Ch 6 Thermochemistry: Energy Flow and Chemical Change PDF

Ch. 6 Thermochemistry: Energy Flow and Chemical Change Chemistry The Molecular Nature of Matter and Change Eighth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribu...

Ch. 6 Thermochemistry: Energy Flow and Chemical Change Chemistry The Molecular Nature of Matter and Change Eighth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Thermodynamics, Thermochemistry Thermodynamics is the study of energy and its transformations. Thermochemistry is a branch of thermodynamics that deals with the heat accompanying chemical changes. ©McGraw-Hill Education. Energy (Physics refresher) Energy is the quantitative property that must be transferred to an object in order to either do work on it, or to heat it, or both. Potential Energy is energy associated to the position of an object while subject to a force. An object (at rest or moving) is said to have potential energy for as long as a force acts on the object. Kinetic Energy is the energy of a moving object in the absence of a force. In the absence of a force, a moving object moves at constant velocity (u): 𝟏 𝑬𝑲 = 𝒎𝒖𝟐 𝟐 Total Energy = Potential Energy + Kinetic Energy ©McGraw-Hill Education. Energy (Physics refresher) Work is the act of converting one type of energy into another type by a force, e.g., potential energy into kinetic energy. The work w done by a constant force of magnitude F that generates a displacement (distance) D of an object is 𝑤 =𝐹∙𝐷 =𝑚∙𝑎∙𝐷 where m is the mass of the object and a is the acceleration. Heat (q), also known as thermal energy, is the energy flow between objects of different initial temperatures in physical contact (thermal contact). Thermal energy is a property proportional to the temperature of an object. ©McGraw-Hill Education. Energy Changes (Physics refresher) There are 2 principles governing energy changes: 1) Lower energy states are more stable and are favored over higher energy states. Therefore, all systems (objects) naturally tend to lower their energy states. 2) Law of Conservation of Energy: “Energy is neither created nor destroyed, it can only be transformed from one type of energy to another.” – Total energy is conserved – and can be converted from one form to another ©McGraw-Hill Education. Potential Energy is Converted to Kinetic Energy A gravitational system. The potential energy gained when a weight is lifted is converted to kinetic energy as the weight falls. The acting force is gravity: 𝐺 = 𝑚𝑔. A lower energy state is more stable. Fig 1.3 ©McGraw-Hill Education. Potential Energy is Converted to Kinetic Energy (2) A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released. The acting force is the elastic force: 𝐹 = −𝑘𝑥 = 𝑚𝑎 Energy is conserved when it is transformed. Fig 1.3 ©McGraw-Hill Education. Potential Energy is Converted to Kinetic Energy A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together. Acting force: electrostatic (coulombic) force Fig 1.3 ©McGraw-Hill Education. Potential Energy is Converted to Kinetic Energy A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car. Acting force: electrostatic and quantum mechanical forces between electrons and nuclei in reactants. Fig 1.3 ©McGraw-Hill Education. Units of Energy The SI unit of energy is the joule (J): – 1 J = 1 kg∙m2/s2 = 1 N·m The calorie was once defined as the quantity of energy needed to raise the temperature of 1 g of water by 1°C. It is a non-SI unit. – 1 cal = 4.184 J The British Thermal Unit (Btu) is often used to rate appliances. One Btu equals the energy required to increase the temperature of 1 lb of water by 1F. It is a non-SI unit. – 1 Btu = 1055 J. Electron Volt: a non-SI unit preferred for energies associated with atoms, molecules and quantum particles. It is the energy an electron acquires when it moves through a potential difference of 1 Volt: 1 𝑒𝑉 = 1.602 × 10−19 𝐽 ©McGraw-Hill Education. Units of Energy 100 W 1W=1 𝑱 𝒔 Power (P) is energy per time unit: 𝑷 = 𝑬 ∆𝒕 1 kWh = 1 kW x 1 h = 1000 W x 3600 s = 3,600,000 J = 3.6 x 106 J = 3.6 MJ ©McGraw-Hill Education. Forms of Energy ©McGraw-Hill Education. Thermal Energy and Temperature Thermal energy is the energy associated with the temperature of an object. – Thermal energy is actually a type of kinetic energy because it arises from the motion of the individual atoms or molecules that make up an object. It is the sum of all kinetic energies of molecules in an object. https://www.youtube.com/watch?v=7fqf7t-fOHI Temperature is an experimental property of an object. It is proportional to the average kinetic energy of all the molecules in the object. Molecules collide continuously in an object, therefore their kinetic energies are continuously averaged. This process creates an equilibrium temperature. ©McGraw-Hill Education. Molecular Speeds at Three Temperatures (5.5 Kinetic Molecular theory refresher) Figure 5.14 ©McGraw-Hill Education. Kinetic Energy and Temperature in an Ideal Gas At a given temperature, all ideal gases in a sample have the same average kinetic energy according to the Kinetic Molecular Theory and Boltzmann statistical distribution: 1 3 𝑅 3 2 𝐸𝐾 = 𝑚𝑢 = 𝑇 = 𝑘𝐵 𝑇 2 2 𝑁𝐴 2 𝑅 𝐽 −23 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡: 𝑘𝐵 = = 1.381 × 10 𝑁𝐴 𝐾 Kinetic energy depends on both the mass and the speed of a particle. At the same T, a heavier gas particle moves more slowly than a lighter one. ©McGraw-Hill Education. Temperature is a SI Base Unit Table 1.2 ©McGraw-Hill Education. Temperature Scales Kelvin (K) – The “absolute temperature scale” begins at absolute zero and has only positive values. Note that the kelvin is not used with the degree sign (o). 