Apchapt 6(thermochemistry)

Chapter 6Chapter 6
EnergyEnergy
ThermochemistryThermochemistry

The Nature of EnergyThe Nature of Energy
The concept of energy is quite familiar, energy itself isThe concept of energy is quite familiar, energy itself is
rather difficult to define precisely.rather difficult to define precisely.
We will defineWe will define energyenergy as the ability to do work or toas the ability to do work or to
produce heat.produce heat.
One of the most important characteristics of energy isOne of the most important characteristics of energy is
that it is conserved.that it is conserved.
Law of conservation of energyLaw of conservation of energy state that: energy can bestate that: energy can be
convert from one form to another but can be neitherconvert from one form to another but can be neither
created nor destroyed.created nor destroyed.
Energy can be classified as:Energy can be classified as:
- potential energy.- potential energy.
- kinetic energy.- kinetic energy.

Potential energy (P.E.):Potential energy (P.E.): energy due to position orenergy due to position or
composition.composition.
e.g.:e.g.: water behind dam has P.E. that can be convert towater behind dam has P.E. that can be convert to
work when water flows down through turbines, creatingwork when water flows down through turbines, creating
electricity.electricity.
e.g.:e.g.: the energy released when gasoline is burned resultthe energy released when gasoline is burned result
from differences in attractive forces between the nucleifrom differences in attractive forces between the nuclei
and electrons in the reactants and products.and electrons in the reactants and products.
Kinetic energy (K.E.):Kinetic energy (K.E.): due to motion of the object.due to motion of the object.
Depend on the mass of the object (m) and its velocity(v)Depend on the mass of the object (m) and its velocity(v)
K.E. = ½ (mvK.E. = ½ (mv22
))
Energy can be convert from one form to anotherEnergy can be convert from one form to another
For example, consider the two balls in the followingFor example, consider the two balls in the following
figurefigure

B
A
P.E. of ball A > P.E. of ball B (because its higher position).
When (A) moves down the hill (P.E. of (A) has decreased
and change to K.E.) and strike (B), part of this K.E. is then
transfer to (B) causing it to raised to higher final position.
Thus P.E. of (B) increased. The final position of (B) has
been increased but lower than the original of position (A).
Both balls in their final position are at rest, so the missing
energy can not be due to their motion.

What happened to the remaining energy?What happened to the remaining energy?
As ball (A) rolls down the hill some of its K.E. isAs ball (A) rolls down the hill some of its K.E. is
transferred to the surface of hill astransferred to the surface of hill as heatheat which is calledwhich is called
frictional heating. Thefrictional heating. The temperaturetemperature of the hill increaseof the hill increase
very slightly.very slightly.
What is the difference between heat and temperature?What is the difference between heat and temperature?
Temperature:Temperature: is a property that reflect the random motionis a property that reflect the random motion
of the particles in a particular substance.of the particles in a particular substance.
Heat:Heat: transfer of energy between two objects due to atransfer of energy between two objects due to a
temperature difference.temperature difference.
From the previous figure: in going from the initial to finalFrom the previous figure: in going from the initial to final
arrangement, ball (B) gain P.E. becausearrangement, ball (B) gain P.E. because workwork waswas
done by ball (A) on ball (B).done by ball (A) on ball (B).
WorkWork is define as: force acting over a distance.is define as: force acting over a distance.

Work is required to raise (B) from its original position toWork is required to raise (B) from its original position to
its final position.its final position.
Part of stored energy as P.E. in (A) has been transferredPart of stored energy as P.E. in (A) has been transferred
through work to (B), increasing the P.E. of (B)through work to (B), increasing the P.E. of (B)
Thus there are two ways to transfer energy:Thus there are two ways to transfer energy:
- Through work- Through work - through heat- through heat
Ball (A) will always lost the same amount of P.E.Ball (A) will always lost the same amount of P.E.
The way that this energy transfer as work or heat dependsThe way that this energy transfer as work or heat depends
on the specific conditions – the pathway.on the specific conditions – the pathway.
e.g.:e.g.: surface of the hill so rough, more frictional heating, lesssurface of the hill so rough, more frictional heating, less
work done on ball (B) and can not move to next level, no workwork done on ball (B) and can not move to next level, no work
donedone
The amount of heat and work will differ. The total energyThe amount of heat and work will differ. The total energy
transferred will be constant. Energy change is independenttransferred will be constant. Energy change is independent
of the pathway, work and heat depend on the pathway.of the pathway, work and heat depend on the pathway.

This brings us to a very important concept: theThis brings us to a very important concept: the statestate
functionfunction oror state propertystate property..
State function: a property of the system that:State function: a property of the system that:
depends only on its present state.depends only on its present state.
does not depend in any way of the system’s past or futuredoes not depend in any way of the system’s past or future
its value does not depend on how the system arrived atits value does not depend on how the system arrived at
the present statethe present state
a change in this function in going from one state toa change in this function in going from one state to
another state is independent of the particular pathwayanother state is independent of the particular pathway
taken between two states.taken between two states.
Energy is a state function, work and heat is not a stateEnergy is a state function, work and heat is not a state
function…. it’s a path functionfunction…. it’s a path function

Chemical EnergyChemical Energy
The ideas we have just illustrated also apply to chemicalThe ideas we have just illustrated also apply to chemical
systems.systems.
To discuss the chemical reactions, the universe is dividedTo discuss the chemical reactions, the universe is divided
into two halves.into two halves.
The system:The system: is the part of the universe on which we wishis the part of the universe on which we wish
to focus our attention.to focus our attention.
The surrounding:The surrounding: include everything else in theinclude everything else in the
universe.universe.
For the reaction:For the reaction:
CHCH4(g)4(g) + 2 O+ 2 O2(g)2(g) COCO2(g)2(g) + 2 H+ 2 H22OO(g)(g) ++
energyenergy
The system: is the reactants and products.The system: is the reactants and products.
The surrounding: everything else … container, room..ect.The surrounding: everything else … container, room..ect.

