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Principles of Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Nature of Energy • Even though Chemistry is the study of matter, energy affects matter. Chapter 6 Thermochemistry • Energy is anything that has the capacity to do work. • Work is a force acting over a distance. 9energy = work = force × distance • Energy can be exchanged between objects through contact. 9collisions http://genchem.chem.wisc.edu/demonstrations/Gen_Chem_Pages/06thermo page/thermochemistry.htm Tro, Principles of Chemistry: A Molecular Approach Energy, Heat, and Work • You can think of energy as a quantity an object can possess. 9or collection of objects • You can think of heat and work as the two different ways that an object can exchange energy with other objects. Tro, Principles of Chemistry: A Molecular Approach 2 Classification of Energy • Kinetic energy is energy of motion or energy that is being transferred. 9Thermal energy is kinetic. 9either out of it, or into it Tro, Principles of Chemistry: A Molecular Approach 3 Tro, Principles of Chemistry: A Molecular Approach 4 Classification of Energy • Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object. 9Energy stored in the structure of a compound is potential. Exchanging Energy • Energy cannot be created • • or destroyed. Energy can be exchanged between objects. Energy can also be transformed from one form to another. 9 heat → light → sound Tro, Principles of Chemistry: A Molecular Approach 5 Some Forms of Energy • electrical 9 kinetic energy associated with the flow of electrical charge • heat or thermal energy 9 kinetic energy associated with molecular motion • light or radiant energy 9 kinetic energy associated with energy transitions in an atom • nuclear 9 potential energy in the nucleus of atoms • chemical 9 potential energy due to the structure of the atoms, the attachment between atoms, the atoms’ relative position to each other in the molecule, or the molecules’ relative positions in the structure Tro, Principles of Chemistry: A Molecular Approach 7 Tro, Principles of Chemistry: A Molecular Approach 6 System and Surroundings • We define the system as the material or process that contains the energy changes we are studying. • We define the surroundings as everything else in the universe. • What we study is the exchange of energy between the system and the surroundings. Surroundings Surroundings System System Tro, Principles of Chemistry: A Molecular Approach 8 Units of Energy • One joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter. 9 1 J = 1 N·m = 1 kg·m2/s2 • One calorie (cal) is the amount of energy needed to raise one gram of water by 1 °C. 9 kcal = energy needed to raise 1000 g of water 1 °C 9 food Calories = kcals = = 9 • The internal energy is the total amount of kinetic and potential energy a system possesses. The change in the internal energy of a system depends only on the amount of energy in the system at the beginning and end. 9 A state function is a mathematical function whose result depends only on the initial and final conditions, not on the process used. 9 ΔE = Efinal – Einitial 9 ΔEreaction = Eproducts – Ereactants Tro, Principles of Chemistry: A Molecular Approach energy change in the system and the surroundings must be zero. 9 ΔEnergyuniverse = 0 = ΔEnergysystem + ΔEnergysurroundings 9 Δ is the symbol that is used to mean change. 4.184 joules (J) (exact) 1000 calories (cal) Internal Energy • • Conservation of energy requires that the total ¾ final amount – initial amount Energy Conversion Factors 1 calorie (cal) 1 kilo calorie (kcal) Tro, Principles of Chemistry: A Molecular Approach Energy Flow and Conservation of Energy 11 Tro, Principles of Chemistry: A Molecular Approach 10 State Function You can travel either of two trails to reach the top of the mountain. One is long and winding; the other is shorter but steeper. Regardless of which trail you take, when you reach the top you will be 10,000 ft. above the base. The height of the mountain is a state function. It depends only on the distance from the base to the peak, not