11.2
Calorimetry
Calorimetry
• The Law of the Conservation of Energy– Energy is neither created nor destroyed in any physical or
Chemical Change
• But to study this requires an isolated system
– Three types of systems
– Open– matter and energy are easily transferred
– Closed– matter can not be transferred but energy is able to
transfer
– Isolated– no matter or energy is able to be transferred
Calorimetry
• Calorimetry is the technological process of
measuring energy changes using an isolated
system called a calorimeter
– Studying the energy loss or gain of a known
system of water as a system undergoes physical or
chemical change
Calorimetry
• Ex – Ice at its freezing point is placed in water.
The temperature of the water is measured
when all the ice is melted.
• As the ice melts it removes energy from the
water
Heat (energy) gained by the ice = Heat (energy) lost by the
water
• Calorimetry can be used to study both physical
changes (heat of melting, heat of freezing, etc) or the
chemical changes (the energy change when one
mole of ammonia is dissolved, the energy change
when 2 grams of octane is burnt)
• We will use it only to study the energy changes in
chemical changes
• So for energy chemical energy changes
H

energy gained or lost
by the chemical system
Q
energy gained or lost
by the water in the
calorimeter
• In a simple calorimeter experiment performed
involving a burning candle and can of water, the
temperature of 100 mL of water is increased from
16.4 to 25.2 when a candle burned forCseveral
C
minutes.
• Calculate the experimental value for the energy given
off by the candle.
c H  Q
c H  mct

J 
 c H  100 g   4.19
  25.2  16.4  C
g C



J
 c H  3687.2  g
C


g
C


 3.68 103 J  3.68kJ
• Notes on symbols
Symbol
Meaning
Q
Total energy change of a system
(calorimeter)
c H
Energy change of a system undergoing
combustion
r H
Energy change of a system undergoing the
chemical reaction shown
 sd H
Energy change of a system undergoing a
simple decomposition reaction
f H
Energy change of a system undergoing a
simple formation reaction
r H m
Energy change of a one mole of a substance
undergoing the chemical reaction shown
• Notes on symbols
Symbol
Meaning
Q
Total energy change of a system
(calorimeter)
c H
Energy change of a system undergoing
combustion
r H
Energy change of a system undergoing the
chemical reaction shown
 sd H
Energy change of a system undergoing a
simple decomposition reaction
f H
Energy change of a system undergoing a
simple formation reaction
r H m
Energy change of a one mole of a substance
undergoing the chemical reaction shown
• Example
• When 50 mL of 1.0 mol/L hydrochloric acid is
neutralized by 75 mL of 1.0 mol/L sodium hydroxide
in a polystyrene cup calorimeter, the temperature of
solution changes from 20.2 to 25.6 . Determine the
C
C
enthalpy change that occurs
in the chemical
system
c H  Q
t  25.6 C  20.2 C
m  50 g  75 g  125 g
J
c  4.19
gC
r H  mct
J
 r H  125 g *4.19
*(25.6  20.2) C
gC
 r H  2828.25J
 r H  2.83kJ
• Is this reaction a endothermic reaction or an
exothermic reaction. Explain?
• What does this mean for the energy stored in the
reactants compared to the energy stored in the
products?
• Molar Enthalpy change of reaction
• Defined as the energy change that takes place when
one mole of a substance takes place
• We do this because scientist often want to compare
the energy to an equation balanced in moles
• So to calculate the molar enthalpy
r H
r H m 
n
 kJ 
 r H m  molar enthalpy 
 of reaction
 mol 
 r H  change of enthalpy  kJ  of reaction
n  number of moles
 r H  n r H m
• Reads as the enthalpy of reaction is equal to the
chemical amount (moles) of substance in the
reaction times the molar enthalpy of reaction for the
substance
• The molar enthalpies are mostly found in text books
or given in the question
• Predict the change in enthalpy due to the
combustion of 10.0g of propane. The molar enthalpy
of combustion for propane is -2043.9 kJ/mol.
 r H  n r H m
 c H  ??
 c H m  2043.9
kJ
mol
n  ??
m  10.0 g
M  ??
m
n
M

kJ
 r H  0.226705...mol * 2043.9
mol
kJ
 r H  463.3643... mol *
mol
 r H  463kJ
10.0 g
44.11g / mol
 0.226705...mol
• What is the energy released when 55.0g of water is
formed from its elements?
• To find the molar enthalpy of formation locate the
table on pg.5 Of the data booklet.
 f H  n f H m
 c H  ??
kJ
 c H m  285.8
mol
n  ??
m  55.0 g
M  18.02 g / mol
m
n
M

