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Thermochemistry
Definitions
• Energy – capacity for doing work or
supplying heat.
• Thermochemistry – study of energy
changes that occur during phase changes
and chem. rxns.
• Chem. Potential Energy – energy stored in
chemical bonds.
Example
5473 kJ/mol
Energy difference.
Lots of energy
stored in bonds!
Little energy stored
in bonds.
Heat
• Represented by q.
• Energy that transfers from one object to
another because of a Temp. difference
between them.
• Heat flows from warm  cool until the
two objects are at the same Temp.
Exothermic vs. Endothermic
• In exothermic processes, the system loses
heat as its surroundings warm up.
– q has a negative value b/c the system is losing
heat.
• In endothermic processes, the system gains
heat as its surroundings cool down.
– q has a positive value b/c the system is gaining
heat.
Potential Energy Diagram of Ice
Melting at 0ºC.
Time 
PotentialEnergy
Ice
Water
Is the melting of ice
an endothermic or
an exothermic
process? How can
you tell?
Measuring Heat Flow
• SI Unit of heat flow: Joule (J)
• Common unit used in chemistry: calorie
(cal)
– Amt. of heat needed to raise 1 gram of water
by 1ºC.
– 1 cal = 4.184 J
• Food Calorie (capital “C”) = 1000 cal, or 1
kilocalorie = 4184 J
What do Calories mean in food?
• 10 grams of sugar has 41 Calories.
– When 10 grams of sugar are burned, 41 kcal
(170 kJ) of energy are released.
– Your body “burns” food for energy.
– In order to use the energy available in 10
grams of sugar, you must do 41 kcal worth of
work.
Heat Capacity
• Amount of heat needed to raise an object’s
temperature by 1°C.
– Depends on the chemical composition and the
mass of the object.
• EXAMPLE: 1 gram of water requires 1
cal to raise its temperature by 1°C.
– 100. g of water require 100. cal to raise the
temp. by 1°C.
Heat Capacity
1 g H2O
10 g H2O
Same temperature change
Specific Heat (c)
• Amt. of heat needed to raise 1 gram of a
substance’s temperature by 1ºC.
– Expressed in J/g ºC, or cal/g ºC
• The higher a substance’s specific heat, the
more energy it takes to heat it.
• Substance’s with low specific heats heat up
and cool down quickly (most metals, e.g.)
Some Specific Heats
Substance Specific Heat
J/gºC cal/g ºC
Water 4.18 1.00
Grain alcohol 2.4 0.58
Ice 2.1 0.50
Steam 1.7 0.40
Chloroform 0.96 0.23
Aluminum 0.90 0.21
Iron 0.46 0.11
Silver 0.24 0.057
Mercury 0.14 0.033
Specific Heat (c)
• c = heat / (mass x change in Temp.)
• c = q / (m x ΔT)
• q = m x c x ΔT
Example Problem
• The temperature of a 95.4-g piece of Cu
increases from 25.0ºC to 48.0ºC when the
Cu absorbs 849 J of heat. What is the
specific heat of Cu?
– SOLUTION: q = m x c x ΔT
• 849 J = (95.4 g) c (48.0ºC – 25.0ºC)
• 849 J = (95.4 g) c (23.0ºC)
• 849 J = (2190 gºC) c
• c = 0.388 J/gºC
• Based on what you know about metals,
does this answer make sense?
Example Problem
• When 435 J of heat is added to 3.4 g of
olive oil at 21ºC, the temperature increases
to 85ºC. What is the specific heat of olive
oil?
– SOLUTION: q = m x c x ΔT
• 435 J = (3.4 g) c (85ºC – 21ºC)
• 435 J = (3.4 g) c (64ºC)
• 435 J = (220 gºC) c
• c = 2.0 J/gºC
Example Problem
• How much heat is required to raise the
temperature of 250.0 g of mercury by
52ºC? The specific heat of mercury is 0.14
J/gºC.
– SOLUTION: q = m x c x ΔT
– q = (250.0 g)(0.14 J/gºC)(52ºC)
– q = 1800 J = 1.8 kJ
Enthalpy Changes
• Enthalpy (H) – the heat content of a system
at constant pressure.
• Enthalpy change (ΔH) – the heat that
enters or leaves a system at constant
pressure.
• q = ΔH
• Neg. ΔH = exothermic process
• Pos. ΔH = endothermic process
Thermochemical Equations
• Enthalpy change can be written as a
reactant or a product.
