Chapter 17 Notes 1 CHAPTER 17 Thermochemistry Thermochemistry The study of the heat changes that occur during chemical reactions and physical changes of state. Chemical Change : new substances created during chemical reaction Physical Change : same substances, different form. 17.1 The Flow of Energy Heat and Work • Energy is the capacity to do work or to supply heat. • Chemical potential energy is energy stored in chemicals because of their compositions. • Light, heat, electricity Energy is stored in the bonds. When our bodies burn glucose, the bonds are broken and energy is released. Glucose Molecule Heat, represented by q, is a form of energy that always flows from a warmer object to a cooler object. Law of conservation of energy States that in any chemical or physical process, energy is neither created nor destroyed. Energy is transferred...
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Chapter 17 Notes
1
CHAPTER 17Thermochemistry
ThermochemistryThe study of the heat changes that occur during chemical reactions and physical changes of state.
Chemical Change: new substances created during chemical reaction
Physical Change: same substances, different form.
17.1 The Flow of EnergyHeat and Work
• Energy is the capacity to do work or to supply heat. • Chemical potential energy is energy stored
in chemicals because of their compositions.• Light, heat, electricity
Energy is stored in the bonds.
When our bodies burn glucose, the bonds are broken and energy is released.
Glucose Molecule
Heat, represented by q, is a form of energy that always flows from a warmer object to a cooler object.
Law of conservation of energyStates that in any chemical or physical process, energy is neither created nor destroyed.
Energy is transferred...
Chapter 17 Notes
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Law of conservation of energy
• when energy is transferred, it is conserved.
• What happens when hot metal is put into cold water?
EXOTHERMIC AND ENDOTHERMIC PROCESSES
• A system is the specific part of the universe on which you focus your attention. (Always the Reactants!)
• The surroundings include everything outside the system. (Products)
• Together, the system and surroundings constitute the universe.
Physical ChangesEndothermic: surroundings cool down• Snow melting or Steam forming• Ice à Liquid à Vapor+ΔH = gaining heat (Heat into system.)
Exothermic : surroundings heat up• Steam condensing or Ice forming• Vapor àLiquid à IceΔH = losing heat (Heat out of system)
System
No Stored Energy Stored EnergyEndothermic
No Stored EnergyStored Energy Exothermic
Physical Change
Chemical Reaction
Endothermic: A + Energy à B + C• Bonds in substance B + C have more stored energy than the bonds in substance A.
Exothermic : A+ B à C + Energy• Bonds in substance A + B have more stored energy than the bonds in substance C.
• Process that absorbs heat from the surroundings is called endothermic. (Positive heat change)• Test tube feels colder
• A process that loses heat to the surroundings is called exothermic. (Negative heat change) • Test tube feels warmer
During a chemical change...
Chapter 17 Notes
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Chemical ChangeNo Stored
Energyin "A"
Stored Energyin B & C
Energyabsorbed from surroundingsA B + C
+
Nothing would happen to "A" if it didn't absorb energy.
Endothermic
No Stored Energyin "C"
Stored Energyin A & B
Energyreleased to surroundings
A + B C +
A & B react because they have stored energy, therefore, the energy is released into the surroundings.
Exothermic
Chemical Change
Chemical Reaction:
• System is the reactants
• Surroundings are the products
Physical Change:
• The system must be defined
• Hot iron in cold water...
• Iron is system=exothermic
• Water is system=endothermic
The calorie is also related to the joule, the SI unit of heat and energy.
• 1 cal = 4.18 J• 1000J = 1kJ• 1 kcal = 4186 J
Units for Measuring Heat Flow
• One calorie is the quantity of heat that raises the temperature of 1g of pure water 1°C.
Dietary Calories
• 1 “C” = 1 kilocalorie = 1000 calories• 10 grams of sugar has 41 calories• Means that when your body burns 10 grams of
sugar, it releases 41 calories of energy.
Chapter 17 Notes
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Heat CapacityThe amount of heat it takes to change an object’s temperature by exactly 1°C.
Depends upon mass and chemical composition.
How much heat can a given mass hold...
Kinetic Energy increases as temperature increases!
Heat Capacity100g of water has twice the heat capacity of 50g of water.
20g of lead has twice the heat capacity of 10g of lead.
Mass is important with regard to heat capacity!!!!
SPECIFIC HEAT CAPACITYThe specific heat capacity, or simply the specific heat, represented by “C”, of a substance is the amount of heat it takes to raise the temperature of 1g of the substance 1°C and is measured in J/(g x ºC).
• Unique for each substance• See table 17.1 on page 508.
SPECIFIC HEAT CAPACITY100g of water has the same specific heat capacity as 5g of water.
500g of aluminum has the same specific heat capacity as 2g of aluminum.
Mass does not matter!!!!
