Zumdahl Ch 6
AP Chemistry
1


From Greek therme (heat); study of
energy changes in chemical reactions
Energy: capacity do work or transfer
heat

Joules (J) or calories (cal); 1 cal = 4.184 J
1.
Kinetic: energy of motion; dependent on
mass & velocity


Applies to motion of large objects &
molecules
Linked to thermal energy (object’s T above 0
K)
2
James Prescott
Joule
(1818-1889)
1
KE  mv
2
2
Potential: stored in “fields” (gravitational and
electrical/magnetic); dependant on position
relative to another object
2.


Applies to large objects where gravity is overriding
force, but not significantly to molecules where gravity is
negligible and electrostatic forces dominate
Associated with chemical energy; stored in arrangement
of atoms or subatomic particles (electrostatic & nuclear
forces, bonding between atoms)
PE  mgh
3





Force: a push or pull on an object
Work: energy transferred to move an object a
certain distance against a force: W = (F)(d)
Heat: energy transferred from a hotter object to a
colder one
Temperature: property that reflects the random
motions (kinetic energies) of particles
State Function/Property: only depends on the
present state, does not depend on past or future.
Pathway independent

Energy is a state function, work and heat are not.
4

System –the part of
the universe on
which you focus
your attention


Closed system: easiest to
study because exchanges
energy with surroundings
but matter is not
exchanged.
Surroundings –
include everything
else in the universe

Universe – System
and the

“0th Law”: 2 systems are in thermal equilibrium
when they are at the same T.


1st Law: Energy can be neither created nor
destroyed, or, energy is conserved.


Thermal equilibrium is achieved when the random
molecular motion of two substances has the same
intensity (and therefore the same T.)
energy gained by surroundings = energy lost by
system
2nd and 3rd Laws: discussed later
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
Includes:




Translational motion
Rotational motion of particles through space
Internal vibrations of particles.
It is difficult to measure all E, so the change in
internal energy (DE) is typically measured:
DE = Efinal - Einitial
DE > 0 Increase in energy of system (gained
from surroundings)
DE < 0 Decrease in energy of system (lost to
surroundings)
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First Law of Thermodynamics

When a system undergoes a chemical or
physical change, the change in internal energy
(E) is equal to the heat (q) added or liberated
from the system plus the work (w) done on or
by the system:
DE = q + w
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q>0
Heat is added to system
q<0
Heat is removed from system
(into surroundings)
w > 0 Work done to system
w < 0 System does work on
surroundings
Always need a quantity and a sign (+/-)
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

Heat flowing into a system from its
surroundings is defined as positive; q has a
positive value
The system gains heat as the surroundings cool
down


Heat flowing out of a system into its
surroundings is defined as negative; q is a
negative value
The system loses heat as the surroundings heat
up.
Calculate ΔE for a system undergoing an
endothermic process in which 15.6 kJ of heat
flows and where 1.4kJ of work is done on the
system.
DE = q + w
q = +15.6 kJ (endothermic)
w = +1.4 kJ (work done on the system)
DE = 15.6 kJ + 1.4 kJ = + 17.0 kJ gained by system

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




P = F/A (F = P x A)
Work = force x distance = F x Δh
Work = F x Δh = P x A x Δh
ΔV = (Vf –Vi) = A x Δh
w = PΔV


Need to have opposite signs b/c when a gas expands
(ΔV is positive), work flows into the surroundings
(w is negative)
w = - PΔV
13

Octane and oxygen gases combust within a closed
cylinder in an engine. The cylinder gives off 1150 J
of heat and a piston is pushed down by 480 J
during the reaction. What is the change in internal
energy of the system?

q is (-) since heat leaves system; w is (-) since work is done
by system. Therefore,
DE = q + w = (-1150 J) + (-480 J) = - 1630 J

1630 J has been liberated from the system (C8H18 and O2)
added to the surroundings (engine, atmosphere, etc.)
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Calculate the work associated with the
expansion of a gas from 46 L to 64 L at a
constant external pressure of 15 atm.
w = - PΔV
P = 15 atm
ΔV = 64 L – 46 L = 18 L
w = - 15 atm x 18 L = -270 Latm
Gas expands, so it does work on the surroundings

1 Latm = 101.3 J
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
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf
E
E
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
Property of a system that is determined by specifying
its condition or its “state”

The value of a state function depends only on its
present state and not on the history of the sample.