1 𝐾 = 1℃ Celsius (oC) – The Celsius scale is based on the freezing (0) and boiling (100) points of water at 1 atmosphere. One degree Celsius equals 1/100 of the difference between these 2 points. This is the temperature scale used most commonly around the world. The Celsius and Kelvin scales use the same size degree although their starting points differ. Fahrenheit (oF) – The Fahrenheit scale is commonly used in the U.S in Meteorology. It is not used in other sciences. The Fahrenheit scale has a different degree size and different zero points than both the Celsius and Kelvin scales. ©McGraw-Hill Education. Thermometry: Freezing and Boiling Points of Water Fig 1.11 ©McGraw-Hill Education. Temperature Difference (Change), ∆T T K = T °C + 273.15 T °C = T K − 273.15 ∆𝑇 𝐾 = ∆𝑇 °𝐶 ∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 1 𝐾 = 1℃ = 1 𝑑𝑒𝑔 ©McGraw-Hill Education. The System and its Surroundings = Universe We assign the term “system” to the material or process in which we are studying the energy changes. In Chemistry, the system is the chemical reaction. We assign the term “surroundings” to everything else with which the system can exchange energy. The system and the surroundings must be in thermal contact. In Chemistry, the surroundings is made of the solvent, the container, the probes, and the nearby air that surrounds the reaction setup. The universe in Thermodynamics (as opposed to the Cosmos) is the name for a system and its surroundings. What we measure in Thermochemistry is the exchange of energy between the reaction (system) and the surroundings. ©McGraw-Hill Education. The Law of Energy Conservation The first law of Thermodynamics states that the total energy of the universe is constant. Energy is conserved, and is neither created nor destroyed. Energy is transferred in the form of heat and/or work. DEuniverse = DEsystem + DEsurroundings = 0 ∆𝑬𝒔𝒚𝒔𝒕𝒆𝒎 = −∆𝑬𝒔𝒖𝒓𝒓𝒐𝒖𝒏𝒅𝒊𝒏𝒈𝒔 In Thermodynamics we are interested in the internal energies of the system and the surroundings, where most of the energy is located, i.e., we ignore the motion of the universe. ©McGraw-Hill Education. Internal Energy The internal energy (E) of a system is the sum of the kinetic and potential energies of all of the particles that compose the system: molecules, electrons in molecules, nuclear particles in the nuclei of the atoms forming molecules. It is difficult and unnecessary to calculate the absolute internal energy of a system. In Thermochemistry we study the change in the internal energy (ΔE) of a system, by measuring and calculating the heat (q) and work (w) it exchanges with the surroundings. ©McGraw-Hill Education. State Function To reach the top of the mountain there are two trails: 1. Long and winding 2. Short but steep Regardless of the trail, when you reach the top, you will be 10,000 ft above the base. A state function is a mathematical function whose result depends only on the initial and final conditions, not on the process (“path”) used. Internal Energy is a state thermodynamic function: ∆𝐸 = 𝐸𝑓𝑖𝑛𝑎𝑙 − 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − 𝐸𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ©McGraw-Hill Education. State and Path Functions in Thermodynamics ©McGraw-Hill Education. Two Different Paths for the Energy Change of a System Even though q and w for the two paths are different, the total DE is the same for both. Figure 6.6 ©McGraw-Hill Education. Transfer of Internal Energy (E) Between a System and its Surroundings. Energy Diagrams DE = Efinal - Einitial = Eproducts - Ereactants Figure 6.2 ©McGraw-Hill Education. Change in Internal Energy, Heat and Work In Thermodynamics we study the change in the internal energy (ΔE) of a system, by measuring and calculating the heat (q) and work (w) it exchanges with surroundings: ∆𝐸 = 𝑞 + 𝑤 ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = −∆𝐸𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 ∆𝐸𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 = 0 𝐼𝑓 ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑞 + 𝑤 𝑡ℎ𝑒𝑛 ∆𝐸𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = − 𝑞 + 𝑤 = −𝑞 − 𝑤 We need a sign convention for q and w! – ΔE > 0 when the object absorbs energy, therefore (q + w) > 0 – ΔE < 0 when the object loses energy, therefore (q + w) < 0 ©McGraw-Hill Education. Table 6.1 Sign Conventions for q, w, and DE We define the sign of the energy change with respect to the system: Energy transferred into the system is positive: +q and +w (Cash in is credit.) Energy transferred out from the system is negative: -q and –w (Cash out is debit.) © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 38 Energy Transfer as Heat Figure 6.3 ©McGraw-Hill Education. Energy Transfer as Work Figure 6.4 ©McGraw-Hill Education. Sample Problem 6.1: Problem and Plan Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released causes the gaseous products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the radiator. If the expanding gases do 451 J of work on the pistons and the system releases 325 J to the surroundings as heat, calculate the change in energy (ΔE) in J, kJ, and kcal. PLAN: We define system and surroundings to choose signs for q and w, and then we calculate ΔE. The system is the reactants and products, and the surroundings are the pistons, the radiator, and the rest of the car. Heat is released by the system, so q is negative. Work is done by the system to push the pistons outward, so w is also negative. We obtain the answer in J and then convert it to kJ and kcal. © McGraw Hill Sample Problem 6.1: Solution SOLUTION: q = − 325 J w = − 451 J ∆E = q+w =−325 J+(−451 J)= − 776 J DE= − 776 J  1 kJ = −0.776 kJ 1000 J DE= − 776 J  1 cal 1 kcal  = −0.185 kcal 4.184 J 1000 cal © McGraw Hill Pressure-Volume Work 1 Pressure-volume work is done when the volume of the system changes in the presence of an external pressure, P. (w = –PDV) Work of expansion: the volume of the system increases if the temperature is raised or if a chemical reaction results in a net increase in the number of moles of gas. The system expands and does work on the surroundings, losing energy. Work of contraction: the volume of the system decreases if the temperature is lowered or if a chemical reaction results in a net decrease in the number of moles of gas. The system contracts and has work done on it by the surroundings, gaining energy as work. © McGraw Hill Pressure-Volume Work 2 An expanding gas pushing back the atmosphere does PV work (w = –P△V) Figure 6.7 © McGraw Hill Work calculation when a gas changes volume force 𝐹 Pressure (p) = = area 𝐴 → 𝐹 =𝑝∙𝐴 Definition of Work in Mechanics (Physics): 𝑤 = 𝐹 ∙ 𝐷 A piston is a disk or short cylinder fitting closely within a tube in which it moves up and down against a liquid or gas, used in an internal combustion engine to derive motion, or in a pump to impart motion. If the cross section of the piston disk is A and the disk moves a distance D, what is change in the gas volume inside the piston? ∆𝑉 = 𝐴 ∙ 𝐷 Without a sign convention, the pressure-volume work is: 𝑤 = 𝐹 ∙ 𝐷 = 𝑝 ∙ 𝐴 ∙ 𝐷 = 𝑝 ∙ ∆𝑉 ©McGraw-Hill Education. Sample Problem 6.2: Calculating Pressure-Volume Work Done by or on a System Calculating Pressure-Volume Work Done by or on a System PROBLEM: A reaction taking place in a container with a piston-cylinder assembly at constant temperature produces a gas, and the volume increases from 125 mL to 652 mL against an external pressure of 988 torr. Calculate the work done (in J; 1 atm⋅L = 101.3 J). PLAN: We are given the external pressure (988 torr) and initial (125 m L) and final volumes (652 m L) and have to find the work done by the gas. We subtract the initial V from the final V to find ΔV and convert it from m L to L. We convert the given P from torr to atm and calculate w. Then we convert the answer from atm⋅L to J. © McGraw Hill Sample Problem 6.2: Solution SOLUTION: Calculating ΔV: ΔV (mL) = Vfinal – Vinitial = 652 mL – 125 mL = 527 mL Converting ΔV from mL to L: DV ( L ) = 527 mL  1L = 0.527 L 1000 mL Converting P from torr to atm: P ( atm ) = 988 torr  1 atm = 1.30 atm 760 torr Calculating w: w(atm∙L)= − P∆𝑉= − (1.30 atm×0.527 L)= − 0.685 atm∙L w ( J ) = −0.685 atm  L  © McGraw Hill 101.3 J = −69.4 J 1 atm  L 6.3 Measuring heat by measuring temperature change: heat capacity of materials 𝑚 𝑞 = 𝐶 ∙ ∆𝑇 = 𝑚 ∙ 𝑐 ∙ ∆𝑇 = ∙ 𝑐 ∙ 𝑀 ∙ ∆𝑇 = 𝑛 ∙ 𝐶𝑚 ∙ ∆𝑇 𝑀 – q = heat lost or gained by the object – DT = Tfinal – Tinitial – C = heat capacity of the object (extensive property) – m = mass in g – c = specific heat capacity of the material (intensive property) – n = amount (mol) – Cm = molar heat capacity The specific heat capacity (c) of a substance is the quantity of heat required to change the temperature of 1 gram of the substance by 1 K. ©McGraw-Hill Education. Extensive and Intensive Properties (refresher) Extensive properties are dependent on the amount of substance present; mass and volume, for example, are extensive properties. – Extensive properties in thermodynamics: internal energy, heat capacity (a calorimeter constant), enthalpy, entropy, Gibbs free energy Intensive properties are independent of the amount of substance; density is an intensive property. – Intensive properties in thermodynamics: temperature, pressure, composition, specific or molar heat capacity, molar enthalpy, molar entropy, molar Gibbs free energy, energy density of a battery, voltage ©McGraw-Hill Education. Specific Heat Capacities (c) of Some Elements, Compounds, and Materials Substance c (J/g·K) Elements Aluminum, Al 0.900 Graphite, C 0.711 Iron, Fe 0.450 Copper, Cu 0.387 Gold, Au 0.129 Compounds Water, H2O(l) 4.184 Ethyl alcohol, C2H5OH(l) 2.46 Ethylene glycol, (CH2OH)2(l) 2.42 Carbon tetrachloride, CCl4(l) 0.862 Materials Wood 1.76 Cement 0.88 Glass 0.84 Granite 0.79 Steel 0.45 Table 6.2 ©McGraw-Hill Education. Sample Problem 6.4: Relating Quantity of Heat and Temperature Change Relating Quantity of Heat and Temperature Change PROBLEM: (a) A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25˚C to 90.˚C? (b) How much heat is needed to raise the temperature of the same mass (125 g) of water from 25˚C to 90.˚C? The specific heat capacity (c) values of Cu and H2O are given in Table 6.2. (c) Explain the reason for the difference in the answers. PLAN: We know the mass (125 g) and c (0.387 J/g⋅K) of Cu and of water (4.184 J/g⋅K) and can find ΔT in ˚C, which equals ΔT in K. We then calculate the heat required for the changes. © McGraw Hill Sample Problem 6.4: Solution SOLUTION: Calculating ΔT for Cu and H2O: ΔT = Tfinal − Tinitial = 90.