Heat Capacity vs. Specific HeatWater has a greater “C” than iron, regardless of the mass. Specific heat only refers to 1g.
Does 100g of Fe have a greater heat capactiy than 5g of water?
(C iron = 0.46 J/g x ºC) (C water = 4.18 J/g x ºC)
Heat Capacity vs. Specific Heat1000g of iron has a greater heat capacity than 5g of water. (C iron = 0.46 J/g x ºC)
(C water = 4.18 J/g x ºC)
1000g Fe
x 0.46 J
5g Water
x 4.18 J
Chapter 17 Notes
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Calculating Heat Change
• q = m x C x T
• ∆ H = m x C x T= change in
CALCULATING SPECIFIC HEAT• Specific Heat = “C”• Heat = “q”• Change in temperature = ∆Τ
heat (Joules or calories)mass (g) x change in temp. (ºC)
Calculating ∆T
• ∆T = Heat m x C
q:
m:
C:
∆T:
Example #1A piece of copper with a mass of 95.4 grams changes temperature from 25°C to 48 °C. The piece of copper absorbs 849 joules during the change. What is the specific heat of copper in J / g x °C? Convert your answer into cal /g x °C.
Example #2How many kilojoules are absorbed by water when 32.0 grams are heated from 25.0 ºC to 80.0 ºC? (Cwater = 4.18 J/g x ºC)
q:
m:
C:
∆T:
Example #3A 181gram chunk of silver has a heat capacity of 42.8 J/ ºC. Calculate the specific heat of silver using the information provided?
q:
m:
C:
∆T:
Chapter 17 Notes
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Example #4A 237 gram piece of iron was at 17 ºC. We added 1246 joules of heat to the iron. What is the new temperature? (C iron = 0.46 J/g x ºC)
q:
m:
C:
∆T:
Extra ExampleA 591.5 gram piece of iron was at 107.5ºC. We removed 5846 joules of heat from the iron. What was the original temperature? (C iron = 0.46 J/g x ºC)
q:
m:
C:
∆T:
You cannot convert ∆T into Fahrenheit.
Is there such a thing as a ∆T?The answer is no…however there is a ∆H.
If temperature goes down, it is a ∆H, but it is a +∆H if the temperature goes up.
17.2 Measuring and Expressing Enthalpy Changes
Calorimeters are devices used to measure the amount of heat absorbed or released during chemical or physical processes.
• Calorimetry is the accurate and precise measurement of the heat change for chemical and physical processes.
For systems at constant pressure, the heat content is the same as a property called the enthalpy, H, of the system.
∆H of the system''+" or ""
Chapter 17 Notes
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EnthalpyOne System vs. Two System
The heat change in one substance, you
determine q
The transfer of heat from one substance to another. You do
not determine q. One temperature increases, the
other decreases.
1 mass1 ∆T 1 specific heat
2 masses2 ∆T 2 specific heats
∆Η = negativeHeat lossTemp decrease
∆Η = positiveHeat GainTemp increase
Hot Iron70.0 °C
Cold Water8.0 °C
Example Problem #5 (One System)How much heat (in joules) is needed to raise 27.0 grams of water from 10.0 °C to 90.0 °C?
Heat (q) = mass x Cwater x ∆Tq:
m:
C:
∆T:
Example Problem #6 (Two different Systems)A 100 gram drinking glass decreased from 30.0 °C to 18.0 °C as 150 grams of water was added. What
was the original temperature of the water? (C glass = 0.5 J/g x ºC) q glass= q water
m x Cglass x ∆T = m x Cwater x ∆TGlass Water
q
mass
C
∆T
Example Problem #7 (Two different Systems)A 130 gram sample of iron is added to 100mL of water at 19.0 °C . The final temperature of the water/iron mixture is to 25.0 °C. What was the original temperature of the iron? (C iron = 0.46 J/g x ºC) (1mL = 1g)
q iron= q waterm x Cwater x ∆T = m x Ciron x ∆T
Iron Water
q
mass
C
∆T
This chart will help on 2 system Problems
Iron Water
TOriginal
TFinal
Chapter 17 Notes
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Extra Example Problem (Two different Systems)A sample of iron has a temperature of 93°C. Its temperature dropped 75°C after placing it in 120mL of water at 6°C. What is the mass of the iron? (C iron = 0.46 J/g x ºC) (1mL = 1g)
q iron = q waterm x Cwater x ∆T = m x Ciron x ∆T
Iron Water
q
mass
C
∆T
THERMOCHEMICAL EQUATIONS• Includes the heat change in equations.• A heat of reaction is the heat change for the equation exactly as it is written.• You can interpret whether the reaction is exothermic or endothermic by looking at the equation.
