TOPIC 8: THERMOCHEMISTRY
Thermochemistry is the branch of chemistry dealing
with determining quantities of heat by measurement
and calculation. Some of these calculations will allow
us to establish indirectly, a quantity of heat that would
be difficult to measure directly.
Some Main Concepts in Thermochemistry
System: The part of the universe
we choose to study.
Surroundings: The parts of the
universe with which the system
interacts.
Systems
groups:
are
classified into
3
Open system: Free transfer of
energy and matter between the
system and its surroundings.
Closed system: Only free transfer
of energy between the system and
its surroundings
Isolated system: No possible
transfer of energy and or matter
between the system and its
surroundings.
Open
system
Closed
system
Isolated
system
Some main Concepts in Thermochemistry
Energy transfer can occur as heat (q) or in several other forms, known
as work (w).
Heat: The form of energy which is transfered from a system with
higher temeperature to a system with low temperature due to the
temperature difference.
Work: The effect that causes the change in position of a body by
exerting a force on it.
Energy: The capacity of a system to perform a work.
Energy transfers occuring as heat and work affect the total amount of
energy contained within a system, its internal energy (E). The
components of internal energy of special interest to us are thermal
energy and chemical energy.
Thermal energy: Energy associated with random molecular motion.
Chemical energy: Energy associated with chemical bonds and
intermolecular forces.
Energy
Kinetic energy (KE): The energy of a moving object. m: mass, v:
velocity
Energy Unit:
1
2
2
Kinetic
kg
m
KE  m v
J ( Jul) 
energy
2
2
s
Potential energy: The stored energy or «energy of position» associated
with forces of attraction or repulsion between objects. The stored
energy arises from the kinetic energy of molecules and atoms.
Potential Energy:
PE  m g h
g: acceleration
of gravity, 9,81
m/s2
h: height of the matter, m
m: mass of the matter,kg
Conservation of Energy
In a closed system energy can be converted from one form into another but it can not
be destroyed(First law of thermodynamics). In other words, the total energy of a
closed system is constant.
In a closed system
E  KE  PE  0
When we release a ball, it is pulled towards the center of
earth by the force of gravity,it falls. Potential energy is
converted into kinetic energy during this fall. When the
falling ball collides with the surface, it reverses direction.
Throghout its rise after collision, the kinetic energy of the
ball decreases while its potential energy increases.(
However, it happens differently in the reality : While the
ball is bouncing on the surface,the energy present as KE
and PE is converted into thermal energy.As the thermal
energy increases , the kinetic energy of the molecules
increases, the temperature of the ball rises. When the ball
slows down and comes to rest, it loses thermal energy to
the surroundings. )
Energy and Temperature
• Thermal energy
– Kinetic energy is in relation to random
molecular motion.
– It is proportional with temperature.
– It is a property of intensity.
• Heat and Work
– Transfer of energy due to the temeperature difference between to systems
is called “heat”
– Transfer of energy occuring as a result of the exertion of a force along a
certain distance is called “work”.
Zero Law of Thermodynamics
If two objects A and B are both in thermal equlibrium with the
thermometer C,that means they are also in thermal equlibrium with
each other.
Heat
Heat transfer between the system and its
surroundings occurs as a result of temperature
difference.
• Heat moves from the hot environment to the
cold environment.
– Temperature is variable.
– Change of the state may occur
The heat flux occuring at constant
temperature is called ISOTHERMAL
PROCESS.
Units of Heat
The quantity of heat energy ,q , depends upon:
How much the temperature is to be changed
The quantity of substance
The nature of substance (type of atoms or molecules)
• Calorie (cal)
– The quantity of heat required to change the
temperature of one gram of water by one
degree Celcius.
• Joule (J)
1 cal = 4.184 J
– SI unit of work and energy
Heat Capacity
• The quantity of heat required to change the
temperature of a system by one degree .
– Molar heat capacity:
• The System is one mole of substance
– Specific heat (capacity), c.
