO2 (g)
51
EXAMPLE: Calculate the standard enthalpy of formation of
CS2 (l) given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ/mol
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ/mol
CO2 (g) + 2SO2 (g)
0 = -1072 kJ/mol
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ/mol
2SO2 (g) DH0rxn = -296.1 kJ/mol x 2
CS2 (l) + 3O2 (g)
0 = +1072 kJ/mol
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ/mol
52
Question #14 :
Calculate the energy involved in the oxidation of
elemental sulfur to sulfur trioxide:
S (s) + 3/2 O2 (g)  SO3 (g)
DHrxn = ?
Given are reactions:
S (s) + O2 (g)  SO2 (g)
DH1 = -296.8 kJ
2 SO2 (g) + O2 (g) 2 SO3 (g) DH2 = -198.4 kJ
Answer:
Question #15 :
What is the enthalpy of reaction, DHrxn, for the formation
of tungsten carbide, WC, from the elements? Given the
equations:
2 W(s) + 3 O2 (g)  2 WO3 (s)
DH = -1680.6 kJ
C(graphite) + O2 (g)  CO2 (g)
DH = -393.5 kJ
2 WC(s) + 5 O2 (g)  2 WO3 (s) + 2 CO2 (g) DH = - 2391.6 kJ
The equation for the production of tungsten carbide
from the elements:
W(s) + C(graphite)
 WC(s)
Question #16 :
To produce silicon (used in semiconductors) from sand,
SiO2, a reaction is used that can be broken down into
three steps:
SiO2 (s) + 2 C (s) → Si (s) + 2 CO (g)
ΔH = 689.9 kJ
Si (s) + 2 Cl2 (g) → SiCl4 (g)
ΔH = – 657.0 kJ
SiCl4 (g) + 2 Mg (s) → 2 MgCl2 (s) + Si (s) ΔH = – 625.6 kJ
a) Write the thermochemical equation for the overall
reaction for the formation of silicon from silicon
dioxide (CO and MgCl2 are the by-products).
b) What is ΔH for the formation of one mole of
silicon?
c) Is the overall reaction exothermic or endothermic?
Masterton 28
Question #17 :
Determine the heat of reaction (kJ) at 298 K for the reaction:
3 NO2(g) + H2O(ℓ)
2 HNO3(ℓ) + NO(g)
Given:
NH3(g) + HNO3(ℓ) NH4NO3(s)
∆Ho = -145.7 kJ/mol
NH4NO3(s)  N2O(g) + 2 H2O(ℓ)
∆Ho = -125.2 kJ/mol
3 NO(g)  N2O(g) + NO2(g)
∆Ho = -155.8 kJ/mol
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(ℓ)
NO(g) + ½ O2(g) NO2(g)
∆Ho= -1169.2 kJ/mol
∆Ho = -56.6 kJ/mol
6.7: Heat of Solution and Heat of Dilution
The heat of solution or enthalpy of solution (DHsoln) is the
heat generated or absorbed when a certain amount of
solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Can be
determined using
calorimeter
Cannot be measured!
Enthalpies of solution for ionic
compounds cover a wide range of
values that are both exo- and
endothermic. These values result
from the hydration of the
individual anions and cations.
6.7
The Solution Process for NaCl
Heat is transferred when a pure
liquid or solid is dissolved in
another liquid (enthalpy of
solution). The solute-solute
interactions are replaced by
solute-solvent interactions.
These effects can be
substantial when an ionic
solid is dissolved in water
and the anion-cation ionic
interactions are replaced by
hydration of the ions
(enthalpy of hydration).
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.7
Heat of Solution.
The heat of solution of a solute
associated with the dissolution
solute in a given amount of
process can be represented
equation.
H2O
HCl (g)  HCl (aq)
is the enthalpy change
of a certain amount of
solvent. The solution
by a thermochemical
∆Hosoln =  75.1 kJ
The energy changes that occur when ionic solids dissolve
in water can be understood by imagining the solution
process to take place in two steps.
We realize that when a compound such as KF dissolves in
water the K+ and F– ions must break away from the crystal
lattice and enter the aqueous phase where they are
surrounded by water molecules.
First, we imagine a step in which the lattice is broken up and the
ions enter the gas phase.
KF(s)  K+(g) + F–(g)
The energy change for breaking up the lattice is called the lattice
energy, U. All ionic crystals have a characteristic lattice energy.
This is the energy required to completely separate all the ions in 1
mole of solid into ions in the gas phase.
For KF(s), U = 821 kJ.
Second, the gaseous K+ and F– ions enter the aqueous phase, where
they are hydrated. During hydration, K+ and F– ions are surrounded
by H2O molecules. Ions in water are attracted to the polar water
molecules, and a quantity of energy is released that is called the
hydration energy, ∆Hhydr.
For K+ and F– ions,
∆Hhydr = – 819 kJ.
According to Hess's law, the overall energy change for
the two steps will be the heat of solution.
∆Hsoln = U + ∆Hhydr
For KF; ∆Hsoln = 821 kJ + ( 819 kJ) = 2 kJ.
K+ (g) + F–(g)
Lattice
energy, U
Heat of
hydration,
∆Hhydr
Hydrated
K+ & F– ions
KF (s)
Heat of
solution,
∆Hsoln
The heat of dilution or enthalpy of dilution (DHdiln) is the
heat change associated with the dilution process.
Enthalpies of dilution are produced when additional
solvent is added to a solution. Strong mineral acids, like
hydrochloric and sulfuric, produce a tremendous amount
of heat when diluted with water. In fact, sufficient heat is
produced to boil the water! A safety rule is to add acid to
water.
Endothermic solution process:
More heat will be absorbed from the
surrounding if additional solvent is added.
Exothermic solution process:
More heat will be released to the
surrounding if additional solvent is added.
Question #18 :
The heat of solution of LiCl is – 37.1 kJ/mol, and the lattice energy of
LiCl (s) is 828.0 kJ/mol. Calculate the total heat of hydration of 1 mol
of gas phase Li+ ions and Cl– ions.
Answer:
Question #19 :
The heat of solution of NaBr is –1.0 kJ/mol. The lattice energy of
NaBr is 735 kJ/mol. Determine the heat of hydration of NaBr and
write the equation for the hydration reaction.
Question #20 :
Given that the heat of solution of LiF is +32 kJ/mol and that the
lattice energy (U) for LiF is 1006 kJ/mol, calculate the heat of
hydration (∆Hhydr) of Li+ and F– ions.