2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > Thermochemistry 17.2 Measuring and Expressing Enthalpy Changes 17.3 Heat in Changes of State 17.4 Calculating Heats of Reaction 17.1 The Flow of Energy > Energy Transformations 2 17.1 The Flow of Energy > Energy Transformations Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 3 Every substance has a certain amount of energy stored inside it. • When you buy gasoline, you are actually buying the stored potential energy it contains. • The controlled explosions of the gasoline in a car’s engine transform the potential energy into useful work, which can be used to propel the car. • The kinds of atoms and the arrangement of the atoms in a substance determine the amount of energy stored in the substance. 5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy Transformations 8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy Transformations • The energy stored in the chemical bonds of a substance is called chemical potential energy. Energy changes occur as either heat transfer or work, or a combination of both. 7 What are the ways in which energy changes can occur? Every substance has a certain amount of energy stored inside it. • Thermochemistry is the study of energy changes that occur during chemical reactions and changes in state. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy Transformations • Energy is the capacity for doing work or supplying heat. • Unlike matter, energy has neither mass nor volume. • Energy is detected only because of its effects. 4 Energy Transformations Lava flowing out of an erupting volcano is very hot. As lava flows, it loses heat and begins to cool slowly. The lava may flow into the ocean, where it cools more rapidly. 17.1 The Flow of Energy Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy Transformations Why does lava cool faster in water than in air? Chapter 17 1 CHEMISTRY & YOU • Heat is also produced, making the car’s engine extremely hot. 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy Transformations Energy changes occur as either heat transfer or work, or a combination of both. • Heat flows spontaneously from a warmer object to a cooler object. • Heat, represented by q, is energy that transfers from one object to another because of a temperature difference between the objects. • If two objects remain in contact, heat will flow from the warmer object to the cooler object until the temperature of both objects is the same. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 9 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 1 2/25/2013 17.1 The Flow of Energy > The energy released when a piece of wood is burned has been stored in the wood as 10 Processes The energy released when a piece of wood is burned has been stored in the wood as A. sunlight. B. heat. B. heat. C. calories. C. calories. D. chemical potential energy. D. chemical potential energy. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Processes 11 Processes • Chemical reactions and changes in physical state generally involve either the absorption or the release of heat. 17.1 The Flow of Energy > Endothermic and Exothermic Processes 12 14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic Processes During any chemical or physical process, the energy of the universe remains unchanged. Processes The law of conservation of energy states that in any chemical or physical process, energy is neither created nor destroyed. 15 Processes During any chemical or physical process, the energy of the universe remains unchanged. Direction of Heat Flow The direction of heat flow is given from the point of view of the system. • Heat is absorbed from the surroundings in an endothermic process. – Heat flowing into a system from its surroundings is defined as positive; q has a positive value. • If the energy of the system decreases during that process, the energy of the surroundings must increase by the same amount. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic • If the energy of the system increases during that process, the energy of the surroundings must decrease by the same amount. 16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic • You can define a system as the part of the universe on which you focus your attention. • Everything else in the universe makes up the surroundings. • Together, the system and its surroundings make up the universe. What happens to the energy of the universe during a chemical or physical process? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic Endothermic and Exothermic Processes Endothermic and Exothermic Processes What happens to the energy of the universe during a chemical or physical process? A. sunlight. 17.1 The Flow of Energy > Endothermic and Exothermic 13 17.1 The Flow of Energy > Endothermic and Exothermic 17.1 The Flow of Energy > 18 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 2 2/25/2013 17.1 The Flow of Energy > Endothermic and Exothermic Processes 17.1 The Flow of Energy > Endothermic and Exothermic Processes In an endothermic process, heat flows into the system from the surroundings. Direction of Heat Flow 17.1 The Flow of Energy > In an exothermic process, heat flows from the system to the surroundings. The direction of heat flow is given from the point of view of the system. • An exothermic process is one that releases heat to its surroundings. – Heat flowing out of a system into its surroundings is defined as negative; q has a negative value. In both cases, energy is conserved. 19 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 21 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 22 23 24 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > Sample Problem 17.1 Recognizing Endothermic and Exothermic Processes On a sunny winter day, the snow on a rooftop begins to melt. As the melted water drips from the roof, it refreezes into icicles. Describe the direction of heat flow as the water freezes. Is this process endothermic or exothermic? 25 26 27 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 3 2/25/2013 17.1 The Flow of Energy > Sample Problem 17.1 17.1 The Flow of Energy > 1 Analyze Identify the relevant concepts. Sample Problem 17.1 17.1 The Flow of Energy > 2 Solve Apply concepts to this situation. • Heat flows from a warmer object to a cooler object. 2 Solve Apply concepts to this situation. First identify the system and the surroundings. Determine the direction of heat flow. • In order for water to freeze, its temperature must decrease. System: water • An endothermic process absorbs heat from the surroundings. Sample Problem 17.1 Surroundings: air • Heat flows out of the water and into the air. • An exothermic process releases heat to the surroundings. First identify the system and surroundings. Then determine the direction of the heat flow. 28 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.1 29 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic Processes 2 Solve Apply concepts to this situation. 30 17.1 The Flow of Energy > Endothermic and Exothermic Processes Units for Measuring Heat Flow Determine if the process is endothermic or exothermic. Heat flow is measured in two common units: • Heat is released from the system to the surroundings. • the calorie Units for Measuring Heat Flow A calorie (cal) is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 C. ° • The word calorie is written with a small c except when referring to the energy contained in food. • the joule • The process is exothermic. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. • The dietary Calorie is written with a capital C. • One dietary Calorie is equal to one kilocalorie, or 1000 calories. 1 Calorie = 1 kilocalorie = 1000 calories 31 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Endothermic and Exothermic Processes 32 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Units for Measuring Heat Flow 33 17.1 The Flow of Energy > Athletes often use instant cold packs to soothe injuries. Many of these packs use the dissociation of ammonium nitrate in water to create a cold-feeling compress. Is this reaction endothermic or exothermic? Why? Athletes often use instant cold packs to soothe injuries. Many of these packs use the dissociation of ammonium nitrate in water to create a cold-feeling compress. Is this reaction endothermic or exothermic? Why? The joule (J) is the SI unit of energy. • One joule of heat raises the temperature of 1 g of pure water 0.2390 C. ° • You can convert between calories and joules using the following relationships: 1 J = 0.2390 cal 34 The instant cold pack feels cold because it removes heat from its surroundings. Therefore, the dissociation of ammonium nitrate in water is endothermic. The system (the cold pack) gains heat as the surroundings lose heat. 4.184 J = 1 cal Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 35 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 36 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 4 2/25/2013 17.1 The Flow of Energy > Heat Capacity and Specific Heat 17.1 The Flow of Energy > Heat Capacity and Specific Heat Heat Capacity and Specific Heat 17.1 The Flow of Energy > Heat Capacity and Specific Heat The heat capacity of an object depends on both its mass and its chemical composition. Heat Capacity and Specific Heat On what factors does the heat capacity of an object depend? On what factors does the heat capacity of an object depend? • The greater the mass of the object, the greater its heat capacity. • The amount of heat needed to increase the temperature of an object exactly 1oC is the heat capacity of that object. 37 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 38 17.1 The Flow of Energy > Heat Capacity and Specific Interpret Data Heat The specific heat capacity, or simply the specific heat, of a substance is the amount of heat it takes to raise the temperature of 1 g of the substance 1 C. ° • Water has a very high specific heat compared with the other substances. • Metals generally have low specific heats. 40 Specific heat J/(g· °C) cal/(g· °C) Liquid water 4.18 1.00 Ethanol 2.4 0.58 Ice 2.1 0.50 Steam 1.9 0.45 Chloroform 0.96 0.23 Aluminum 0.90 0.21 Iron 0.46 0.11 Silver 0.24 0.057 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Heat Capacity and Specific Heat Specific Heat of Water When a freshly baked apple pie comes out of the oven, both the filling and the crust are at the same temperature. • The filling, which is mostly water, has a higher specific heat than the crust. 39 Heat Specific Heat of Water Water in lakes and oceans absorbs heat from the air on hot days and releases it back into the air on cool days. • This property of water is responsible for moderate climates in coastal areas. 41 17.1 The Flow of Energy > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CHEMISTRY & YOU Heat will flow from the lava to the surroundings until the lava and surroundings are at the same temperature. Air has a smaller specific heat than water. Why would lava then cool more quickly in water than in air? 42 CHEMISTRY & YOU Heat will flow from the lava to the surroundings until the lava and surroundings are at the same temperature. Air has a smaller specific heat than water. Why would lava then cool more quickly in water than in air? Water requires more energy to raise its temperature than air. Therefore, lava in contact with water loses more heat energy than lava in contact with air, allowing it to cool more quickly. • This release of heat is why you have to be careful not to burn your tongue when eating hot apple pie. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > • In order to cool down, the filling must give off a lot of heat. 43 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Heat Capacity and Specific Specific Heat of Water Just as it takes a lot of heat to raise the temperature of water, water also releases a lot of heat as it cools. Specific Heats of Some Common Substances Substance Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. • A massive steel cable requires more heat to raise its temperature by 1oC than a steel nail does. 44 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 45 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 5 2/25/2013 17.1 The Flow of Energy > Heat Capacity and Specific Heat 17.1 The Flow of Energy > Heat Capacity and Specific Heat Calculating Specific Heat Calculating Specific Heat To calculate the specific heat (C) of a substance, you divide the heat input by the mass of the substance times the temperature change. q C= = m × ∆T q C= m × ∆T = 17.1 The Flow of Energy > Calculating the Specific Heat of a Substance heat (J or cal) mass (g) × change in temperature ( C) The temperature of a 95.4-g piece of copper increases from 25.0 C to 48.0 C when the copper absorbs 849 J of heat. What is the specific heat of copper? ° • m is mass. • ∆T is the change in temperature. ∆T = Tf – Ti ° ° • q is heat, expressed in terms of joules or calories. heat (J or cal) mass (g) × change in temperature (oC) Sample Problem 17.2 ° • The units of specific heat are either J/(g· C) or cal/(g· C). 46 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.2 47 ° Sample Problem 17.2 48 • Substitute the known quantities into the equation to calculate the unknown value CCu. q CCu = m × ∆T Cu ∆T = (48.0 C – 48.0 C) = 23.0 C q = 849 J Sample Problem 17.2 2 Calculate Solve for the unknown. • Start with the equation for specific heat. ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Use the known values and the definition of specific heat. ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 1 Analyze List the knowns and the unknown. KNOWNS mCu = 95.4 g ° 849 J CCu = 95.4 g × 23.0oC J/(g· C) ° = 0.387 UNKNOWN ° C = ? J/(g· C) 49 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.2 50 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 51 17.1 The Flow of Energy > ° ° The specific heat of ethanol is 2.4 J/(g· C). A sample of ethanol absorbs 676 J of heat, and the temperature rises from 22 C to 64 C. What is the mass of ethanol in the sample? 3 Evaluate Does the result make sense? ° • Remember that liquid water has a specific heat of 4.18 J/(g· C). ° ° • Metals have specific heats lower than water. C= q m × ∆T • Thus, the calculated value of 0.387 J/(g· C) seems reasonable. m= 676 J 2.4 J/(g· C) × (64 C – 22 C) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 53 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. ° ° The specific heat of ethanol is 2.4 J/(g· C). A sample of ethanol absorbs 676 J of heat, and the temperature rises from 22 C to 64 C. What is the mass of ethanol in the sample? ° 52 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 54 m= ° ° ° q C × ∆T = 6.7 g ethanol Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 6 2/25/2013 17.1 The Flow of Energy > Key Concepts & Key Equation 17.1 The Flow of Energy > Glossary Terms • thermochemistry: the study of energy changes that occur during chemical reactions and changes in state Energy changes occur as either heat transfer or work, or a combination of both. q m × ∆T Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > BIG IDEA • endothermic process: a process that absorbs heat from the surroundings • heat (q): energy that transfers from one object to another because of a temperature difference between the objects The heat capacity of an object depends on both its mass and its chemical composition. C= • law of conservation of energy: in any chemical or physical process, energy is neither created nor destroyed • chemical potential energy: energy stored in chemical bonds During any chemical or physical process, the energy of the universe remains unchanged. 