When the reaction evolve heat, it is said to beWhen the reaction evolve heat, it is said to be ExothermicExothermic
reactions (exo- mean “out of ’’). Energy flowreactions (exo- mean “out of ’’). Energy flow out of theout of the
systemsystem. e.g.: combustion of CH. e.g.: combustion of CH44
Endothermic reactions :Endothermic reactions : absorb energy from theabsorb energy from the
surroundings. Energy flowsurroundings. Energy flow into a systeminto a system..
Where does the energy come from in an exothermic reactions?Where does the energy come from in an exothermic reactions?
It come from the difference in P.E. between the products andIt come from the difference in P.E. between the products and
the reactants.the reactants.
In an exothermic reactions, some of the P.E. stored in theIn an exothermic reactions, some of the P.E. stored in the
chemical bonds is being converted to thermal energy viachemical bonds is being converted to thermal energy via
heat.heat.
In an endothermic reactions, the situation is reversed. EnergyIn an endothermic reactions, the situation is reversed. Energy
that flows into the system as heat is used to increase the P.E.that flows into the system as heat is used to increase the P.E.
of the products.of the products.
This is shown in the following figures for exo- and endothermicThis is shown in the following figures for exo- and endothermic

System
Surroundings
Energy released to surrounding as heat
∆E <0
Potentialenergy
∆(P.E.)
2 mol O2
1 mol CH4
Reactants
2 mol H2O
1 mol CO2
(products)
CH + 2O CO + 2H O + Heat4 2 2 2→

System
Surroundings
Energy absorbed from the surroundings
∆E >0
N + O 2NO2 2 + heat →Potentialenergy
∆(P.E.)
2 mol NO
(Products)
1 mol N2
1 mol O2
(Reactants)

The study of energy and its interconversions is calledThe study of energy and its interconversions is called
thermodynamics.thermodynamics.
The law of conservation of energy is called theThe law of conservation of energy is called the first lawfirst law
of thermodynamicsof thermodynamics and is stated as follows:and is stated as follows: the energythe energy
of the universe is constant.of the universe is constant.
TheThe internal energy (E)internal energy (E) of a system can be define as:of a system can be define as: thethe
sum of the K.E. and P.E. of all “particles” in the system.sum of the K.E. and P.E. of all “particles” in the system.
The internal energy can be changed by a flow ofThe internal energy can be changed by a flow of work,work,
heat, or bothheat, or both. That is:. That is:
∆∆E = q + wE = q + w
∆E: change in the system’s internal energy
q: the heat.
w: work

Every thermodynamic quantity has three parts.Every thermodynamic quantity has three parts.
1.1. A unit ( Joules of calories).A unit ( Joules of calories).
2.2. A number: giving the magnitude of the change.A number: giving the magnitude of the change.
3.3. A sign: indicate the direction of the flow.A sign: indicate the direction of the flow.
For exothermic process:For exothermic process:
q:q: negative, -ve sign indicates that the system’snegative, -ve sign indicates that the system’s
energy is decreasing.energy is decreasing. Heat flow out the systemHeat flow out the system
w: negative, system does work on the surrounding.w: negative, system does work on the surrounding.
energy flow out of the system.energy flow out of the system.
For endothermic process:For endothermic process:
q: positive, +ve sign indicates that the system’sq: positive, +ve sign indicates that the system’s
energy is increasing. Heat flow into the system.energy is increasing. Heat flow into the system.
w: positive, work done on the system by thew: positive, work done on the system by the
surrounding. Energy flow into of the system.surrounding. Energy flow into of the system.

SummarySummary
Heat given off is negative.Heat given off is negative.
Heat absorbed is positive.Heat absorbed is positive.
Work done by system on surroundings isWork done by system on surroundings is
positive.positive.
Work done on system by surroundings isWork done on system by surroundings is
negative.negative.
q , w : positive: energy enter system.q , w : positive: energy enter system.
q , w : negative: energy leaves system.q , w : negative: energy leaves system.

Example:Example: A system released 125 kJ of heat while 104 kJA system released 125 kJ of heat while 104 kJ
of work is done on it, calculate ∆Eof work is done on it, calculate ∆Esys.sys. , ∆E, ∆Esurr.surr.
q = - 125 kJ w = + 104 kJq = - 125 kJ w = + 104 kJ
∆∆EEsys.sys. = q + w = - 125 kJ + 104 kJ = - 21 kJ= q + w = - 125 kJ + 104 kJ = - 21 kJ
∆∆EEsurr.surr. = + 21 kJ= + 21 kJ

A common type of work associated with chemical processA common type of work associated with chemical process
is work done by a gas (through expansion) or work doneis work done by a gas (through expansion) or work done
on a gas (through compression).on a gas (through compression).
Suppose we have a gas inSuppose we have a gas in
a cylindrical containera cylindrical container
with movable piston aswith movable piston as
shown in this figure whereshown in this figure where
F: force acting on pistonF: force acting on piston
of area (A)of area (A)
Pressure is define as forcePressure is define as force
per unit areaper unit area
P = F/AP = F/A
∆h ∆h
∆V
Initial state Final state
Area = A
P = F/A
P = F/A