T & E are state functions.
Consider 50 g of water at 25°C: EH2O does not depend on
how the water got to be 25°C (whether it was ice that
melted or steam that condensed or…)

Work (w) and heat (q) are not state functions because
the ratio of q and w are dependent on the scenario.
Consider the combustion of gasoline in a car engine vs.
burning in the open.
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

Measurement of heat flow
Heat capacity, C: amount of heat required to raise T
of an object by 1 K (or 1 °C)
q = C DT

Specific heat (or specific heat capacity, c): heat
capacity of 1 g of a substance
q = m c DT
Ex: How much energy is required to heat 40.0 g of
iron (c = 0.45 J/(g °C) from 0.0ºC to 100.0ºC?
q = m c DT = (40.0 g)(0.45 J/(g °C))(100.0 – 0.0 ºC)
= 1800 J
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


Calorimetry is an experimental technique used to
measure the heat transferred in a physical or
chemical process.
The apparatus used in this procedure is called as a
“Calorimeter”. There are two types of calorimetersConstant pressure (coffee cup) and constant volume
(bomb calorimeter). We will discuss constant
pressure calorimeter in detail.
Constant Pressure Calorimeter: The coffee cup
calorimeter is an example of this type of
calorimeter. The system in this case is the
“contents” of the calorimeter and the surroundings
are cup and the immediate surroundings.
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During the rxn:
qrxn + q solution = 0

where qrxn is the heat gained or lost in the
chemical reaction and qsolutionis the heat gained
or lost by solution. Heat exchange in this
system (qrxn), is equal to enthalpy change.
Assuming no heat transfer takes place between
the system and surroundings, qrxn + q solution =
0
chemlab.truman.edu
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

Since most reactions occur in containers open to
the air, w is often negligible. If a reaction produces
a gas, the gas must do work to expand against the
atmosphere. This mechanical work of expansion is
called PV (pressure-volume) work.
Enthalpy (H): change in the heat content (qp) of a
reaction at constant pressure
H = E + PV
DH = DE + PDV (at constant P)
DH = (qp + w) + (-w)
DH = qp
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Sign conventions
DH > 0
Heat is gained from surroundings
+ DH in endothermic reaction

DH < 0
Heat is released to surroundings
- DH in exothermic reaction
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chemlab.truman.edu
23
Sample Problem #1:
200. ml of 0.400 M HCl is mixed with the same
amount and molarity of NaOH solution, inside a
coffee-cup calorimeter. The temperature of the
solutions before mixing was 25.10 oC, and 27.78 oC
after mixing and letting the reaction occur. Find the
molar enthalpy of the neutralization of the acid,
assuming the densities of all solutions are 1.00 g/ml
and their specific heat capacities are 4.18 J/(g * K).
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Sample Problem #2:
1. 0.500g of magnesium chips are placed in a
coffee-cup calorimeter and 100.0 ml of 1.00 M HCl
is added to it. The reaction that occurs is:
Mg (s) + 2HCl (aq)  H2 (g) + MgCl2 (aq)
The temperature of the solution increases from
22.2oC to 44.8 oC. What’s the enthalpy change for
the reaction, per mole of Mg? (Assume specific
heat capacity of solution is 4.18 J/(g * °C) and
the density of the HCl solution is 1.00 g/ml.)
Ans. -7.60 x 105 J/mol Mg
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Sample Problem #3:
1. 5.00 g of a metal is heated in an oven to 100°C. It is
then placed in 100. g of H2O at 25.0°C. The final temp
of the solution is 30.0 °C. Calculate the specific heat of
the metal.
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
Also called heat of reaction:
1.
Enthalpy is an extensive property (depends on amounts
of reactants involved).
Ex: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
DHrxn = - 890. kJ

Combustion of 1 mol CH4 produces 890. kJ
… of 2 mol CH4 → (2)(-890. kJ) = -1780 kJ
What is the DH of the combustion of 100. g CH4?
 100. g CH4  1 mol CH4  - 890. kJ 


 5550 kJ





1

 16.0426 g CH4  1 mol CH 4 
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2.
DHreaction = - DHreverse reaction
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
DH = - 890. kJ
CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g)
DH = +890. kJ
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3.
DHrxn depends on states of reactants and products.
CO2 (g) + 2 H2O (g) CH4 (g) + 2 O2 (g) DH = 802 kJ
2 H2O (l)  2 H2O (g)
DH = 88 kJ
So:
CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g) DH = 890. kJ
CH4(g) + 2 O2(g)
CH4 (g) + 2 O2 (g)
802
kJ CO (g) + 2 H O (g)
2
2
2 H2O (g)
88
2 H2O (l)
kJ
890.
kJ
CO2(g) + 2 H2O(l)

If a rxn is carried out in a series of steps,
DHrxn =  (DHsteps) = DH1 + DH2 + DH3 + …Germain Hess
(1802-1850)
Ex. What is DHrxn of the combustion of propane?
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
3 C (s) + 4 H2 (g)  C3 H8 (g)
C3H8 (g)  3 C (s) + 4 H2 (g)
DH1 = -103.85 kJ
DH1 = +103.85 kJ
3[C (s) + O2 (g)  CO2 (g) ]
DH2 = 3( -393.5 kJ )
4[H2 (g) + ½ O2 (g)  H2O (l) ]
DH3 = 4( -285.8 kJ )
DHrxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ
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