55 17.1 The Flow of Energy > Glossary Terms • exothermic process: a process that releases heat to its surroundings • system: a part of the universe on which you focus your attention • heat capacity: the amount of heat needed to increase the temperature of an object exactly 1 C • surroundings: everything in the universe outside the system • specific heat: the amount of heat needed to increase the temperature of 1 g of a substance 1 C; also called specific heat capacity 56 17.1 The Flow of Energy > Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. BIG IDEA ° ° 57 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Matter and Energy • During a chemical or physical process, the energy of the universe is conserved. – If energy is absorbed by the system in a chemical or physical process, the same amount of energy is released by the surroundings. – Conversely, if energy is released by the system, the same amount of energy is absorbed by the surroundings. 58 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Solar hot water system installed on low cost housing in the Kouga Local Municipality, South Africa 59 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 60 17.1 The Flow of Energy > Thermochemistry Remember: The concept of specific heat allows you to measure heat flow in chemical and physical processes. 17.1 The Flow of Energy 17.2 Measuring and Expressing Enthalpy Changes 17.3 Heat in Changes of State 17.4 Calculating Heats of Reaction 61 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 62 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CHEMISTRY & YOU How can you measure the amount of heat released when a match burns? Chapter 17 END OF 17.1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 63 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 7 2/25/2013 17.1 The Flow of Energy > Calorimetry 17.1 The Flow of Energy > Calorimetry 17.1 The Flow of Energy > Calorimetry Calorimetry is the measurement of the heat flow into or out of a system for chemical and physical processes. Calorimetry How can you measure the change in enthalpy of a reaction? Calorimetry is the measurement of the heat flow into or out of a system for chemical and physical processes. • In a calorimetry experiment involving an endothermic process, the heat absorbed by the system is equal to the heat released by its surroundings. • In an exothermic process, the heat released by the system is equal to the heat absorbed by its surroundings. 64 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry 65 17.1 The Flow of Energy > Calorimetry Calorimetry is the measurement of the heat flow into or out of a system for chemical and physical processes. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry 66 Constant-Pressure Calorimeters The enthalpy (H) of a system accounts for the heat flow of the system at constant pressure. Foam cups can be used as simple calorimeters because they do not let much heat in or out. • The heat absorbed or released by a reaction at constant pressure is the same as the change in enthalpy, symbolized as ∆H. • Most chemical reactions and physical changes carried out in the laboratory are open to the atmosphere and thus occur at constant pressure. 68 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry Constant-Pressure Calorimeters The value of ∆H of a reaction can be determined by measuring the heat flow of the reaction at constant pressure. • In this textbook, the terms heat and enthalpy change are used interchangeably. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry Constant-Pressure Calorimeters • The insulated device used to measure the absorption or release of heat in chemical or physical processes is called a calorimeter. 67 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 69 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry Constant-Pressure Calorimeters Constant-Pressure Calorimeters • To measure the enthalpy change for a reaction in aqueous solution in a foam cup calorimeter, dissolve the reacting chemicals (the system) in known volumes of water (the surroundings). • Measure the initial temperature of each solution, and mix the solutions in the foam cup. • After the reaction is complete, measure the final temperature of the mixed solutions. • In other words, q = ∆H. 70 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 71 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 72 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 8 2/25/2013 17.1 The Flow of Energy > Calorimetry 17.1 The Flow of Energy > Calorimetry Constant-Pressure Calorimeters Constant-Pressure Calorimeters You can calculate the heat absorbed or released by the surroundings (qsurr) using the formula for the specific heat, the initial and final temperatures, and the heat capacity of water. Constant-Pressure Calorimeters The heat absorbed by the surroundings is equal to, but has the opposite sign of, the heat released by the system. qsurr = m × C × ∆T • m is the mass of the water. • C is the specific heat of water. qsurr = m × C × ∆T 73 17.1 The Flow of Energy > Calorimetry qsurr = –qsys • ∆T = Tf – Ti Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry 74 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry Constant-Pressure Calorimeters Constant-Volume Calorimeters The enthalpy change for the reaction (∆H) can be written as follows: Calorimetry experiments can also be performed at a constant volume using a device called a bomb calorimeter. 75 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry Constant-Volume Calorimeters In a bomb calorimeter, a sample of a compound is burned in a constant-volume chamber in the presence of oxygen at high pressure. qsys = ∆H = –qsurr = –m × C × ∆T • The sign of ∆H is positive for an endothermic reaction and negative for an exothermic reaction. 76 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calorimetry 77 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 78 17.1 The Flow of Energy > 17.1 The Flow of Energy > 80 81 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Constant-Volume Calorimeters • The heat that is released warms the water surrounding the chamber. • By measuring the temperature increase of the water, it is possible to calculate the quantity of heat released during the combustion reaction. 