Work is defined as a force acting over a distance.Work is defined as a force acting over a distance.
w= force x distance = F x ∆hw= force x distance = F x ∆h
P = F/ A or F = P x AP = F/ A or F = P x A
w= (P x A x ∆ hw= (P x A x ∆ h
Volume of the cylinder (V) = A x hVolume of the cylinder (V) = A x h
w = P x ∆Vw = P x ∆V
where ∆V = final volume – initial volumewhere ∆V = final volume – initial volume
P: the external pressureP: the external pressure
units of work is liter - atm (L-atm)units of work is liter - atm (L-atm)
to change this unit to joule (J) (1.0 L. atm. = 101.3 J)to change this unit to joule (J) (1.0 L. atm. = 101.3 J)
If the volume of a gas increases, the system has done workIf the volume of a gas increases, the system has done work
on the surroundings. work is negative.on the surroundings. work is negative.
If the volume of a gas decreases, the surroundings hasIf the volume of a gas decreases, the surroundings has
done work on the system. work is positivedone work on the system. work is positive
Volume of the cylinder

Example: A)Example: A) What amount of work is done when 15 LWhat amount of work is done when 15 L
of gas is expanded to 25 L at 2.4 atm pressure?of gas is expanded to 25 L at 2.4 atm pressure?
B)B) If 2.36 J of heat are absorbed by the gas above. whatIf 2.36 J of heat are absorbed by the gas above. what
is the change in energy?is the change in energy?
C)C) How much heat would it take to change the gasHow much heat would it take to change the gas
without changing the internal energy of the gas?without changing the internal energy of the gas?
A)A) w = P ∆V = 2.4 atm. x (25 – 15) = 24 L. atm.w = P ∆V = 2.4 atm. x (25 – 15) = 24 L. atm.
w = - 24 L. atm. x 101 (J/L. atm) = - 2431.2 Jw = - 24 L. atm. x 101 (J/L. atm) = - 2431.2 J
B) ∆E = w + q = - 2431.2 + 2.36 J = - 2428.84 JB) ∆E = w + q = - 2431.2 + 2.36 J = - 2428.84 J
C) ∆E = 0C) ∆E = 0
q = w so q = + 2431.2 Jq = w so q = + 2431.2 J

Enthalpy and CalorimetryEnthalpy and Calorimetry
EnthalpyEnthalpy
A less familiar property of a system is its enthalpy,A less familiar property of a system is its enthalpy,
abbreviated H, which is define asabbreviated H, which is define as
H = E + PVH = E + PV
E: internal energyE: internal energy P: pressure of the systemP: pressure of the system
V: volume of the systemV: volume of the system
Since (E, P, V) are all state functions,Since (E, P, V) are all state functions, enthalpy is also aenthalpy is also a
state function.state function.
What exactly is enthalpy?What exactly is enthalpy?
To answer this question, consider a process carried outTo answer this question, consider a process carried out
at constant pressure, the only work allowed is pressure-at constant pressure, the only work allowed is pressure-
volume work (w = -P∆V)volume work (w = -P∆V)

Under these conditions:
∆E = qp + w = qp - P∆V
qp = ∆E + P ∆V (qp: heat at constant pressure)
Since : H = E + PV then
change in H = (change in E) + (change in PV)
∆H = ∆E + ∆(PV)
∆H = ∆E + P∆V (since P is constant)
identical
So : ∆H = qp
At constant pressure, the change in the enthalpy ∆H of
the system is equal to the energy flow as heat.

This means that for a reaction studied at constantThis means that for a reaction studied at constant
pressure, the flow of heat is measured as ∆Hpressure, the flow of heat is measured as ∆H
∆∆H = HH = Hproductsproducts – H– Hreactantsreactants
HHproductsproducts >> HHreactantsreactants , ∆H will be positive,, ∆H will be positive,
heat absorbed by the system,heat absorbed by the system,
the reaction is endothermicthe reaction is endothermic
HHproductsproducts < H< Hreactantsreactants , ∆H will be negative,, ∆H will be negative,
heat evolved by the system,heat evolved by the system,
the reaction is exothermicthe reaction is exothermic

Example:Example: consider the following reaction:consider the following reaction:
CHCH4(g)4(g) + 2 O+ 2 O2(g)2(g) COCO2(g)2(g) + 2 H+ 2 H22OO(ℓ)(ℓ) ∆H= -981kJ∆H= -981kJ
Calculate the enthalpy change for each of the followingCalculate the enthalpy change for each of the following
cases: A) 1.0 gcases: A) 1.0 g CHCH44 of is burned in enoughof is burned in enough OO22
B) 1x10B) 1x1033
LL CHCH44 of gas at 740 torr (0.97 atm.)atof gas at 740 torr (0.97 atm.)at
25ºC is burned in excess25ºC is burned in excess OO22
A) mole of CA) mole of CHH44 = 1.0 g/ (16g/mol) = 0.0625 mol= 1.0 g/ (16g/mol) = 0.0625 mol
from the equation: 1 mol Cfrom the equation: 1 mol CHH44 evolve 981 kJevolve 981 kJ
qqpp = ∆H = (- 981kJ/mol) x 0.0625 mol = -55.69 kJ= ∆H = (- 981kJ/mol) x 0.0625 mol = -55.69 kJ
B) n =PV/RT = 0.97 atm.x1x10B) n =PV/RT = 0.97 atm.x1x1033
L/(0.0821 L.atm/mol K)298KL/(0.0821 L.atm/mol K)298K
n = 39.65 moln = 39.65 mol
∆∆H = (- 891 kJ/mol) x 39.65 mol = - 35325.6 kJH = (- 891 kJ/mol) x 39.65 mol = - 35325.6 kJ