– Reactant  endothermic
– Product  exothermic
• Example: The reaction of calcium oxide
with water is exothermic.
– It produces 65.2 kJ of heat per mole of CaO
reacted.
– CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ
An Exothermic Reaction
CaO(s) + H2O(l)
Ca(OH)2(s)
ΔH = -65.2 kJ
CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ
Thermochemical Equations
2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g)
2NaHCO3(s)
Na2CO3(s) + H2O(g) + CO2(g)
ΔH = +129 kJ
Thermochemical Equations and
Stoichiometry
• You can use thermochemical equations in
stoichiometry.
– How much heat energy is produced when 55.0 grams
of ethanol is burned completely?
• C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + 1300. kJ
• Given: 55.0 g C2H5OH
• Want: kJ
• Conversion factors:
– 1 mol C2H5OH produces 1300. kJ when burned
– 1 mol C2H5OH = 46.07 g/mol
55.0 g C2H5OH
OHHCg46.07
OHHCmol1
x
52
52
OHHCmol1
kJ1300.
x
52
= 1550 kJ
Thermochemical Equations and
Stoichiometry
• 0.500 grams of methane gas are burned completely beneath a
container that holds 100. grams of water, originally at 20.0º. If
all of the heat from the combustion reaction goes into the
water, what will the water’s final temperature be?
– CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 803 kJ
– First find out how much total heat is released.
– Given: 0.500 g CH4(g)
– Want: kJ
– Conversion factors:
• 1 mol CH4 = 16.05 g CH4
• 1 mol CH4 produces 803 kJ when completely burned
0.500 g CH4
4
4
CHg16.05
CHmol1
x
4CHmol1
kJ803
x = 25.0 kJ
Thermochemical Equations and
Stoichiometry
• The combustion of 5.00 grams of methane releases 250. kJ of heat.
– Now we’ll calculate how hot the water in the container will get if it
absorbs all of the heat.
– First convert 25.0 kJ to J
– q = m x c x ∆T
– 2.50x104
J = (100. g) (4.18 J/gºC) ∆T
– 2.50x104
J = (418 J/ºC) ∆T
∆T = 59.8ºC
– The water will get 59.8ºC warmer.
– The final temperature will be 20.0ºC + 59.8ºC = 79.8ºC.
25.0 kJ
kJ1
J1000
x = 2.50x104
J

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Thermochemistry

  • 2. Definitions • Energy – capacity for doing work or supplying heat. • Thermochemistry – study of energy changes that occur during phase changes and chem. rxns. • Chem. Potential Energy – energy stored in chemical bonds.
  • 3. Example 5473 kJ/mol Energy difference. Lots of energy stored in bonds! Little energy stored in bonds.
  • 4. Heat • Represented by q. • Energy that transfers from one object to another because of a Temp. difference between them. • Heat flows from warm  cool until the two objects are at the same Temp.
  • 5. Exothermic vs. Endothermic • In exothermic processes, the system loses heat as its surroundings warm up. – q has a negative value b/c the system is losing heat. • In endothermic processes, the system gains heat as its surroundings cool down. – q has a positive value b/c the system is gaining heat.
  • 6. Potential Energy Diagram of Ice Melting at 0ºC. Time  PotentialEnergy Ice Water Is the melting of ice an endothermic or an exothermic process? How can you tell?
  • 7. Measuring Heat Flow • SI Unit of heat flow: Joule (J) • Common unit used in chemistry: calorie (cal) – Amt. of heat needed to raise 1 gram of water by 1ºC. – 1 cal = 4.184 J • Food Calorie (capital “C”) = 1000 cal, or 1 kilocalorie = 4184 J
  • 8. What do Calories mean in food? • 10 grams of sugar has 41 Calories. – When 10 grams of sugar are burned, 41 kcal (170 kJ) of energy are released. – Your body “burns” food for energy. – In order to use the energy available in 10 grams of sugar, you must do 41 kcal worth of work.
  • 9. Heat Capacity • Amount of heat needed to raise an object’s temperature by 1°C. – Depends on the chemical composition and the mass of the object. • EXAMPLE: 1 gram of water requires 1 cal to raise its temperature by 1°C. – 100. g of water require 100. cal to raise the temp. by 1°C.
  • 10. Heat Capacity 1 g H2O 10 g H2O Same temperature change
  • 11. Specific Heat (c) • Amt. of heat needed to raise 1 gram of a substance’s temperature by 1ºC. – Expressed in J/g ºC, or cal/g ºC • The higher a substance’s specific heat, the more energy it takes to heat it. • Substance’s with low specific heats heat up and cool down quickly (most metals, e.g.)