79 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 9 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 82 83 84 17.1 The Flow of Energy > CHEMISTRY & YOU 17.1 The Flow of Energy > What type of calorimeter would you use to measure the heat released when a match burns? Describe the experiment and how you would calculate the heat released. CHEMISTRY & YOU 17.1 The Flow of Energy > Enthalpy Change in a Calorimetry Experiment What type of calorimeter would you use to measure the heat released when a match burns? Describe the experiment and how you would calculate the heat released. When 25.0 mL of water containing 0.025 mol HCl at 25.0 C is added to 25.0 mL of water containing 0.025 mol NaOH at 25.0 C in a foam-cup calorimeter, a reaction occurs. Calculate the enthalpy change (in kJ) during this reaction if the highest temperature observed is 32.0 C. Assume that the densities of the solutions are 1.00 g/mL and the volume of the final solution is equal to the sum of the volumes of the reacting solutions. ° A constant-volume, or bomb, calorimeter would be used to measure the heat released when a match burns. The match would be ignited in the chamber. By measuring the temperature increase in the water and using the equation q = –m × C × ∆T , the heat released, q, can be calculated. 85 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.3 86 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 1 Analyze List the knowns and the unknown. Sample Problem 17.3 KNOWNS 87 ° Sample Problem 17.3 2 Calculate Solve for the unknown. First calculate the total mass of the water. ∆H = ? kJ Vfinal = VHCl + VNaOH • Use ∆H = –qsurr = –m × C × ∆T to solve for ∆H. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > UNKNOWN Cwater = 4.18 J/(g· C) • You must also calculate ∆T. ° ° 1 Analyze List the knowns and the unknown. • Use dimensional analysis to determine the mass of the water. Sample Problem 17.3 mwater = 50.0 mL × = 25.0 mL + 25.0 mL = 50.0 mL ° ° 1.00 g = 50.0 g 1 mL Ti = 25.0 C Tf = 32.0 C Assume that the densities of the solutions are 1.00 g/mL to find the total mass of the water. densitysolution = 1.00 g/mL 88 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 89 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 90 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 10 2/25/2013 17.1 The Flow of Energy > Sample Problem 17.3 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Now calculate ∆T. ° ° Sample Problem 17.3 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. ° ∆T = Tf – Ti = 32.0 C – 25.0 C = 7.0 C 3 Evaluate Does the result make sense? Use the values for mwater, Cwater, and ∆T to calculate ∆H. • The temperature of the solution increases, which means that the reaction is exothermic, and thus the sign of ∆H should be negative. ∆H = –qsurr = –mwater × Cwater × ∆T • About 4 J of heat raises the temperature of 1 g of water 1 C, so 200 J of heat is required to raise 50 g of water 1 C. Raising the temperature of 50 g of water 7 C requires about 1400 J, or 1.4 kJ. ° ° ° ° = –(50.0 g)(4.18J/(g·oC))(7.0 C) = –1500 J = –1.5 kJ 91 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 92 ° Use the relationship 1 kJ = 1000 J to convert your answer from J to kJ. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > The initial temperature of the water in a constant-pressure calorimeter is 24 C. A reaction takes place in the calorimeter, and the temperature rises to 87 C. The calorimeter contains 367 g of water, which has a specific heat of 4.18 J/(g· C). Calculate the enthalpy change during this reaction. ° • This estimated answer is very close to the calculated value of ∆H. 93 ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations The initial temperature of the water in a constant-pressure calorimeter is 24 C. A reaction takes place in the calorimeter, and the temperature rises to 87 C. The calorimeter contains 367 g of water, which has a specific heat of 4.18 J/(g· C). Calculate the enthalpy change during this reaction. ° Sample Problem 17.3 ° ° ∆H = –m × C × ∆T ° Thermochemical Equations How can you express the enthalpy change for a reaction in a chemical equation? ° = –367 g × 4.18 J/(g· C) × (87 C – 24 C) ° 94 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations 95 = –97000 J = –97 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations 96 17.1 The Flow of Energy > Thermochemical Equations In the equation describing the exothermic reaction of calcium oxide and water, the enthalpy change can be considered a product. In a chemical equation, the enthalpy change for the reaction can be written as either a reactant or a product. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. A chemical equation that includes the enthalpy change is called a thermochemical equation. CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ Calcium oxide is one of the components of cement. 97 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 98 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 99 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 11 2/25/2013 17.1 The Flow of Energy > Thermochemical Equations 17.1 The Flow of Energy > Thermochemical Equations Heats of Reaction The heat of reaction is the enthalpy change for the chemical equation exactly as it is written. Heats of Reaction Each mole of calcium oxide and water that reacts to form calcium hydroxide produces 65.2 kJ of heat. • Heats of reaction are reported as ∆H. CaO(s) + H2O(l) → Ca(OH)2(s) • The physical state of the reactants and products must also be given. ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations 101 Baking soda (sodium bicarbonate) decomposes when it is heated. This process is endothermic. 2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) + CO2(g) The carbon dioxide released in the reaction causes muffins to rise while baking. ∆H = –65.2 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations Heats of Reaction 2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) + CO2(g) Remember that ∆H is positive for endothermic reactions. Therefore, you can write the reaction as follows: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Heats of Reaction • In exothermic processes, the chemical potential energy of the reactants is higher than the chemical potential energy of the products. • The standard conditions are that the reaction is carried out at 101.3 kPa (1 atm) and 25 C. 100 17.1 The Flow of Energy > Thermochemical Equations 102 17.1 The Flow of Energy > Thermochemical Equations Heats of Reaction Heats of Reaction The amount of heat released or absorbed during a reaction depends on the number of moles of the reactant involved. To see why the physical state of the reactants and products must be stated, compare the following two equations. H2O(l) → H2(g) + • The decomposition of 2 mol of sodium bicarbonate requires 85 kJ of heat. ∆H = 85 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. H2O(g) → H2(g) + 1 2 1 2 O2(g) O2(g) • Therefore, the decomposition of 4 mol of the same substance would require twice as much heat, or 170 kJ. 103 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations 104 17.1 The Flow of Energy > Heats of Reaction To see why the physical state of the reactants and products must be stated, compare the following two equations. H2O(l) → H2(g) + H2O(g) → H2(g) + 1 2 1 2 O2(g) O2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 17.4 105 Calculate the amount of heat (in kJ) required to decompose 2.24 mol NaHCO3(s). ∆H = 85 kJ for 2 mol NaHCO3 ° 106 UNKNOWN ∆H = ? kJ for 2.24 mol NaHCO3 ∆H = 44.0 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 17.4 KNOWNS amount of NaHCO3(s) that decomposes = 2.24 mol • The vaporization of 1 mol of liquid water to water vapor at 25 C requires 44.0 kJ of heat. H2O(l) → H2O(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 1 Analyze List the knowns and the unknown. Use the thermochemical equation to write a conversion factor relating kJ of heat and moles of NaHCO3. Then use the conversion factor to determine ∆H for 2.24 mol NaHCO3. 