PRINCIPLES OF CHEMISTRY I

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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 5
DR. AUGUSTINE OFORI AGYEMAN Assistant professor
of chemistry Department of natural
sciences Clayton state university
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CHAPTER 5 THERMOCHEMISTRY
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THERMOCHEMISTRY
- The study of the relationship between heat and
chemical reactions
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ENERGY
- The ability to do work or to transfer heat -
Energy is necessary for life humans, plants,
animals, cars - Forms of energy are
interconvertible Two major categories - Kinetic
energy - Potential energy
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ENERGY
Kinetic Energy (Ek) - Energy of motion - Atoms
and molecules possess kinetic energy since they
have mass and are in motion
m mass of the object (kg) v velocity (speed)
of the object (m/s) - Units The joule (J), 1 J
1 kg-m2/s2, 4.184 J 1 cal
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ENERGY
Calculate the kinetic energy of an object of mass
38 g, which is moving with a constant velocity of
54 m/s. (a) in joules and (b) in calories
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ENERGY
Potential Energy (Ep) - Energy by virtue of its
position relative to other objects - Arises when
there is a force operating on an object
Ep mgh
m mass of the object (kg) g gravitational
constant (9.8 m/s2) h height of the object
relative to a reference point (m) - Units The
joule (J), 1 J 1 kg-m2/s2, 4.184 J 1 cal
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ENERGY
A 25-g marble is thrown upward and travels
through a vertical distance of 10.0 m from the
ground. Calculate the potential energy of the
marble at a height of 5.0 m from the ground.
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ENERGY
Electrostatic Potential Energy (Eel) - Due to
interactions between charged particles
? constant of proportionality (8.99 x 109
J-m/C2) C coulomb, a unit of electrical
charge Q1 and Q2 electrical charges on two
interacting objects d distance between the two
objects
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ENERGY
Electrostatic Potential Energy (Eel) - Due to
interactions between charged particles - For
molecular-level objects, Q1 and Q2 are on the
order of magnitude of electron charge (1.60 x
10-19 C) - Same sign Q1 and Q2 (both positive or
both negative) causes repulsion and Eel is
positive - Opposite signs Q1 and Q2 (one positive
and one negative) cause attraction and Eel is
negative - The lower the energy of a system, the
more stable the system - Opposite charges
interact more strongly and the system is more
stable
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ENERGY
Work (w) - The energy transferred when a force
moves an object - The product of force (F) and
distance (d) through which the object moves
w F x d
Force - Any kind of push or pull exerted on an
object
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ENERGY
Heat (q) - Energy used to cause the temperature
of an object to change - A form of energy
necessary to change the temperature of a
substance Chemical Energy - Potential energy
resulting from forces that hold atoms together
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SYSTEM AND SURROUNDINGS
System - The limited and well-defined portion of
the universe under study Surroundings -
Everything else in the universe Studying energy
changes in a chemical reaction - The reactants
and products make up the system - The reaction
container and everything else make up the
surroundings
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SYSTEM AND SURROUNDINGS
Open System - Matter and energy can be exchanged
with the surroundings (water boiling on a stove
without a lid) Closed System - Energy but not
matter can be exchanged with the surroundings
(two reactants in a closed cylinder reacting to
produce energy) Isolated System - Neither matter
nor energy can be exchanged with the
surroundings (insulated flask containing hot
tea)
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INTERNAL ENERGY (E)
- Sum of all potential and kinetic energies of
all components - Change in internal energy
final energy minus initial energy ?E Efinal -
Einitial - Energy can neither be created nor
destroyed - Energy is conserved
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INTERNAL ENERGY (E)
?E Efinal - Einitial If Efinal gt Einitial ?E
is positive and system has gained energy from its
surroundings If Efinal lt Einitial ?E is
negative and system has lost energy to its
surroundings
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INTERNAL ENERGY (E)
Energy Diagram
Efinal
Einitial
?E lt 0
Internal energy, E
Internal energy, E
?E gt 0
Energy lost to surroundings
Energy gained from surroundings
Einitial
Efinal
E of system decreases
E of system increases
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INTERNAL ENERGY (E)
?E q w q heat added to or liberated from a
system w work done on or by a system Internal
energy of a system increases when - Heat is
added to the system from surroundings (positive