on how you arrive there! Tro, Principles of Chemistry: A Molecular Approach 12 Energy Flow Energy Diagrams “graphical” way of showing the direction of energy flow during a process. • If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be +. initial energy added ΔE = + Internal Energy • initial final energy removed ΔE = – 13 • • Surroundings ΔE + System ΔE – 14 Energy Flow in a Chemical Reaction • When energy flows into a system, it must all come from the surroundings. When energy flows into a system, ΔEsystem is +. When energy flows out of the surroundings, ΔEsurroundings is –. therefore: ΔEsystem = –ΔEsurroundings system, it must all flow into the surroundings. When energy flows out of a system, ΔEsystem is –. When energy flows into the surroundings, ΔEsurroundings is +. therefore: –ΔEsystem = ΔEsurroundings Tro, Principles of Chemistry: A Molecular Approach Energy Flow • • • • If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be –. Tro, Principles of Chemistry: A Molecular Approach • When energy flows out of a final • The total amount of internal energy Surroundings ΔE – System ΔE + in 1 mol of C(s) and 1 mole of O2(g) is greater than the internal energy in 1 mole of CO2(g). 9 at the same temperature and pressure • In the reaction C(s) + O2(g) → CO2(g), there will be a net release of energy into the surroundings. Surroundings 9 −ΔEreaction = ΔEsurroundings 9 ΔEreaction = −ΔEsurroundings 15 C(s), O2(g) energy energy absorbed released CO2(g) ΔE ΔErxn rxn==–+ • In the reaction CO2(g) → C(s) + O2(g), there will be an absorption of energy from the surroundings into the reaction. Tro, Principles of Chemistry: A Molecular Approach Internal Energy Internal Energy • Energy diagrams are a Tro, Principles of Chemistry: A Molecular Approach 16 System C+O CO C + 2O→ 2 → CO2 Energy Exchange Energy Exchange • Energy is exchanged between the system and surroundings through heat and work. 9 q = heat (thermal) energy 9 w = work energy 9 q and w are NOT state functions; their value depends on the process. ΔE = q + w • Energy is exchanged between the system and surroundings through either heat exchange or work being done. Tro, Principles of Chemistry: A Molecular Approach 17 Heat and Work • On a smooth table, most of the kinetic energy is transferred from the first ball to the second—with a small amount lost through friction. Tro, Principles of Chemistry: A Molecular Approach 18 Example 6.1 If burning fuel in a potato cannon gives 855 J of kinetic energy to the potato and releases 1422 J of heat from the cannon, what is the change in internal energy of the fuel? Given: Find: qpotato = 855 J, wpotato = 1422 J ΔEfuel, J Concept Plan: Relationships: qsystem = −qsurroundings, wsystem = −wsurroundings, ΔE = q + w Solution: Check: Tro, Principles of Chemistry: A Molecular Approach 19 The unit and sign are correct. Tro, Principles of Chemistry: A Molecular Approach 20 Heat Exchange • Heat is the exchange of thermal energy between the system and surroundings. • It occurs when the system and surroundings have a difference in temperature. • Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature. 9thermal equilibrium Tro, Principles of Chemistry: A Molecular Approach Quantity of Heat Energy Absorbed “Heat Capacity” • When a system absorbs heat, its temperature increases. • The increase in temperature is directly proportional to the amount of heat absorbed. • The proportionality constant is called the heat capacity, C. 9 Units of C are J/°C . q = C × ΔT • The heat capacity of an object depends on its mass. 9 200 g of water requires twice as much heat to raise its temperature by 1 °C as does 100 g of water. • The heat capacity of an object depends on the type of material. 