2NaHCO3(s) + 85 kJ → Na2CO3(s) + H2O(l) CO2(g) ∆H = 285.8 kJ ∆H = 241.8 kJ difference = 44.0 kJ 17.1 The Flow of Energy > Using the Heat of Reaction to Calculate Enthalpy Change ∆H = 241.8 kJ difference = 44.0 kJ ∆H = 285.8 kJ 107 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 108 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 12 2/25/2013 17.1 The Flow of Energy > Sample Problem 17.4 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Sample Problem 17.4 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Write the conversion factor relating kJ of heat and moles of NaHCO3. 3 Evaluate Does the result make sense? Using dimensional analysis, solve for ∆H. • The 85 kJ in the thermochemical equation refers to the decomposition of 2 mol NaHCO3(s). 85 kJ ∆H = 2.24 mol NaHCO3(s) × 2 mol NaHCO (s) 3 = 95 kJ 85 kJ 2 mol NaHCO3(s) • Therefore, the decomposition of 2.24 mol should absorb more heat than 85 kJ. The thermochemical equation indicates that 85 kJ are needed to decompose 2 mol NaHCO3(s). 109 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations • The answer of 95 kJ is consistent with this estimate. 110 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemical Equations 111 °C CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + 890 kJ • This is an exothermic reaction. • The heat of combustion (∆H) for this reaction is –890 kJ per mole of methane burned. 17.1 The Flow of Energy > 113 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Which of the following thermochemical equations represents an endothermic reaction? A. Cgraphite(s) + 2 kJ A. Cgraphite(s) + 2 kJ B. 2H2(g) + O2(g) Cdiamond(s) B. 2H2(g) + O2(g) 2H2O + 483.6 kJ Substance Formula Hydrogen H2(g) Carbon C(s, graphite) ∆H (kJ/mol) –286 –394 Methane CH4(g) Acetylene C2H2(g) –1300 Ethanol C2H6O(l) –1368 –890 Propane C3H8(g) Glucose C6H12O6(s) –2808 Octane C8H18(l) –5471 Sucrose C12H22O11(s) –5645 –2220 114 Like other heats of reaction, heats of combustion are reported as the enthalpy changes when the reactions are carried out at 101.3 kPa and 25 C. ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Which of the following thermochemical equations represents an endothermic reaction? Interpret Data Heats of Combustion at 25 • Burning 1 mol of methane releases 890 kJ of heat. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Heats of Combustion Small amounts of natural gas within crude oil are burned off at oil refineries. Heats of Combustion The heat of combustion is the heat of reaction for the complete burning of one mole of a substance. 112 Sample Problem 17.4 Key Concepts & Key Equation The value of ∆H of a reaction can be determined by measuring the heat flow of the reaction at a constant pressure. In a chemical equation, the enthalpy change for the reaction can be written as either a reaction or a product. Cdiamond(s) 2H2O + 483.6 kJ qsys = ∆H = –qsurr = –m × C × ∆T 115 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 116 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 117 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 13 2/25/2013 17.1 The Flow of Energy > Glossary Terms 17.1 The Flow of Energy > The heat of reaction or process can be determined experimentally through calorimetry. • heat of combustion: the heat of reaction for the complete burning of one mole of a substance • enthalpy (H): the heat content of a system at constant pressure 17.1 The Flow of Energy > 119 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 120 When your body heats up, you start to sweat. The evaporation of sweat is your body’s way of cooling itself to a normal temperature. 17.1 The Flow of Energy 17.2 Measuring and Expressing Enthalpy Changes 17.3 Heat in Changes of State 17.4 Calculating Heats of Reaction Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Heats of Fusion and 17.1 The Flow of Energy > Solidification 122 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Heats of Fusion and 17.1 The Flow of Energy > Solidification Heats of Fusion and Solidification 123 Heats of Fusion and • The heat absorbed by one mole of a solid substance as it melts to a liquid at constant temperature is the molar heat of fusion (∆Hfus). • The gain of heat causes a change of state instead of a change in temperature. • The molar heat of solidification (∆Hsolid) is the heat lost when one mole of a liquid substance solidifies at a constant temperature. • The temperature of the substance undergoing the change remains constant. 124 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 125 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Solidification All solids absorb heat as they melt to become liquids. What is the relationship between molar heat of fusion and molar heat of solidification? CHEMISTRY & YOU Why does sweating help cool you off? Thermochemistry END OF 17.2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Chapter 17 121 BIG IDEA Matter and Energy • heat of reaction: the enthalpy change for a chemical equation exactly as it is written • calorimeter: an insulated device used to measure the absorption or release of heat in chemical or physical processes Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > • thermochemical equation: a chemical equation that includes the enthalpy change • calorimetry: the precise measurement of heat flow out of a system for chemical and physical processes 118 Glossary Terms 126 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 14 2/25/2013 Heats of Fusion and 17.1 The Flow of Energy > Solidification Heats of Fusion and 17.1 The Flow of Energy > Solidification ° ° ∆Hfus = –∆Hsolid Sample Problem 17.5 ° Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.5 129 Sample Problem 17.5 2 Calculate Solve for the unknown. Start by expressing ∆Hfus as a conversion factor. Express the molar mass of ice as a conversion factor. 1 mol H2O(s) 6.01 kJ UNKNOWN mice = ? g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. • Convert moles of ice to grams of ice. KNOWNS Initial and final temperature are 0 C ° ∆Hsolid = –6.01 kJ/mol 128 1 Analyze List the knowns and the unknown. • Find the number of moles of ice that can be melted by the addition of 2.25 kJ of heat. How many grams of ice at 0 C will melt if 2.25 kJ of heat are added? ∆Hfus = 6.01 kJ/mol H2O(l) → H2O(s) 17.1 The Flow of Energy > Sample Problem 17.5 Using the Heat of Fusion in PhaseChange Calculations ° H2O(s) → H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. ° • The melting of 1 mol of ice at 0 C to 1 mol of liquid water at 0 C requires the absorption of 6.01 kJ of heat. • The conversion of 1 mol of liquid water at 0 C to 1 mol of ice at 0 C releases 6.01 kJ of heat. The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies. 127 17.1 The Flow of Energy > 18.0 g H2O(s) 1 mol H2O(s) Use the thermochemical equation ∆Hfus = 6.01 kJ/mol H2O(s) + 6.01 kJ → H2O(l). ∆H = 2.25 kJ 130 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.5 131 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Sample Problem 17.5 132 17.1 The Flow of Energy > 3 Evaluate Does the result make sense? Calculate the amount of heat absorbed to liquefy 15.0 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol. • To melt 1 mol of ice, 6.01 kJ of energy is required. Multiply the known enthalpy change by the conversion factors. • Only about one-third of this amount of heat (roughly 2 kJ) is available. 