• The system is 1 g of substance
– Heat capacity, C
• Mass x specific heat .
q = mcT
q = CT
Conservation of Energy
• In interactions between a system and its
surroundings, the total energy remains constantenergy is neither created nor destroyed.
• Applied to the exchange of heat, this means :
qsystem + qsurroundings = 0
qsystem = -qsurroundings
Thus, heat lost by a system is gained by its
surroundings, and vice versa.
Experimental Determination of Specific
Heats
Lead
The transfer of energy, as heat, from the lead to the cooler water causes the
temperature of the lead decrease and that of the water increase, to the point
where both the lead and water are at the same temperature. If we consider lead
to be the system and water as the surroundings, we can write :
qlead = -qwater
Experimental Determination of Specific
Heats
qlead = -qwater
q water = mcT = (50.0 g water)(4.184 J/g water °C)(28.8 - 22.0)°C
q water = 1.4x103 J
q lead = -1.4x103 J = mcT = (150.0 g lead)(c)(28.8 - 100.0)°C
c lead = 0.13 Jg-1°C-1
Heats of Reaction and Calorimetry
• Chemical Energy.
– Type of energy related to the internal energy of the
system.
– The energy which occurs as a result of chemical
reactions. The batteries and accumulators are the
vehicles which convert the chemical energy into
electrical energy. The accumulation of electrical energy
in batteries is performed by chemical methods.
Chemical energy can be also converted into the
mechanical, heat and light energy.
• Heat of reaction, qrxn.
– The quantity of heat exchanged between a system and
its surroundings when a chemical reaction occurs
within a system at constant temperature and pressure.
Heat of Reaction
• Exothermic reaction.
CaO(s) + H2O(l) → Ca(OH)2 (aq)
– Gives off heat to the surroundings, qrxn < 0.
– Temperature increase in the system (a)
• Endothermic reaction.
– Gain of heat from the surroundings, qrxn > 0.
• Temperature decrease in the system (b)
• Calorimeter
– Device for measuring quantities of heat
Ba(OH)2·8H2O + 2NH4Cl(s)→BaCl2(s) + 2 NH3(aq) + 8 H2O(l)
Bomb Calorimetry
The type of calorimeter shown in the figure is called bomb calorimetry. It is
ideally suited for measuring the heat evolved in combustion reaction. It is an
isolated system from its surroundings
When the combustion reaction occurs,
chemical energy is converted to thermal
energy and temperature of the system
rises.The heat of reaction qrxn is the
quantity of heat that the system would have
to lose to its surroundings to be restored to
its initial temperature and pressure.
qrxn = -qcal
qcal = qbomb + qwater + qwire +…
qcal = miciT = CcalT
Heat capacity of calorimeter
heat
Example
Determination of heat of reaction by the bomb calorimeter
The combustion of 1,010 g sucrose, C12 H22 O11 , in a bomb
calorimeter causes the temperature rise from 24,92˚ to
28,33˚C. The heat capacity of the calorimeter assembly is
4,90 kJ/ ˚C
(a) What is the heat of combustion of sucrose (kJ/mol) ?
(b) Verify the claim of sugar producers that one teaspoon of
sugar (4,8 g) contains only 19 calories .
Solution
Example 7-3
Calculate qcal
qcal = CT = (4,90 kJ/°C)(28,33-24,92)°C = (4,90)(3,41) kJ
= 16,7 kJ
Calculate qrxn
qrxn = -qcal = -16,7 kJ
in 1,01 g sample of sugar
Solution
Example 7-3
Conversion of the unit ofqrxn as kJ/mol :
-16,7 kJ
qrxn = -qcal =
= -16,5 kJ/g
1,010 g
qrxn
343,3 g
= -16,5 kJ/g
1,00 mol
= -5,65 x 103 kJ/mol
(a)
(b) Verification of the claim that a teaspon of sugar(4,8 g) has 19 Calories
For a teaspoon
qrxn
4,8 g 1,00 kal
)= -19 cal/teaspoon (b)
= (-16,5 kJ/g)(
)(
4,184 J
1 ts
The Coffee-Cup Calorimeter
• Easy to handle .