CalorimetryCalorimetry
Calorimetry: is the science of measuring heat. It based onCalorimetry: is the science of measuring heat. It based on
observing the temperature change when body absorbedobserving the temperature change when body absorbed
or released energy as heator released energy as heat
The device used experimentally to determine the heatThe device used experimentally to determine the heat
associated with chemical reaction is called aassociated with chemical reaction is called a
calorimeter.calorimeter.
Substances respond differently to being heated, this
property is measured by the heat capacity, C, which
defined as:
C = (heat absorbed/ increase in temp.) = (J/ºC)
Heat capacity is the energy required to raise the
temperature of a substance by one degree Celsius.
Energy required depend on the amount of substance
present.

In defining the heat capacity of a substance, the amount
of substance must be specified.
If the heat capacity is given per gram of substance, it is
called specific heat capacity, and its unit is (J/ºC. g)
If the heat capacity is given per mol of substance, it is
called molar heat capacity, and its unit is (J/ºC. mol)
The specific heat capacities of some common substances
are given in the
following table:
substance Specific heat capacity
(J/ºC.g)
H2O(ℓ( .4 18
H2O(s( .2 03
Al(s( .0 89
Fe(s( .0 45
Hg(ℓ( .0 14
C(s( .0 71

Two kinds of calorimeters are usedTwo kinds of calorimeters are used
1. Constant pressure calorimeter (called a coffee-cup1. Constant pressure calorimeter (called a coffee-cup
calorimeter)calorimeter)
-- Since theSince the pressure remain constantpressure remain constant
(atmospheric pressure) during the(atmospheric pressure) during the
process, it is used to determineprocess, it is used to determine
the enthalpy,the enthalpy, ∆H∆H(heat) of the(heat) of the
reactionreaction
- C= heat absorbed/ ∆T = ∆H/ ∆T- C= heat absorbed/ ∆T = ∆H/ ∆T
- specific heat capacity = C/mass- specific heat capacity = C/mass
- heat =∆H∆H=specific heat x mass x ∆T

Example:Example: The specific heat of graphite is 0.71 J/The specific heat of graphite is 0.71 J/ºC.ºC.g.g.
Calculate the energy needed to raise the temperature ofCalculate the energy needed to raise the temperature of
75 kg of graphite from 21º C to 75ºC.75 kg of graphite from 21º C to 75ºC.
heat = specific heat x mass x ∆T
heat = 0.71 J/ºC.g x 75x10J/ºC.g x 75x1033
g x (75-21)ºCg x (75-21)ºC
heat = 2875500 J = 2875.5 kJheat = 2875500 J = 2875.5 kJ

Example:Example: If 50.0 ml of 1.0 M HCl solution at 25ºC mixed withIf 50.0 ml of 1.0 M HCl solution at 25ºC mixed with
1.0 M NaOH solution also at 25ºC in calorimeter, after the1.0 M NaOH solution also at 25ºC in calorimeter, after the
reactants are mixed by stirring, the temperature is observed toreactants are mixed by stirring, the temperature is observed to
increased to 31.9 ºC. Calculate the heat of neutralization of theincreased to 31.9 ºC. Calculate the heat of neutralization of the
reaction. HCl + NaOHreaction. HCl + NaOH HH22O + NaClO + NaCl
Assuming that:Assuming that: specific heat capacity of solution=4.18specific heat capacity of solution=4.18 J/ºC.gJ/ºC.g
density of the solution = 1.0g/mldensity of the solution = 1.0g/ml
mass of solution = 100 ml x 1 g/mlmass of solution = 100 ml x 1 g/ml
energy released by the reaction = energy absorbed by solutionenergy released by the reaction = energy absorbed by solution
Heat released by the reaction = specific heat x (mass)solution x ∆T
heat = 4.18 J/ºC.g x 100 g x (31.9 - 25)ºCJ/ºC.g x 100 g x (31.9 - 25)ºC
heat = 2.9 x 10heat = 2.9 x 1033
JJ (heat released when 0.05 mol H(heat released when 0.05 mol H22O produced)O produced)
heat (kJ/mol Hheat (kJ/mol H22O) = 2.9 x 10O) = 2.9 x 1033
J/0.05 mol = 58 kJ/molJ/0.05 mol = 58 kJ/mol

Example:Example: A 46.2 g sample of copper is heated to 95.4ºCA 46.2 g sample of copper is heated to 95.4ºC
and then placed in a calorimeter containing 75.0 g ofand then placed in a calorimeter containing 75.0 g of
water at 19.6ºC. The final temperature of both the waterwater at 19.6ºC. The final temperature of both the water
and the copper is 21.8ºC. What is the specific heat ofand the copper is 21.8ºC. What is the specific heat of
copper?copper?
heat absorbed by water = heat lost by the metalheat absorbed by water = heat lost by the metal
(specific heat x mass x ∆T)water = (specific heat x mass x∆T)metal
4.18 J/ºC.g x 75.0g x (21.8 – 19.6) ºC =
specific heat x 46.2 g x (95.4 – 21.8) ºC
(specific heat)metal = 0.203 J/ºC.g

