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Chapter ? 1 Thermochemistry Prentice Hall © 2005 Chapter Six 2 Energy • Energy is the capacity to do work (to displace or move matter). • Energy literally means “work within”; however, an object does not contain work. • Potential energy is energy of position or composition. • Kinetic energy is the energy of motion. Ek = ½ mv2 Energy has the units of joules (J or kg . m2/s2) Prentice Hall © 2005 Chapter Six 3 Potential Energy and Kinetic Energy At what point in each bounce is the potential energy of the ball at a maximum? Prentice Hall © 2005 Chapter Six 4 Thermochemistry: Basic Terms • Thermochemistry is the study of energy changes that occur during chemical reactions. • System: the part of the universe being studied. • Surroundings: the rest of the universe. Prentice Hall © 2005 Chapter Six 5 Types of Systems • Open: energy and matter can be exchanged with the surroundings. • Closed: energy can be exchanged with the surroundings, matter cannot. • Isolated: neither energy nor matter can be exchanged with the surroundings. Prentice Hall © 2005 A closed system; energy (not matter) can be exchanged. After the lid of the jar is unscrewed, which kind of system is it? Chapter Six 6 Internal Energy (E) • Internal energy (E) is the total energy contained within a system • Part of E is kinetic energy (from molecular motion) – Translational motion, rotational motion, vibrational motion. – Collectively, these are sometimes called thermal energy • Part of E is potential energy – Intermolecular and intramolecular forces of attraction, locations of atoms and of bonds. – Collectively these are sometimes called chemical energy Prentice Hall © 2005 Chapter Six 7 Heat (q) • Technically speaking, heat is not “energy.” • Heat is energy transfer between a system and its surroundings, caused by a temperature difference. • Thermal equilibrium occurs when the system and surroundings reach the same temperature and heat transfer stops. Prentice Hall © 2005 More energetic molecules … … transfer energy to less energetic molecules. How do the root-mean-square speeds of the Ar atoms and the N2 molecules compare at the point of thermal equilibrium? Chapter Six Exchange of Heat between System and Surroundings 8 • When heat is absorbed by the system from the surroundings, the process is endothermic. Prentice Hall © 2005 Chapter Six Exchange of Heat between System and Surroundings 9 • 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. Prentice Hall © 2005 Chapter Six 10 Work (w) • Like heat, work is an energy transfer between a system and its surroundings. • Unlike heat, work is caused by a force moving through a distance (heat is caused by a temperature difference). • A negative quantity of work signifies that the system loses energy. • A positive quantity of work signifies that the system gains energy. • There is no such thing as “negative energy” nor “positive energy”; the sign of work (or heat) signifies the direction of energy flow. Prentice Hall © 2005 Chapter Six 11 Pressure-Volume Work For now we will consider only pressure-volume work. work (w) = –PDV Prentice Hall © 2005 How would the magnitude of DV compare to the original gas volume if the two weights (initial and final) were identical? Chapter Six 12 What is work? • • • • • • Work is a force acting over a distance. w= F x Dd P = F/ area d = V/area w= (P x area) x D (V/area)= PDV Work can be calculated by multiplying pressure by the change in volume at constant pressure. • units of liter - atm L-atm Prentice Hall © 2005 Chapter Six 13 Work needs a sign • If the volume of a gas increases, the system has done work on the surroundings. • work is negative • w = - PDV • Expanding work is negative. • Contracting, surroundings do work on the system w is positive. • 1 L atm = 101.3 J Prentice Hall © 2005 Chapter Six 14 Examples • What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure? • If 2.36 J of heat are absorbed by the gas above. what is the change in energy? Prentice Hall © 2005 Chapter Six 15 Same rules for heat and work • Heat given off is negative. • Heat absorbed is positive. • Work done by system on surroundings is negative. • Work done on system by surroundings is positive. Prentice Hall © 2005 Chapter Six 16 State Functions • The state of a system: its exact condition at a fixed instant. • State is determined by the kinds and amounts of matter present, the structure of this matter at the molecular level, and the prevailing pressure and temperature. • A state function is a property that has a unique value that depends only the present state of a system, and does not depend on how the state was reached (does not depend on the history of the system). • Law of Conservation of Energy – in a physical or chemical change, energy can be exchanged between a system and its surroundings, but no energy can be created or destroyed. Prentice Hall © 2005 Chapter Six 17 State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its DE is the same. – But q and w are different in the two cases. Prentice Hall © 2005 Chapter Six 18 First Law of Thermodynamics • “Energy cannot be created or destroyed.” • Inference: the internal energy change of a system is simply the difference between its final and initial states: DE = Efinal – Einitial • Additional inference: if energy change occurs only as heat (q) and/or work (w), then: DE = q + w Prentice Hall © 2005 Chapter Six 19 First Law: Sign Convention • Energy entering a system carries a positive sign: – heat absorbed by the system, or – work done on the system • Energy leaving a system carries a negative sign – heat given off by the system – work done by the system Prentice Hall © 2005 Chapter Six 20 Direction • Every energy measurement has three parts. 1. A unit ( Joules of calories). 2. A number how many. 3. and a sign to tell direction. • negative - exothermic • positive- endothermic Prentice Hall © 2005 Chapter Six 21 Surroundings System Energy DE <0 Prentice Hall © 2005 Chapter Six 22 Surroundings System Energy DE >0 Prentice Hall © 2005 Chapter Six 23 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, DE = q + w. Prentice Hall © 2005 Chapter Six 24 Example 6.1 A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? Example 6.2: A Conceptual Example The internal energy of a fixed quantity of an ideal gas depends only on its temperature. If a sample of an ideal gas is allowed to expand against a constant pressure at a constant temperature, (a) what is DU for the gas? (b) Does the gas do work? (c) Is any heat exchanged with the surroundings? Prentice Hall © 2005 Chapter Six 25 DE, q, w, and Their Signs Prentice Hall © 2005 © 2009, Prentice-Hall, Chapter Six Inc. 26 Heats of Reaction (qrxn) • qrxn is the quantity of heat exchanged between a reaction system and its surroundings. • An exothermic reaction gives off heat – In an isolated system, the temperature increases. – The system goes from higher to lower energy; qrxn is negative. • An endothermic reaction absorbs heat – In an isolated system, the temperature decreases. – The system goes from lower to higher energy; qrxn is positive. Prentice Hall © 2005 Chapter Six 27 Conceptualizing an Exothermic Reaction Surroundings are at 25 °C 25 °C Typical situation: some heat is released to the surroundings, some heat is absorbed by the solution. Hypothetical situation: all heat is instantly released to the surroundings. Heat = qrxn Prentice Hall © 2005 32.2 °C 35.4 °C In an isolated system, all heat is absorbed by the solution. Maximum temperature rise. Chapter Six 28 Internal Energy Change at Constant Volume • For a system where the reaction is carried out at constant volume, DV = 0 and DE = qV. • All the thermal energy produced by conversion from chemical energy is released as heat; no P-V work is done. Prentice Hall © 2005 Chapter Six 29 lnternal Energy Change at Constant Pressure • For a system where the reaction is carried out at constant pressure, DE = qP – PDV or DE + PDV = qP • Most of the thermal energy is released as heat. • Some work is done to expand the system against the surroundings (push back the atmosphere). Prentice Hall © 2005 Chapter Six 30 Calorimetry • We measure heat flow using calorimetry. • A calorimeter is a device used to make this measurement. • A “coffee cup” calorimeter may be used for measuring heat involving solutions. A “bomb” calorimeter is used to find heat of combustion; the “bomb” contains oxygen and a sample of the material to be burned. Prentice Hall © 2005 Chapter Six 31 Calorimetry n Measuring heat. n Use a calorimeter. n Two kinds n Constant pressure calorimeter (called a coffee cup calorimeter) n heat capacity for a material, C is calculated n C= heat absorbed/ DT = DH/ DT n specific heat capacity = C/mass Prentice Hall © 2005 Chapter Six 32 Calorimetry, Heat Capacity, Specific Heat • Heat evolved in a reaction is absorbed by the calorimeter and its contents. • In a calorimeter we measure the temperature change of water or a solution to determine the heat absorbed or evolved by a reaction. • The heat capacity (C) of a system is the quantity of heat required to change the temperature of the system by 1 °C. C = q/DT (units are J/°C) • Molar heat capacity is the heat capacity of one mole of a substance. • The specific heat (s) is the heat capacity of one gram of a pure substance (or