– An isolated system composed of styrofoam
cup.
– Reactants are mixed up in the cup
– Temperature difference is measured at the end
of reaction.
– System is at constant pressure.
qrxn = -qcal
qrxn = - Ccal T
WORK
In thermodynamics work means the transfer of energy between the system and
its surroundings due to an external makroscopic force.
• Apart from transfer of heat, some chemical processes
may do work.(the expansion or compression of gases)
• The gas pushes the atmosphere.
• The volume changes.
• Pressure-volume work
Pressure-Volume Work
Work  force dis tance
Fext
P

w Fx
A
w  Fext  x
Fext  P  A
w  P  A  x
F  Fext
V  Vf  Vi
w   P  V
Unit of work
L  atm  101,3 J
Since work is done by the system , it has a negative sign - .
If the volume increases, work has - sign.
Example 7-3
Calculation of Pressure-Volume Work.
What is the work done in joules, when 0,100 mol He(g) at
298K expands from an initial pressure of Pi= 2,4 atm to a final
pressure Pf= 1,3 atm?
We assume that He behaves as an ideal gas:
Vi = nRT/Pi
= (0,100 mol)(0,08201 L atm mol-1 K-1)(298K)/(2,40 atm)
= 1,02 L
Vf = nRT/Pf= 1,88 L
V = 1.88-1.02 L = 0.86 L
Example 7-3
Calculation of work done by the system
w = -PV
= -(1.30 atm)(0.86 L)(
= -1.1 x 102 J
101 J
)
1 L atm
Conversion factor:
8.3145 J/mol K ≡ 0.082057 L atm/mol K
1 ≡ 101.33 J/L atm
Pressure-Volume Work
Example: What is the quantity of work, in
joules, done by the gas in the figure next
if it expands against a constant pressure
of 0,980 atm and the change in
Volume(ΔV) is 25 L .
w   P  V
w  0,980atm  25 L
101,3 J
w  24,5 L atm 
1 L atm
w  2,48 kJ
First Law of Thermodynamics
• Internal Energy, U.
– The TOTAL energy of the system (potential and
kinetic).
•Translational kinetic energy.
•Rotations.
•Bond vibrations.
•Intermolecular attractions
•Chemical bonds.
•Electrons.
The First Law of Thermodynamics
• A system keeps the energy only in form of
internal energy.
– A system does not contain energy in the form of
heat or work.
– Heat and work are the means by which the
system exchanges energy with its
surroundings.
– Heat and work only exist during a change.
U = q + w
• Law of Conservation of energy
– The energy of an isolated system is constant.
Uisolated = 0
The First Law of Thermodynamics
Any energy entering the
system carries + sign.Thus,if
heat is absorbed
by the
system, q>0 .If work is done
on the system ,w>0.
Any energy leaving the system
carries a – sign.Thus if heat is
given off by the system , q<0.
If work is done by the system,
w<0.
System
If, on balance more energy
enters the system than leaves,
ΔU is positive.If more energy
leaves than enters, ΔU is
negative.
Functions of State
• Any property having a unique value when the
state of the system is defined is called function
of state.
• Example:
•
•
•
•
The state of water at 293,15 K and 1,00 atm is specific
In this state d = 0.99820 g/mL
Density just depends on the state of the system.
It does not depend on how it is reached.
Functions of State
State 2
– Can not be measured.
– We do not need to know
the real value.
• The U between U2 and
U1 have a unique value.
– It can be measured
easily.
Internal energy
• U is a function of state.
State 1
tot
The value of the internal energy is the value of heat energy given off
from the surroundings to the system to reach from the state U1 to U2
Functions depending on the path to be
followed
• Heat and work are not functions of state!
