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Chapter 6 Thermochemistry Chapter 6 Chapter 6 Thermochemistry 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 Chemical Hand Warmers The Nature of Energy: Key Definitions The First Law of Thermodynamics: There is no Free Lunch Quantifying Heat and Work Measuring DE for Chemical Reactions: Constant – Volume Calorimetry Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Constant – Pressure Calorimetry: Measuring DHrxn Relationships Involving DHrxn Determining Enthalpies of Reaction from Standard Enthalpies of Formation Energy Use and the Environment 2 Chapter 6 Thermochemistry Matter vs Energy • Chapters 1 – 5 have mainly been concerned with matter – – • How we describe it How we measure it This chapter describes energy – – How we describe it How we measure it 3 Section 6.1 Chemical Hand Warmers Thermochemistry • The study of the relationships between matter and energy is a field called thermochemistry • 4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s) + heat – Oxidation of iron • This reaction takes place inside chemical heat packs • Notice heat is written as a product 4 Section 6.1 Chemical Hand Warmers Thermochemistry • Another familiar reaction we have seen • 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) + heat – Oxidation of octane • This reaction also generates heat • The heat is used to do work – we will talk about this in this chapter 5 Section 6.2 The Nature of Energy: Key Definitions The Transfer of Energy: Heat and Work • Energy is something an object has • Heat and work are ways energy is transferred • When you hold a warm object in you hand, energy is transferred from the object to your hand in the form of heat • When a ball on a pool table hits another ball energy is transferred from the first ball to the second ball in the form of work 6 Section 6.2 The Nature of Energy: Key Definitions Potential vs Kinetic Energy • Potential energy is stored energy. Energy of position. – The water in a reservoir behind a dam, an automobile poised to coast downhill, and a coiled spring have potential energy waiting to be released. • Kinetic energy is the energy of motion. – When the water falls over the dam and turns a turbine, when the car rolls downhill, or when the spring uncoils and makes the hands on a clock move, the potential energy in each is converted to kinetic energy. 7 Section 6.2 The Nature of Energy: Key Definitions Potential vs Kinetic Energy in Chemistry • Chemical Potential energy is also stored energy – the energy stored in chemical bonds – – – – CO2 + H2O Low Ep More stable Lower energy sugars hi Ep less stable higher energy • Chemical Kinetic energy is the energy of motion of the particles – what we measure when we measure temperature 8 Section 6.2 The Nature of Energy: Key Definitions The Law of Conservation of Energy • Law of conservation of energy – energy can be neither created nor destroyed – Can assume different forms • Energy from iron and oxygen in our heat pack example becomes heat • Energy from octane becomes heat that is converted to work • Energy from the sun is converted to complex sugars in plants 9 Section 6.2 The Nature of Energy: Key Definitions The Law of Conservation of Energy • Law of conservation of energy – energy can be neither created nor destroyed – Can be transferred from one object to another • In this chapter we are going to learn how to quantitatively measure the transfer of energy as either heat or work 10 Section 6.2 The Nature of Energy: Key Definitions System vs Surroundings • Energy is exchanged between the system under investigation and the surroundings • System – part of the universe on which we wish to focus attention – Kind of a weird definition can be just the chemicals reacting or can include the beaker and liquids in it • Surroundings – everything with which the system exchanges energy • The definition isn’t really the most important thing – Understanding how they exchange Energy is important 11 Section 6.2 The Nature of Energy: Key Definitions Exothermic vs Endothermic Reaction • Exothermic Reaction – – • Endothermic Reaction – – • Heat flows out of the system to the surroundings Amount lost by system exactly the same as amount gained by surroundings * Heat flows is into a system from the surroundings. Amount gained by system is exactly the same as amount lost by surroundings* * What law makes this true?* 12 Section 6.3 (preview) The First Law of thermodynamics: There is no Free Lunch Heat and Work: Pathways to Energy Change • A system can exchange energy with its surrounding through heat, work, or both. • The change in Energy is the sum of the heat exchanged and work done DE = q + w 13 Section 6.2 The Nature of Energy: Key Definitions Units of Energy • In Chapter 5 we saw that Kinetic Energy was defined as ½ mv2 where m = mass and v = velocity 2 The SI unit of Energy is mv 2 m = kg 1J s • 1 J = the joule • Another common unit is the calorie (cal) – 1 cal = 4.184 Joules • The Calorie we see on nutritional labels = 1000 cal = kilocal 14 Section 6.2 The Nature of Energy: Key Definitions Energy Conversion Factors • Don’t memorize these! • Be able to use them. 15 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Internal Energy • The Internal Energy of a system