homogeneous mixture). s = C/m = q/(mDT) q = s m DT Prentice Hall © 2005 Chapter Six 33 Calorimetry n molar heat capacity = C/moles n heat = specific heat x m x DT n heat = molar heat x moles x DT n Make the units work and you’ve done the problem right. n A coffee cup calorimeter measures DH. n An insulated cup, full of water. n The specific heat of water is 1 cal/gºC n Heat of reaction= DH = sh x mass x DT Prentice Hall © 2005 Chapter Six 34 Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity. Prentice Hall © 2005 Chapter Six 35 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. Prentice Hall © 2005 Chapter Six 36 Heat Capacity and Specific Heat Specific heat, then, is Specific heat = s= Prentice Hall © 2005 heat transferred mass temperature change q m DT © 2009, Prentice-Hall, Chapter Six Inc. 37 More on Specific Heat • • • • • • q = mass x specific heat x DT If DT is positive (temperature increases), q is positive and heat is gained by the system. If DT is negative (temperature decreases), q is negative and heat is lost by the system. The calorie, while not an SI unit, is still used to some extent. Water has a specific heat of 1 cal/(g oC). 4.184 J = 1 cal One food calorie (Cal or kcal) is actually equal to 1000 cal. Prentice Hall © 2005 Chapter Six 38 Many metals have low specific heats. The specific heat of water is higher than that of almost any other substance. Prentice Hall © 2005 Chapter Six 39 Heat Capacity: A Thought Experiment • Place an empty iron pot weighing 5 lb on the burner of a stove. • Place an iron pot weighing 1 lb and containing 4 lb water on a second identical burner (same total mass). • Turn on both burners. Wait five minutes. • Which pot handle can you grab with your bare hand? • Iron has a lower specific heat than does water. It takes less heat to “warm up” iron than it does water. Prentice Hall © 2005 Chapter Six 40 Example Calculate the heat capacity of an aluminum block that must absorb 629 J of heat from its surroundings in order for its temperature to rise from 22 °C to 145 °C. Prentice Hall © 2005 Chapter Six 41 Example 6.7 How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g of water from 25.0 to 100.0 °C? Example 6.8 What will be the final temperature if a 5.00-g silver ring at 37.0 °C gives off 25.0 J of heat to its surroundings? Use the specific heat of silver listed in Table 6.1. Prentice Hall © 2005 Chapter Six 42 Examples n The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. n A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper? Prentice Hall © 2005 Chapter Six 43 Calorimetry n Constant volume calorimeter is called a bomb calorimeter. n Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water. n The heat capacity of the calorimeter is known and tested. n Since DV = 0, PDV = 0, DE = q Prentice Hall © 2005 Chapter Six 44 Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Prentice Hall © 2005 Chapter Six 45 Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure DH for the reaction with this equation: q = m s DT Prentice Hall © 2005 Chapter Six 46 Bomb Calorimeter n thermometer n stirrer n full of water n ignition wire n Steel bomb n sample Prentice Hall © 2005 Chapter Six 47 Properties n intensive properties not related to the amount of substance. n density, specific heat, temperature. n Extensive property - does depend on the amount of stuff. n Heat capacity, mass, heat from a reaction. Prentice Hall © 2005 Chapter Six 48 Enthalpy n Symbol is H n Change in enthalpy is DH n delta H n If heat is released the heat content of the products is lower n DH is negative (exothermic) n If heat is absorbed the heat content of the products is higher n DH is positive (endothermic) 48 Prentice Hall © 2005 Chapter Six 49 Enthalpy n Since DE = q + w and w = -PDV, we can substitute these into the enthalpy expression: DH = DE + PDV DH = (q+w) − w DH = q n So, at constant pressure, the change in enthalpy is the heat gained or lost. Prentice Hall © 2005 © 2009, Prentice-Hall, Chapter Six Inc. 50 Enthalpy and Enthalpy Change Enthalpy is the sum of the internal energy and the pressure-volume product of a system: H = E + PV For a process carried out at constant pressure, qP = DE + PDV so qP = DH The evolved H2 pushes back the atmosphere; work is done at constant pressure. Mg + 2 HCl MgCl2 + H2 Most reactions occur at constant pressure, so for most reactions, the heat evolved equals the enthalpy change. Prentice Hall © 2005 Chapter Six 51 Properties of Enthalpy • Enthalpy is an extensive property. – It depends on how much of the substance is present. • Since E, P, and V are all state functions, enthalpy H must be a state function also. • Enthalpy changes