22. Constant volume calorimeter is called a bomb. Constant volume calorimeter is called a bomb
calorimeter.calorimeter.
Material is put in a container with pure oxygen. WiresMaterial is put in a container with pure oxygen. Wires
are used to start the combustion. The container is putare used to start the combustion. The container is put
into a container of water.into a container of water.
The heat capacity of the calorimeter is known andThe heat capacity of the calorimeter is known and
tested.tested.
∆∆E = q + w = q + p ∆VE = q + w = q + p ∆V
Since ∆V = 0, P∆V = 0, ∆E = qSince ∆V = 0, P∆V = 0, ∆E = qvv

Bomb CalorimeterBomb Calorimeter
thermometerthermometer
stirrerstirrer
full of waterfull of water
ignition wireignition wire
Steel bombSteel bomb
samplesample

Example:Example: A 1.65 g of Mg (molar mass = 24.3 g/mol) wasA 1.65 g of Mg (molar mass = 24.3 g/mol) was
oxidized to MgO in calorimeter with heat capacity =oxidized to MgO in calorimeter with heat capacity =
6.27 kJ/ºC, if the temperature increase from 21.3ºC to6.27 kJ/ºC, if the temperature increase from 21.3ºC to
28.56ºC. Calculate ∆E in kJ for the reaction:28.56ºC. Calculate ∆E in kJ for the reaction:
2 Mg + O2 Mg + O22 MgOMgO
Heat (kJ) = heat capacity x ∆THeat (kJ) = heat capacity x ∆T
Heat (kJ) = 6.27Heat (kJ) = 6.27 kJ/ºC x (28.56 – 21.3)kJ/ºC x (28.56 – 21.3) ºCºC
= 45.52 kJ= 45.52 kJ (heat released when 1.65g Mg oxidized or(heat released when 1.65g Mg oxidized or
0.0679 mol Mg)0.0679 mol Mg)
Moles of Mg = 1.65 g/ (24.3g/mol) = 0.0679 molMoles of Mg = 1.65 g/ (24.3g/mol) = 0.0679 mol
heat for the above reaction (2 mol Mg) =heat for the above reaction (2 mol Mg) =
45.52 kJ x (2 mol Mg/ 0.0679 mol Mg)45.52 kJ x (2 mol Mg/ 0.0679 mol Mg)
= 1.34 x 10= 1.34 x 1033
kJkJ

Example:Example: A) The combustion of 0.1584 g benzoic acidA) The combustion of 0.1584 g benzoic acid
increase the temperature of bomb calorimeter by 2.54ºC,increase the temperature of bomb calorimeter by 2.54ºC,
calculate the heat capacity of calorimeter (energy released bycalculate the heat capacity of calorimeter (energy released by
combustion of benzoic acid = 26.42 kJ/g)combustion of benzoic acid = 26.42 kJ/g)
B) A 0.2130 g of vanillin (molar mass = 152 g/mol) is burned inB) A 0.2130 g of vanillin (molar mass = 152 g/mol) is burned in
the same calorimeter and the temperature increased bythe same calorimeter and the temperature increased by
3.25ºC. Calculate the heat of combustion per gram of3.25ºC. Calculate the heat of combustion per gram of
vanillin…… per molvanillin…… per mol
A)A) (Heat capacity)(Heat capacity)calorimetercalorimeter = (26.42kJ/g) x 0.1584 g / 2.54ºC= (26.42kJ/g) x 0.1584 g / 2.54ºC
= 1.65 kJ/ºC= 1.65 kJ/ºC
B)B) ∆E = q∆E = qvv = heat capacity x ∆T == heat capacity x ∆T = --1.65 kJ/ºC x 3.25ºC1.65 kJ/ºC x 3.25ºC
= - 5.36 kJ= - 5.36 kJ
= - 5.36 kJ/0.2130 g = - 25.18 kJ/g= - 5.36 kJ/0.2130 g = - 25.18 kJ/g
= - 5.36 kJ/ 0.0014 mol = - 3828.57 kJ/mol= - 5.36 kJ/ 0.0014 mol = - 3828.57 kJ/mol

Hess’s LawHess’s Law
Enthalpy is a state function.Enthalpy is a state function.
It is independent of the path.It is independent of the path.
This is mean:This is mean: in going from a particular set ofin going from a particular set of
reactants to a particular set of products, the change inreactants to a particular set of products, the change in
enthalpy is the same whether the reaction takes placeenthalpy is the same whether the reaction takes place
in one step or in a series of steps.in one step or in a series of steps. This principle isThis principle is
known asknown as Hess’s lawHess’s law
We can add equations to come up with the desiredWe can add equations to come up with the desired
final product, and add the ∆H.final product, and add the ∆H.

