1-Energy and Chemistry_rev14Sep2024-2.pdf

I. Energy and Chemistry A. Chemical Energy from the Combustion of Fuels The Hindenburg disaster. The Hindenburg, a German airship filled with hydrogen gas, was destroyed in a spectacular fire at Lakehurst, New Jersey, in 1937. Outline TheFirst  The FirstLaw Law...

I. Energy and Chemistry A. Chemical Energy from the Combustion of Fuels The Hindenburg disaster. The Hindenburg, a German airship filled with hydrogen gas, was destroyed in a spectacular fire at Lakehurst, New Jersey, in 1937. Outline TheFirst  The FirstLaw Lawof of Thermodynamics Thermodynamics Enthalpy  Enthalpy Enthalpiesof of  Enthalpies Reaction Reaction Calorimetry  Calorimetry Hess’sLaw  Hess’s Law Enthalpiesof of  Enthalpies Formation Formation Fuels  Fuels A Look Ahead… We begin by studying the nature and different types of energy, which, in principle, are interconvertible. Next, we build up our vocabulary in learning thermochemistry, which is the study of heat change in chemical reactions. We see that the vast majority of reactions are either endothermic (absorbing heat) or exothermic (releasing heat). A Look Ahead… We learn that thermochemistry is part of a broader subject called the first law of thermodynamics, which is based on the law of conservation of energy. We see that the change in internal energy can be expressed in terms of the changes in heat and work done by a system. We then become acquainted with a new term for energy, called enthalpy, whose change applies to processes carried out under constant-pressure conditions. Discover that the enthalpy change associated with a given chemical reaction is equal to the enthalpies of the products minus the enthalpies of the reactants. This quantity is directly proportional to the amount of reactant consumed in the reaction.  Energy Energy is the ability to do work or transfer heat. – Energy used to cause an object that has mass to move is called work. – Energy used to cause the temperature of an object to rise is called heat. Thermochemistry © 2012 Pearson Education, Inc. Energy is the capacity to do work. Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position e.g. Chemical energy: associated with relative positions and arrangements within a given Thermochemistry substance 6  Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 Ek =  mv2 2 Thermochemistry © 2012 Pearson Education, Inc. Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. The most important form of potential energy in molecules is electrostatic potential energy, Eel: K Q 1Q 2 Eel = d Thermochemistry © 2012 Pearson Education, Inc.  Units of Energy The SI unit of energy is the joule (J): kg m2 1 J = 1  s2 An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = 4.184 J Thermochemistry © 2012 Pearson Education, Inc. Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open closed isolated Thermochemistry Exchange: mass & energy 10 energy nothing  Thermochemistry © 2012 Pearson Education, Inc. Answer: Open system. Humans exchange matter and energy with their surroundings. Thermochemistry © 2012 Pearson Education, Inc. Definitions: System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston). Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  ANSWER: The number of molecules will change because three reactant molecules (2 H2 and 1 O2) are needed to make two product (H2O) molecules. However, in a closed system like this one the total mass will not change. Thermochemistry © 2012 Pearson Education, Inc.  Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy Thermochemistry 16  Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects. Thermochemistry © 2012 Pearson Education, Inc.  Heat Units Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 °C (or 1 kelvin). The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. Thermochemistry  Heat Units The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. – It is named in honor of the English physicist James Prescott Joule. 1 calorie = 4.184 joules Thermochemistry Law of conservation of energy: During a chemical or physical change, energy can be neither created nor destroyed, although its form can change. Thermochemistry  Conversion of Energy Energy can be converted from one type to another. For example, the cyclist in Figure 5.2 has potential energy as she sits on top of the hill. Thermochemistry © 2012 Pearson Education, Inc.  Conversion of Energy As she coasts down the hill, her potential energy is converted to kinetic energy. At the bottom, all the potential energy she had at the top of the hill is now kinetic energy. Thermochemistry © 2012 Pearson Education, Inc.  First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Thermochemistry © 2012 Pearson Education, Inc.  Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Thermochemistry © 2012 Pearson Education, Inc.  