Transcript Thermochemistry

Thermochemistry
• When energy is changed from one form to
another, all of the energy can be accounted for.
This is called the Law of Conservation of Energy.
• Heat is energy. If heat is released by a chemical
system, an equal amount of heat will be absorbed
by the surroundings
• (For example) - When your body breaks
down sugar completely, the body
releases the same amount of heat as
compared to burning the same amount
of sugar in a flame.
• Heat is transferred due to a temperature
difference. Faster molecules speed up the
slower molecules. This slows the faster
molecules.
• In an endothermic process, the system absorbs
heat from the surroundings. A process that
absorbs heat is a(n) endothermic process.
• In an exothermic process, the system
gives off heat to the surroundings.
• If you were to touch the flask in which
an endothermic reaction was occurring,
the flask would probably feel warmer
than before the reaction started.
• In an exothermic reaction, the energy
stored in the chemical bonds of the
reactants is greater than the energy
stored in the bonds of the products.
• As perspiration evaporates from your
skin, your body is cooled.
• With respect to your body, this process
is said to be exothermic.
• A calorie is the quantity of heat that raises the
temperature of 1 gram of pure water by 1C.
• 1 Calorie = 4.18 kJ of heat
• How many kJ of energy can be released
by a banana containing 150 Cal?
(1 Calorie = 4.18 kJ)
4.18 kJ
150 Cal 
 630 kJ
1 Cal
• An object's heat capacity is the amount
of heat required to change the
temperature of an object by exactly
1°C.
• The specific heat capacity or specific heat is the
heat divided by the mass and temperature change.
• Where c = specific heat Q = heat m = mass and
T = change of temperature
• The temperature of a 10. g sample of metal
changed from 25°C to 50°C when it
absorbed 500. J of heat. What is the specific
heat of this sample?
• c=?
• m = 10. g
Q
• T = 50°C-25°C = 25°C c 
• Q = 500. J
mT
500. J
Q
J
c

 2.0
mT 10. g  25C
g C
• Determine the specific heat of a
material if an 18 g sample absorbed
75 J as it was heated from 15°C to
40.°C.
• c=?
• m = 18 g
• T = 40.°C-15°C = 25°C
• Q = 75 J
75 J
Q
J
c

 0.17
mT 18 g  25C
g C
• What is the specific heat of a substance
if 2000. cal are required to raise the
temperature of a 300. g sample by
20.C?
• c=?
• Q = 2000. cal
• m = 300. g
• T = 20.C
2000. cal
Q
cal
c

 0.33
mT 300. g  20.C
g C
• What is the amount of heat required to
raise the temperature of 200.0 g of
aluminum by 10.C?
• (specific heat of aluminum = 0.21
cal/gC)
• Q=?
• m = 200. g
• c = 0.21 cal/gC
• T = 10.C
• Q = mcT
• Q = (200. g)(0.21 cal/gC)( 10.C) = 420 cal
• Water has the highest specific heat of
all common substances.
cal
c = 1.00
g°C
• The symbol H stands for the heat of
reaction for a chemical reaction.
• This is also called the change in
enthalpy. H = Q
• The heat content of a system is equal
to the enthalpy only for a system that is
at constant pressure.
• Calorimetry depends on the law of
conservation of energy.
• If 27.0 mL of water containing HCl is
mixed with 28.0 mL of water containing
NaOH in a calorimeter such that the
initial temperature of each solution was
24.0°C and the final temperature of the
mixture is 33.0°C, how much heat (in
kJ) is released in the reaction? Assume
that the densities of the solutions are
1.00 g/mL.
•
•
•
•
H = ?
m = 27.0 g + 28.0 g = 55.0 g
T = 33.0°C - 24.0°C = 9.0°C
c = 0.00418 kJ/g°C
 0.00418kJ 
9.0  C  2.1 kJ
H  mcT  55.0 g 
gC 

• A lead mass is heated and placed in a
foam cup calorimeter containing
40.0 mL of water at 17.0°C. The water
reaches a temperature of 20.0°C. How
many joules of heat were released by
the lead?
•
•
•
•
H = ?
m = 40.0 g
T = 20.0°C - 17.0°C = 3.0°C
c = 4.18 J/g°C
 4.18 J 
3.0  C  502 J
H  mcT  40.0 g 
 gC 
• Fusion = solid  liquid
• Solidification = liquid  solid
• Molar heat of fusion –
–heat required to melt a mole = Hfusion
• Molar heat of solidification –
–heat required to freeze a mole =Hsolid
•
Hfusion = -Hsolid
• Vaporization is liquid  vapor
• Condensation is vapor  liquid
• Molar heat of vaporization –
– heat required to vaporize a mole = Hvap
• Molar heat of condensation –
– heat required to condense a mole = Hcond
•
Hvap = -Hcond
• Given the equation
I2(s) + 62.4 kJ  I2(g)
H = +62.4 kJ
• How much heat, in kJ, is
released when 108 g of
water at O°C freezes to ice
at O°C if Hsolid
for water = -6.01 kJ/mol?
1 mol H 2O
6.01 kJ
108 g 

 36.1 kJ
18 g H 2O 1 mol H 2O
• How much heat is released in
the condensation of 27.0 g of
steam at 100°C to water at
100°C if
Hcond for water = -40.7 kJ/mol?
1 mol H 2O
40.7 kJ
27.0 g 

 61.0 kJ
18 g H 2O 1 mol H 2O
• How many grams of ice at 0°C
can be melted into water at 0°C
by the addition of 75.0 kJ of
heat?
Hfus for water = 6.01 kJ/mol
18 g H 2O 1 mol H 2O
75.0 kJ

 225 g
1 mol H 2O
6.01 kJ
• The heat of solution is the
amount of heat absorbed or
released when a solid
dissolves.
• If the molar heat of solution
of NaOH is -445.1 kJ/mol,
how much heat (in kJ) will
be released if 80.00 g of
NaOH are dissolved in
water?
1 mol NaO H
445.1 kJ
80.00 g NaO H

 890.2 kJ
40. g NaO H 1 mol NaO H
• Hess's law makes it possible to
calculate H for complicated chemical
reactions.
• Hess’s Law – If you add 2 or more
thermochemical equations to give a
final equation, then you also ADD the
heats of reactions to give the final heat
of reaction.
• Given the equation:
• C2H4 (g) + 3O2(g) + 2H2O(l) + 1411 kJ
• How much heat is released
when 8.00 g of O2 react?
1 mol O 2
1411kJ
8.00 g O 2 

 118 kJ
32. g O 2 3 mol NaO H
• What is the heat of reaction
(H) for the combustion
(with O2) of benzene, C6H6 to
form carbon dioxide and
water? Write the balanced
equation for the reaction.
• Standard heats of formation:
• C6H6 = +48.50 kJ
• O2 = 0.0 kJ
• CO2 = -393.5 kJ
• H20 = -285.8 kJ
• 2 C6H6 + 15 O2  12 CO2 + 6 H2O
• H + (2) (48.50 kJ) + (15)(0.0 kJ)
= (12) (-393.5 kJ) + (6) (-285.8 kJ)
• H = -3266.9 kJ