is the sum of all the kinetic and potential energies of the particles that compose the system. • Potential energy is the energy stored in the chemical bonds • Kinetic energy is the energy of the motion of the particles 16 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Internal Energy is a State Function • A state function is a quantity whose value depends only on the state of the system not on the pathway the system arrived in that state. • What does this mean? 17 Section 6.3 The First Law of thermodynamics: There is no Free Lunch What is a state function • The change in altitude is a state function. • The pathway is not 18 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Change in Internal Energy • Just like a change in altitude a change in internal energy is a state function. DE = Efinal – Einitial • In a chemical system the reactants are the initial state (ground level) and the products are the final state (peak) DE = Eproducts – Ereactants 19 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Change in Internal Energy • For the reaction C (s) + O2 (g) CO2 (g) • We can represent the DE like this • Since DE = Eproducts – Ereactants DE is negative 20 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Change in Internal Energy • If the system (the reaction) loses energy DE is negative • Where does the energy go? • Energy has to be conserved (1st Law of TD) • Energy is released to the surroundings 21 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Change in Internal Energy • For the reverse reaction CO2 (g) C (s) + O2(g) • The reaction diagram would look like this • Since DE = Eproducts – Ereactants DE is positive 22 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Change in Internal Energy • If the system (the reaction) gains energy DE is positive • Energy is absorbed from the surroundings 23 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Concept Check Hydrogen gas and oxygen gas react explosively to form water. 2H 2 (g) + O2 (g) 2H 2O(g) + energy (heat) Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? 24 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Solution Hydrogen gas and oxygen gas react explosively to form water. 2H 2 (g) + O2 (g) 2H 2O(g) + energy (heat) Which is lower in energy: a mixture of hydrogen and oxygen gases, or water? Water is lower in energy because a lot of energy was released when hydrogen and oxygen gases reacted. 25 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Summarizing Energy Flow • In both cases the DEsys is the opposite of the DEsurr DEsys = – DEsurr • If reactants have higher internal energy than products DEsys is negative energy flows out of the system • If reactants have lower internal energy than products DEsys is positive and energy flows into the system 26 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Heat and Work: Pathways to Energy Change • A system can exchange energy with its surrounding through heat, work, or both. • The change in Internal Energy is the sum of the heat exchanged and work done DE = q + w 27 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Heat and Work: Pathways to Energy Change • When we looked at the change in altitude on the mountain we saw that the two pathways were very different. – The change in altitude was a state function – The pathways were not state functions • Same situation for heat and work DE = heat + work = q + w DE is a state function – q and w are not 28 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Heat and Work: Pathways to Energy Change DE is a state function – q and w are not • Think about a gallon of gas – Burning in an open container (lots of heat, not much work) – Moving a car 40 miles (lots of work, less heat) • Same change in DE but q and w are completely different depending on the path 29 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Sign conventions for DE, q and w DE with no subscript always refers to DE of the system – Negative sign means the system loses energy • What about heat and work? • Signs are defined from the point of view of the system. – – – – If the system does work w is negative If the system loses heat q is negative If work on done on the system w is positive If the system gains heat q is positive 30 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Concept Check Determine the value of DE for the combustion of 1 gallon of gasoline in a car engine q = – 8.89 x 104 kJ w = – 3.81 x 103 kJ 31 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Solution Determine the value of DE for the combustion of 1 gallon of gasoline in a car engine DE = q + w = – 8.89 x 104 kJ + – 3.81 x 103 kJ DE = – 9.27 x 104 kJ 32 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Concept Check In this problem for the determination of the value of DE for the combustion of 1 gallon of gasoline in a car engine, why are the values of q and w both negative. q = – 8.89 x 104 kJ w = – 3.81 x 103 kJ 33 Section 6.3 The First Law of thermodynamics: There is no Free Lunch Solution In this problem for the determination of the value of DE for the combustion of 1 gallon of gasoline in a car engine, why are the values of q and w both negative. q = – 8.89 x 104 kJ combustion produces heat which leaves the system w = – 3.81 x 103 kJ work is done on the surroundings 34 Section 6.4 Quantifying Heat and Work How do we measure Heat and Work • How do we measure the amount of heat gained or released or the amount of work done on a system, or performed by a system? • We measure changes in temperature and volume. • Then use a bunch of equations. • Lets do heat first. 