have unique values. DH = qP Prentice Hall © 2005 Two logs on a fire give off twice as much heat as does one log. Enthalpy change depends only on the initial and final states. In a chemical reaction we call the initial state the ____ and the final state the ____. Chapter Six 52 Enthalpy Diagrams • Values of DH are measured experimentally. • Negative values indicate exothermic reactions. • Positive values indicate endothermic reactions. A decrease in enthalpy during the reaction; DH is negative. Prentice Hall © 2005 An increase in enthalpy during the reaction; DH is positive. Chapter Six 53 Enthalpy of Reaction The change in enthalpy, DH, is the enthalpy of the products minus the enthalpy of the reactants: DH = Hproducts − Hreactants Prentice Hall © 2005 Chapter Six 54 Enthalpy of Reaction This quantity, DH, is called the enthalpy of reaction, or the heat of reaction. Prentice Hall © 2005 Chapter Six 55 The Truth about Enthalpy 1. Enthalpy is an extensive property. 2. DH for a reaction in the forward direction is equal in size, but opposite in sign, to DH for the reverse reaction. 3. DH for a reaction depends on the state of the products and the state of the reactants. Prentice Hall © 2005 © 2009, Prentice-Hall, Chapter Six Inc. 56 Reversing a Reaction • DH changes sign when a process is reversed. • Therefore, a cyclic process has the value DH = 0. Same magnitude; different signs. Prentice Hall © 2005 Chapter Six 57 Example 6.3 Given the equation (a) H2(g) + I2(s) 2 HI(g) DH = +52.96 kJ calculate DH for the reaction (b) HI(g) ½ H2(g) + ½ I2(s). Example 6.4 The complete combustion of liquid octane, C8H18, to produce gaseous carbon dioxide and liquid water at 25 °C and at a constant pressure gives off 47.9 kJ of heat per gram of octane. Write a chemical equation to represent this information. Prentice Hall © 2005 Chapter Six ΔH in Stoichiometric Calculations 58 • For problem-solving, heat evolved (exothermic reaction) can be thought of as a product. Heat absorbed (endothermic reaction) can be thought of as a reactant. • We can generate conversion factors involving DH. • For example, the reaction: H2(g) + Cl2(g) 2 HCl(g) DH = –184.6 kJ can be used to write: –184.6 kJ ———— 1 mol H2 Prentice Hall © 2005 –184.6 kJ ———— 1 mol Cl2 –184.6 kJ ———— 2 mol HCl Chapter Six 59 Example 6.5 What is the enthalpy change associated with the formation of 5.67 mol HCl(g) in this reaction? H2(g) + Cl2(g) 2 HCl(g) Prentice Hall © 2005 DH = –184.6 kJ Chapter Six 60 Measuring Enthalpy Changes for Chemical Reactions For a reaction carried out in a calorimeter, the heat evolved by a reaction is absorbed by the calorimeter and its contents. qrxn = – qcalorimeter qcalorimeter = mass x specific heat x DT By measuring the temperature change that occurs in a calorimeter, and using the specific heat and mass of the contents, the heat evolved (or absorbed) by a reaction can be determined and the enthalpy change calculated. Prentice Hall © 2005 Chapter Six 61 Example 6.11 A 50.0-mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction. Example 6.12 Express the result of Example 6.11 for molar amounts of the reactants and products. That is, determine the value of DH that should be written in the equation for the neutralization reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Prentice Hall © 2005 DH = ? Chapter Six Bomb Calorimetry: Reactions at Constant Volume 62 • Some reactions, such as combustion, cannot be carried out in a coffee-cup calorimeter. • In a bomb calorimeter, a sample of known mass is placed in a heavywalled “bomb,” which is then pressurized with oxygen. • Since the reaction is carried out at constant volume, –qrxn = qcalorimeter = DE … but in many cases the value of DE is a good approximation of DH. Prentice Hall © 2005 Chapter Six 63 Hess’s Law of Constant Heat Summation • Some reactions cannot be carried out “as written.” • Consider the reaction: C(graphite) + ½ O2(g) CO(g). • If we burned 1 mol C in ½ mol O2, both CO and CO2 would probably form. Some C might be left over. However … Prentice Hall © 2005 Chapter Six 64 Hess’s Law of Constant Heat Summation • … enthalpy change is a state function. • The enthalpy change of a reaction is the same whether the reaction is carried out in one step or through a number of steps. • Hess’s Law: If an equation can be expressed as the sum of two or more other equations, the enthalpy change for the desired equation is the sum of the enthalpy changes of the other equations. Prentice Hall © 2005 Chapter Six 65 Hess’s Law n Enthalpy is a state function. n It is independent of the path. n We can add equations to to come up with the desired final product, and add the DH n Two