CHAPTER 17 Thermochemistry

Transcription

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 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... 1

2 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 Changes No Stored Energy Endothermic Stored Energy Endothermic: surroundings cool down Snow melting or Steam forming Ice Liquid Vapor +ΔH = gaining heat (Heat into system.) Exothermic: surroundings heat up Steam condensing or Ice forming Vapor Liquid Ice ΔH = losing heat (Heat out of system) System Stored Energy Exothermic No Stored Energy Physical Change Chemical Reaction Endothermic: A + Energy B + C Bonds in substance B + C have more stored energy than the bonds in substance A. Exothermic : A+ B C + Energy Bonds in substance A + B have more stored energy than the bonds in substance C. During a chemical change... 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 2

3 A No Stored Energy in "A" + Endothermic Energy absorbed from surroundings Chemical Change B + C Stored Energy in B & C Nothing would happen to "A" if it didn't absorb energy. A + B C + Stored Energy in A & B No Stored Energy in "C" Exothermic Energy released to surroundings Chemical Change A & B react because they have stored energy, therefore, the energy is released into the surroundings. 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 1000 J = 1 kj 1 kcal = 4186 J Units for Measuring Heat Flow One calorie is the quantity of heat that raises the temperature of 1 g 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. 3

4 Heat Capacity The 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... Heat Capacity 100 g of water has twice the heat capacity of 50 g of water. 20 g of lead has twice the heat capacity of 10 g of lead. Mass is important with regard to heat capacity!!!! Kinetic Energy increases as temperature increases! SPECIFIC HEAT CAPACITY The 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 1 g 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 CAPACITY 100 g of water has the same specific heat capacity as 5 g of water. 500 g of aluminum has the same specific heat capacity as 2 g of aluminum. Mass does not matter!!!! Heat Capacity vs. Specific Heat Water has a greater C than iron, regardless of the mass. Specific heat only refers to 1 g. (C iron = 0.46 J/g x ºC) (C water = 4.18 J/g x ºC) Does 100 g of Fe have a greater heat capactiy than 5 g of water? Heat Capacity vs. Specific Heat 1000 g of iron has a greater heat capacity than 5 g of water g Fe x 0.46 J (C iron = 0.46 J/g x ºC) (C water = 4.18 J/g x ºC) 5 g Water x 4.18 J 4

5 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 Example #1 A 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 #2 How many kilojoules are absorbed by water when 32.0 grams are heated from 25.0 ºC to 80.0 ºC? (C water = 4.18 J/g x ºC) Example #3 A 181gram chunk of silver has a heat capacity of 42.8 J/ ºC. Calculate the specific heat of silver using the information provided? 5

6 Example #4 A 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) Extra Example A gram piece of iron was at 107.5ºC. We removed5846 joules of heat from the iron. What was the original temperature? (C iron = 0.46 J/g x ºC) 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 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. ''+" or " " H of the system 6

7 Enthalpy One System vs. Two System The heat change in one substance, you determine q 1 mass 1 T 1 specific heat The transfer of heat from one substance to another. You do not determine q. One temperature increases, the other decreases. 2 masses 2 T 2 specific heats Hot Iron 70.0 C Cold Water 8.0 C Η = negative Heat loss Temp decrease Η = positive Heat Gain Temp increase 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 C water x 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 q mass C Glass Water m x C glass x T = m x C water x T T Example Problem #7 (Two different Systems) A 130 gram sample of iron is added to 100 ml 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) (1 ml = 1 g) This chart will help on 2 system Problems Iron Water Iron Water q iron = q water T Original q mass m x C water x T = m x C iron x T T Final C T 7

8 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 120 ml of water at 6 C. What is the mass of the iron? (C iron = 0.46 J/g x ºC) (1 ml = 1 g) q mass C T Iron Water q iron = q water m x C water x T = m x C iron x 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 system C (s) + 2S (s) + Energy CS 2 (l) Positive H (Heat Change) Exothermic: system surroundings Mg (s) + 2HCl (aq) MgCl 2 (aq) + H 2 (g) + Energy Negative H (Heat Change) Example Problem #8 How much heat is released when combining 25.0 ml of 0.5 molar HCl (aq) with 25.0 ml of 0.5 molar NaOH (aq)? During the reaction, there was a temperature change of 7 ºC. (Hint: aqueous is water) (1 ml = 1 g) HCl+NaOH H 2 O+NaCl+Energy Example Problem #9 Using the equation below, calculate the amount of heat (in kj) released by burning 24.0 g of CH 4. CH 4 + 2O 2 CO 2 + 2H 2 O kj ( H = 890kJ) 8

9 Example Problem #10 Using the equation below, calculate the amount of heat (in kj) required to break down g of KClO 3. 2KClO kj 2KCl + 3O 2 ( H = kj) 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 #11 How much energy (in kj) is absorbed when 25.0 grams of water increases from water at 40 ºC to steam at 120 ºC? C steam = 1.70 J/g x ºC Liquid Stea 40.7 kj/mol Phase Change Diagram 150 ºC 100 ºC C water = 4.18 J/g x ºC Solid Liquid: 6.01 kj/mol C ice = 2.10 J/g x ºC 50 ºC 0 ºC Temp Rises Phase Change Diagram C steam = 1.70 J/g x ºC Steam Liquid: 40.7 kj/mol C water = 4.18 J/g x ºC Liquid Solid: 6.01 kj/mol C ice = 2.10 J/g x ºC 50 ºC 0 ºC 150 ºC 100 ºC Temp Lowers 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. H fus = 6.01 kj/mol (Water) 9

10 H fus Example Problem How much heat is gained (in kj) when 57.6 g of ice is melted completely? H 2 O (s) H 2 O (l) H fus = 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) H solid Example Problem How much heat is lost (in kj) when 48.6 g of water completely freezes? H 2 O (l) H 2 O (s) H solid = 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 Problem A 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? 10

11 C steam = 1.70 J/g x ºC Phase Change Diagram 150 ºC ºC 0 ºC Liquid Stea 40.7 kj/mol C water = 4.18 J/g x ºC 100 ºC 2. 0 ºC 0 ºC 3. 0 ºC 100 ºC Solid Liquid: 6.01 kj/mol C ice = 2.10 J/g x ºC 50 ºC 0 ºC Temp Rises ºC 100 ºC ºC 125 ºC Do all phase change problems need 5 steps? Phase Change Example Problem A 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? Phase Change Diagram ºC 100 ºC C steam = 1.70 J/g x ºC 150 ºC ºC 100 ºC Steam Liquid: 40.7 kj/mol 100 ºC ºC 0 ºC C water = 4.18 J/g x ºC 4. 0 ºC 0 ºC Liquid Solid: 6.01 kj/mol C ice = 2.10 J/g x ºC 50 ºC 0 ºC Temp Lowers 5. 0 ºC 7 ºC 11

12 Example Problem A 56.8 g 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 Moles KJ H soln Example Problem When 5 g of solid NaOH is dissolved in water, how much heat is released? H soln = kj/mol Potential Test Questions CH 4 + 2O 2 CO 2 + 2H 2 O kj Is this exothermic or endothermic? What is the H? How many moles of CH 4 are needed to release 890 kj? How many moles of O 2 are needed to release 890 kj? Energy released by 3.12 mol of CH 4? Energy released by 78.3 g of O 2? 12