Internal Energy By definition, the change in internal energy, ∆E, is the final energy of the system minus the initial energy of the system: ∆E = Efinal − Einitial Thermochemistry © 2012 Pearson Education, Inc.  Changes in Internal Energy If ∆E > 0, Efinal > Einitial – Therefore, the system absorbed energy from the surroundings. – This energy change is called endergonic. Thermochemistry © 2012 Pearson Education, Inc.  Changes in Internal Energy If ∆E < 0, Efinal < Einitial – Therefore, the system released energy to the surroundings. – This energy change is called exergonic. Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: If Efinal = Einitial, then ΔE=0 Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  ANSWER: Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is, ∆E = q + w. (mathematical statement of the first law) Thermochemistry © 2012 Pearson Education, Inc. ∆E, q, w, and Their Signs Thermochemistry © 2012 Pearson Education, Inc.  Work Done On the System w=Fxd w = -P ∆V ∆V > 0 F PxV= x d3 = F x d = w -P∆V < 0 d2 wsys < 0 Work is not a state function. ∆w = wfinal - winitial Thermochemistry initial 34 final  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: The system does work on the surroundings to move the piston upward, so w < 0. Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: No. If ΔV is zero, then the expression w = -PΔV is also zero. Thermochemistry © 2012 Pearson Education, Inc. A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? Thermochemistry 39 A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm? w = -P ∆V (a) ∆V = 5.4 L – 1.6 L = 3.8 L P = 0 atm W = -0 atm x 3.8 L = 0 L atm = 0 joules (b) ∆V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm w = -3.7 atm x 3.8 L = -14.1 L atm 101.3 J = -1430 J w = -14.1 L atm x 1L atm Thermochemistry 40  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: 0.69 L-atm = 70 J Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  ANSWER: ΔE = 50 J + (-85 J) = -35 J Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. Answer: +55 J Thermochemistry © 2012 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. Thermochemistry © 2012 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic. Thermochemistry © 2012 Pearson Education, Inc. Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l) Thermochemistry 52 Schematic of Exothermic and Endothermic Processes Thermochemistry 53  Thermochemistry © 2012 Pearson Education, Inc.  Answer: Endothermic Thermochemistry © 2012 Pearson Education, Inc.  Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature ∆E = Efinal - Einitial ∆P = Pfinal - Pinitial ∆V = Vfinal - Vinitial ∆T = Tfinal - Tinitial Potential energy of hiker 1 and hiker 2 is the same even though they took Thermochemistry different paths. 56  State Functions Usually, we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. Thermochemistry © 2012 Pearson Education, Inc.  State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. – In the system depicted in Figure 5.9, the water could have reached room temperature from either direction. Thermochemistry © 2012 Pearson Education, Inc.  State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so, ∆E depends only on Einitial and Efinal. Thermochemistry © 2012 Pearson Education, Inc. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its ∆E is the same. – But q and w are different in the two cases. Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  ANSWER: The battery is doing work on the surroundings, so w < 0. Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: Your weight Thermochemistry © 2012 Pearson Education, Inc.  Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV Thermochemistry © 2012 Pearson Education, Inc.  Enthalpy Change When the system changes at constant pressure, the change in enthalpy, ∆H, is ∆H = ∆(E + PV) This can be written ∆H = ∆E + P∆V Thermochemistry © 2012 Pearson Education, Inc.  Enthalpy Change Since ∆E = q + w and w = −P∆V, we can substitute these into the enthalpy expression: ∆H = ∆E + P∆V ∆H = (q + w) − w ∆H = q So, at constant pressure, the change in enthalpy is the heat gained or lost. Thermochemistry © 2012 Pearson Education, Inc. Endothermicity and Exothermicity A process is endothermic when ∆H is positive. Thermochemistry © 2012 Pearson Education, Inc. Endothermicity and Exothermicity A process is endothermic when ∆H is positive. A process is exothermic when ∆H is negative. Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Enthalpy of Reaction The change in enthalpy, ∆H, is the enthalpy of the products minus the enthalpy of the reactants: ∆H = Hproducts − Hreactants Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: ΔH is positive; the fact that the flask (part of the surroundings) gets cold means that the system is absorbing heat, meaning that q is positive. (See Figure 5.8.) Because the process occurs at P constant pressure, q = ΔH. p Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: In order to solidify, the gold must cool to below its melting temperature. It cools by transferring heat to its surroundings. The air around the sample would feel hot because heat is transferred to it from the molten gold, meaning the process is exothermic. (You may notice that solidification of a liquid is the reverse of the melting we analyzed in the exercise. As we will see, reversing the direction of a process changes the sign of the heat transferred.) Thermochemistry © 2012 Pearson Education, Inc.  Enthalpy of Reaction This quantity, ∆H, is called the enthalpy of reaction, or the heat of reaction. Thermochemistry © 2012 Pearson Education, Inc.  The Truth about Enthalpy 1. Enthalpy is an extensive property. 2. ∆H for a reaction in the forward direction is equal in size, but opposite in sign, to ∆H for the reverse reaction. 3. ∆H for a reaction depends on the state of the products and the state of the reactants. Thermochemistry © 2012 Pearson Education, Inc. The Truth about Enthalpy Thermochemistry © 2012 Pearson Education, Inc. The Truth about Enthalpy Thermochemistry © 2012 Pearson Education, Inc. The Truth about Enthalpy Thermochemistry © 2012 Pearson Education, Inc. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. ∆H = H (products) – H (reactants) ∆H = heat given off or absorbed during a reaction at constant pressure Hproducts > Hreactants Hproducts < Hreactants ∆H > 0 Thermochemistry ∆H < 0 82  Thermochemical Equations Is ∆H negative or positive? System absorbs heat Endothermic ∆H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) ∆H = 6.01 kJ/molThermochemistry 83  Thermochemical Equations Is ∆H negative or positive? System gives off heat Exothermic ∆H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H = -890.4 Thermochemistry kJ/mol 84  Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) ∆H = 6.01 kJ/mol If you reverse a reaction, the sign of ∆H changes H2O (l) H2O (s) ∆H = -6.01 kJ/mol If you multiply both sides of the equation by a factor n, then ∆H must change by the same factor n. 2H2O (s) 2H2O (l) ∆H = 2 x 6.01 = 12.0 kJ Thermochemistry 85  Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) ∆H = 6.01 kJ/mol H2O (l) H2O (g) ∆H = 44.0 kJ/mol How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) ∆H = -3013 kJ/mol 1 mol P4 3013 kJ 266 g P4 x x = 6470 kJ 123.9 g P4 1 mol P4 Thermochemistry 86  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc. ANSWER: -14.4 kJ Thermochemistry © 2012 Pearson Education, Inc.  Thermochemistry © 2012 Pearson Education, Inc.  A Comparison of ∆H and ∆E 2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) ∆H = -367.5 kJ/mol ∆E = ∆H - P∆V At 25 oC, 1 mole H2 = 24.5 L at 1 atm P∆V = 1 atm x 24.5 L = 2.5 kJ ∆E = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol Thermochemistry Sodium reacting with water to form 95 hydrogen gas.  Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity. Thermochemistry © 2012 Pearson Education, Inc.  Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 °C). Thermochemistry © 2012 Pearson Education, Inc.  Heat Capacity and Specific Heat Specific heat, then, is heat transferred Specific heat = mass × temperature change q c= m × ∆T Thermochemistry © 2012 Pearson Education, Inc.  Heat Capacity The heat capacity (C) of a body of matter is the quantity of heat (q) it absorbs or releases when it experiences a temperature change (ΔT) of 1 °C (or 1 kelvin): Heat capacity is an extensive property. For example, the heat capacity of a large cast iron pan is greater than the heat capacity of a small cast iron pan.Thermochemistry  Specific Heat Capacity The specific heat capacity (c) of a substance, commonly called its “specific heat,” is the quantity of heat required to raise the temperature of 1 gram of a substance by 1 °C (or 1 kelvin): Specific heat capacity is an intensive property—it only depends on the identity of the substance, not the amount. For example, the specific heat capacity of a large and small cast iron pan are identical. Thermochemistry  Figure 9.8 Due to its larger mass, a large frying pan has a larger heat capacity than a small frying pan. Because they are made of the same material, both frying pans have the same specific heat. (credit: Mark Blaser) Table 9.1 Specific Heats of Common Substances at 25 °C and 1 bar Substance Symbol (state) Specific Heat (J/g °C) helium He(g) 5.193 water H2O(l) 4.184 ethanol C2H6O(l) 2.376 ice H2O(s) 2.093 (at −10 °C) water vapor H2O(g) 1.864 nitrogen N2(g) 1.040 air 1.007 oxygen O2(g) 0.918 aluminum Al(s) 0.897 carbon dioxide CO2(g) 0.853 argon Ar(g) 0.522 iron Fe(s) 0.449 copper Cu(s) 0.385 lead Pb(s) 0.130 gold Au(s) 0.129 Thermochemistry silicon Si(s) 0.712  Calculating Heat The amount of heat, q, entering or leaving a substance can be calculated: If a substance gains thermal energy, Tfinal > Tinitial, then the value of q is positive. If a substance loses thermal energy, Tfinal < Tinitial, then the value of q is negative. Thermochemistry  Example 9.1 A flask containing 8.0 × 102 g of water is heated, and the temperature of the water increases from 21 °C to 85 °C. How much heat did the water absorb? Thermochemistry  Calorimetry One technique that can be used to measure the amount of heat involved in a chemical or physical process is known as calorimetry. Calorimetry is used to measure the amount of heat transferred to or from a substance. The heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat. Thermochemistry  System and Surroundings System: the substance or substances undergoing the chemical or physical change Surroundings: all other matter, including components of the measurement apparatus, that serve to either provide heat to the system or absorb heat from the system Thermochemistry  Calorimeter A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. Many processes measured in a calorimeter occur in solution. If the reaction is exothermic, the heat produced by the reaction is absorbed by the solution. If the reaction is endothermic, the heat required for the reaction to occur is provided by the solution. Thermochemistry  Figure 9.11 In a calorimetric determination, either (a) an exothermic process occurs and heat, q, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, q, is positive, indicating that thermal energy is transferred from the surroundings to the system.  Calorimeter The calorimeter must be well insulated in order to prevent the transfer of heat with the outside environment. To obtain accurate results, all heat exchange must take place only between the system and the surroundings. Coffee-cup calorimeters (constructed of polystyrene cups) are often used in general chemistry labs. Commercial calorimeters of better design are used in industry and for research. Thermochemistry Figure 9.12 A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture.  Figure 9.13 Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research.  Calorimetry Principles Consider a hot piece of metal (M) and cool water (W). If the metal is placed in the water within a calorimeter, heat will transfer from M to W. The temperature of M will decrease. The temperature of W will increase. Eventually, thermal equilibrium will be reached and both objects will have the same temperature. Thermochemistry  Calorimetry Principles Since this is done in a calorimeter, the heat exchange is only between M and W. The net change in heat is zero. Rearranging shows that the heat gained by M is equal to the heat lost by W. The heat of both substances is equal in magnitude but opposite in sign. Thermochemistry  Figure 9.14 In a simple calorimetry process, (a) heat, q, is transferred from the hot metal, M, to the cool water, W, until (b) both are at the same temperature.  Example 9.3 A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured to be 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table 9.1) and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the outside environment). Thermochemistry  Example 9.3 The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which must of been initially at some higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the outside environment, then heat given off by rebar = −heat taken in by water, or: Thermochemistry  Example 9.3 Since we know how heat is related to other measurable quantities, we have: Letting f = final and i = initial, in expanded form, this becomes: crebar × mrebar × (T f , rebar − Ti , rebar ) = −cwater × mwater × (T f , water − Ti , water ) Thermochemistry  Example 9.3 The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields: ( 0.449 J / g ⋅ C ) ( 360 g ) ( 42.7 C − T   i , rebar )= − ( 4.184 J / g ⋅ C ) ( 425 g ) ( 42.7 C − 24.0 C )    Solving this equation gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C. Thermochemistry  Calorimetry Principles When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. Because energy is neither created nor destroyed during a chemical reaction, the heat produced or consumed by the reaction (the “system”), qreaction, plus the heat absorbed or lost by the solution (the “surroundings”), qsolution, must add up to zero. Thermochemistry  Calorimetry Principles 0 qreaction + qsolution = This means that the amount of heat produced or consumed by the reaction equals the amount of heat absorbed or lost by the solution: qreaction = − qsolution This concept lies at the heart of all calorimetry problems and calculations. Thermochemistry  Example 9.5 When 50.0 mL of 1.00 M HCl(aq) and 50.0 mL of 1.00 M NaOH(aq), both at 22.0 °C, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximate amount of heat produced by this reaction? HCl(aq) + NaOH(aq) ⟶ NaCl(aq) + H2O(l) Thermochemistry  Example 9.5 To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solution temperature reaches 28.9 °C. The heat given off by the reaction is equal to that taken in by the solution. Therefore: qreaction = − qsolution Thermochemistry  Example 9.5 Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change: qsolution = csolution × msolution × ∆Tsolution Thermochemistry  Example 9.5 To proceed with this calculation, we need to make a few reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and density values. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 × 102 g (two significant figures). The specific heat of water is 4.184 J/g °C, so we use this value for the specific heat of the solution. Substituting these values gives: qsolution = ( 4.184 J/g × o C )(1.0 × 102 g )( 28.9 o C − 22.0 o C ) = 2.9 × 103 J Thermochemistry  Example 9.5 Finally, since we are trying to find the heat of the reaction, we have: qreaction = − qsolution = − 2.9 × 103 J The negative sign indicates that the reaction is exothermic. The reaction produces 2.9 kJ of heat. Thermochemistry  Another Example The addition of 3.15 g of Ba(OH)2・8H2O to a solution of 1.52 g of NH4SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Thermochemistry  Another Example Dissolving 3.0 g of CaCl2(s) in 150.0 g of water in a calorimeter (Figure 9.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? Thermochemistry  Figure 9.15 Chemical hand warmers produce heat that warms your hand on a cold day. In this one, you can see the metal disc that initiates the exothermic precipitation reaction. (credit: modification of work by Science Buddies TV/YouTube)  Figure 9.16 An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water. The cold pack then removes thermal energy from your body.  Bomb Calorimeter A bomb calorimeter is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. This type of calorimeter consists of a robust steel container (the “bomb”) that contains the reactants and is itself submerged in water. The energy produced by the reaction is trapped in the steel bomb and the surrounding water. Thermochemistry  Bomb Calorimetry Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Thermochemistry © 2012 Pearson Education, Inc.  Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ∆E, not ∆H. For most reactions, the difference is very small. Thermochemistry © 2012 Pearson Education, Inc.  Figure 9.17 (a) A bomb calorimeter is used to measure heat produced by reactions involving gaseous reactants or products, such as combustion. (b) The reactants are contained in the gas-tight “bomb,” which is submerged in water and surrounded by insulating materials. (credit a: modification of work by “Harbor1”/Wikimedia commons)  Whole-Body Calorimeter Whole-body calorimeters of various designs are large enough to hold a human being. These calorimeters are used to measure the metabolism of individuals under different conditions: – environmental conditions – dietary regimes – health conditions, such as diabetes A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods. Thermochemistry  Figure 9.18 (a) Macaroni and cheese contain energy in the form of the macronutrients in the food. (b) The food’s nutritional information is shown on the package label. In the US, the energy content is given in Calories (per serving); the rest of the world usually uses kilojoules. (credit a: modification of work by “Rex Roof”/Flickr)  Bomb Calorimetry Example When a 0.740-g sample of trinitrotoluene (TNT), C7H5N2O6, is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?  Hess’s Law There are two ways to determine the amount of heat involved in a chemical change: – Measure it experimentally. – Calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements experimentally. For these reactions, the amount of heat involved must be calculated, usually using Hess’s law. Thermochemistry  Hess’s Law Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function. Enthalpy changes depend only on where a chemical process starts and ends, not on the path it takes from start to finish. Thermochemistry  Hess’s Law For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written: C(s) + O2(g) ⟶ CO2(g) ΔH° = −394 kJ Thermochemistry  Hess’s Law In the two-step process: – First carbon monoxide is formed. – Then, carbon monoxide reacts further to form carbon dioxide. Step 1: C(s) + ½ O2(g) ⟶ CO(g) ΔH° = −111 kJ Step 2: CO(g) + ½ O2(g) ⟶ CO2(g) ΔH° = −283 kJ Sum: C(s) + ½ O2(g) + CO(g) + ½ O2(g) ⟶ CO(g) + CO2(g) Net Change: C(s) + O2(g) ⟶ CO2(g) Thermochemistry  Hess’s Law According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. The enthalpy change of the entire two-step reaction: ΔH o = - 111 kJ + - 283 kJ = - 394 kJ We see that ΔH of the overall reaction is the same whether it occurs in one step or two. Thermochemistry  Figure 9.24 The formation of CO2(g) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.  