9 1000 J of heat energy will raise the temperature of 100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C. 21 Tro, Principles of Chemistry: A Molecular Approach 22 “Specific Heat Capacity” • measure of a substance’s intrinsic ability to absorb heat • The specific heat capacity (Cs) is the amount of heat energy required to raise the temperature of one gram of a substance 1 °C. 9units are J/(g·°C) Tro, Principles of Chemistry: A Molecular Approach 23 Quantifying Heat Energy • The heat capacity of an object is proportional to its mass and the specific heat of the material. • So we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object. heat = (mass) × (specific heat capacity) × (temp. change) q = (m) × (Cs) × (ΔT) Tro, Principles of Chemistry: A Molecular Approach 24 Example 6.2 How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from –8.0 °C to 37.0 °C? Sort information Given: T1 = –8.0 °C, T2 = 37.0 °C, m = 3.10 g Practice—Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C. q, J Find: Conceptual Plan: Strategize Relationships: q = m · Cs · ΔT Cs = 0.385 J/g (Table 6.4) Follow the conceptual plan to solve the problem. Check Solution: Check: The unit and sign are correct. Answer: 6.2 x 102 J Tro, Principles of Chemistry: A Molecular Approach 25 Pressure–Volume Work • PV work is work that is the result of a volume • • change against an external pressure. When gases expand, ΔV is +, but the system is doing work on the surroundings, so wgas is –. As long as the external pressure is kept constant, Tro, Principles of Chemistry: A Molecular Approach 26 Example 6.3 If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done? Given: Find: V1 = 0.100 L, V2 = 1.85 L, P = 1.00 atm w, J Conceptual Plan: Relationships: 101.3 J = 1 atm·L Solution: –work = external pressure × change in volume w = –PΔV 9 To convert the units to joules, use 101.3 J = 1 atm·L. Check: The unit and sign are correct. Tro, Principles of Chemistry: A Molecular Approach 27 Tro, Principles of Chemistry: A Molecular Approach 28 Exchanging Energy between System and Surroundings • exchange of heat energy q = mass × specific heat × ΔTemperature • exchange of work w = –Pressure × ΔVolume Measuring ΔE, Calorimetry at Constant Volume • Since ΔE = q + w, we can determine ΔE by measuring q and w. • In practice, it is easiest to do a process in such a way that there is no change in volume, w = 0. 9 at constant volume, ΔEsystem = qsystem • In practice, it is not possible to observe the temperature • changes of the individual chemicals involved in a reaction. So instead, we use an insulated, controlled surroundings and measure the temperature change in it. The surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water. qsurroundings = qcalorimeter = –qsystem –ΔEreaction = qcal = Ccal × ΔT Tro, Principles of Chemistry: A Molecular Approach 29 Bomb Calorimeter • used to measure Tro, Principles of Chemistry: A Molecular Approach Example 6.4 When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If Ccal = 4.90 kJ/°C, find ΔE for burning 1 mole. Given: ΔE because it is a constant volume system 30 1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C, 2.9506 × 10−3 mol C12H22O11, ΔT = 3.41 °C, Conceptual Plan: Relationships: Ccal, ΔT qcal qrxn qcal = Ccal × ΔT = –qrxn Solution: Tro, Principles of Chemistry: A Molecular Approach 31 Tro, Principles of Chemistry: A Molecular Approach 32 Ccal = 4.90 kJ/°C qrxn, mol ΔE Practice—When 1.22 g of HC7H5O2 (MM 122.12) is burned in a bomb calorimeter, the temperature rises from 20.27 °C to 22.67 °C. If ΔE for the reaction is −3.23 × 103 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? Enthalpy • The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume. 9 H is a state function. H = E + PV • The enthalpy change, ΔH, of a reaction is the heat evolved in a reaction at constant pressure. ΔHreaction = qreaction at constant pressure • Usually ΔH and ΔE are similar in value; the difference is largest for reactions that produce or use large quantities of gas. Answer: 13.4 kJ/ oC Tro, Principles of Chemistry: A Molecular Approach 33 Tro, Principles of Chemistry: A Molecular Approach 34 Molecular View of Exothermic Reactions Endothermic and Exothermic Reactions • For an exothermic reaction, the outside • When ΔH is –, heat is being released by the system. • Reactions that release heat are called exothermic • • • • • • reactions. When ΔH is +, heat is being absorbed by the system. Reactions that absorb heat are called endothermic