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A.P. Chemistry


Chapter 6

Thermochemistry

1
st

Law of Thermodynamics
-

total energy of the
universe is constant (p. 244)

Definitions and State Functions


System
-

a system is the part of the universe that
is being studied. In Chemistry, this is often just a
particular reaction (p. 243)


Surroundings
-

everything else in the universe
outside of the system (p. 243)


Open System
-

one where both energy and matter
can be transferred to and from the surroundings


Closed system
-

one where only energy can be
transferred to and from the surroundings



Isolated System
-

one where neither energy nor
matter can be transferred to and from the
surroundings


State Functions
-

those that depend only upon
the initial and final state of a substance. Examples
are
/
\
H,
/
\
S, and
/
\
G (p. 243)


Heat
-

the symbol used is q (p. 245)


Enthalpy
-

in
thermochemistry
, every substance is
said to have a heat content or enthalpy (H). Most
reactions involve an enthalpy change
/
\
H which
measures the heat transferred (p. 248) (Under
constant pressure, q =
/
\
H)


Problem:

Calculate the work done on the system when 6.0 L of a gas is
compressed to 1.0 L by a constant external pressure of 2.0 atm
.




(1.0 x 10
3

J; obtaining a positive value for work means that work is done on
the system by the surroundings. A positive work value means the system
gains energy.)

Problem:

A gas, initially at a pressure of 10.0
atm

and having a volume of 5.0
L is allowed to expand at constant temperature against a constant external
pressure of 4.0
atm

until the new volume is 12.5 L. Calculate the work
done by the gas on the surroundings.











(w =
-
30 L
-
atm

=
-
3.0 x 10
3

J)






Problem
: A gas is allowed to expand at constant
temperature from a volume of 10.0 L to 20.0 L
against an external pressure of 1.0 atm. If the
gas also absorbs 250 J of heat from the
surroundings. What are the values of q, w, and
/
\
E?




(q = 250 J

w

=
-
1.0 x 10
3

J

/
\
E = q + w =
-
750 J)

Energy is measured in joules (J). Another unit of
energy you may come across is the calorie
(cal).



1.000 calorie = 4.184 J


1kilocalorie = 1000 calories = 1 Cal


1 kilojoules = 1000 joules

Example
: Convert 1 joule to calories


1 J x 1 calorie/4.184 J =


Specific Heat Capacity
The specific heat capacity of a
substance is defined as the amount of energy required
to change the temperature of one gram of a substance
by one degree Celsius. A useful relationship to use is




Energy Change = (specific heat
capacity)(mass)(temperature change)





= mc
/
\
T (
/
\
T =
Temp.
final



temp.
initial
)

The unit for specific heat capacity is usually
J/
g
o
C
. The
specific heat for water is designated as
c
p
, and is
4.184
J/
g
o
C
, or
1 cal/
g
o
C
. Generally, specific heat capacity has
the symbol of
c

for other substances.


Example
: The specific heat capacity for
aluminum is 0.900 J/g
-
o
C

Calculate the energy needed to raise the
temperature of 8.50 x 10
2

gram block of
aluminum from 22.8
o
C to 94.6
o
C.


q = (8.50 x 10
2

g)(0.900 J/g
o
C
)(94.6
o
C


22.8
o
C) =


Example
: Calculate the molar heat capacity of
aluminum.


0.900 J/g
o
C

x 27 g/ mol =


Enthalpy (p. 248)


In
thermochemistry
, every substance is said to have a heat content or
enthalpy. Enthalpy is given the symbol,
H
. Most reactions involve an
enthalpy change,
/
\
H


(delta H), where:





/
\
H =
H
products


H
reactants





/
\
H
o
reaction

=
n
p
/
\
H
o
f
products
-

n
p
/
\
H
o
f
reactants



(
n = # of moles
)


(Data for
/
\
H
o
f
is in Appendix 4, p. A21
-
A24
)


Enthalpy is a molar dependent function
.


Example
: problem #33. 283.