35 Section 6.4 Quantifying Heat and Work Heat • Heat is the exchange of thermal energy between a system and its surroundings caused by a temperature difference. • Heat and temperature are not the same thing. • Although sometimes we talk about them as if they were. • Temperature is a measure of thermal Energy. • Heat is a transfer of thermal Energy – Heat is not a substance contained by an object, although we often talk of heat as if this were true. 36 Section 6.4 Quantifying Heat and Work Temperature Change and Heat Capacity • Substances respond differently to being heated Think of a pool vs a car door on a hot day Metal absorbs heat fast Water absorbs heat slowly • The Heat capacity (C) is a measure of this property C heat absorbed increase in tempera ture 37 Section 6.4 Quantifying Heat and Work Temperature Change and Heat Capacity heat absorbed C increase in tempera ture • Metal absorbs heat fast • Has a low heat capacity • Water absorbs heat slowly • Has a high heat capacity 38 Section 6.4 Quantifying Heat and Work Heat Capacity • Specific heat capacity (Cs): The energy required to raise the temperature of one gram of a substance by one degree Celsius. • Molar heat capacity (Cm): The energy required to raise the temperature of one mole of substance by one degree Celsius. 39 Section 6.4 Quantifying Heat and Work Specific Heat Capacities • Specific Heat Capacities (per g) and Molar heat capacities (per mole) are called intensive properties • Intensive properties of matter depend on the identity of the matter – not the amount 40 Section 6.4 Quantifying Heat and Work Heat Calculations • Specific Heat Capacity is used to determine the relationship between the amount of heat added to a system and the corresponding change in temperature. DT is always Tfinal - Tinitial 41 Section 6.4 Quantifying Heat and Work Learning Check A piece of iron with a mass of 75.0 g at 125.0 °C is allowed to cool to room temperature of 25.0 °C . Determine the magnitude and sign of q. The specific heat capacity of iron is 0.449 J/°C g. 42 Section 6.4 Quantifying Heat and Work Solution A piece of iron with a mass of 75.0 g at 125.0 °C is allowed to cool to room temperature of 25.0 °C . Determine the magnitude and sign of q. The specific heat capacity of iron is 0.449 J/°C g. q = m × Cs × ΔT m = 75.0 g Cs = 0.449 J/°C g. DT = 25.0 – 125.0 = -100.0°C 0.449 J q = 75.0 g x x - 100.0 C = - 3.37x10 3 J Cg 43 Section 6.4 Quantifying Heat and Work Learning Check The specific heat capacity of silver is 0.24 J/°C g. a.Calculate the energy required to raise the temperature of 150.0g Ag from 273 K to 298 K. b.Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0°C (called the molar heat capacity of silver). 44 Section 6.4 Quantifying Heat and Work Solution The specific heat capacity of silver is 0.24 J/°C °g. a.Calculate the energy required to raise the temperature of 150.0g Ag from 273 K to 298 K. q = m × Cs × ΔT m = 150.0 g Cs = 0.24 J/°C g. ΔT = 298 – 273 = 25°C 0.24 J q = 150.0 g x o x 25 oC = 9.0x10 2 J Cg 45 Section 6.4 Quantifying Heat and Work Solution The speciﬁc heat capacity of silver is 0.24 J/°C °g. b. Calculate the energy required to raise the temperature of 1.0 mol Ag by 1.0°C (called the molar heat capacity of silver). J 107.9 g J 0.24 x = 26 gC mol mol C 46 Section 6.4 Quantifying Heat and Work Thermal Energy Transfer • If two substances at different temperatures are combined the heat lost by one substance is absorbed by the other • qsys = – qsurr • For example if we place a piece of hot metal in a beaker of water • qmetal = – qwater • The negative sign here just signifies that the heat is moving in the opposite direction 47 Section 6.4 Quantifying Heat and Work Thermal Energy Transfer • How do we use this in calculations • qmetal = – qwater • mmetal × Csmetal × ΔTmetal = – mwater × Cswater × ΔTwater • There are two types of these problems • 6.3 in the book – unnecessarily complicated – you are not responsible for any calculations that ask you for final temperature of both substances • Other type gives you all variables except one 48 Section 6.4 Quantifying Heat and Work Learning Check • An iron bar at 65.5°C is placed in a beaker of water at 25.0 °C. If the mass of the water is 65.0 g and the final temperature at thermal equilibrium is 32.5 °C. What is the mass of the iron bar? The specific heat capacity of water is 4.184 J/ °C g. 49 Section 6.4 Quantifying Heat and Work Solution • An iron bar at 65.5°C is placed in a beaker of water at 25.0 °C. If the mass of the water is 65.0 g and the final temperature at thermal equilibrium is 32.5 °C. What is the mass of the iron bar? • qmetal = – qwater • mmetal × Csmetal × ΔTmetal = – mwater × Cswater × ΔTwater J mmetal 65.0g4.18 o 7.5oC mwater Cs water DTwater g C = 140 g = J Csiron DTiron o 0.449 o –33.0 C g C 50 Section 6.4 Quantifying Heat and Work Work • We know that energy transfer can occur via heat or work • We have seen how to determine the amount of heat transfer by measuring the change in temperature • Now we will look at how to determine the amount of work by measuring the change in volume 51 Section 6.4 Quantifying Heat and Work Work: Pressure-Volume Work • Chemical reactions can do several different types of work • We are only going to consider pressure-volume (or PV) work. • Work is a force (F) acting through a distance. • PV work occurs when the force is caused by a volume change against an external pressure. – Like in the engine of a car – when the pistons are pushed outward against the external atmospheric pressure. 