˚C − 25˚C = 65˚C = 65 K = 65 deg (a) Calculating q for Cu: q = c  mass ( g )  DT = 0.387 J g  K × 125 g × 65 K = 3.14×103 J (b) Calculating q for H2O: q = c  mass ( g )  DT = 4.184 J g  K × 125 g × 65 K = 3.40×104 J (c) The heat required to raise the temperature of the water if 10.8 times the amount needed to heat an equal mass of copper because the specific heat capacity of water is 10.8 times the specific heat capacity of copper. © McGraw Hill Calorimetry. Bomb calorimeter Calorimetry is used to measure thermal energy exchanged between reaction and surroundings. A bomb calorimeter has a constant volume. It is used to measure ΔE for all reactions in which at least one reactant is in the gas phase, e.g., combustion reactions [O2(g)]. The heat capacity of the calorimeter is the amount of heat absorbed by the calorimeter for each degree rise in temperature and is called the calorimeter constant. – Ccal, kJ/°C © 2017 Pearson Education, Inc. Bomb calorimeter © 2017 Pearson Education, Inc. Measuring DE: Calorimetry at Constant Volume How does the Bomb Calorimeter work? In practice, we cannot expose the thermometer (or any other temperature sensor – thermocouple, thermistor, etc.) to the combustion reaction because it will be damaged. Instead, we measure the temperature change in the surroundings (water and the inner guts of the calorimeter). We limit the surroundings to the inner, sealed bomb, water, probes and stirrer by building a vacuum mantle in the walls of the outer container. The outer vessel cannot exchange heat with the air surrounding it because vacuum is the perfect thermal insulator. Thermodynamic changes of a system with no exchange of heat with the surroundings are called adiabatic transformations (or adiabatic paths). © 2017 Pearson Education, Inc. Measuring DE: Calorimetry at Constant Volume In a bomb calorimeter: 𝒘 = 𝟎 (𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒗𝒐𝒍𝒖𝒎𝒆 𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕) → ∆𝑬𝒓𝒙𝒏(𝒔𝒚𝒔) = 𝒒 + 𝒘 = 𝒒𝑽 (𝒒𝑽 𝒊𝒔 𝒉𝒆𝒂𝒕 𝒂𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒗𝒐𝒍𝒖𝒎𝒆) qsurroundings = qadiabatic calorimeter = ─qsystem= −∆𝑬𝒓𝒙𝒏 ∆𝑬𝒓𝒙𝒏 = −𝒒𝒄𝒂𝒍𝒐𝒓𝒊𝒎𝒆𝒕𝒆𝒓 ∆𝑬𝒓𝒙𝒏 = −𝑪𝒄𝒂𝒍 ∆𝑻 𝑰𝒇 ∆𝑻 > 𝟎 𝐭𝐡𝐞𝐧 ∆𝑬𝒓𝒙𝒏 < 𝟎 𝑰𝒇 ∆𝑻 < 𝟎 𝐭𝐡𝐞𝐧 ∆𝑬𝒓𝒙𝒏 > 𝟎 © 2017 Pearson Education, Inc. Sample Problem 6.7 – Problem and Plan Calculating the Heat of a Combustion Reaction PROBLEM: Oxyacetylene torches produce such high temperatures that they are often used to weld and cut metal. When 2.50 g of acetylene (C2H2) is burned in a bomb calorimeter with a heat capacity of 10.85 kJ/K, the temperature increases from 23.488°C to 34.988°C. What is ΔE (in kJ/mol) for this combustion reaction? PLAN: To find the heat released during the combustion of 2.50 g of acetylene, we find ΔT, multiply it by the given heat capacity of the calorimeter (10.85 kJ/K), and change the sign since −qrxn = qcalorimeter. We use the molar mass of acetylene to calculate the heat released by 1 mole of acetylene (ΔE). ©McGraw-Hill Education. Sample Problem 6.7 – Solution SOLUTION: Finding ΔT: ΔT = Tfinal − Tinitial = 34.988°C − 23.488°C = 11.500°C = 11.500 K Calculating the heat absorbed by the calorimeter: qcalorimeter = Ccalorimeter × ΔT = 10.85 kJ/K × 11.500 K = 124.8 kJ Therefore, qrxn = −124.8 kJ for 2.50 g of acetylene. ∆E = ©McGraw-Hill Education. −124.8 kJ 2.50 g C2 H2 ∗ 26.04 g C2 H2 1 mol C2 H2 = 1.30 ∗ 103 kJ C H mol 2 2 Practice Problem 6.50 Calculating the Heat of a Combustion Reaction PROBLEM: A chemist places 1.750 g of ethanol, C2H6O, in a bomb calorimeter with a heat capacity of 12.05 kJ/K. The sample is burned and the temperature of the calorimeter increases by 4.287C. Calculate ΔE for the combustion of ethanol in kJ/mol. ©McGraw-Hill Education. Practice Problem 6.7B – Problem and Plan Calculating the Heat of a Combustion Reaction PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department for Consumer Affairs places one serving in a bomb calorimeter and burns it in O2. The initial temperature is 21.862C, and the temperature rises to 26.799C. If the heat capacity of the calorimeter is 8.151 kJ/K, is the manufacturer’s claim correct? PLAN: To find the heat released during the combustion of one serving of dessert, we find ΔT, multiply it by the given heat capacity of the calorimeter (8.151 kJ/K), and change the sign since −qrxn = qcalorimeter. At the end, we have to convert the unit for qrxn from kJ to Cal. ©McGraw-Hill Education. Practice Problem 6.7B – Solution SOLUTION: Finding ΔT: ΔT = Tfinal − Tinitial = 26.799°C − 21.862°C = 4.937°C = 4.937 K Calculating the heat absorbed by the calorimeter (surroundings): 𝑞𝑠𝑢𝑟𝑟 = 𝑞𝑐𝑎𝑙 = 𝐶𝑐𝑎𝑙 × ∆𝑇 = 8.151 𝑘𝐽 𝐾 × 4.937 𝐾 = 40.241 𝑘𝐽 Therefore: 𝑞𝑠𝑦𝑠 = 𝑞𝑟𝑥𝑛 = −𝑞𝑠𝑢𝑟𝑟 = −𝑞𝑐𝑎𝑙 = −40.241 𝑘𝐽/𝑠𝑒𝑟𝑣𝑖𝑛𝑔 𝑞𝑟𝑥𝑛 ©McGraw-Hill Education. 