q) - Work is done on the system by surroundings
(positive w)
w
q
system
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INTERNAL ENERGY (E)
?E q w q heat added to or liberated from a
system w work done on or by a system Internal
energy of a system decreases when - Heat is lost
by the system to the surroundings (negative q) -
Work is done by the system on the surroundings
(negative w)
-w
-q
system
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INTERNAL ENERGY (E)
Endothermic Process - Process in which system
absorbs heat (endo- means into) - Heat flows
into system from its surroundings (melting of
ice - the reason why it feels cold) - Heat is a
reactant and ?E is positive N2(g) O2(g)
heat ? 2NO(g)
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INTERNAL ENERGY (E)
Exothermic Process - Process in which system
loses heat - Heat flows out of the system (exo-
means out of) (combustion of gasoline) -
Heat is a product and ?E is negative CH4(g)
2O2(g) ? CO2(g) 2H2O(l) heat
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INTERNAL ENERGY (E)
State Function - Property that depends on
initial and final states of the system - Does
not depend on path or how a change occurs -
Internal Energy depends on initial and final
states - Internal Energy is a state function -
q and w, on the other hand, are not state
functions
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INTERNAL ENERGY (E)
Internal energy is influenced by -
Temperature - Pressure - Total quantity of
matter - Internal Energy is an extensive
property
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INTERNAL ENERGY (E)
Calculate ?E for a system absorbing 58 kJ of heat
from its surroundings while doing 19 kJ of work
on the surroundings. State whether it is an
endothermic or an exothermic process q 58 kJ
(heat is added to the system from surroundings) w
-19 kJ (work is done by the system on the
surroundings) ?E q w ?E ( 58 19) kJ
39 kJ Endothermic
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ENTHALPY (H)
- Heat flow in processes occurring at constant
pressure - Only work-pressure (P-V work) are
performed At constant pressure w - P?V w
work p pressure ?V change in volume Vfinal
- Vinitial
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ENTHALPY (H)
Expansion of Volume - ?V is a positive quantity
and w is a negative quantity - Energy leaves the
system as work - Work is done by the system on
the surroundings Compression of Volume - ?V is
a negative quantity and w is a positive
quantity - Energy enters the system as work -
Work is done on the system by the surroundings
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ENTHALPY (H)
Calculate the work associated with the expansion
of a gas from 32 L to 58 L at a constant pressure
of 12 atm w - P?V w - (12 atm)(58 L - 32
L) - 310 L.atm Gas expands hence work is done
by system on surroundings
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ENTHALPY (H)
Calculate the work associated with the
compression of a gas from 58 L to 32 L at a
constant pressure of 12 atm w - P?V w -
(12 atm)(32 L - 58 L) 310 L.atm Gas compresses
hence work is done on system by surroundings
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ENTHALPY (H)
H E PV H, E, P, and V are all state
functions Change in Enthalpy ?H ?(E PV)
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ENTHALPY CHANGE (?H)
Change in Enthalpy at Constant Pressure ?H ?E
P?V ?H (qp w) - w qp qp heat at
constant pressure ?E q w P?V - w
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ENTHALPY CHANGE (?H)
?H qp Change in enthalpy heat gained or lost
at constant pressure Positive ?H - System gains
heat from the surroundings - Endothermic
process Negative ?H - System releases heat to
the surroundings - Exothermic process
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ENTHALPY CHANGE (?H)
?H Hfinal - Hinitial - Enthalpy change is a
state function - Enthalpy is an extensive
property
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ENTHALPY CHANGE (?H)
Standard Enthalpy Change (?Ho) - When all
reactants and products are in their standard
states Standard State - Pure form of a
substance at standard temperature and pressure
(STP) Conditions of STP - Standard
temperature 273 K or 0 oC - Standard pressure
1.00 atm (101.325 kPa or 100 kPa)
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ENTHALPY CHANGE (?H)

• Enthalpy of Reaction (?Hrxn)
• ?H accompanying a chemical reaction
• Enthalpy of Formation (?Hf)
• - ?H for forming a substance from its component
  elements
• Enthalpy of Vaporization
• - ?H for converting liquids to gases
• Enthalpy of Fusion
• - ?H for melting solids
• Enthalpy of Combustion
• - ?H for combusting a substance in oxygen

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ENTHALPY OF REACTION (?Hrxn)
- Heat of reaction - Enthalpy change
accompanying a chemical reaction ?Hrxn
Hproducts Hreactants ?Horxn standard
enthalpy of reaction ?Horxn Hoproducts
Horeactants Thermochemical Equation - A chemical
equation for which ?H is given
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ENTHALPY OF REACTION (?Hrxn)
Consider combustion of methane CH4(g) 2O2(g) ?