55.0 g
18.02 g / mol
 3.05216...mol
kJ
 f H  3.05216...mol * 285.8
mol
kJ
 r H  872.3085461... mol *
mol
 r H  872kJ
Calorimetry Problems
• Since the basic principle in Calorimetry is that the
energy gained or lost by the water came from or left
the system (The chemical reaction)
• So
r H  Q
Energy gained/lost
by the chemical system
n r H m  mct
Energy gained/lost
by the water
• In a research laboratory, scientist perform an
experiment to compare the combustion of ethanol to
the combustion of octane.
• When 3.50g of ethanol is combusted in a
sophisticated calorimeter the temperature of 3.63L
19.88 C to 26.18 C
of water rises from
• Determine the molar enthalpy of Combustion for the
ethanol
water
ethanol
n  ???
J
gC
m  3.63kg
m  3.50 g
t  6.30 C
 c H m  ????
M  46.09
c  4.19
g
mol
c H  Q
nc H m  mct
3.50 g
J
3
 c H m  3.63 10 g * 4.19
*6.30 C
g
gC
46.09
mol
0.07593..
g
g
mol
 c H m  95821.11 g *
J
g C
* C
0.07593..mol * c H m  95821.11J
95821.11J
c H m 
0.07593..mol
J
 c H m  1261827.132
mol
MJ
 c H m  1.26
mol
• Since the temperature of water increased, the reaction is
exothermic. To indicate that we use a negative sign to show
that the system is losing energy (the reactants have more
energy than the reactants)
MJ
 c H m  1.26
mol
• Determine the final temperature of the calorimeter
from the last question containing 3.63L at 21.25 if
C
3.50 g of octane was burnt in its combustion
chamber. The molar enthalpy of combustion of
octane is
 c H m  5470.1
kJ
mol
c H  Q
Octane
 c H m  5470.1
kJ
mol
water
nc H m  mct
n  ???
m  3.50 g
g
M  114.26
mol
J
gC
m  3.63kg
c  4.19
t  ??? C
3.50 g
114.26
g
mol
167.559...
*5470.1
g
g
*
kJ
J
 3.63kg * 4.19
t
mol
gC
kJ
J
 15.2097k g *
* t
mol
g C
mol
167.559...kJ
 t
kJ
15.2097
C
tf
 11.01662185 C
 25.25 C  11.01662185 C
t f  36.27 C
• Calculate the experimental molar enthalpy when
1.50g of ethanol undergoes combustion in a precise
calorimeter and the temperature of the 1.00L of
water in the calorimeter increases from 22.3 C to 31.3 C
• What assumptions are made in this experiment?
• List the manipulated, controlled and responding
variables in this experiment
Ethanol
m  1.50 g
 c H m  ???
 c H  ??
M  ??
H  Q
nc H m  mct
me
 c H m  mwct
M
Water
m  1.00 103 g
c  ??
t  (31.3 C  22.3 C )  9.0 C
mwct
c H m 
me
M

J 
3
1.00 10 g   4.19 g C  (31.3 C  22.3 C )


c H m 
1.50 g
g
46.08 mol

J 
 (31.3 C  22.3 C )
1.00 10 g  4.19


g
C


c H m 
1.50 g

3

46.08
g
mol
J
 c H m  1158451
mol
kJ
 c H m  1.16 10
mol
3
• Is the combustion of ethanol an exothermic process
or an endothermic process. Explain.
• Determine the experimental error of the experiment
if the accepted value for the combustion of ethanol
is -1366.8kJ/mol
Exp Error 
Exp Error 
expermental value -accepted value
100%
accepted value
kJ
kJ
-1366.8
mol
mol 100%
kJ
1366.8
mol
1.16 103
 15.1%
Homework
•
•
•
•
Pg. 494
#3, 4, 5, 6, 7
Pg. 519
#16, 17, 18