1 mol H2O(s) 18.0 g H2O(s) × 6.01 kJ 1 mol H2O(s) = 6.74 g H2O(s) mice = 2.25 kJ × Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. • So, only about one-third mol of ice, or 18.0 g/3 = 6 g, should melt. • This estimate is close to the calculated answer. 133 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 134 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 135 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 15 2/25/2013 Heats of Vaporization and 17.1 The Flow of Energy > 17.1 The Flow of Energy > Condensation Calculate the amount of heat absorbed to liquefy 15.6 g of methanol (CH4O) at its melting point. The molar heat of fusion for methanol is 3.16 kJ/mol. Heats of Vaporization and Condensation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Interpret Data 137 Heats of Vaporization and ∆Hfus (kJ/mol) ∆Hvap (kJ/mol) 5.66 23.3 Ethanol (C2H6O) 4.93 38.6 Hydrogen (H2) 0.12 Methanol (CH4O) 3.22 Oxygen (O2) 0.44 Water (H2O) 6.01 139 138 Heats of Vaporization and The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses. ∆Hvap = –∆Hcond • The molar heat of condensation (∆Hcond) is the amount of heat released when one mole of vapor condenses at its normal boiling point. 0.90 6.82 40.7 CHEMISTRY & YOU 140 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Explain why the evaporation of sweat off your body helps cool you off. CHEMISTRY & YOU 141 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Interpret Graphs A heating curve graphically describes the enthalpy changes that take place during phase changes. Explain why the evaporation of sweat off your body helps cool you off. 143 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Energy is required to vaporize (or evaporate) a liquid into a gas. When liquid sweat absorbs energy from your skin, the temperature of your skin decreases. 142 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Condensation • When a vapor condenses, heat is released. 35.2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > • The amount of heat required to vaporize one mole of a given liquid at a constant temperature is called its molar heat of vaporization (∆Hvap). Condensation is the exact opposite of vaporization. Heats of Physical Change Ammonia (NH3) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Condensation This table lists the molar heats of vaporization for several substances at their normal boiling point. Substance A liquid that absorbs heat at its boiling point becomes a vapor. What is the relationship between molar heat of vaporization and molar heat of condensation? 1 mol 3.16 kJ ∆H = 15.6 g CH4O × 32.05 g CH O × 1 mol 4 = 1.54 kJ 136 Heats of Vaporization and 17.1 The Flow of Energy > Condensation Remember: The temperature of a substance remains constant during a change of state. 144 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 16 2/25/2013 17.1 The Flow of Energy > Sample Problem 17.6 17.1 The Flow of Energy > Using the Heat of Vaporization in PhaseChange Calculations 17.1 The Flow of Energy > 1 Analyze List the knowns and the unknown. • First convert grams of water to moles of water. How much heat (in kJ) is absorbed when 24.8 g H2O(l) at 100 C and 101.3 kPa is converted to H2O(g) at 100 C? ° Sample Problem 17.6 2 Calculate Solve for the unknown. Start by expressing the molar mass of water as a conversion factor. • Then find the amount of heat that is absorbed when the liquid is converted to steam. ° Sample Problem 17.6 1 mol H2O(l) 18.0 g H2O(l) KNOWNS UNKNOWN Initial and final conditions ∆H = ? kJ are 100 C and 101.3 kPa ° Mass of liquid water converted to steam = 24.8 g ∆Hvap = 40.7 kJ/mol 145 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.6 146 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Sample Problem 17.6 147 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Express ∆Hvap as a conversion factor. ∆H = 24.8 g H2O(l) × • Knowing that the molar mass of water is 18.0 g/mol, 24.8 g H2O(l) can be estimated to be somewhat less than 1.5 mol H2O. 1 mol H2O(l) 40.7 kJ × 18.0 g H2O(l) 1 mol H2O(l) • The calculated enthalpy change should be a little less than 1.5 mol × 40 kJ/mol = 60 kJ, and it is. = 56.1 kJ Use the thermochemical equation Sample Problem 17.6 3 Evaluate Does the result make sense? Multiply the mass of water in grams by the conversion factors. 40.7 kJ 1 mol H2O(l) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. H2O(l) + 40.7 kJ → H2O(g). 148 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 149 17.1 The Flow of Energy > The molar heat of condensation of a substance is the same, in magnitude, as which of the following? 151 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The molar heat of condensation of a substance is the same, in magnitude, as which of the following? A. molar heat of fusion B. molar heat of vaporization B. molar heat of vaporization C. molar heat of solidification C. molar heat of solidification D. molar heat of formation D. molar heat of formation 152 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Heat of Solution A. molar heat of fusion Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 150 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Heat of Solution What thermochemical changes can occur when a solution forms? 153 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17 2/25/2013 17.1 The Flow of Energy > Heat of Solution 17.1 The Flow of Energy > Heat of Solution During the formation of a solution, heat is either released or absorbed. 17.1 The Flow of Energy > Heat of Solution A practical application of an exothermic dissolution process is a hot pack. During the formation of a solution, heat is either released or absorbed. • In a hot pack, calcium chloride, CaCl2(s), mixes with water, producing heat. • The enthalpy change caused by the dissolution of one mole of substance is the molar heat of solution (∆Hsoln). CaCl2(s) → Ca2+(aq) + 2Cl–(aq) ∆Hsoln = –82.8 kJ/mol 154 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Heat of Solution 155 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > The dissolution of ammonium nitrate, NH4NO3(s), is an example of an endothermic process. Sample Problem 17.7 156 17.1 The Flow of Energy > Calculating the Enthalpy Change in Solution Formation • The cold pack shown here contains solid ammonium nitrate crystals and water. • The solution process absorbs energy from the surroundings. Sample Problem 17.7 1 Analyze List the knowns and the unknown. Use the heat of solution for the dissolution of NaOH(s) in water to solve for the amount of heat released (∆H). How much heat (in kJ) is released when 2.50 mol NaOH(s) is dissolved in water? • Once the solute dissolves, the pack becomes cold. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. KNOWNS ∆Hsoln = –44.5 kJ/mol NH4NO3(s) → NH4+(aq) + NO3–(aq) amount of NaOH(s) dissolved = 2.50 mol ∆Hsoln = 25.7 kJ/mol UNKNOWN ∆H = ? kJ 157 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.7 158 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Sample Problem 17.7 159 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Start by expressing ∆Hsoln as a conversion factor. ∆H = 2.50 mol NaOH(s) × Sample Problem 17.7 3 Evaluate Does the result make sense? • ∆H is 2.5 times greater than ∆Hsoln, as it should be. Multiply the number of moles by the conversion factor. –44.5 kJ 1 mol NaOH(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. • Also, ∆H should be negative, as the dissolution of NaOH(s) in water is exothermic. –44.5 kJ 1 mol NaOH(s) = –111 kJ Use the thermochemical equation NaOH(s) → Na+(aq) + OH–(aq) + 44.5 kJ/mol. 160 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 161 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 162 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 18 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if ∆Hsoln = 25.7 kJ/mol? 17.1 The Flow of Energy > How much heat (in kJ) is absorbed when 50.0 g of NH4NO3(s) are dissolved in water if ∆Hsoln = 25.7 kJ/mol? 1 mol ∆H = 50.0 g NH4NO3 × 80.04 g NH NO × 4 3 = 16.1 kJ Key Concepts The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies; that is, ∆Hfus = –∆Hsolid. The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses; that is, ∆Hvap = –∆Hcond. 25.7 kJ 1 mol During the formation of a solution, heat is either released or absorbed. 163 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Glossary Terms 164 17.1 The Flow of Energy > • molar heat of fusion (∆Hfus): the amount of heat absorbed by one mole of a solid substance as it melts to a liquid at a constant temperature 17.1 The Flow of Energy > 167 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Thermochemistry END OF 17.3 CHEMISTRY & YOU 168 17.4 Calculating Heats of Reaction 170 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Hess’s Law Hess’s Law How can you calculate the heat of reaction when it cannot be directly measured? Diamonds are gemstones composed of carbon. Over a time period of millions and millions of years, diamond will break down into graphite, which is another form of carbon. 17.1 The Flow of Energy 17.2 Measuring and Expressing Enthalpy Changes 17.3 Heat in Changes of State Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > How much heat is released when a diamond changes into graphite? Chapter 17 169 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. • molar heat of solution (∆Hsoln): the enthalpy change caused by the dissolution of one mole of a substance • molar heat of vaporization (∆Hvap): the amount of heat absorbed by one mole of a liquid as it vaporizes at a constant temperature Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Glossary Terms 165 • molar heat of condensation (∆Hcond): the amount of heat released by one mole of a vapor as it condenses to a liquid at a constant temperature • molar heat of solidification (∆Hsolid): the amount of heat lost by one mole of a liquid as it solidifies at a constant temperature 166 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 171 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 19 2/25/2013 17.1 The Flow of Energy > Hess’s Law 17.1 The Flow of Energy > Hess’s Law 17.1 The Flow of Energy > Hess’s Law C(s, diamond) → C(s, graphite) Hess’s law of heat summation states that if you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. 172 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Hess’s Law Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations. 173 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Hess’s Law C(s, diamond) → C(s, graphite) a. C(s, graphite) + O2(g) → CO2(g) ∆H = –393.5 kJ b. C(s, diamond) + O2(g) → CO2(g) ∆H = –395.4 kJ 174 ∆H = –393.5 kJ b. C(s, diamond) + O2(g) → CO2(g) ∆H = –395.4 kJ b. C(s, diamond) + O2(g) → CO2(g) c. CO2(g) → C(s, graphite) + O2(g) C(s, diamond) → C(s, graphite) ∆H = –395.4 kJ ∆H = 393.5 kJ If you also add the values of ∆H for equations b and c, you get the heat of reaction for this conversion. If you add equations b and c, you get the equation for the conversion of diamond to graphite. Write equation a in reverse to give: ∆H = 393.5 kJ When you reverse a reaction, you must also change the sign of ∆H. C(s, diamond) + O2(g) → CO2(g) ∆H = –395.4 kJ C(s, diamond) + O2(g) → CO2(g) ∆H = –395.4 kJ CO2(g) → C(s, graphite) + O2(g) ∆H = CO2(g) → C(s, graphite) + O2(g) ∆H = 393.5 kJ C(s, diamond) → C(s, graphite) ∆H = –1.9 kJ 393.5 kJ C(s, diamond) → C(s, graphite) 175 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Hess’s Law 176 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > C(s, diamond) + O2(g) → CO2(g) ∆H = –395.4 kJ CO2(g) → C(s, graphite) + O2(g) ∆H = 393.5 kJ C(s, diamond) → C(s, graphite) ∆H = –1.9 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Hess’s Law C(s, diamond) → C(s, graphite) a. C(s, graphite) + O2(g) → CO2(g) c. CO2(g) → C(s, graphite) + O2(g) Although the enthalpy change for this reaction cannot be measured directly, you can use Hess’s law to find the enthalpy change for the conversion of diamond to graphite by using the following combustion reactions. CHEMISTRY & YOU 177 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > How can you determine ∆H for the conversion of diamond to graphite without performing the reaction? CHEMISTRY & YOU How can you determine ∆H for the conversion of diamond to graphite without performing the reaction? You can use Hess’s law by adding thermochemical equations in which the enthalpy changes are known and whose sum will result in an equation for the conversion of diamond to graphite. 178 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 179 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 180 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 20 2/25/2013 17.1 The Flow of Energy > Hess’s Law 17.1 The Flow of Energy > Hess’s Law 17.1 The Flow of Energy > Hess’s Law C(s, graphite) + O2(g) → CO2(g) Another case where Hess’s law is useful is when reactions yield products in addition to the product of interest. You can calculate the desired enthalpy change by using Hess’s law and the following two reactions that can be carried out in the laboratory: • Suppose you want to determine the enthalpy change for the formation of carbon monoxide from its elements. C(s, graphite)+ 12 O2(g) → CO(g) ∆H = ? • Carrying out the reaction in the laboratory as written is virtually impossible. 181 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > C(s, graphite) + O2(g) → CO2(g) ∆H = –393.5 kJ CO2(g) → CO(g) + 12 O2(g) ∆H = C(s, graphite)+ 21 O2(g) → CO(g) ∆H = –110.5 kJ 182 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > A. heats of fusion for each of the compounds in the reaction A. heats of fusion for each of the compounds in the reaction B. two other reactions with known heats of reaction B. two other reactions with known heats of reaction C. specific heat capacities for each compound in the reaction C. specific heat capacities for each compound in the reaction Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Standard Heats of Formation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Standard Heats of Formation ° The standard heat of formation (∆Hf ) of a compound is the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. • Scientists specify a common set of conditions as a reference point. ° • Thus, ∆H ° = 0 for the diatomic molecules • These conditions, called the standard state, refer to the stable form of a substance at 25 C and 101.3 kPa. • The ∆Hf of a free element in its standard state is arbitrarily set at zero. ° f H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and I2(s). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 283.0 kJ 183 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation Standard Heats of Formation How can you calculate the heat of reaction when it cannot be directly measured? D. density for each compound in the reaction 185 Enthalpy changes generally depend on the conditions of the process. 