CO2(g) 2H2O(l) ?H -890 kJ ?H is an
extensive property (depends on amount) From
balanced equation 1 mol CH4 reacts with 2 mol O2
to release 890 kJ of heat Similarly 3 mol CH4
reacts with 6 mol O2 to release (3 x 890 kJ) of
heat
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ENTHALPY OF REACTION (?Hrxn)
For the reverse reaction ?H is equal in magnitude
but opposite in sign CO2(g) 2H2O(l) ? CH4(g)
2O2(g) ?H 890 kJ ?H depends on state of
the reactants and products For instance,
enthalpy of H2O(g) gt enthalpy of H2O(l) Heat is
absorbed when converting from liquid to gas
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ENTHALPY OF REACTION (?Hrxn)
Consider the following reaction CH4(g) 2O2(g) ?
CO2(g) 2H2O(l) ?H -890 kJ/mol Calculate the
enthalpy change if 2.0 g methane is burned in
excess oxygen When 1 mol of methane is burned,
890 kJ of heat is released
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CALORIMETRY
- Measurement of heat flow - Device used to
measure heat flow is the calorimeter
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CALORIMETRY
Heat Capacity - Amount of heat required to raise
the temperature of a substance by 1 K (or 1 oC)
Units cal/K (cal/oC) or J/K (J/oC) 1 cal
4.184 J cal calorie and J Joule - Greater
heat capacity requires greater amount of heat to
produce a given temperature increase
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CALORIMETRY
Specific Heat Capacity (Cs) - The quantity of
heat energy necessary to raise the temperature
of 1 gram of a substance by 1 K (or 1 oC)
Units cal/g.K (cal/g.oC) or J/g.K
(J/g.oC) Calorie - The amount of heat energy
needed to raise the temperature of 1 gram of
water by 1 Kelvin (or 1 degree Celsius) Heat
mass(g) of solution x specific heat of
solution x ?T q mCs ?T
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CALORIMETRY
Constant-Pressure - Heat lost by reaction
(qrxn) heat gained by solution (qsoln) - qrxn
qsoln qsoln mass(g) of solution x
specific heat of solution x ?T - Specific
heat of dilute aqueous solutions are
approximately equal to that of water (4.18
J/g.K)
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CALORIMETRY
Constant-Volume (Bomb Calorimetry) - Mostly
used to study combustion reactions - Heat
capacity of the calorimeter (Ccal) is first
determined using a substance that releases a
known quantity of heat qrxn - Ccal x ?T
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CALORIMETRY
Calculate the amount of heat needed to increase
the temperature of 140 g of a solution from 28 oC
to 72 oC. Take the specific heat capacity of the
solution as that of water, 4.18 J/g.K
qsoln mCs ?T ?T (72 - 28) oC 44 oC 44
K q (4.18 J/g.K)(140 g)(44 K) 26,000 J
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CALORIMETRY
When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of
0.100 M HCl are mixed in a constant-pressure
calorimeter, the temperature of the mixture
increases from 22.20 oC to 23.11 oC as per the
following reaction AgNO3(aq) HCl(aq) ?