Formation: a reaction that describes a substance
formed from its elements
Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g)  2 NH4NO3 (s)
Standard enthalpy of formation (DHf): forms 1
mole of compound from its elements in their
standard state (at 298 K)
2 C (graphite) + 3 H2 (g) + ½ O2 (g)  C2H5OH (l)
DHf = - 277.7 kJ

DHf of the most stable form of any element equals zero.
H2, N2 , O2 , F2 , Cl2 (g)
Br2 (l), Hg (l)
C (graphite), P4 (s, white), S8 (s), I2 (s)
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32
The complete
(long) way to
calculate the
ENTHALPY of any
reaction is to take
apart the reactant
molecules piece by
piece to elements,
and then build the
product molecules
from all of the
pieces, bond by
bond.
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A GREAT SHORTCUT exists however:
To estimate the
enthalpy of a
reaction ΔRH ° we
need only to know
the enthalpies of
formation of the
reactants and the
products.
(we calculate the
large RED arrow by
the difference
between the known
small GREEN arrows)
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An example: The combustion of liquid benzene :
Firstly, we need to define the overall reaction, and
balance it to 1 mole of the molecule we’re interested in :
1 C6H6 (l) + 15/2 O2 (g)  6 CO2 (g) + 3 H2O (l)
= ???
ΔRH ° =
ΔfH ° { products }
-
ΔRH °
ΔfH ° { reactants }
Secondly, we need to look up the ΔfH ° values
corresponding to both the reactants and the products,
and multiply by the moles:
ΔfH ° { products }
= 6 ( - 394 kJ/mol ) + 3 (
- 286 kJ/mol )
= - 3222 kJ/mol
to form the products
ΔfH ° { reactants } = ( + 49 kJ/mol )
+
15/2 ( zero )
=
+ 49
kJ/mol to form the reactants
----------------------------------------------Overall reaction ΔRH ° = - 3268 kJ/mol
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36
Enthalpies of combustion ΔCH ° are also commonly
tabulated :
( converted into enthalpies per gram, for
convenience )
methane
octane
dodecane
methanol
ΔCH °
ΔCH °
ΔCH °
ΔCH °
=
=
=
=
- 55 kJ/g
- 48 kJ/g
- 51 kJ/g
- 23 kJ/g
glucose
carbohydrates
tristearin
ΔCH ° = - 16 kJ/g
ΔCH ° ~ - 18 kJ/g
ΔCH ° = - 38 kJ/g
human beings at age 20 need around 10,000 kJ per
day of energy from combustions (~12,000 kJ for men,
ΔH0rxn   n  ΔH0f (products) -  n  ΔH0f (reactants )
Ex. Combustion of propane:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Given:
Compound
C3H8 (g)
CO2 (g)
H2O (l)
H2O (g)
DHrxn (kJ/mol)
-103.85
-393.5
-285.8
-241.82
DHrxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)]
= - 2219.8 kJ
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
Amount of energy required to break a particular bond
between two elements in gaseous state. Given in kJ/mol.
Remember, breaking a bond always requires energy!


Bond enthalpy indicates the “strength” of a bond.
Bond enthalpies can be used to figure out DHrxn .
Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ?
 1 C-H & 1 Cl-Cl bond are broken (per mole)
 1 C-Cl & 1 H-Cl bond are formed (per mole)
DHrxn ≈  (Hbonds broken) -  (Hbonds formed)
Note: this is the “opposite” of Hess’ Law where
DHrxn = DHproducts – Dhreactants
Bond Enthalpy link
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Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) DHrxn = ?
Bond
C-H
H-Cl
C-C
Ave DH/mol
413
431
348
Bond
Ave DH/mol
Cl-Cl
242
C-Cl
328
C=C
614
DHrxn ≈  (Hbonds broken) -  (Hbonds formed)
DHrxn ≈ [(1(413) + 1(242)] – [1(328) + 1(431)]
DHrxn ≈ -104 kJ/mol
DHrxn = -99.8 kJ/mol (actual)
Note:
2 C-C
2(348) = 696 kJ
≠ 1 C=C
≠ 614 kJ
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*CH3(g) + H(g) + 2 Cl(g)
H
Absorb E,
break 1 C-H
and 1 Cl-Cl
bond
CH4(g) + Cl2(g)
DHrxn
Release E,
form 1 C-Cl
and 1 H-Cl
bond
CH3Cl (g) + HCl (g)
DHrxn =  (Hbonds broken) +  (- Hbonds formed)
DHrxn =  (Hbonds broken) -  (Hbonds formed)
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