• The values of heat and work depend upon the path
to be followed for a change in the system.
• 0,1 mol He
298 K, 2,40 atm (State 1)
B
(1,02 L)
298 K, 1,80 atm
A↓
(1,36 L)
298 K, 1,30 atm (State 2) C
(1,88 L)
wBC = (-1,80 atm)(1,36-1,02)L – (1,30 atm)(1,88-1,36)L
= -0,61 L atm – 0,68 L atm = -1,3 L atm
= -1,3 x 102 J
in contrast;
2
Heats of Reaction and Entalpy Change:
U and H
Reactants → Products
Ui
Uf
U = Uf- Ui
U = qrxn + w
In a system at a constant volume(Bomb
calorimeter):
w = -PV= 0
U = qrxn + 0 = qrxn = qv
However, lots of chemical reactions occur in the earth
under constant pressure!
What is the relationship between qp and qv ?
Heat of Reactions
Internal energy
First state
First state
qV = qP + w
Final state
Final state
Heat of reactions
qV = qP + w
w = - PV and U = qv :
U = qP - PV
P,V,U are functions of state
qP = U + PV
H = U + PV
Enthalpy
H = Hf – Hi = U + PV
At constant pressure and
temperature:
H = U + PV = qP
Comparison of heat of reactions
2 CO(g) + O2(g) → 2CO2(g)
qP
= -566 kJ/mol
= H
heat
Constant
volume
PV = P(Vf – Vi)
= RT(nf – ni)
= -2,5 kJ
U = H - PV
Constant
pressure
heat
= -563,5 kJ/mol
= qV
Example
• The heat of combustion at constant volume of
CH4 (g) is measured in a bomb calorimeter at
25˚C and is found to be – 885,389 J/mol. What
is the enthalpy change,ΔH ?
• Solution:
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (l)
ΔU= - 885,389 J
Δn= Σnproducts - Σnreactants 1-(2+1)= - 2 (Note that,
the mole number of H2O is not included in the
calculation!!!Since volume change of liquids and
solids are too small, it is here neglected.
Example
• ΔH= ΔU + PΔV
• ΔH= ΔU + ΔnRT
• ΔH= - 885,389 -2x 8,314 J/molK x 298,15K
• ΔH=- 885,389 kJ- 4,957 kJ
• ΔH=- 890,346 kJ
Note that the value measured by the bomb
calorimeter is equal to ΔU!!!
Example
• For the reaction ;
B2H6(g) + 3 O2(g)
B2O3(s) + 3 H2O(l)
ΔU is – 2143,2 kJ
a) Calculate ΔH for the reaction at 25 ˚C
b) Determine the value of standard enthalpy
of formation of B2H6(g).
For B2O3 (s) ΔHf˚=-1264,0 kJ/mol and for
H2O(l) ΔHf˚=-285,9 kJ/mol
Solve the problem by yourself!
Enthalpy during the phase transitions
Molar enthalpy of vaporization:
H2O (l) → H2O(g)
H = 44,0 kJ ; 298 K
Molar enthalpy of melting :
H2O (s) → H2O(l)
H = 6,01 kJ ; 273,15 K
Heat energy during the phase transition:
qp= n*Hphchange
Example 7-3
Enthalpy change during phase transition
Calculate the molar heat of vaporization of 50,0 g sample of
the water whose temperature is risen from 25,0°C to 100°C .
cwater= 4,184 J/g °C, Hvap = 44,0 kJ/mol
Think about the solution in two steps: First step: Increase in
the temperature of the water. Second step: The vaporization
Hvap = 44,0 kJ/mol
Solution:
qP = mcH2OT + nHvap
50,0 g
= (50,0 g)(4,184 J/g °C)(100-25,0)°C +
x 44,0 kJ/mol
18,0 g/mol
= 15,69 kJ + 122 kJ = 137,69 kJ
Standard Enthalpies of Formation
• Standard state is the pure substance at 1 atm
pressure at the temperature of interest.