Hess’s Law and Enthalpies of Formation We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: – (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by – (ii) recombinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). Thermochemistry  Hess’s Law and Enthalpies of Formation The standard enthalpy change of the overall reaction, therefore, is equal to: – (ii) the sum of the standard enthalpies of formation of all the products – plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows: o D H reaction = å n ´ D H of (products) - å n ´ D H of (reactants) With ∑ representing “the sum of” and n representing the stoichiometric coefficients. Thermochemistry  Example 9.15 What is the standard enthalpy change for the reaction: 3NO2(g) + H2O(l) ⟶ 2HNO3(aq) + NO(g) ΔH° = ? Use the special form of Hess’s law given previously and the enthalpy of formation values found in Appendix G: ΔH oreaction = ∑ n × ΔH o f (products) − ∑ n × ΔH o f (reactants)  −207.4 kJ   +90.2 kJ   =  2 mol HNO3 (aq)×  + 1 mol NO(g) ×   mol   mol   +33.2 kJ   −285.8 kJ   −  3 mol NO 2 (g) ×  + 1 mol H 2 O(l) ×   mol   mol  = − 138.4 kJ Thermochemistry  Standard Enthalpy of Formation Standard enthalpy of formation (ΔH°f) is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from its constituent free elements in their most stable states under standard state conditions. For example, ΔH°f of CO2(g) is −393.5 kJ/mol. C(s) + O2(g) ⟶ CO2(g) ΔH°f = ΔH° = −393.5kJ By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions. Thermochemistry Standard Enthalpies of Formation Standard enthalpies of formation, ∆Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). Thermochemistry © 2012 Pearson Education, Inc.  Example 9.12 Write the heat of formation reaction equations for: (a) C2H5OH(l) (b) Ca3(PO4)2(s) Remembering that ΔH°f reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have: (a) 2C(s,graphite) + 3H2(g) + ½ O2(g) ⟶ C2H5OH(l) (b) 3Ca(s) + ½ P4(s) + 4O2(g) ⟶ Ca3(PO4)2(s) Note: The standard state of carbon is graphite, and phosphorus exists as P4. Thermochemistry  Calculation of ∆H C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Imagine this as occurring in three steps: C3H8(g) → 3C(graphite) + 4H2(g) Thermochemistry © 2012 Pearson Education, Inc.  Calculation of ∆H C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Imagine this as occurring in three steps: C3H8(g) → 3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) →3CO2(g) Thermochemistry © 2012 Pearson Education, Inc.  Calculation of ∆H C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Imagine this as occurring in three steps: C3H8(g) → 3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) →3CO2(g) 4H2(g) + 2O2(g) → 4H2O(l) Thermochemistry © 2012 Pearson Education, Inc.  Calculation of ∆H C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) The sum of these equations is C3H8(g) → 3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) →3CO2(g) 4H2(g) + 2O2(g) → 4H2O(l) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) Thermochemistry © 2012 Pearson Education, Inc.  Calculation of ∆H We can use Hess’s law in this way: ∆H = Σn∆Hf,products – Σm∆Hf°,reactants where n and m are the stoichiometric coefficients. Thermochemistry © 2012 Pearson Education, Inc.  Calculation of ∆H C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)] = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ Thermochemistry © 2012 Pearson Education, Inc. Using Bond Energies to Calculate Approximate Enthalpy Changes Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH4, 1660 kJ, is equal to the standard enthalpy change of the reaction: The average C–H bond energy, DC–H, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The 415 kJ/mol value is the average, not the exact value required to break any one bond. Thermochemistry Thermochemistry Using Bond Energies to Calculate Approximate Enthalpy Changes Example 9.16 Methanol, CH3OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H2, from which methanol can be produced. Using the bond energies in Table 9.4, calculate the approximate enthalpy change, ΔH, for the reaction here: Thermochemistry Using Bond Energies to Calculate Approximate Enthalpy Changes Example 9.16 Methanol, CH3OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H2, from which methanol can be produced. Using the bond energies in Table 9.4, calculate the approximate enthalpy change, ΔH, for the reaction here: Thermochemistry Thermochemistry Thermochemistry  Check Your Learning Ethyl alcohol, CH3CH2OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction: Thermochemistry  Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats. Thermochemistry © 2012 Pearson Education, Inc. Thermochemistry  Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels. Thermochemistry © 2012 Pearson Education, Inc. Thermochemistry Thermochemistry Thermochemistry