reactions. Chemical heat packs contain iron filings that are oxidized in an exothermic reaction. Your hands get warm because the released heat of the reaction is absorbed by your hands. Chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process. Your hands get cold because they are giving away your heat to the reaction. Tro, Principles of Chemistry: A Molecular Approach 35 • • temperature rises due to release of thermal energy. This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat. During the course of a reaction, old bonds are broken and new bonds made. The products of the reaction have less chemical potential energy than the reactants. The difference in energy is released as heat. Tro, Principles of Chemistry: A Molecular Approach 36 Molecular View of Endothermic Reactions • In an endothermic reaction, the surrounding • • • temperature drops due to absorption of some of its thermal energy. During the course of a reaction, old bonds are broken and new bonds made. The products of the reaction have more chemical potential energy than the reactants. To acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products. Tro, Principles of Chemistry: A Molecular Approach 37 Example 6.6 How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)? Given: Find: Enthalpy of Reaction • The enthalpy change in a chemical reaction is an extensive property. 9 The more reactants you use, the larger the enthalpy change. • By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written. • For the reaction: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) Tro, Principles of Chemistry: A Molecular Approach 38 Practice—How much heat is evolved when a 0.483-g diamond is burned? (ΔHcombustion = –395.4 kJ/mol C) 13.2 kg C3H8 q, kJ/mol Conceptual Plan: Relationships: 1 kg = 1000 g, 1 mol C3H8 = –2044 kJ, Molar Mass = 44.09 g/mol Solution: Check: The sign is correct and the value is reasonable. Tro, Principles of Chemistry: A Molecular Approach 39 ΔH = –2044 kJ Answer: -15.9 kJ Tro, Principles of Chemistry: A Molecular Approach 40 Measuring ΔH Calorimetry at Constant Pressure • Reactions done in aqueous solution are at constant pressure. 9 open to the atmosphere • The calorimeter is often nested foam cups containing the solution. Example 6.7 What is ΔHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacting in 100.0 mL of solution changes the temperature from 25.6 °C to 32.8 °C? (assume d sol = 1.00 g/mL and specific heat of the solution is 4.18 J/g·°C) Given: Find: Conceptual Plan: 0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C d = 1.00 g/mL, Cs = 4.18 J/g·°C ΔH, kJ/mol Cs, ΔT, m qsol’n qrxn qrxn, mol ΔH Relationships: Solution: qreaction = –qsolution = –(masssolution × Cs, solution × ΔT) y ΔHreaction = qconstant pressure = qreaction step2 9 To get ΔHreaction per mol, divide by the number of moles. Step 1 Tro, Principles of Chemistry: A Molecular Approach 41 Practice—What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00-mL sample of 0.250 M HCl(aq) is added to a 50.00-mL sample of 0.250 NaOH(aq) if the initial temperature is 19.50 °C and the ΔHrxn is −57.2 kJ/mol NaOH? (Assume the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.18 J/g·°C.) Tro, Principles of Chemistry: A Molecular Approach 42 Relationships Involving ΔHrxn Rule 1 When a reaction is multiplied by a factor, ΔHrxn is multiplied by that factor. 9because ΔHrxn is extensive C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ 2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = 2(−393.5 kJ) = −787.0 kJ Rule 2 If a reaction is reversed, then the sign of ΔH is reversed. ΔH = +393.5 kJ CO2(g) → C(s) + O2(g) Answer: 21.21 oC Tro, Principles of Chemistry: A Molecular Approach 43 Tro, Principles of Chemistry: A Molecular Approach 44 Relationships Involving ΔHrxn Hess’s Law • If a reaction can Given the following information: ΔH° = −206 kJ Cu(s) + Cl2(g) → CuCl2(s) 2 Cu(s) + Cl2(g) → 2 CuCl(s) ΔH° = −36 kJ be expressed as a series of steps, then the ΔHrxn for the overall reaction is the sum of the heats of reaction for each step. Tro, Principles of Chemistry: A Molecular