• Standard molar enthalpy of formation(molar heat of
formation), Hf
– The difference in enthalpy between one mole of a
compound in its standard state and its elements in their
most stable forms and standard states.
We will use enthalpies of formation to perform a
variety of calculations. The first thing we have to do is
to write a chemical calculation and then to sketch an
enthalpy diagram
Temperature values must be given with H° !!
Enthalpy Diagrams
Reactants
Endothermic
reactants
Reactants
Enthalpy
Enthalpy
Products
Products
Exothermic
reactants
Indirect Determination of H :
Hess’s Law
• H is an Extensive Property .
– Enthalpy change is directly proportional to the amounts
of substances in a system.
N2(g) + O2(g) → 2 NO(g)
H = +180.50 kJ
½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ
(we divide all coeeficients and H value by two)
• H changes sign when a process is reversed.
NO(g) → ½N2(g) + ½O2(g)
H = -90.25 kJ
Hess’s Law
• Hess’s Law of Constant Heat Summation:
– If a process occurs in stages or steps (even if only
hypothetically) the enthalpy change for the overall(net)
process (Hnet )is the sum of the enthalpy changes for the
individual steps (H).
½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
H = +33.18 kJ
Hess’s Law
Enthalpy
½N2(g) + ½O2(g) → NO(g)
NO(g) + ½O2(g) → NO2(g)
½N2(g) + O2(g) → NO2(g)
Standard Enthalpy of Formation
Hf °
• We assign enthalpies of zero to the elements in
their most stable forms when in the standard state.
• Most stable forms of the elements are the ones
indicated below
• Na(s), H2(g), N2(g), O2(g), C(graphite), Br2(l)
Formation Enthalpy
Standard Enthalpies of Formation
Enthalpy of Formation
(formaldehyde)
Standard Enthalpies of Formation
Pozitive Enthalpies
of Formation
Standard Enthalpies
of Formation
Negative Enthalpies
of Formation
Standard Enthalpies of Reaction
• If the reactants and products of a reaction
are all in their standard states, we call the
enthalpy change for a reaction standard
enthalpy of reaction.
• ΔHrxn0
• Instead of standard enthalpy of reaction the
term is also frequently used:
– Standard heat of reaction
Standard Enthalpy of Reaction
Enthalpy
Standard Enthalpy of Formation
Htot = -2Hf °NaHCO3+ Hf °Na2CO3+ Hf °CO2 + Hf °H2O
Standard Enthalpy of Formation
•
Since enthalpy is a state of function , it is independent on
the pathway!!
• The overall(net) enthalpy change of the reaction is equal to
the sum of all enthalpy changes of the steps in the whole
reaction.
Hrxn = H°dec+ H°f
TABLE 8.1 Some Standard Enthalpies of Formation at
298 K
SUBSTANCE
ΔHf˚,298 K
kJ/mol
SUBSTANCE
ΔHf˚,298K
kJ/mol
CO(g)
-110,5
HF(g)
-271,1
CO2(g)
-393,5
HI(g)
26,48
CH4(g)
-74,81
H2O(g)
-241,8
C2H2(g)
226,7
H2O(l)
-285,8
C2H4(g)
52,26
H2S(g)
-20,63
C2H6(g)
-84,68
NH3(g)
-46,11
C3H8(g)
-103,8
NO(g)
90,25
C4H10(g)
-125,7
N2O(g)
82,05
CH3OH(l)
-238,7
NO2(g)
33,18
C2H5OH(l)
-277,7
N2O4(g)
9,16
HBr(g)
-36,40
SO2(g)
-296,8
HCl(g)
-92,31
SO3(g)
-395,7
Standard Enthalpy of Formation
Decomposition
Product
Formation
Enthalpy
Enthalpy
Formation
Decomposition
Reactant
Overall
Overall
Products
Reactants
Endothermic reaction
Exotermic reaction
Hrxn = ΣυpHf ° products- Συr Hf°reactants
Determination of the enthalpy change of
reaction
The enthalpy change of the combustion reaction of C6H6 is -6535
kJ. Determine the molar enthalpy of formation of C6H6?