Hess’s law can be illustrated in the following example:Hess’s law can be illustrated in the following example:
NN2(g)2(g) + 2 O+ 2 O2(g)2(g) 2 NO2 NO2(g)2(g) ∆H∆H11 = 68 kJ= 68 kJ
The reaction also can also be carried out in two distinctThe reaction also can also be carried out in two distinct
steps:steps:
NN2(g)2(g) + O+ O2(g)2(g) 2 NO2 NO(g)(g) ∆H∆H22 = 180 kJ= 180 kJ
2 NO2 NO(g)(g) + O+ O2(g)2(g) 2 NO2 NO2(g)2(g) ∆H∆H33 = -112 kJ= -112 kJ
Net reaction:Net reaction:
NN2(g)2(g) + 2 O+ 2 O2(g)2(g) 2 NO2 NO2(g)2(g) ∆H∆H11=68 kJ = ∆H=68 kJ = ∆H22 +∆H+∆H33
Two characteristics ofTwo characteristics of ∆H for a reaction:∆H for a reaction:
If the reaction is reversed the sign of ∆H is also reversedIf the reaction is reversed the sign of ∆H is also reversed
If the coefficient in a balanced reaction is multiplied by anIf the coefficient in a balanced reaction is multiplied by an
integer, the value of ∆H is multiply by the same integerinteger, the value of ∆H is multiply by the same integer

N2 , 2O2
O2 , NO2
68 kJ
NO2
180 kJ
-112 kJ
H(kJ)

Example:Example: Given the following data:Given the following data:
a)a) 2 NH2 NH3(g)3(g) NN2(g)2(g) + 3 H+ 3 H2(g)2(g) ∆H = 92 kJ∆H = 92 kJ
b)b) 2 H2 H2(g)2(g) + O+ O2(g)2(g) 2 H2 H22OO(g)(g) ∆H = - 484 kJ∆H = - 484 kJ
CalculateCalculate ∆∆H for the reaction:H for the reaction:
2 N2 N2(g)2(g) + 6 H+ 6 H22OO(g)(g) 3 O3 O2(g)2(g) + 4 NH+ 4 NH3(g)3(g)
Solution:Solution:
Reverse equation (a) and multiply it by the coefficient (2)Reverse equation (a) and multiply it by the coefficient (2)
Reverse equation (b) and multiply it by the coefficient (3)Reverse equation (b) and multiply it by the coefficient (3)
22 NN2(g)2(g) + 6H+ 6H2(g)2(g) 4NH4NH3(g)3(g) ∆H = -184 kJ∆H = -184 kJ
6 H6 H22OO(g)(g) 6 H6 H2(g)2(g) + 3 O+ 3 O2(g)2(g) ∆H = + 1452 kJ∆H = + 1452 kJ
2 N2 N2(g)2(g) + 6 H+ 6 H22OO(g)(g) 4NH4NH3(g)3(g) ++ 3 O3 O2(g)2(g)
∆∆H = + 1268 kJH = + 1268 kJ

Standard EnthalpyStandard Enthalpy
The enthalpy change for a reaction at standardThe enthalpy change for a reaction at standard
conditions (25ºC, 1 atm , 1 M solutions)conditions (25ºC, 1 atm , 1 M solutions)
Symbol ∆HºSymbol ∆Hº
When using Hess’s Law, work by adding theWhen using Hess’s Law, work by adding the
equations up to make it look like the answer.equations up to make it look like the answer.
The other parts will cancel out.The other parts will cancel out.

Example:Example: Given the following thermochemical equation:Given the following thermochemical equation:
a) Ca) C22HH2(g)2(g) + 5/2 O+ 5/2 O2(g)2(g) → 2 CO→ 2 CO2(g)2(g) + H+ H22OO(ℓ)(ℓ) ∆Hº = - 1300 kJ∆Hº = - 1300 kJ
b)b) CC(s)(s) + O+ O2(g)2(g) → CO→ CO2(g)2(g) ∆Hº = - 394 kJ∆Hº = - 394 kJ
c)c) HH2(g)2(g) + ½ O+ ½ O2(g)2(g) → H→ H22OO(ℓ)(ℓ) ∆Hº = - 286 kJ∆Hº = - 286 kJ
Calculate ∆Hº for the reaction:Calculate ∆Hº for the reaction:
2 C2 C(s)(s) + H+ H2(g)2(g) → C→ C22HH4(g)4(g)
Solution:Solution: Reverse equation (a). Multiply equation (b) by theReverse equation (a). Multiply equation (b) by the
coefficient 2. equation (c) remain as it is.coefficient 2. equation (c) remain as it is.
2 CO2 CO2(g)2(g) + H+ H22OO(ℓ)(ℓ) → C→ C22HH2(g)2(g) + 5/2 O+ 5/2 O2(g)2(g) ∆Hº = +1300 kJ∆Hº = +1300 kJ
2 C2 C(s)(s) + 2 O+ 2 O2(g)2(g) → 2 CO→ 2 CO2(g)2(g) ∆Hº = - 788 kJ∆Hº = - 788 kJ
HH2(g)2(g) + ½ O+ ½ O2(g)2(g) → H→ H22OO(ℓ)(ℓ) ∆Hº = - 286 kJ∆Hº = - 286 kJ
2 C2 C(s)(s) + H+ H2(g)2(g) → C→ C22HH4(g)4(g) ∆Hº = +229 kJ∆Hº = +229 kJ