52 Section 6.4 Quantifying Heat and Work Work: Pressure-Volume Work • So we define work as Force x Distance or F x D • W=FxD • In chapter 5 we defined pressure P = F/A which we can rearrange to F = P x A • Substitute (P x A) for F into the definition for work and we get • w = P x A x D (pressure x area x distance) 53 Section 6.4 Quantifying Heat and Work Work: Pressure-Volume Work • w = P x A x D (pressure x area x distance) • w = P x A x Dh (distance between initial and final state • w = PDV (where A x Dh = DV) 54 Section 6.4 Quantifying Heat and Work Work: Pressure-Volume Work • • • • So Pressure volume work is defined w = PDV The only thing missing is the sign As the volume of the cylinder expands, work is done on the surroundings, so the sign of the work must be negative according to the way we defined work and heat, the only way to accomplish this is to include the negative sign in the formula • w = – PDV 55 Section 6.4 Quantifying Heat and Work Concept Check Which of the following performs more work? a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L. 56 Section 6.4 Quantifying Heat and Work Solution Which of the following performs more work? a)A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L. w = – (2 atm)(4.0 – 1.0) = – 6 L·atm b) A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L w = – (3 atm)(3.0 – 1.0) = – 6 L·atm They both perform the same amount of work. 57 Section 6.4 Quantifying Heat and Work Concept Check A balloon is being inﬂated to its full extent by heating the air inside it. The volume of the balloon changes from 4.00 x106 L to 4.50 x106 L. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate the work done in this process in Joules. (To convert between L atm and J, use 1 L atm =101.3 J.) 58 Section 6.4 Quantifying Heat and Work Solution A balloon is being inﬂated to its full extent by heating the air inside it. The volume of the balloon changes from 4.00 x106 L to 4.50 x106 L. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate the work done in this process in Joules. (To convert between L atm and J, use 1 L atm =101.3 J.) w = PDV = 1.0 atm x 4.50 x 10 6 L - 4.00 x 10 6 L = - 5.0 x 10 5 L atm 101.3 J w = 5.0x10 L atmx = 5.1x10 7 J L atm 5 59 Section 6.4 Quantifying Heat and Work Concept Check The balloon in the previous example was expanded by heating the air inside it by the addition of 1.3 x108J of energy as heat. Calculate DE for the process. 60 Section 6.4 Quantifying Heat and Work Solution DE = q+w Heat = 1.3 x108J (given in problem) Work = – 5.1 x 107 J (previous problem) DE = q + w = 1.3x10 J + 5.1x10 J 8x10 J 8 7 7 61 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Measuring DE - Calorimetry • We measure heat evolved in a chemical reaction using calorimetry • In calorimetry we measure the thermal energy (heat) exchanged between the reaction (the system) and the surroundings by measuring the change in temperature • Two types of Calorimetry – Constant Volume – measures DE – Constant Pressure – measures DH (see slide 86) 62 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Measuring DE - Calorimetry • We are going to look at Constant Volume Calorimetry first • Measure of DE • Change in Internal Energy of a Reaction – Remember Internal Energy is the sum of all the kinetic and potential energy. 63 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Measuring DE • So we know that systems exchange energy with their surroundings via heat and work. 64 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Measuring DE • Change in internal energy during a reaction is a sum of both the heat and work. DE = q + w • We could determine q and w separately and then add them to determine DE. • It would be easier though if we could force all the change in internal energy to manifest as heat flow (q) with no work done. Only 1 measurement. 65 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Constant Volume Calorimetry How do we force all of the DE to be heat flow (q) Well DE = q + w And w = – PDV So DE = q – PDV So if we carry out the reaction at constant volume then DV = 0 and w = 0 so DE = q • This is written as DErxn = qv • • • • • 66 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Constant Volume Calorimetry • How do we measure DErxn = qv • Device called a constant volume (or bomb)calorimeter. – Sample is burned in oxygen gas = system – Surroundings are the calorimeter. • Measure heat flow by change in temperature 67 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Constant Volume Calorimetry • DErxn = qv • Temperature change measured in the bomb calorimeter is converted to DE using • qcal = Ccal x ΔT C = heat capacity of the calorimeter (J/°C) DT = change in temperature (°C) The