Problem:

The
thermochemical

equation for the combustion of propane is:


C
3
H
8
(g) + 5O
2
(g)


3 CO
2
(g) + 4H
2
O(l)
/
\
H
o

rxn

=
-
2220 kJ/mol

How many kJ of heat are released when 0.50 mol of propane reacts?

How much heat is released when 88.2 g of propane reacts?








(q =
-
1.1 x 10
3

J; q =
-
4440 kJ)

Problem:

Determining the Calorimeter Constant

The combustion of benzoic acid is often used as a standard source of heat for calibrating combustion
bomb calorimeters. The heat of combustion of benzoic acid has been accurately determined to be
26.42 kJ/g. When 0.8000 g of benzoic acid was burned in a calorimeter containing 950 g of water, a
temperature rise of 4.08
oC

was observed. What is the heat capacity of the bomb calorimeter (the
calorimeter constant)?













(
C
cal


= 1.21 x 10
3

J/
o
C
)

Enthalpy Level Diagrams


Endothermic
-

/
\
H is positive, enthalpy of products > enthalpy of reactants.
Energy must be put into the reaction for it to occur. (beaker gets cold)(p.
243)






















Exothermic
-

/
\
H is negative, enthalpy of products < enthalpy of reactants.
Energy is released from the reaction as it occurs. (beaker gets hot)(p. 243)






When discussing the enthalpy of a substance, it is
necessary to state the conditions under which the
enthalpy is measured. Usually enthalpy changes
are stated under standard conditions (p. 260,
table), i.e.,


Gases at 1
atm

pressure


All solutions at unit concentration (1 M)


All substances in their normal physical state at
standard temperature and pressure


Standard temperature (273 K)



A reaction which is
spontaneous

will tend toward
conditions of lower enthalpy, more negative
values of
/
\
H.


Standard Enthalpy of Formation (
/
\
H
f
o
)(p. 260)
-

defined
as the enthalpy change when one mole of a substance
is formed from its elements, in their standard form,
under standard conditions.



2C(graphite) + 3H
2
(g) + ½ O
2
(g)


C
2
H
5
OH(l)
/
\
H
f
o

=
-
279 kJ/mol



Practice:

CH
3
Br(l
)

CH
3
COC
2
H
5
(l)

NaNO
3
(s)



/
\
H
f
o

may be negative or positive. If the enthalpy change
is negative, then the energy is released and the
reaction is exothermic. If it is positive, then energy is
taken in and it is endothermic
.

Example
: Problem #61, 285







Example
: problem #67, p. 286





Problem:

Using
/
\
H
o
f

values in the back of your textbook, calculate the
standard enthalpy change for the incomplete combustion of
ethane:


C
2
H
6
(g) + 5/2 O
2
(g)


2CO(g) + 3 H
2
O(l)



(
/
\
H
rxn

=
-
993.7 kJ/mol)(note: this problem and the values used to
solve it were obtained from another book; the values may vary
slightly.)




Standard Enthalpy of Combustion(
/
\
H
c
o
)
-

defined as the
enthalpy change when one mole of a substance is
completely burned in oxygen under standard
conditions. Since energy is released in such a reaction,
/
\
H
c
o

will usually be negative.





C
2
H
6
(g) + 3 ½ O
2
(g)


2CO
2
(g) + 3H
2
O(l)



/
\
H
c
o

=
-
1565 kJ/mol



Remember that the combustion of hydrocarbons or
organic molecules in air (O
2
) will produce carbon
dioxide and water. This becomes more complicated for
other compounds.






Hess’s Law(p. 256)
-

the enthalpy change during a
reaction depends only on the nature of the
reactants and products and is independent of the
route taken.

In algebraically manipulating the equations for
Hess’s Law,

If a reaction is reversed, the sign of
/
\
H is also
reversed.

The magnitude of
/
\
H is directly proportional to the
quantities of reactants and products in a reaction.
If the coefficients in a balanced reaction are
multiplied by an integer, then the value of
/
\
H is
multiplied by the same integer.


Example
: problem #53, p. 285