1000 𝐽 1 𝑐𝑎𝑙 1 𝐶𝑎𝑙 = −40.241 𝑘𝐽 × × × = −9.618 𝐶𝑎𝑙/𝑠𝑒𝑟𝑣𝑖𝑛𝑔 1 𝑘𝐽 4.184 𝐽 1000 𝑐𝑎𝑙 Calorimetry at Constant Pressure Dewar calorimeters Figure 6.9. Styrofoam cup calorimeter ©McGraw-Hill Education. Calorimetry at Constant Pressure Most reactions in Chemistry happen in solution, liquid or solid state. We cannot constrain liquid and solid states to undergo thermal changes at constant volume because it would require extremely high pressures, not easily attainable in laboratories. Instead, we measure the thermal effects at constant pressure, i.e., in baric equilibrium with the surrounding air. Thermodynamic changes at constant pressure are called isobaric transformations (or isobaric paths) and are indicated by subscript P. ©McGraw-Hill Education. Reaction Heat at Constant Pressure (qp) 1. The heat at constant pressure, qp, is larger than the heat at constant volume, qV (the heat measured in a bomb calorimeter is different from the heat in a Dewar calorimeter): ∆𝐸 = 𝑞𝑃 + 𝑤𝑃 𝑤𝑃 = −𝑃𝑒𝑥𝑡 ∆𝑉 ≠ 0 → 𝑞𝑃 = ∆𝐸 − 𝑤𝑃 𝑞𝑉 = ∆𝐸 𝑤𝑉 = 0 → 𝑞𝑃 = 𝑞𝑉 − 𝑤𝑃 = 𝑞𝑉 + 𝑃𝑒𝑥𝑡 ∆𝑉 𝑜𝑟 𝑞𝑃 ≠ 𝑞𝑉 2. In a constant-pressure calorimeter, the heat released by the reaction (system) is transferred to the reaction products, solvent and the Dewar (surroundings). Here, the heat capacity of the empty calorimeter must be determined in a separate measurement by heating pure water in the Dewar with a calibrated amount of energy. ©McGraw-Hill Education. Calculation of Heat at Constant Pressure (qp) 𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛: 𝑞𝑠𝑦𝑠(𝑟𝑥𝑛) = −𝑞𝑠𝑢𝑟𝑟 𝑞𝑟𝑥𝑛 = − 𝑚𝑓𝑖𝑛𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑓𝑖𝑛𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 + 𝐶𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟 ∙ ∆𝑇 𝑚𝑓𝑖𝑛𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = ෍ 𝑚𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 Ccalorimeter and cfinal solution need to be measured in separate experiments. Calorimetry at constant pressure is not trivial. In problems, often Ccalorimeter is ignored (assumed 0) and cfinal.solution is assumed 4.184 J/(g K) for pure water. ©McGraw-Hill Education. Sample Problem 6.5 – Problem and Plan Determining the Specific Heat Capacity of a Solid PROBLEM: You heat 22.05 g of a solid in a test tube to 100.00°C and then add the solid to 50.00 g of water in a coffee-cup calorimeter. The water temperature changes from 25.10°C to 28.49°C. Find the specific heat capacity of the solid. PLAN: This type of problem calls for idealized (or simplified) calorimetry. The universe here is limited to the hot and cold objects, and it excludes the calorimeter. We cannot ignore the calorimeter heat capacity in heat measurements. We are given the masses of the solid (22.05 g) and of H2O (50.00 g), and we can find the temperature changes of the water and of the solid by subtracting the given values, always using Tfinal − Tinitial. We set the heat released by the solid (−qsolid) equal to the heat absorbed by the water (qH2O). The specific heat of water is known, and we solve for csolid. ©McGraw-Hill Education. Sample Problem 6.5 – Solution SOLUTION: Finding ΔTsolid and ΔTH2O: Tinitial (solid) = 100.00°C; Tinitial (H2O) = 25.10°C; Tfinal (solid and H2O) = 28.49°C ΔTH2O = Tfinal − Tinitial = 28.49°C − 25.10°C = 3.39°C = 3.39 K ΔTsolid = Tfinal − Tinitial = 28.49°C − 100.00°C = −71.51°C = −71.51 K Setting up the heat exchange equation (calorimetry equation): 𝑞𝑠𝑦𝑠 = −𝑞𝑠𝑢𝑟𝑟 𝑚𝑠𝑜𝑙𝑖𝑑 𝑐𝑠𝑜𝑙𝑖𝑑 ∆𝑇𝑠𝑜𝑙𝑖𝑑 = −𝑚𝐻2𝑂 𝑐𝐻2𝑂 ∆𝑇𝐻2𝑂 Solving for csolid: csolid = − ©McGraw-Hill Education. (cH2O ∗massH2O ∗∆TH2O ) masssolid ∗∆Tsolid =− 4.184 J ∗50.00 g ∗3.39 K gK 22.05 g ∗ −71.51 K = 0.450 J gK Enthalpy: A New Thermodynamical Function for Change at Constant Pressure ΔE = q + w To determine ΔE, both heat and work must be measured. Most reactions are carried out in open vessels, i.e., at constant pressure. At constant pressure, work is 𝑤 = −𝑃∆𝑉 and heat is qP: ∆𝐸 = 𝑞𝑃 − 𝑃∆𝑉 𝑂𝑅 𝑞𝑝 = ∆𝐸 + 𝑃∆𝑉 We define the enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume: H = E + PV Enthalpy is a state function because both E and PV are state functions. © McGraw Hill Change in Enthalpy at constant pressure We cannot measure the absolute value of the enthalpy because we cannot measure the absolute value of E. However, we can measure the change in enthalpy, ΔH, just like we can measure ΔE. How? If: H = E + PV then 𝐻𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐻𝑓𝑖𝑛𝑎𝑙 = 𝐸𝑓𝑖𝑛𝑎𝑙 + 𝑃𝑉𝑓𝑖𝑛𝑎𝑙 𝐻𝑓𝑖𝑛𝑎𝑙 − 𝐻𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸𝑓𝑖𝑛𝑎𝑙 − 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 + 𝑃 𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 or ∆𝐻 = ∆𝐸 + 𝑃∆𝑉 but ∆𝐸 = 𝑞 + 𝑤 = 𝑞𝑃 − 𝑃∆𝑉 → 𝑞𝑝 = ∆𝐸 + 𝑃∆𝑉 The enthalpy change, DH, of a reaction is the heat involved in a reaction at constant pressure: ∆𝐻 = 𝑞𝑃 © 2017 Pearson Education, Inc. Change in Enthalpy at constant volume If: H = E + PV, derive the change in enthalpy, ΔH, associated with a bomb calorimeter. The answer should be: ∆𝐻 = 𝑞𝑉 + 𝑉∆𝑃 The change in pressure during reaction in a bomb calorimeter must be measured. A pressure gauge is needed! Or we assume that gases are ideal: 𝑉∆𝑃 = ∆𝑛𝑔𝑎𝑠 𝑅𝑇 © 2017 Pearson Education, Inc. Change in Enthalpy for a System Definition of Enthalpy: 𝐻 = 𝐸 + 𝑃𝑉 𝑃𝑉 𝑖𝑠 𝑎𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑒𝑟𝑚 (a work) Change in Enthalpy of a system for an isobaric transformation: ∆𝐻 = ∆𝐸 + 𝑃∆𝑉 = 𝑞𝑃 Change in Enthalpy of a system for an isochoric transformation: ∆𝐻 = ∆𝐸 + 𝑉∆𝑃 = 𝑞𝑉 + 𝑉∆𝑃 Change in Enthalpy of a system for a general transformation: ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) Just like internal energy, enthalpy is an extensive property, i.e., it depends on the mass of the system undergoing transformation. It is also a state function. ©McGraw-Hill Education. DH as a Measure of DE Usually DH and DE are similar in value; the difference is largest for reactions that produce or use large quantities of gas. DH is