Approach Practice—Hess’s Law Calculate the ΔH° for the reaction below: Cu(s) + CuCl2(s) → 2 CuCl(s) ΔH° = ? kJ 45 Tro, Principles of Chemistry: A Molecular Approach Practice—Hess’s Law Given the following information: Cu(s) + Cl2(g) → CuCl2(s) 2 Cu(s) + Cl2(g) → 2 CuCl(s) ΔH° = −206 kJ ΔH° = −36 kJ Calculate the ΔH° for the reaction below: Cu(s) + CuCl2(s) → 2 CuCl(s) ΔH° = ? kJ 46 Practice—Hess’s Law Given the following information: 2 NO(g) + O2(g) → 2 NO2(g) 2 N2(g) + 5 O2(g) + 2 H2O(l) → 4 HNO3(aq) N2(g) + O2(g) → 2 NO(g) ΔH° = −116 kJ ΔH° = −256 kJ ΔH° = +183 kJ Calculate the ΔH° for the reaction below: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH° = ? Solution: reverse the 1st equation and keep the second the same CuCl2(s) → Cu(s) + Cl2(g) 2 Cu(s) + Cl2(g) → 2 CuCl(s) Cu(s) + CuCl2(s) → 2 CuCl(s) ΔH° = +206 kJ ΔH° = −36 kJ ΔH° = +170. kJ Answer: ΔH = -137kJ Tro, Principles of Chemistry: A Molecular Approach 47 Tro, Principles of Chemistry: A Molecular Approach 48 Standard Conditions Formation Reactions • The standard state is the state of a material at a defined set of conditions. 9 pure gas at exactly 1 atm pressure 9 pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest ¾ usually 25 °C 9 substance in a solution with concentration 1 M • The standard enthalpy change, ΔH°, is the enthalpy • change when all reactants and products are in their standard states. The standard enthalpy of formation, ΔHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements. 9 The elements must be in their standard states. 9 The ΔHf° for a pure element in its standard state = 0 kJ/mol. • reactions of elements in their standard state to form 1 mole of a pure compound 9If you are not sure what the standard state of an element is, find the form in Appendix IIB that has a ΔHf° = 0. 9Since the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions. ¾ by definition Tro, Principles of Chemistry: A Molecular Approach 49 Writing Formation Reactions— Write the formation reaction for CO(g). • The formation reaction is the reaction between the • elements in the compound, which are C and O. C + O → CO(g) The elements must be in their standard state. 9 There are several forms of solid C, but the one with ΔHf° = 0 is graphite. 9 Oxygen’s standard state is the diatomic gas. Tro, Principles of Chemistry: A Molecular Approach 50 Write the formation reactions for the following. CO2(g) C(s, graphite) + O2(g) → CO2(g) Al2(SO4)3(s) C(s, graphite) + O2(g) → CO(g) • The equation must be balanced, but the coefficient of the product compound must be 1. H2O (g) 9 Use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient. C(s, graphite) + ½ O2(g) → CO(g) Tro, Principles of Chemistry: A Molecular Approach 51 Tro, Principles of Chemistry: A Molecular Approach 52 Calculating Standard Enthalpy Change for a Reaction • Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products. • The ΔH° for the reaction is then the sum of the ΔHf° for the component reactions. ΔH°reaction = Σ n ΔHf°(products) – Σ n ΔHf°(reactants) 9Σ means sum. 9n is the coefficient of the reaction. Tro, Principles of Chemistry: A Molecular Approach 53 Example—Calculate the enthalpy change in the reaction. 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) Step 2 ΔH°reaction = Σ n ΔHf°(products) – Σ n ΔHf°(reactants) Example—Calculate the enthalpy change in the reaction. 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) Step 1: Write formation reactions for each compound and determine the ΔHf° for each. 2 C(s, gr) + H2(g) → C2H2(g) ΔHf° = +227.4 kJ/mol C(s, gr) + O2(g) → CO2(g) ΔHf° = –393.5 kJ/mol H2(g) + ½ O2(g) → H2O(l) ΔHf° = –285.8 kJ/mol Tro, Principles of Chemistry: A Molecular Approach 54 Practice—Calculate the ΔH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions. CaCO3(s) → CaO(s) + O2(g) Material ΔHf°, kJ/mol CaCO3(s) −1207.6 O2(g) 0 CaO(s) −634.9 ΔHrxn = [(4•ΔHCO2 + 2•ΔHH2O) – (2•ΔHC2H2 + 5•ΔHO2)] ΔHrxn = [(4•(–393.5) + 2•(–285.8)) – (2•(+227.4) + 5•(0))] ΔHrxn = −2600.4 kJ Tro, Principles of Chemistry: A Molecular Approach 55 Tro, Principles of Chemistry: A Molecular Approach 56