  • 12. Some Specific Heats Substance Specific Heat J/gºC cal/g ºC Water 4.18 1.00 Grain alcohol 2.4 0.58 Ice 2.1 0.50 Steam 1.7 0.40 Chloroform 0.96 0.23 Aluminum 0.90 0.21 Iron 0.46 0.11 Silver 0.24 0.057 Mercury 0.14 0.033
  • 13. Specific Heat (c) • c = heat / (mass x change in Temp.) • c = q / (m x ΔT) • q = m x c x ΔT
  • 14. Example Problem • The temperature of a 95.4-g piece of Cu increases from 25.0ºC to 48.0ºC when the Cu absorbs 849 J of heat. What is the specific heat of Cu? – SOLUTION: q = m x c x ΔT • 849 J = (95.4 g) c (48.0ºC – 25.0ºC) • 849 J = (95.4 g) c (23.0ºC) • 849 J = (2190 gºC) c • c = 0.388 J/gºC • Based on what you know about metals, does this answer make sense?
  • 15. Example Problem • When 435 J of heat is added to 3.4 g of olive oil at 21ºC, the temperature increases to 85ºC. What is the specific heat of olive oil? – SOLUTION: q = m x c x ΔT • 435 J = (3.4 g) c (85ºC – 21ºC) • 435 J = (3.4 g) c (64ºC) • 435 J = (220 gºC) c • c = 2.0 J/gºC
  • 16. Example Problem • How much heat is required to raise the temperature of 250.0 g of mercury by 52ºC? The specific heat of mercury is 0.14 J/gºC. – SOLUTION: q = m x c x ΔT – q = (250.0 g)(0.14 J/gºC)(52ºC) – q = 1800 J = 1.8 kJ
  • 17. Enthalpy Changes • Enthalpy (H) – the heat content of a system at constant pressure. • Enthalpy change (ΔH) – the heat that enters or leaves a system at constant pressure. • q = ΔH • Neg. ΔH = exothermic process • Pos. ΔH = endothermic process
  • 18. Thermochemical Equations • Enthalpy change can be written as a reactant or a product. – Reactant  endothermic – Product  exothermic • Example: The reaction of calcium oxide with water is exothermic. – It produces 65.2 kJ of heat per mole of CaO reacted. – CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ
  • 19. An Exothermic Reaction CaO(s) + H2O(l) Ca(OH)2(s) ΔH = -65.2 kJ CaO(s) + H2O(l)  Ca(OH)2(s) + 62.5 kJ
  • 20. Thermochemical Equations 2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) ΔH = +129 kJ
  • 21. Thermochemical Equations and Stoichiometry • You can use thermochemical equations in stoichiometry. – How much heat energy is produced when 55.0 grams of ethanol is burned completely? • C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g) + 1300. kJ • Given: 55.0 g C2H5OH • Want: kJ • Conversion factors: – 1 mol C2H5OH produces 1300. kJ when burned – 1 mol C2H5OH = 46.07 g/mol 55.0 g C2H5OH OHHCg46.07 OHHCmol1 x 52 52 OHHCmol1 kJ1300. x 52 = 1550 kJ
  • 22. Thermochemical Equations and Stoichiometry • 0.500 grams of methane gas are burned completely beneath a container that holds 100. grams of water, originally at 20.0º. If all of the heat from the combustion reaction goes into the water, what will the water’s final temperature be? – CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 803 kJ – First find out how much total heat is released. – Given: 0.500 g CH4(g) – Want: kJ – Conversion factors: • 1 mol CH4 = 16.05 g CH4 • 1 mol CH4 produces 803 kJ when completely burned 0.500 g CH4 4 4 CHg16.05 CHmol1 x 4CHmol1 kJ803 x = 25.0 kJ
  • 23. Thermochemical Equations and Stoichiometry • The combustion of 5.00 grams of methane releases 250. kJ of heat. – Now we’ll calculate how hot the water in the container will get if it absorbs all of the heat. – First convert 25.0 kJ to J – q = m x c x ∆T – 2.50x104 J = (100. g) (4.18 J/gºC) ∆T – 2.50x104 J = (418 J/ºC) ∆T ∆T = 59.8ºC – The water will get 59.8ºC warmer. – The final temperature will be 20.0ºC + 59.8ºC = 79.8ºC. 25.0 kJ kJ1 J1000 x = 2.50x104 J