calorimeter absorbs heat so the sign of the qrxn is reversed to reflect that the system (the sample) released heat • qrxn = – qcal 68 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Learning Check When a 1.50 g sample of methane or hydrogen gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C for methane and 14.3°C for hydrogen. Calculate qrxn for methane and hydrogen. The bomb calorimeter has heat capacity of 11.3 kJ/°C. 69 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Solution When a 1.50 g sample of methane or hydrogen gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C for methane and 14.3°C for hydrogen. Calculate qrxn for methane and hydrogen. The bomb calorimeter has heat capacity of 11.3 kJ/°C. 11.3 kJ For Methane q cal = 7.3 C x o = 82 kJ C q rxn = q cal so q rxn = 82 kJ 11.3 kJ For Hydrogen q cal = 14.3 oC x o = 162 kJ C q rxn = q cal so q rxn = 162 kJ o 70 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry DErxn per mole • We just calculated the heat of reaction (in Joules) based on temperature change and heat capacity of the calorimeter DErxn can also be expressed per mole of reactant • Divide qrxn/moles of reactant 71 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Learning Check Calculate DErxn/mole for methane and hydrogen from the last problem. 72 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Solution Calculate DErxn/mole for methane and hydrogen from the last problem. For Methane 82 kJ 16.04 g x = - 880 kJ/mol 1.50 g mol 162 kJ 2.016 g = x = - 218 kJ/mol 1.50 g mol q rxn = 82 kJ DE rxn = For Hydrogen q rxn = 162 kJ DE rxn 73 Section 6.5 Measuring DE for Chemical Reactions: Constant-Volume Calorimetry Two Types of Calorimetry • On a previous slide we saw • Two types of Calorimetry – Constant Volume – measures DE – Constant Pressure – measures DH • What is DH DH is a change in Enthalpy • What is Enthalpy? 74 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure What is Enthalpy • • Enthalpy (H) is the total energy of a system Enthalpy is the sum of internal energy and a quantity PV – – • • Internal energy (E) - energy required to create a system PV - energy required to make room for the system it by displacing its environment and establishing its volume and pressure. So H = E + PV Enthalpy is a State function just like Internal Energy 75 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure What is Enthalpy • • • • So H = E + PV Enthalpy represents both internal energy and a quantity PV So a change in Enthalpy represents a change in internal energy and a change in PV DH = DE +D(PV) How do we measure Enthalpy? 76 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure How do we measure Enthalpy? • • • • • • When a reaction occurs in a sealed bomb calorimeter all the energy exchanged is in the form of heat Most chemical reactions in the lab don’t occur this way – they occur in open containers on the bench. Energy is exchanged as both heat and work In reality though, most of the time we are not really interested in the small amount of work the reaction does expanding against the atmosphere What we are really interested in is the heat exchanged How do we determine this value? 77 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Change in Enthalpy is the Heat Flow at Constant Pressure • If H = E + PV • And DH = DE +D(PV) – • So DH = DE +PDV – • • • • At constant P – the only change is volume And we have already defined ΔE = q + w So DH = q + w +PDV and –PDV = w so So DH = q + w – w And DH = qp So Enthalpy is a measure of heat exchange only! 78 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Enthalpy vs Internal Energy DH and DE can seem quite similar • They both represent state functions DE is a measure of all the energy (heat and work) exchanged with the surroundings DH is a measure of only the heat exchanged under conditions of constant pressure 79 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Summarizing Enthalpy •The value of DH is the amount of heat absorbed or released under conditions of constant pressure – Endothermic Reaction DH will be positive – heat absorbed • Reaction absorbs heat from the surroundings • Feels cool – Exothermic Reaction DH will be negative – heat released • Reaction gives off heat to the surroundings • Feels warm 80 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Learning Check Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DH = –2044 kJ For this reaction do the products or the reactants have the higher enthalpy? Is this reaction exothermic or endothermic? 