rules n If the reaction is reversed the sign of DH is changed n If the reaction is multiplied, so is DH Prentice Hall © 2005 Chapter Six 66 Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps, DH for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” Prentice Hall © 2005 Chapter Six 67 Hess’s Law Because DH is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. Prentice Hall © 2005 Chapter Six 68 Example 6.14 Calculate the enthalpy change for reaction (a) given the data in equations (b), (c), and (d). (a) 2 C(graphite) + 2 H2(g) C2H4(g) DH = ? (b) C(graphite) + O2(g) CO2(g) DH = –393.5 kJ (c) C2H4(g) + 3 O2 2 CO2(g) + 2 H2O(l) DH = –1410.9 kJ (d) H2(g) + ½ O2 H2O(l) DH = –285.8 kJ Prentice Hall © 2005 Chapter Six 69 Standard Enthalpies of Formation • It would be convenient to be able to use the simple relationship ΔH = Hproducts – Hreactants to determine enthalpy changes. • Although we don’t know absolute values of enthalpy, we don’t need them; we can use a relative scale. • We define the standard state of a substance as the state of the pure substance at 1 atm pressure and the temperature of interest (usually 25 °C). • The standard enthalpy change (ΔH°) for a reaction is the enthalpy change in which reactants and products are in their standard states. • The standard enthalpy of formation (ΔHf°) for a reaction is the enthalpy change that occurs when 1 mol of a substance is formed from its component elements in their standard states. Prentice Hall © 2005 Chapter Six 70 Standard Enthalpies of Formation n Hess’s Law is much more useful if you know lots of reactions. n Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states. n Standard states are 1 atm, 1M and 25ºC n For an element it is 0 Prentice Hall © 2005 Chapter Six 71 Standard Enthalpies of Formation Standard enthalpies of formation, DHf°, are measured under standard conditions (25 °C and 1.00 atm pressure). Prentice Hall © 2005 Chapter Six 72 Standard Enthalpy of Formation When we say “The standard enthalpy of formation of CH3OH(l) is –238.7 kJ”, we are saying that the reaction: C(graphite) + 2 H2(g) + ½ O2(g) CH3OH(l) has a value of ΔH of –238.7 kJ. We can treat ΔHf° values as though they were absolute enthalpies, to determine enthalpy changes for reactions. Question: What is ΔHf° for an element in its standard state [such as O2(g)]? Hint: since the reactants are the same as the products … Prentice Hall © 2005 Chapter Six 73 Calculations Based on Standard Enthalpies of Formation • • • DH°rxn = Snp x DHf°(products) – Snr x DHf°(reactants) The symbol S signifies the summation of several terms. The symbol n signifies the stoichiometric coefficient used in front of a chemical symbol or formula. In other words … 1. Add all of the values for DHf° of the products. 2. Add all of the values for DHf° of the reactants. 3. Subtract #2 from #1 (This is usually much easier than using Hess’s Law!) Prentice Hall © 2005 Chapter Six 74 Example 6.15 Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds. One reaction for producing synthesis gas is 3 CH4(g) + 2 H2O(l) + CO2(g) 4 CO(g) + 8 H2(g) ΔH° = ? Use standard enthalpies of formation from Table 6.2 to calculate the standard enthalpy change for this reaction. Example 6.16 The combustion of isopropyl alcohol, common rubbing alcohol, is represented by the equation 2 (CH3)2CHOH(l) + 9 O2(g) 6 CO2(g) + 8 H2O(l) ΔH° = –4011 kJ Use this equation and data from Table 6.2 to establish the standard enthalpy of formation for isopropyl alcohol. Example 6.17: A Conceptual Example Without performing a calculation, determine which of these two substances should yield the greater quantity of heat per mole upon complete combustion: ethane, C2H6(g), or ethanol, CH3CH2OH(l). Prentice Hall © 2005 Chapter Six 75 Ionic Reactions in Solution • We can apply thermochemical concepts to reactions in ionic solution by arbitrarily assigning an enthalpy of formation of zero to H+(aq). Prentice Hall © 2005 Chapter Six 76 Example 6.18 H+(aq) + OH–(aq) H2O(l) ΔH° = –55.8 kJ Use the net ionic equation just given, together with ΔHf° = 0 for H+(aq), to obtain ΔHf° for OH–(aq). Prentice Hall © 2005 Chapter Six 77 Looking Ahead • A reaction that occurs (by itself) when the reactants are brought together under the appropriate conditions is said to be spontaneous. • A discussion of entropy is needed to fully understand the concept of spontaneity, and will be discussed in Chapter 17. • A spontaneous reaction isn’t necessarily fast (rusting; diamond graphite; etc. are slow). • The difference between the tendency of a reaction to occur and the rate at which a reaction occurs will be discussed in Chapter 13. Prentice Hall © 2005 Chapter Six