Example:Example: Given the following thermochemical equation:Given the following thermochemical equation:
a) Ha) H2(g)2(g) + 1/2 O+ 1/2 O2(g)2(g) → H→ H22OO(ℓ)(ℓ) ∆Hº = - 285.8 kJ∆Hº = - 285.8 kJ
b)b) NN22OO5(g)5(g) ++ HH22OO(ℓ)(ℓ) → 2 HNO→ 2 HNO3(3(ℓℓ)) ∆Hº = - 76.6 kJ∆Hº = - 76.6 kJ
c)c) 1/2 N1/2 N2(g)2(g) + 3/2 O+ 3/2 O2(g)2(g) + 1/2+ 1/2 HH2(g)2(g) →→ HNOHNO3(ℓ)3(ℓ) ∆Hº = - 174.1 kJ∆Hº = - 174.1 kJ
Calculate ∆Hº for the reaction:Calculate ∆Hº for the reaction:
22 NN2(g)2(g) + 5 O+ 5 O2(g)2(g) → N→ N22OO5(g)5(g)
Solution:Solution: Reverse equation (a), and multiply it by 2.Reverse equation (a), and multiply it by 2. ReverseReverse
equationequation (b),(b), multiply it bymultiply it by 2. multiply equation (c) by 42. multiply equation (c) by 4
22 /H/H22OO(ℓ)(ℓ) → H→ H2(g)2(g) + 1/2 O+ 1/2 O2(g)2(g) ∆Hº =+∆Hº =+ 285.8285.8 kJkJ
2/2/ 2 HNO2 HNO3(ℓ)3(ℓ) →→ NN22OO5(g)5(g) + H+ H22OO(ℓ)(ℓ) ∆Hº = + 76.6 kJ∆Hº = + 76.6 kJ
4/ ½ N4/ ½ N2(g)2(g) + 3/2 O+ 3/2 O2(g)2(g) + 1/2 H+ 1/2 H2(g)2(g) → HNO→ HNO3(ℓ)3(ℓ) ∆Hº = - 174.1 kJ∆Hº = - 174.1 kJ
2 N2 N2(g)2(g) + 5 O+ 5 O2(g)2(g) → N→ N22OO5(g)5(g) ∆Hº = +28.8 kJ∆Hº = +28.8 kJ

Standard Enthalpies of FormationStandard Enthalpies of Formation
Standard enthalpy of formationStandard enthalpy of formation (∆H(∆Hƒƒ
ºº
):): is the heatis the heat
change that results whenchange that results when one moleone mole of a compound isof a compound is
formed from itsformed from its elementselements in their standard states.in their standard states.
Standard states are 1 atm, 1M and 25ºCStandard states are 1 atm, 1M and 25ºC
For an elements:For an elements: ∆H∆Hƒƒ
ºº
== 00
∆∆HHƒƒ
ºº
(O(O22) = 0) = 0
The formation of NOThe formation of NO22 from its elements in their standardfrom its elements in their standard
states:states:
½ N½ N2(g)2(g) + O+ O2(g)2(g) → NO→ NO2(g)2(g) ∆H∆Hƒƒº = 34 kJ/molº = 34 kJ/mol
TheThe ∆H∆Hƒƒº values for common substances are presentº values for common substances are present
in certain tables (in certain tables (∆H∆Hƒƒ
ºº
for some substances are shownfor some substances are shown
in the following table)in the following table)..
∆Hƒ
º
(C, graphite) = 0

Write the equation for the formation of methanolWrite the equation for the formation of methanol
CHCH33OH.OH.
CC(s)(s) + 2 H+ 2 H2(g)2(g) + ½ O+ ½ O2(g)2(g) → CH→ CH33OHOH(ℓ)(ℓ) ∆Hº∆Hºƒƒ = -239kJ/mol.= -239kJ/mol.
Several important characteristics of the definition of ∆HºSeveral important characteristics of the definition of ∆Hºƒƒ ::
The reaction is written so that elements are in theirThe reaction is written so that elements are in their
standard states.standard states.
1.0 mol of product is formed so ∆Hº1.0 mol of product is formed so ∆Hºƒƒ are always givenare always given
in kJ/mol.in kJ/mol.
For most compounds ∆HºFor most compounds ∆Hºƒƒ is negative, because you areis negative, because you are
making bonds; making bonds is exothermic process.making bonds; making bonds is exothermic process.

Enthalpy is a state function, so we can invoke Hess’s law andEnthalpy is a state function, so we can invoke Hess’s law and
choose any convenient pathway from reactants to products andchoose any convenient pathway from reactants to products and
then sum the enthalpy changes along the chosen pathway.then sum the enthalpy changes along the chosen pathway.
A convenient pathway involve the use of ∆HºA convenient pathway involve the use of ∆Hºƒƒ ..
For exampleFor example we will calculate ∆Hº for the combustion of CHwe will calculate ∆Hº for the combustion of CH44::
CHCH4(g)4(g) + 2 O+ 2 O2(g)2(g) → CO→ CO2(g)2(g) + 2 H+ 2 H22OO(ℓ)(ℓ)
By using ∆HºBy using ∆Hºƒƒ for each compound in the reactionfor each compound in the reaction
1)1) CC(s)(s) + 2 H+ 2 H2(g)2(g) → CH→ CH4(g)4(g) ∆Hº∆Hºƒƒ = - 75 kJ/mol= - 75 kJ/mol
2)2) CC(s)(s) + O+ O2(g)2(g) → CO→ CO2(g)2(g) ∆Hº∆Hºƒƒ = - 394 kJ/mol= - 394 kJ/mol
3)3) HH2(g)2(g) + ½ O+ ½ O2(g)2(g) → H→ H22OO(g)(g) ∆Hº∆Hºƒƒ = - 286 kJ/mol= - 286 kJ/mol
Reverse equation (1), multiply equation (3) by the coefficient 2Reverse equation (1), multiply equation (3) by the coefficient 2
The reactions become:The reactions become:

1)1) CHCH4(g)4(g) →→CC(s)(s) + 2 H+ 2 H2(g)2(g) ∆Hº∆Hºƒƒ = + 75 kJ/mol= + 75 kJ/mol
2)2) CC(s)(s) + O+ O2(g)2(g) → CO→ CO2(g)2(g) ∆Hº∆Hºƒƒ = - 394 kJ/mol= - 394 kJ/mol
3)3) 2 / H2 / H2(g)2(g) + ½ O+ ½ O2(g)2(g) → H→ H22OO(g)(g) ∆Hº∆Hºƒƒ = - 286 kJ/mol= - 286 kJ/mol
CHCH4(g)4(g) + 2 O+ 2 O2(g)2(g) → CO→ CO2(g)2(g) + 2 H+ 2 H22OO(ℓ)(ℓ)
∆∆Hº = 75 + (- 394) + (-286 x 2) = - 891 kJ/molHº = 75 + (- 394) + (-286 x 2) = - 891 kJ/mol
∆∆HºHºreactionreaction = - ∆Hº= - ∆Hºƒƒ for CHfor CH44 + ∆Hº+ ∆Hºƒƒ for COfor CO22+∆Hº+∆Hºƒƒ for Hfor H22OO
reactantsreactants productsproducts
So:So:
∆∆HH00
== ΣΣ nnpp ∆Hº∆Hºƒƒ (products) –(products) – ΣΣ nnrr ∆Hº∆Hºƒƒ (reactants)(reactants)
n: number of moles of each substance in the balancen: number of moles of each substance in the balance
equationequation

Example:Example: Calculate the enthalpy (∆Hº) for the followingCalculate the enthalpy (∆Hº) for the following
reactions:reactions:
a) 4 NHa) 4 NH3(g)3(g) + 5O+ 5O2(g)2(g) → 4 NO→ 4 NO(g)(g) + 6H+ 6H22OO(g)(g)
b) 3 NOb) 3 NO2(g)2(g) + H+ H22OO(ℓ)(ℓ) → 2 HNO→ 2 HNO3(aq)3(aq) + NO+ NO(g)(g)
∆∆HH00
reactionreaction == Σ ∆Σ ∆HH00
(products) –(products) – Σ ∆Σ ∆HH00
(reactants)(reactants)
a)a) ∆∆HH00
rxnrxn== [6 ∆Hº[6 ∆Hºƒƒ HH22O + 4 ∆HºO + 4 ∆Hºƒƒ NO] – [5∆HºNO] – [5∆Hºƒƒ OO22 + 4 ∆Hº+ 4 ∆HºƒƒNHNH33]]
= [6 x (-242) + 4 x 90] – [5 x 0 + 4 x (- 46)= [6 x (-242) + 4 x 90] – [5 x 0 + 4 x (- 46)
= - 908 kJ= - 908 kJ
b)b) ∆∆HH00
rxnrxn== [2 ∆Hº[2 ∆Hºƒƒ HNOHNO33 + ∆Hº+ ∆Hºƒƒ NO] – [∆HºNO] – [∆Hºƒƒ HH22O + 3 ∆HºO + 3 ∆Hºƒƒ NONO22]]
= [2 x (- 207) + 90 ] – [ - 286 + 3 x 34]= [2 x (- 207) + 90 ] – [ - 286 + 3 x 34]
= - 140 kJ= - 140 kJ

Example:
Benzene (C6H6) burns in air to produce carbon dioxide
and liquid water and librates 3271 kJ at 25ºC and 1 atm.
Calculate ∆Hºƒ of benzene in kJ/mol.
2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(l)
∆H0
rxn ∆H0
(products)f= Σ ∆H0
(reactants)fΣ-
∆H0
rxn 6∆H0
(H2O)f12 ∆H0
(CO2)f= [ + ] - 2∆H0
(C6H6)f[ ]
- 3271 = [ 12 × -393.5 + 6 × -285.8 ] – [2∆H0
f (C6H6)]
6.6
∆H0
f = + 49 kJ

Keep in mind the following key concepts when doingKeep in mind the following key concepts when doing
enthalpy calculations:enthalpy calculations:
When a reaction is reversed, the magnitude of ∆HWhen a reaction is reversed, the magnitude of ∆H
remains the same, but its sign changes.remains the same, but its sign changes.
When the balanced equation for a reaction is multipliedWhen the balanced equation for a reaction is multiplied
by an integer, the value of ∆H for that reaction must beby an integer, the value of ∆H for that reaction must be
multiply by the same integer.multiply by the same integer.
The change in enthalpy for a given reaction can beThe change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of thecalculated from the enthalpies of formation of the
reactants and products.reactants and products.
∆∆HH00
== ΣΣ nnpp ∆Hº∆Hºƒƒ (products) –(products) – ΣΣ nnrr ∆Hº∆Hºƒƒ (reactants(reactants
Elements in their standard states are not included in theElements in their standard states are not included in the
∆H∆Hreactionreaction calculations. That is,calculations. That is, ∆Hº∆Hºƒƒ for an element in itsfor an element in its
standard state is zero.standard state is zero.

Apchapt 6(thermochemistry)

More Related Content

What's hot (15)

Similar to Apchapt 6(thermochemistry) (20)

More from علي علي (6)

Recently uploaded (20)

Apchapt 6(thermochemistry)