the change in heat for a system at constant pressure. qP = DE + PDV = DH DH ≈ DE – for reactions that do not involve gases – for reactions in the gas phase in which the total amount (mol) of gas does not change – for reactions in which qP is much larger than PDV, even if the total mol of gas does change. ©McGraw-Hill Education. Enthalpy of Exothermic and Endothermic Processes An exothermic process releases heat, so ∆𝐻 < 0 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑞𝑃 < 0. An endothermic process absorbs heat, so ∆𝐻 > 0. Figure 6.8 ©McGraw-Hill Education. Sample Problem 6.3 – Problem and Plan Drawing Enthalpy Diagrams and Determining the Sign of ΔH PROBLEM: In each of the following cases, determine the sign of ΔH, state whether the reaction is exothermic or endothermic, and draw an enthalpy diagram: 1 2 a) H2(g) + O2(g) → H2O(l) + 285.8 kJ b) 40.7 kJ + H2O (l) → H2O (g) PLAN: In each equation, we note whether heat is a “product” (exothermic; ΔH < 0) or a “reactant” (endothermic; ΔH > 0). For exothermic reactions, reactants are above products on the enthalpy diagram; for endothermic reactions, reactants are below products. The ΔH arrow always points from reactants to products. ©McGraw-Hill Education. Sample Problem 6.3 – Solution SOLUTION: a) Heat is a product (on the right), so ΔH < 0 and the reaction is exothermic. b) Heat is a reactant (on the left), so ΔH > 0 and the reaction is endothermic. Heat is absorbed during the reaction. ©McGraw-Hill Education. Sample Problem 6.6 – Determining the Enthalpy Change of an Aqueous Reaction Determining the Enthalpy Change of an Aqueous Reaction PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.00°C and add 25.0 mL of 0.500 M HCl, also at 25.00°C. After stirring, the final temperature is 27.21°C. [Assume that the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/mL) and specific heat capacity of pure water.] a) Calculate qsoln (in J). b) Calculate the ΔH for the reaction in kJ/(mol H2O). MY COMMENTS: 1) Again, the heat capacity of the calorimeter is ignored. 2) Volumes are not additive because the intermolecular forces change during mixing and reaction, so the densities will change, too. Only masses are additive. Measuring volumes is convenient, but masses is what we really need. 3) A 0.500 M solution is not dilute enough to be likened to pure water. The final solution contains NaCl (reaction product) and NaOH (excess reactant). So, c = 4.184 J/(g K) is only an approximation for convenience. 4) Calorimetry measurements done right are not trivial. ©McGraw-Hill Education. Sample Problem 6.6 – Plan PLAN: a) The solution is the surroundings, and as the reaction takes place, heat flows into the solution. To find qsoln, we need the mass of solution, the change in temperature, and the specific heat capacity. We know the volumes (25.0 mL and 50.0 mL), so we find their masses with the given density (1.00 g/mL). Then, to find qsoln, we multiply the total mass by the given c (4.184 J/g⋅K) and the change in T, which we find from Tfinal − Tinitial. b) The heat of the surroundings is qsoln, and it is the negative of the heat of the reaction (qrxn), which equals ΔH. And dividing qrxn by the amount (mol) of water formed in the reaction gives ΔH in kJ per mole of water formed. To calculate the amount (mol) of water formed, we write the balanced equation for the acid-base reaction and use the volumes and the concentrations (0.500 M) to find amount (mol) of each reactant (H+ and OH−). Since the amounts of two reactants are given, we determine which is limiting, that is, which gives less product (H2O). ©McGraw-Hill Education. Sample Problem 6.6 – Solution SOLUTION: – Finding masssoln and ΔTsoln: Total mass of solution = (25.0 mL + 50.0 mL) * 1.00 g/mL = 75.0 g ∆T = 27.21 ℃ - 25.00 ℃ = 2.21 ℃ = 2.21 K – Finding qsoln: qsoln = csoln × masssoln × ΔTsoln = (4.184 J/g⋅K)(75.0 g)(2.21 K) = 693 J ©McGraw-Hill Education. Sample Problem 6.6 – Solution, Cont’d SOLUTION: – Writing the balanced equation: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) – Finding amounts (mol) of reactants: Amount of HCl = 0.500 mol HCl/L * 0.0250 L = 0.0125 mol HCl Amount of NaOH = 0.500 mol NaOH/L * 0.0500 L = 0.0250 mol NaOH – Finding the amount (mol) of product: All the coefficients in the equation are 1, which means that the amount (mol) of reactant yields that amount of product. Therefore, HCl is limiting because it yields less product: 0.0125 mol of H2O. – Finding ΔH: Heat absorbed by the solution was released by the reaction; that is, qsoln = −qrxn = 693 J ∆H ©McGraw-Hill Education. so qrxn = −693 J kJ 𝑞𝑟𝑥𝑛 1 kJ −693 J 1 kJ 𝑘𝐽 = ∗ = ∗ = −55.4 mol mol 𝐻2 𝑂 1000 J 0.0125 mol H2 O 1000 J 𝑚𝑜𝑙 𝐻2 𝑂 Stoichiometry of Thermochemical Equations A thermochemical equation is a balanced equation that includes ΔHrxn and the phase states of the chemicals participating in the reaction: 2𝐻2 𝑂 𝑙 → 2𝐻2 𝑔 + 𝑂2 𝑔 ∆𝐻𝑟𝑥𝑛 = 572 𝑘𝐽 – The sign of ΔH indicates whether the reaction is exothermic (ΔH < 0) or endothermic (ΔH > 0). A forward reaction has the opposite sign of the reverse reaction: 2𝐻2 𝑔 + 𝑂2 𝑔 → 2𝐻2 𝑂 𝑙 ∆𝐻𝑟𝑥𝑛 = −572 𝑘𝐽 The magnitude of ΔH is proportional to the amount of substance. The coefficients of the balanced equation need not be the lowest integers: 1 𝐻2 𝑔 + 𝑂2 𝑔 → 𝐻2 𝑂 𝑙 ∆𝐻𝑟𝑥𝑛 = −286 𝑘𝐽 2 1 𝑏𝑎𝑟,298 𝐾 Data are tabulated as kJ/mol. Example: ∆𝐻𝑟𝑥𝑛 = −286 ©McGraw-Hill Education. 