(ΔHCO2(g)=-393,5 kJ/mol ; ΔHH2O(l)= -285,8 kJ/mol
°
Ionic Reactions in Solutions
• Many chemical reactions in aqueous solution are
best thought of as reactions between ions and
represented by net ionic equations.
• To calculate the net enthalpy change of
reaction,we need to know the the enthalpy of
formation data for individual ions.
• However, we can not create ions of a single type
in a chemical reaction.
• We must choose a particular ion to which we
assign an enthalpy of formation of zero in aqueous
solutions. We then compare the enthalpies of
formation of other ions to this reference ion.
• ΔH0(H+) (aq) = 0
Table 7.3 Enthalpies of Formation of Ions
in Aqueous Solutions
Example
• Calculating the enthalpy change in an ionic reaction
• Given that ΔHf˚(BaSO4 (s))= 1473 kJ/mol,what is the
enthalpy change for the precipitation of barium sulfate?
• Solution:
• Ba2+ (aq) + SO42- (aq)
BaSO4(s)
• The enthalpy of formation of BaSO4(s) is given and those
of Ba2+(aq) and SO42-(aq) are found in Table 7.3
• ΔH˚=ΔHf˚(BaSO4 (s))- ΔHf˚(Ba2+ (aq))-ΔHf˚(SO42- (aq))
ENTROPY and FREE ENERGY
• THE MEANING OF SPONTANEOUS CHANGE
• Spontaneity: A spontaneous process is a process that
occurs in a system left to itself
• A nonspontaneous process will not occur unless some
external action is continiously applied.
• Examples: a) the reaction of NaOH(aq) and HCl(aq) is
• a neutralization process and spontaneous
• b) the electrolysis of liquid water is
• a nonspontaneous process, since electric current is
required to decompose liquid water into its elements
• c)The melting of an ice cube is
• spontaneous above the melting point(0˚C) ,
• but it is nonspontaneous below the melting point
ENTROPY and FREE ENERGY
• Entropy: The thermodynamic property
related to the degree of disorder in a system,
designated by the symbol S.
• The greater the degree of randomness in a
system , the greater its entropy.
• The entropy change, ΔS, is the difference
in entropy between two states and also has a
unique value.
• A(g) + B(g)
mixture of A(g) and B(g)
• ΔS=Smix of gases-(SA(g)+ SB(g)) > 0
ENTROPY AND FREE ENERGY
• The increase in disorder (ΔS>0) outweighs the fact that
heat must be absorbed(ΔH>0) and the process is
spontaneous.
• Entropy increases are expected when
• Pure liquids or liquid solutions are formed from solids
• Gases are formed, from either solids or liquids
• The number of molecules of gases increases as a result of
chemical reaction
• The temperature of a substance increases.(Increased
temperature means increased molecular motion (either
vibrational motion of atoms or ions in a solid, or
translational motion of molecules in a liquid or gas))
ENTROPY and FREE ENERGY
• Ice melting in a warm
room is a common way of
increasing entropy
ENTROPY and FREE ENERGY
ENTROPY and FREE ENERGY
•
ENTROPY and FREE ENERGY
•
CASE ΔH
ΔS
ΔG
RESULT
EXAMPLE
1
-
+
-
Spontaneous at all temperature
2 N2O(g)
2
N2(g)+ O2(g)
2
-
-
+
Spontaneous at low temperature
Nonspontaneous at high
temperature
+
-
Nonspontaneous at low temperature
+
Nonspontaneous at all temperature
3
4
+
+
+
-
Spontaneous at high temperature
H2O(l)
H2O(s)
2 NH3(g)
N2(g)+3 H2(g)
3 O2(g)
2O3(g)
Evaluating Entropy and Entropy Changes
•