AgCl(s) HNO3(aq) Calculate ?H (kJ/mol) for the
reaction if the combined solution has a mass of
100.0 g and a specific heat of 4.18 J/g.oC
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CALORIMETRY
?T (23.11 - 22.20) oC 0.91 oC - qrxn
qsoln Solution temperature increases so qrxn is
negative (exothermic) qrxn - (4.18
J/g.oC)(100.0 g)(0.91 oC) - 380 J - 0.38
kJ mol AgNO3 (0.100 M)(0.0500 L) 0.00500
mol ?H (-0.38 kJ)/(0.00500 mol) -76 kJ/mol
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CALORIMETRY
When a 0.5865-g sample of lactic acid (HC3H5O3)
is burned in a bomb calorimeter whose heat
capacity is 4.812 kJ/oC, the temperature
increases from 23.10 oC to 24.95 oC. Calculate
the heat of combustion of lactic acid (a) per
gram and (b) per mole ?T (24.95 - 23.10) oC
1.85 oC qrxn - Ccal x ?T - (4.812
kJ/oC)(1.85 oC) - 8.90 kJ (a) ?H (8.90
kJ)/(0.5865 g) - 15.2 kJ/g (b) ?H (15.2
kJ/g)(90.09 g/mol) - 1370 kJ/mol
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HESSS LAW
If a reaction is carried out in a series of steps
- Enthalpy change for the overall reaction will
equal the sum of the enthalpy changes for the
individual steps - Overall ?H is independent of
the number of steps - Overall ?H is independent
of the reaction path - Useful for calculating
energy changes that are very difficult to
measure
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HESSS LAW
Useful Hints - Work backward from the required
reaction, manipulating reactants and products
of any given reactions - Reverse any reactions
as needed - Multiply reactions by appropriate
factors as needed - Use trial and error but
allow the final reaction to guide you
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HESSS LAW
To determine ?H for A ? 2C, given A ? 2B ?HA C
? B ?HC Reverse second reaction and multiply
by 2 A ? 2B ?HA 2B ? 2C -2?HC A ? 2C ?H
?HA -2?HC
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HESSS LAW
Calculate the ?H for the conversion of graphite
to diamond Cgraphite(s) ? Cdiamond(s) using the
following combustion reactions Cgraphite(s)
O2(g) ? CO2(g) ?H -394 kJ Cdiamond(s) O2(g)
? CO2(g) ?H -396 kJ Reverse the second
reaction and sum the 2 reactions Cgraphite(s)
O2(g) ? CO2(g) ?H -394 kJ CO2(g) ?
Cdiamond(s) O2(g) ?H 396
kJ Cgraphite(s) ? Cdiamond(s) ?H 2 kJ
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ENTHALPY OF FORMATION (?Hf)
- ?H for forming a substance from its component
elements Standard Enthalpy of Formation (?Hfo)
- ?Hf when all substances are in their standard
states - Most stable form of elements are used
for elements existing in more than one form
under standard conditions (O2 is used for
oxygen, H2 is used for hydrogen) - ?Hfo of the
most stable form of any element is zero
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ENTHALPY OF FORMATION (?Hf)
n and m are stoichiometric coefficients
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ENTHALPY OF FORMATION (?Hf)
Calculate the standard change in enthalpy for the
thermite reaction in which a mixture of powdered
aluminum and iron(III) oxide is ignited with a
magnesium fuse as shown below 2Al(s) Fe2O3(s) ?
Al2O3(s) 2Fe(s) ?Hfo for Fe2O3(s) -826
kJ/mol ?Hfo for Al2O3(s) -1676 kJ/mol ?Hfo for
Al(s) ?Hfo for Fe(s) 0
?Hfo for Al2O3(s) - ?Hfo for Fe2O3(s) -1676 -
(-826 kJ) -850 kJ
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SOURCES OF ENERGY
Foods C6H12O6(s) 6O2(g) ? 6CO2(g)
6H2O(l) ?Ho -2803 kJ Glucose from
carbohydrate reacts with oxygen to produce
carbon dioxide water and energy in our
bodies
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SOURCES OF ENERGY
Fuels Fossil Fuels - Carbon or hydrocarbons
found in the earths crust - Major sources of
energy Coal (solid) - High molecular weight
hydrocarbons (most abundant) Petroleum (liquid)
- Composed of many compounds and mostly
hydrocarbons Natural Gas - Composed of gaseous
hydrocarbons
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SOURCES OF ENERGY
Fuels Fossil Fuels - Carbon or hydrocarbons
found in the earths crust - Major sources of
energy Hydrocarbons - Compounds of carbon and
hydrogen - The greater the percentage of carbon
and hydrogen in a fuel, the higher its fuel value
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SOURCES OF ENERGY
Nuclear Energy - generated from splitting or
fusion of the nuclei of atoms Nonrenewable
Sources of Energy - Limited resources and rate of
consumption is much greater than rate of
regeneration - Fossil fuel and nuclear energy
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SOURCES OF ENERGY
Renewable Sources of Energy - Unlimited
resources and are inexhaustible Solar energy
from the sun Wind energy from wind mills
Geothermal energy from heat stored in the earth
Biomass from crops and biological waste
Hydroelectric energy from rivers