𝑘𝐽 𝑚𝑜𝑙 𝐻2 𝑂(𝑙) Relationship Between Amount and Heat Energy Transfer The value of ΔH can be used in a calculation in the same way as a mole ratio. ©McGraw-Hill Education. Sample Problem 6.8 – Problem Using the Enthalpy Change of a Reaction (ΔH) to Find the Amount of a Substance PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be written as: ∆ 3 Al2O3(s) → 2Al(s) + O2(g) 2 ∆H = 1676 kJ If aluminum is produced this way, how many grams of aluminum can form when 1.000×103 kJ (1.000 MJ) of heat is transferred? ©McGraw-Hill Education. Sample Problem 6.8 – Plan PLAN: From the balanced equation and the enthalpy change, we see that 1676 kJ of heat is thermochemically equivalent to 2 mol of Al. With this equivalent quantity, we convert the given number of kJ to amount (mol) formed and then convert amount to mass in g. ∆ 3 Al2O3(s) → 2Al(s) + O2(g) 2 ©McGraw-Hill Education. ∆H = 1676 kJ Sample Problem 6.8 – Solution SOLUTION: Combining steps to convert from heat transferred to mass of Al: m (g) of Al = (1.000 x 103 kJ) x ©McGraw-Hill Education. 2 mol 𝐴𝑙 1676 kJ x 26.98 g Al = 1 mol Al 32.20 g Al Hess’s Law Some reactions are difficult to run in a calorimeter (extreme conditions, biochemical process, etc.). The enthalpy of reaction can be calculated from the enthalpies of reaction of other reactions that involve common reactants and products with the extreme condition reaction by using the Hess’s law. Hess’s law states that the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps. Hess’s law applies to phase changes as well. DHoverall = DH1 + DH2 + ………. + DHn DH for an overall reaction can be calculated if the DH values for the individual steps are known. Hess’s law is a corollary of the 1st law of thermodynamics and enthalpy as state function. ©McGraw-Hill Education. Calculating DH For An Overall Process Identify the target equation, the step whose DH is unknown. – Note the amount of each reactant and product. Manipulate each equation with known DH values so that the target amount of each substance is on the correct side of the chemical equation. – Change the sign of DH when you reverse an equation. – Multiply amount (mol) and DH by the same factor. Add the manipulated equations and their resulting DH values to get the target equation and its DH. – ©McGraw-Hill Education. All substances except those in the target equation must cancel. (All chemicals present in various individual steps but not in the target equation are called spectators.) Sample Problem 6.9 – Problem and Plan Using Hess’s Law to Calculate an Unknown ΔH PROBLEM: Two pollutants that form in automobile exhaust are CO and NO. An environmental chemist must convert these pollutants to less harmful gases through the following: CO (g) + NO (g) → CO2 (g) + ½N2 (g) DH = ? Given the following information, calculate the unknown DH: – Equation A: CO (g) + ½O2 (g) → CO2 (g) DHA = –283.0 kJ – Equation B: N2 (g) + O2 (g) → 2NO (g) DHB = 180.6 kJ PLAN: We note the amount (mol) of each substance and on which side each appears in the target equation. We adjust equations A and/or B and their ΔH values as needed and add them together to obtain the target equation and the unknown ΔH. ©McGraw-Hill Education. Sample Problem 6.9 – Solution SOLUTION: Multiply Equation B by ½ and reverse it: 1 2 NO (g) → ½N2 (g) + ½O2 (g) ∆𝐻 = − × 180.6 𝑘𝐽 = −90.3 𝑘𝐽 Add the manipulated equations together: – Equation A: CO (g) + ½O2 (g) – ½ Reversed Equation B: NO (g) → CO2 (g) → ½N2 (g) + ½O2 (g) RESULT: CO (g) + NO (g) → CO2 (g) + ½ N2 (g) DHrxn = - 283.0 kJ + (- 90.3) kJ = –373.3 kJ ©McGraw-Hill Education. DH = –283.0 kJ DH = –90.3 kJ Hess’s Law Practice Problem Find ΔHrxn for the reaction: 𝑁2 𝑂 𝑔 + 𝑁𝑂2 (𝑔) → 3𝑁𝑂(𝑔) Use these reactions with known ΔH’s: 2𝑁𝑂 𝑔 + 𝑂2 𝑔 → 2𝑁𝑂2 𝑔 ∆𝐻 = −113.1 𝑘𝐽 𝑁2 𝑔 + 𝑂2 𝑔 → 2𝑁𝑂 𝑔 ∆𝐻 = +182.6 𝑘𝐽 2𝑁2 𝑂 𝑔 → 2𝑁2 𝑔 + 𝑂2 𝑔 ©McGraw-Hill Education. ∆𝐻 = −163.2 𝑘𝐽 Standard States for Thermochemical Equations Values of the ΔHrxn for a given equation vary somewhat with the pressure and temperature conditions. The standard states for tabulating thermodynamic data are: 1) For a gas: 1 bar (very close to 1 atm: 1 atm = 1.01325 bar) and 298 K. 2) For a pure substance: the most stable form at 1 bar and 298 K. Examples: Br2(l), Hg(l), C(s,graphite), S(s,rhombic), O2(g) 3) For a liquid solution: 1 M concentration, 1 bar, 298 K The symbol for the change in enthalpy at standard conditions is ∆𝐻 °. ° The standard enthalpy of a reaction, ∆𝐻𝑟𝑥𝑛 , is the enthalpy change when all the reactants and products are in their standard states. ©McGraw-Hill Education. Formation Equations, Standard Enthalpy of Formation In a formation thermochemical equation, 1 mol of compound forms from its elements. The standard enthalpy of formation, ∆𝐻𝑓° , is the enthalpy of the formation reaction. Examples: 𝐶 𝑠, 𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒 + 2𝐻2 𝑔 → 𝐶𝐻4 𝑔 ∆𝐻𝑓° = −74.9 𝑘𝐽 1 𝑁𝑎 𝑠 + 𝐶𝑙2 𝑔 → 𝑁𝑎𝐶𝑙 𝑠 ∆𝐻𝑓° = −411.1 𝑘𝐽 2 For an element in its standard state ∆𝐻𝑓° = 0. Standard state for elements means: the correct phase at 1 bar and 298 K, the correct molecular structure (not individual atoms, e.g., Cl2(g)), the correct allotrope (e.g., C(graphite) not C(diamond)). ©McGraw-Hill Education. Selected Standard Enthalpies of Formation at 25°C (298K) See Appendix B Formula Δ Hf ° (kJ/mol) Δ Hf ° (kJ/mol) Br2(l) 0 Br2(g) 30.9 Cl2(g) 0 Cl(g) 121.0 HCl(g) −92.3 Hydrogen Calcium Ca(s) 