81 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Solution Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DH = –2044 kJ For this reaction do the products or the reactants have the higher enthalpy? The ΔH is negative therefore the products have less enthalpy than the reactants. Is this reaction exothermic or endothermic? Exothermic 82 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Endothermic and Exothermic Processes: A Molecular View • An exothermic reaction releases energy as heat • Where does that energy come from? • Two forms of energy kinetic and potential. – Kinetic is energy of motion – reflected in temperature. – Exothermic reaction can’t be drawing from the pool of kinetic energy – otherwise the temperature would go down. • In an exothermic reaction the reaction feels warm • So the energy must be coming from the stored potential energy in the bonds of the reactants 83 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure How do we measure Enthalpy • • • • • Total enthalpy (H) cannot be measured directly We can only measure a change in Enthalpy (DH) And we do this by measuring heat flow (DT) We measure a change in enthalpy by measuring DT at constant pressure Then we convert the change in temperature to an amount of heat 84 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure How do we measure Enthalpy vs Internal Energy • The value of DE for a reaction is the amount of heat absorbed or released under constant volume. – Bomb Calorimeter DE = qv • The value of DH for a reaction is the amount of heat absorbed or released under constant pressure. – Coffee Cup Calorimeter DH = qp 85 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Coffee-Cup Calorimeter • Constant Pressure Calorimeter • The reaction takes place in solution inside the inner cup. • Change in temperature of the solution is measured. 86 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Coffee-Cup Calorimeter • Temperature change measured in the coffeecup calorimeter is converted to DH using • qsoln = msoln x Cs,soln x ΔT msoln = mass of solution Cs,soln = heat capacity of the solution inside the calorimeter (J/g°C) DT = change in temperature (°C) • qrxn = – qsoln 87 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Learning Check • Magnesium metal reacts with hydrochloric acid according to the following equation • Mg(s) + 2HCl(g) MgCl2 (aq) + H2 (g) • 0.316 g of magnesium metal is combined with HCl to a final volume of 200.0 mL and completely reacts. The temperature of the solution changes from 25.6°C to 32.8°C. The density of the solution is 1.00 g/mL and the Cs,soln = 4.18 J/g°C. Determine qrxn. 88 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Solution • Mg(s) + 2HCl(g) MgCl2 (aq) + H2 (g) • 0.316 g of magnesium metal is combined with HCl to a final volume of 200.0 mL and completely reacts. The temperature of the solution changes from 25.6°C to 32.8°C. The density of the solution is 1.00 g/mL and the Cs,soln = 4.18 J/g°C. Determine qrxn. q so ln = msoln x C s,soln x DT = 200.0 g x 4.18 J o 3 x 7.2 C = 6.0x 10 J o gC q rxn = – 6.0x 10 3 J 89 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn DHrxn per mole DHrxn can also be expressed per mole of reactant • Divide qrxn/moles of reactant 90 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Learning Check • Find DHrxn/mole for the previous problem. 91 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Solution • Find DHrxn/mole for the previous problem. qrxn 6.0x 10 3 J 24.31 g J DHrxn = x = 4.6 x 10 5 mole Mg 0.316 g Mg mole mole 92 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Conceptual Connection • Lighters are usually fueled by butane (C4H10). When 1 mole of butane burns at constant pressure it produces 2658 kJ of heat and does 3kJ of work. What are the values of DH and DE for the combustion of one mole of butane. 93 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Solution • Lighters are usually fueled by butane (C4H10). When 1 mole of butane burns at constant pressure it produces 2658 kJ of heat and does 3kJ of work. What are the values of DH and DE for the combustion of one mole of butane. • w and q are both negative – work is done on the surroundings and heat is lost to surroundings DE = q + w = – 2661 kJ DH = q = – 2658 kJ 94 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Conceptual Connection • The same reaction with exactly the same amount of reactants is conducted in a bomb calorimeter and a coffee cup calorimeter. In one measurement qrxn = – 12.5 kJ and in the other qrxn = – 11.8 kJ. Which value was determined in the bomb calorimeter? (Assume the reaction has a positive DV in the coffee-cup calorimeter) 95 Section 6.7 Constant-Pressure Calorimetry: Measuring DHrxn Solution • The same reaction with exactly the same amount of reactants is conducted in a bomb calorimeter and a coffee cup calorimeter. In one measurement qrxn = – 12.5 kJ and in the other qrxn = – 11.8 kJ. Which value was determined in the bomb calorimeter? (Assume the reaction has a positive DV in the coffee-cup calorimeter) • The larger value 12.5 was measured in the bomb calorimeter which measures DE (heat and work). The coffee cup calorimeter measures DH (heat only). 96 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Stoichiometry Involving DH: Thermochemical Equations • Enthalpy change is also called enthalpy of reaction or heat of reaction. • Extensive property – depends on amount of material undergoing the reaction – The amount of heat generated depends on the amount of reactant or product • Consider the combustion of propane 97 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Example Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DHrxn = –2044 kJ This means that 2044 kJ of energy is released for every mol of propane or every 5 moles of oxygen reacted 98 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Example Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DHrxn = –2044 kJ Calculate DH in which 5.00 g of propane is burned in excess oxygen at constant pressure. 