187 ∆H = –110.5 kJ 17.1 The Flow of Energy > According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following? D. density for each compound in the reaction ∆H = C(s, graphite)+ 12 O2(g) → CO(g) 283.0 kJ According to Hess’s law, it is possible to calculate an unknown heat of reaction by using which of the following? 184 ∆H = –393.5 kJ CO2(g) → CO(g) + 12 O2(g) 188 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 186 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Interpret Data Standard Heats of Formation (∆Hf ° Substance ∆Hf (kJ/mol) Substance Al2O3(s) –1676.0 F2(g) Br2(g) 30.91 Fe(s) Br2(l) 0.0 Fe2O3(s) C(s, diamond) 1.9 H2(g) C(s, graphite) CH4(g) NO2(g) NaCl(s) 0.0 0.0 O3(g) P(s, white) –187.8 I2(g) –1207.0 I2(s) –635.1 N2(g) 0.0 NH3(g) 62.4 0.0 33.85 –411.2 142.0 –241.8 –285.8 H2O2(l) 90.37 O2(g) H2O(g) –110.5 189 0.0 –822.1 H2O(l) –393.5 Cl2(g) f (kJ/mol) NO(g) 0.0 CO2(g) CaO(s) 0.0 –74.86 CO(g) CaCO3(s) °) at 25°C and 101.3 kPa ° Substance ∆H ° ∆Hf (kJ/mol) P(s, red) S(s, rhombic) S(s, monoclinic) 0.0 –18.4 0.0 0.30 0.0 SO2(g) –296.8 –46.19 SO3(g) –395.7 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 21 2/25/2013 Standard Heats of Formation 17.1 The Flow of Energy > Standard Heats of Formation 17.1 The Flow of Energy > For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. Standard Heats of Formation 17.1 The Flow of Energy > For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. • Such an enthalpy change is called the standard heat of reaction (∆H ). • The standard heat of reaction is the difference between the standard heats of formation of all the reactants and products. ° ° = ∆H °(products) – ∆H °(reactants) ∆H 190 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Standard Heats of Formation 17.1 The Flow of Energy > 191 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Sample Problem 17.8 Balance the equation of the reaction of CO(g) with O2(g) to form CO2(g). Then determine ∆H using the standard heats of formation of the reactants and products. ° KNOWNS ∆Hf CO(g) = –110.5 kJ/mol ° ∆H °O (g) = 0 kJ/mol (free element) ∆H °CO (g) = –393.5 kJ/mol f • Notice that water has a lower enthalpy than the elements from which it is formed. 193 17.1 The Flow of Energy > Sample Problem 17.8 194 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. ° of the product in a similar way. Find ∆Hf °(reactants) = 2 mol CO(g) × ∆H °CO(g) + 1 mol O (g) × °O (g) –110.5 kJ 2 = 2 mol CO(g) × ∆Hf ∆Hf 2 0 kJ 2 mol CO(g)+ 1 mol O2(g) × 1 mol O2(g) = –221.0 kJ Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 197 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. °(products) = 2 mol CO (g) × °CO (g) 2 2 –393.5 kJ = 2 mol CO2(g) × 1 mol CO (g) 2 = –787.0 kJ Remember to take into account the number of moles of each reactant and product. 196 Sample Problem 17.8 2 Calculate Solve for the unknown. ° of all the reactants. f Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Find and add ∆Hf ∆Hf ∆Hf ° 2 195 2 Calculate Solve for the unknown. First write the balanced equation. 2CO(g) + O2(g) → 2CO2(g) Sample Problem 17.8 UNKNOWN ∆H = ? kJ 2 f Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 17.8 1 Analyze List the knowns and the unknown. ° • The enthalpy difference between the reactants and products, –285.8 kJ/mol, is the standard heat of formation of liquid water from the gases hydrogen and oxygen. f Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calculating the Standard Heat of Reaction What is the standard heat of reaction (∆H ) for the reaction of CO(g) with O2(g) to form CO2(g)? This enthalpy diagram shows the standard heat of formation of water. f 192 Remember to take into account the number of moles of each reactant and product. 198 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 22 2/25/2013 17.1 The Flow of Energy > Sample Problem 17.8 17.1 The Flow of Energy > 2 Calculate Solve for the unknown. Calculate ∆H ° for the reaction. f Standard heats of formation are used to calculate the enthalpy change for the reaction of carbon monoxide and oxygen. ° • The ∆H is negative, so the reaction is exothermic. • 2CO(g) + O2(g) → 2CO2(g) f • This outcome makes sense because combustion reactions always release heat. = (–787.0 kJ) – (–221.0 kJ) • The diagram shows the difference between ∆Hf (product) and ∆Hf (reactants) after taking into account the number of moles of each. ° ° = –566.0 kJ 199 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > 200 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calculate the standard heat of reaction for the following: CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g) CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g) °(CH (g)) = –74.86 kJ/mol °(C(s, diamond)) = 1.9 kJ/mol ∆H °(HCl(g)) = –92.3 kJ/mol °(CH (g)) = –74.86 kJ/mol °(C(s, diamond)) = 1.9 kJ/mol ∆H °(HCl(g)) = –92.3 kJ/mol ∆Hf 4 ∆Hf 201 4 f ∆Hf °(reactants) = [1 mol CH (g) × ∆H °CH (g)] + [1 mol Cl ∆Hf ∆Hf °(products) = [1 mol C(s) × ∆H °C(s, diamond)] + [4 mol HCl × °HCl(g)] 4 f 4 2 × ∆Hf For a reaction that occurs at standard conditions, you can calculate the heat of reaction by using standard heats of formation. °Cl (g)] 2 = –74.86 kJ + 0.0 kJ = –74.86 kJ 202 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Glossary Terms ° f °(products) – ∆H °(reactants) = –367.3 kJ – (–74.86 kJ) = – f Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > BIG IDEA ° = ∆H °(products) – ∆H °(reactants) ∆H = 1.9 kJ + (4 × –92.3 kJ) = –367.3 kJ ∆H = ∆Hf 292.4 kJ 203 Key Concepts & Key Equation Hess’s law allows you to determine the heat of reaction indirectly by using the known heats of reaction of two or more thermochemical equations. ∆Hf f Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Calculate the standard heat of reaction for the following: ∆Hf Standard Heats of Formation 17.1 The Flow of Energy > 3 Evaluate Does the result make sense? ° = ∆H °(products) – ∆H °(reactants) ∆H Sample Problem 17.8 f f 204 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 17.1 The Flow of Energy > Matter and Energy • Hess’s law of heat summation: if you add two or more thermochemical equations to give a final equation, then you also add the heats of reaction to give the final heat of reaction The heat of reaction can be calculated by using the known heats of reaction of two or more thermochemical equations or by using standard heats of formation. ° • standard heat of formation (∆Hf ): the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25 C END OF 17.4 ° 205 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 206 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 207 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 23 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 208 209 210 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 211 212 213 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 214 215 216 24 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 217 218 219 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 220 221 222 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 223 224 225 25 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 226 227 228 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 229 230 231 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 232 233 234 26 2/25/2013 17.1 The Flow of Energy > 17.1 The Flow of Energy > 17.1 The Flow of Energy > 235 236 237 17.1 The Flow of Energy > 17.1 The Flow of Energy > 238 239 27
© Copyright 2024