• Reactants are system, products are surroundings
• The physical state of the reactants and products in a thermochemical reaction must be stated.• The heat of combustion is the heat of reaction for the complete burning of one mole of a substance.• ∆H = Heat of Reaction
Endothermic: surroundings à systemC (s) + 2S (s) + Energy à CS2 (l)
• Positive ∆H (Heat Change)
Exothermic: system à surroundingsMg (s) + 2HCl (aq) à MgCl2 (aq) + H2 (g)+ Energy• Negative ∆H (Heat Change)
Example Problem #8How much heat is released when combining 25.0mL of 0.5 molar HCl (aq) with 25.0mL of 0.5 molar NaOH (aq)? During the reaction, there was a temperature change of 7 ºC. (Hint: aqueous is water) (1mL = 1g)
HCl+NaOH à H2O+NaCl+Energyq:
m:
C:
∆T:
Example Problem #9Using the equation below, calculate the amount of heat (in kJ) released by burning 24.0g of CH4.
CH4 + 2O2 àCO2 + 2H2O + 890 kJ (∆H = 890kJ)
Chapter 17 Notes
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Example Problem #10Using the equation below, calculate the amount of heat (in kJ) required to break down 152.0g of KClO3.
2KClO3 + 84.9kJ è 2KCl + 3O2 (∆H = +84.9kJ)
17.3 HEAT IN CHANGES OF STATE• The energy required for a phase change is different than the energy required to raise the temperature of a substance 1ºC.
Example Problem #11How much energy (in kJ) is absorbed when 25.0grams of water increases from water at 40 ºC to steam at 120 ºC?
Phase Change Diagram
50 ºC
0 ºC
100 ºC
150 ºC
TempRises
C ice = 2.10 J/g x ºC
C water = 4.18 J/g x ºC
C steam = 1.70 J/g x ºC
Solid à Liquid: 6.01 kJ/mol
Liquid à Steam: 40.7 kJ/mol
C ice = 2.10 J/g x ºC
C water = 4.18 J/g x ºC
C steam = 1.70 J/g x ºC
Liquid à Solid: 6.01 kJ/mol
Steam à Liquid: 40.7 kJ/mol
50 ºC
0 ºC
100 ºC
150 ºC
TempLowers
Phase Change Diagram • The heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature is the molar heat of fusion. • ∆ Hfus= 6.01 kj/mol (Water)
Chapter 17 Notes
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∆ Hfus Example ProblemHow much heat is gained (in kJ) when 57.6g of ice is melted completely?• H2O (s) à H2O (l) • ∆Hfus = 6.01 kJ/mol
• The heat lost when one mole of a liquid changes to a solid at a constant temperature is the molar heat of solidification.• ∆ H solid = 6.01 kj/mol (Water)
∆ Hsolid Example ProblemHow much heat is lost (in kJ) when 48.6g of water completely freezes?• H2O (l) à H2O (s) • ∆Hsolid = 6.01 kJ/mol
• The condensation of 1 mol of vapor releases heat as the molar heat of condensation, • ∆ H cond = 40.7 kj/mol (Water)
• The vaporization of 1 mol of vapor absorbs heat as the molar heat of vaporization.• ∆ H vap = 40.7 kj/mol (Water)
Phase Change Example ProblemA 50.0 gram piece of ice at 125 ºC changes into steam at 125 ºC. How much energy (in kj) is absorbed during this process?
Chapter 17 Notes
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Phase Change Diagram
50 ºC
0 ºC
100 ºC
150 ºC
TempRises
C ice = 2.10 J/g x ºC
C water = 4.18 J/g x ºC
C steam = 1.70 J/g x ºC
Solid à Liquid: 6.01 kJ/mol
Liquid à Steam: 40.7 kJ/mol
1. Ice @ 125 ºC ice @ 0 ºC
2. Ice @ 0 ºC water @ 0 ºC
3. Water @ 0 ºC water @ 100 ºC
4. Water @ 100 ºC steam @ 100 ºC
5. Steam @ 100 ºC steam @ 125 ºC
Do all phase change problems need 5 steps? Phase Change Example ProblemA 39.0 gram sample of steam at 112 ºC changes into ice at 7 ºC. How much energy (in kj) is released during this process?
Example ProblemA 56.8g piece of ice is exposed to 2.0 kJ of heat. Calculate the amount of ice (in grams) that should melt?
Molar Heat of Solution• The heat change caused by dissolution (dissolving) of one mole of substance is the molar heat of solution.• Grams à Moles à KJ• ∆ H soln
Example Problem• When 5g of solid NaOH is dissolved in water, how much heat is released?• ∆ H soln = 445.1 kj/mol
Potential Test Questions
CH4 + 2O2 àCO2 + 2H2O + 890 kJ• Is this exothermic or endothermic?• What is the ∆H?• How many moles of CH4 are needed to release 890kJ?• How many moles of O2 are needed to release 890kJ?• Energy released by 3.12 mol of CH4?• Energy released by 78.3g of O2?