1 mol C 3H 8 2044 kJ 5.00 g C 3H 8 x x = – 232 kJ 44.09 g C 3H 8 mol C 3H 8 99 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Learning Check Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DHrxn = –2044 kJ Calculate DH in which 25.0 g of water is released when propane is burned in excess oxygen at constant pressure. 100 Section 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure Solution Consider the combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DHrxn = –2044 kJ Calculate DH in which 25.0 g of water is released when propane is burned in excess oxygen at constant pressure. 1 mol H 2O 1 mol C 3H 8 2044 kJ 25.0 g H 2O x 18.02 g H 2O x 4 mol H 2O x mol C 3H 8 = – 709 kJ 101 Section 6.8 Relationships Involving DHrxn Characteristics of Enthalpy Changes DHrxn • The change in enthalpy for a reaction (DHrxn) is associated with a particular reaction • If we change the characteristics of the the reaction then DHrxn changes too • Lets look at a couple of examples 102 Section 6.8 Relationships Involving DHrxn Characteristics of Enthalpy Changes 1. If a chemical reaction is multiplied by some factor then DHrxn is multiplied by the dame factor. A + 2B C DH1 Multiply by 2 2A + 4B 2C DH2 = DH1 x 2 103 Section 6.8 Relationships Involving DHrxn Characteristics of Enthalpy Changes 2. If a reaction is reversed, then DHrxn changes sign. A + 2B C DH1 Reverse reaction C A + 2B DH2 = – DH1 104 Section 6.8 Relationships Involving DHrxn 3. If a chemical reaction can expressed as the sum of a series of steps, then DHrxn for the overall equation is the sum of the heats of reactions for each step. A + 2B C DH1 C 2D DH2 ––––––––––––––––––– A + 2B 2D DH3 = DH1 + DH2 105 Section 6.8 Relationships Involving DHrxn Hess’s Law • This last relationship is called Hess’s Law 106 Section 6.8 Relationships Involving DHrxn Example of Hess’s Law • N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ This reaction also can be carried out in two distinct steps, with enthalpy changes designated by DH2 and DH3. N2(g) + O2(g) → 2NO(g) DH2 = 180 kJ 2NO(g) + O2(g) → 2NO2(g) DH3 = – 112 kJ N2(g) + 2O2(g) → 2NO2(g) DH2 + DH3 = 68 kJ 107 Section 6.8 Relationships Involving DHrxn Example • Consider the following data: 1 3 NH3 (g) N2 (g) H2 (g) 2 2 2 H2 (g) O2 (g) 2 H2O(g) • DH = 46 kJ DH = 484 kJ Calculate ΔH for the reaction 2 N2 (g) 6 H2O(g) 3 O2 (g) 4 NH3 (g) 108 Section 6.8 Relationships Involving DHrxn Example 1 3 NH3 (g) N2 (g) H2 (g) 2 2 2 H2 (g) O2 (g) 2 H2O(g) DH = 46 kJ DH = 484 kJ Desired reaction: 2 N2 (g) 6 H2O(g) 3 O2 (g) 4 NH3 (g) Reverse the two reactions: 1 3 N2 (g) H2 (g) NH3 (g) DH = 46 kJ 2 H2O(g) 2 H2 (g) O2 (g) DH = +484 kJ 2 2 109 Section 6.8 Relationships Involving DHrxn Example • Multiply reactions to give the correct numbers of reactants and products: 1 3 4 N2 (g) H2 (g) NH3 (g) 2 2 3 2 H2O(g) 2 H2 (g) O2 (g) • 4 DH = 46 kJ 3 DH = +484 kJ Desired reaction: 2 N2 (g) 6 H2O(g) 3 O2 (g) 4 NH3 (g) 110 Section 6.8 Relationships Involving DHrxn Example • • Final reactions: 2 N2 (g) 6 H2 (g) 4 NH3 (g) DH = 184 kJ 6 H2O(g) 6 H2 (g) 3 O2 (g) DH = +1452 kJ Desired reaction: 2 N2 (g) 6 H2O(g) 3 O2 (g) 4 NH3 (g) DH = +1268 kJ 111 Section 6.8 Relationships Involving DHrxn Problem-Solving Strategy • • • Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal. Reverse any reactions as needed to give the required reactants and products. Multiply reactions to give the correct numbers of reactants and products. 112 Section 6.8 Relationships Involving DHrxn Why? • Why on earth would we calculate DH this way? 113 Section 6.8 Relationships Involving DHrxn Learning Check Two forms of carbon are graphite, the soft, black, slippery material used in “lead" pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), calculate ΔH for the conversion of graphite to diamond. Cgraphite (s) Cdiamond (s) Cgraphite (s) + O2 (g) CO 2 (g) DH = 394 kJ Cdiamond (s) + O2 (g) CO2 (g) DH = 396 kJ 114 Section 6.8 Relationships Involving DHrxn Solution Two forms of carbon are graphite, the soft, black, slippery material used in “lead" pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), calculate ΔH for the conversion of graphite to diamond. Cgraphite (s) Cdiamond (s) Cgraphite (s) + O2 (g) CO 2 (g) DH = 394 kJ CO2 (g) Cdiamond (s) + O2 (g) Cgraphite (s) Cdiamond (s) DH = 396 kJ DH = 2 kJ The ΔH for the conversion of graphite to diamond is 2 kJ/mole graphite 115 Section 6.8 Relationships Involving DHrxn Learning Check • Find DHrxn for the reaction • N2O(g) + NO2(g) 3NO(g) • Using • 2 NO(g) + O2(g) 2 NO2(g) • N2(g) + O2(g) 2 NO(g) • 2 N2O(g) 2 N2(g) + O2 (g) DH = – 113.1 kJ DH = 182.6 kJ DH = –163.2 kJ 116 Section 6.8 Relationships Involving DHrxn Solution • Desired Reaction N2O(g) + NO2(g) 3NO(g) • Reverse eq 1, multiply eq 1 and 3 by 1/2 • ½[2NO2(g) 2 NO(g) + O2(g)] ½[DH = + 113.1 kJ] • N2(g) + O2(g) 2 NO(g) DH = 182.6 kJ • ½[2 N2O(g) 2 N2(g) + O2(g)] ½[DH = –163.2 kJ] 117 Section 6.8 Relationships Involving DHrxn Solution • Desired Reaction N2O(g) + NO2(g) 3NO(g) • Reverse eq 1, multiply eq 1 and 3 by 1/2 • NO2(g) NO(g) + ½ O2(g) • N2(g) + O2(g) 2 NO(g) • N2O(g) N2(g) + ½ O2(g) ½[DH = + 113.1 kJ] DH = 182.6 kJ ½[DH = –163.2 kJ] 118 Section 6.8 Relationships Involving DHrxn Solution • Desired Reaction N2O(g) + NO2(g) 3NO(g) • Reverse eq 1, multiply eq 1 and 3 by 1/2 • NO2(g) NO(g) + ½ O2(g) • N2(g) + O2(g) 2 NO(g) • N2O(g) N2(g) + ½ O2(g) • ½ + 113.1 = 56.55 kJ 182.6 kJ ½ –163.2 kJ = –81.6 kJ 157.6 kJ 119 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation How many ways are there to determine DHrxn? • So far we have looked at two different ways to determine DHrxn • Calorimetry – coffee cup calorimeter – qp = DHrxn • Hess’s Law – where we infer DHrxn by knowing the values of DHrxn of our reaction from other reactions with known values • Now we will look at a third way 120 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes • The third method to determine DHrxn uses tables of Enthalpies (or Heats of Formation) • This method determines the amount of heat required to make all the reactants and products in something called the “standard state” and then compares them to each other. • So lets talk about standard state. 121 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes DH is the change in enthalpy for a chemical reaction – The difference in enthalpy between the products and reactants DH = Hproducts – Hreactants • So the difference in enthalpy is an absolute value (like the difference in altitude) • But enthalpy itself (like altitude) is a relative quantity – defined relative to some standard (such as sea level in the case of altitude) 122 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes • So what are the standards that we use when talk about enthalpy • Well there are three – but they are related • The standard state • The standard enthalpy change (DH°) • The standard enthalpy of formation (DHf°) 123 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes • Standard State • For a gas – pure gas at 1 atm of pressure • For a liquid or solid – pure substance in its most stable form at 1 atm of pressure and 298 K (25 °C) • For a solution – 1 M solution 124 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes • Standard Enthalpy Change (DH°) • The change in enthalpy for a process when all reactants and products are in their standard states. • Standard Enthalpy of Formation (DHf°) • For a pure compound – the change in enthalpy when 1 mole of the compound forms from its constituent elements in their standard states. • For a pure element in its standard state DHf°= 0 125 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard States and Standard Enthalpy Changes • When we assign a value of 0 for DHf° to a pure element in its standard state it is the same thing as assigning sea level an altitude of 0. • The interesting thing to notice about the table of enthalpies (or heats) of formation for compounds is that most of the values are negative – Below sea level • This actually means that compounds have less enthalpy (more stable) that the original elements that comprised them. 126 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 127 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Standard Enthalpy of Formation (DHf°) We can use DHf° to calculate DH for a reaction Add up all the DHf° for the products – Because the are being formed • Subtract all the DHf° of the reactants – Because they are being broken down • Pay attention to stoichiometry of equation • • DHrxn° = SnpDHf° (products) – SnpDHf° (reactants) 128 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Example Calculate DH° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: DHf° (kJ/mol) Na(s) 0 H2O(l) –286 NaOH(aq) –470 H2(g) 0 •[2(–470) + 0] – [0 + 2(–286)] = –368 kJ DH = –368 kJ 129 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Learning Check Calculate DH° for the following reaction: 4 NH3 (g) + 7O2(g) → 4 NO2(g) + 6H2O(l) Use the Table on the next slide. 130 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 131 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Solution Calculate DH° for the following reaction: 4 NH3 (g) + 7O2(g) → 4 NO2(g) + 6H2O(l) [4(34) + 6(–286)] – [4(–46) + 0] = –1396 kJ DH = –1396 kJ 132 Section 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation Problem-Solving Strategy: Enthalpy Calculations 1. When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes. 2. When the balanced equation for a reaction is multiplied by an integer, the value of ΔH for that reaction must be multiplied by the same integer. 3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: DH°rxn = SnpDHf° (products) - SnrHf° (reactants) 4. Elements in their standard states are not included in the DHreaction calculations because DHf° for an element in its standard state is zero. 133 Section 6.10 Energy Use and the Environment Energy Consumption • Combustion of Fossil fuels is Highly Exothermic • Coal C (s) + O2 (g) CO2 (g) DHrxn° = – 393.5 kJ • Natural Gas CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) DHrxn° = – 802.3 kJ • Petroleum C8H18 (l) + 25/2 O2 (g) 8 CO2 (g) + 9 H2O (g) DHrxn° = – 5074.1 kJ 134 Section 6.10 Energy Use and the Environment Environmental Problems Associated with Fossil Fuel Use • Non renewable • Produce greenhouse gases (CO2, H2O) • Produce other gases from side reactions due to impurities – Sulfur oxides – contribute to acid rain – Nitrogen oxides – contribute to SMOG – Ozone – ground level ozone is a dangerous pollutant 135

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