0 CaO(s) −635.1 −1206.9 Carbon C(graphite) 0 C(diamond) 1.9 Δ Hf ° (kJ/mol) H2(g) 0 H(g) 218.0 Mercury Hg(l) 0 Hg(g) 61.3 N2(g) CO2(g) −393.5 NH3(g) −45.9 CH4(g) −74.9 NO(g) 90.3 HCN(g) CS2(l) Table 6.3 −238.6 135 87.9 Ag(s) 0 0 −127.0 Sodium Na(s) 0 Na(g) 107.8 NaCl(s) Nitrogen −110.5 CH3OH(l) Silver AgCl(s) CO(g) ©McGraw-Hill Education. Formula Chlorine Bromine CaCO3(s) Formula −411.1 Sulfur S8(rhombic) 0 S8(monoclinic) 0.3 Oxygen O2(g) 0 O3(g) 143 H2O(g) −241.8 H2O(l) −285.8 SO2(g) −296.8 SO3(g) −396.0 Standard Enthalpy of Formation: details Most formation reactions cannot be run in one step, therefore their enthalpy of reaction cannot be measured in a calorimeter. Most standard enthalpies of formation are calculated from accessible thermochemical equations using Hess’s law. For an element in its standard state ∆𝐻𝑓° = 0 by international agreement, for the following reasons: – Formation of an element from itself is not a reaction. – We cannot measure absolute enthalpies (H), we can only measure changes in enthalpy (ΔH). Elements are set as relative zero enthalpies in Thermodynamics. – Compared to elements, most chemical compounds have a negative standard enthalpy of formation. ©McGraw-Hill Education. Sample Problem 6.10 – Problem and Plan Writing Formation Equations PROBLEM: Write a balanced formation equation for each of the following and include the value of ΔHf° : (a) AgCl(s); (b) CaCO3(s); (c) HCN(g). PLAN: We write the elements as the reactants and 1 mol of the compound as the product, being sure all substances are in their standard states. Then, we balance the equations, keeping a coefficient of 1 for the product, and find the Δ H f ° values in Appendix B. ©McGraw-Hill Education. Sample Problem 6.10 – Solution SOLUTION: a) Ag (s) + ½Cl2 (g) → AgCl (s) DH°f = –127.0 kJ b) Ca (s) + C(graphite) + 3/2 O2 (g) → CaCO3 (s) DH°f = –1206.9 kJ c) ½H2 (g) + C(graphite) + ½N2 (g) → HCN (g) ©McGraw-Hill Education. DH°f = 135 kJ Determining DH°rxn from DH°f Values Any reaction can be thought of as happening in 2 steps: 1. The decomposition of reactants to elements. This step would be the reverse of the formation reactions of reactants. 2. The re-combination of elements to form products. This would be the formation reactions of products. ° ° ° ∆𝐻𝑟𝑥𝑛 = ෍ 𝑚∆𝐻𝑓(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ෍ 𝑛∆𝐻𝑓(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) Figure 6.12 ©McGraw-Hill Education. Sample Problem 6.11 – Problem and Plan Calculating ΔH°rxn from ΔH°f Values PROBLEM: Nitric acid is used to make many products, including fertilizers, dyes, and explosives. The first step in its production is the oxidation of ammonia: 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g) Calculate DH°rxn from DH°f values. PLAN: Use the DH°f values from Appendix B and apply the equation: DHrxn = SmDH°f (products) – SnDH°f (reactants) ©McGraw-Hill Education. Sample Problem 6.11 – Solution SOLUTION: DHrxn = SmDHºf (products) – SnDHºf (reactants) DHrxn = [4(DHºf of NO(g) + 6(DHºf of H2O(g)] – [4(DHºf of NH3(g) + 5(DHºf of O2(g)] = (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) – [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] = –906 kJ DHrxn = –906 kJ ©McGraw-Hill Education. 9.4. Bond Energy and Enthalpy of Reaction Chemistry The Molecular Nature of Matter and Change Eighth Edition Martin S. Silberberg and Patricia G. Amateis ©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Bond Energy The bond energy (BE) is the energy needed to overcome the attraction between the nuclei and the shared electrons. The stronger the bond the higher the bond energy. Bond Energy (also known as the bond dissociation energy, D) is the standard enthalpy change for breaking a bond in 1 mol of gaseous molecules (to 1 mol of each gaseous individual atoms or radicals): 𝑘𝐽 𝐶𝑂 𝑔 → 𝐶 𝑔 + 𝑂 𝑔 𝐷 𝐶 ≡ 𝑂 = 1070 𝑚𝑜𝑙 𝑘𝐽 𝐻2 𝑂 𝑔 → 𝐻 𝑔 + 𝑂𝐻 𝑔 𝐷 𝑂 − 𝐻 = 467 𝑚𝑜𝑙 As bond dissociation is always an endothermic process, the bond energy is always a positive number. ©McGraw-Hill Education. Average Bond Energies (kj/mol) and Lengths (pm) ©McGraw-Hill Education. Bond Energies and DH°rxn The heat released or absorbed during a chemical change is due to differences between the bond energies of reactants and products: 1. First, the heat is absorbed by reactants to break the existing bonds (ΔH > 0). 2. Second, heat is released by products in forming new bonds. DH°rxn = SDH°reactant bonds broken+ SDH°product bonds formed ©McGraw-Hill Education. Why would we calculate ∆𝑯𝒓𝒙𝒏 from Bond Energies? When the ∆𝐻𝑟𝑥𝑛 cannot be measured in a calorimeter for various reasons: Reaction is too slow kinetically Reaction can never be carried out in a calorimeter because it requires extreme conditions (high T, high p, catalyst, etc.) Not enough thermochemical data to use Hess’s Law It is a proposed or newly isolated chemical (synthesis reaction is not yet known) One or more of the reactants cannot be isolated in pure form for calorimetric studies (e.g., biochemical compounds) © 2017 Pearson Education, Inc. DH°rxn from bond energies is an estimated value: within 10% of the actual value The actual bond energy depends on the surrounding atoms in molecule and other (quantum mechanical) factors (e.g., type of molecular orbital, hybridization, conjugation, etc.) Tabulated values are average bond energies. For example, the D(C-H) value is an average for alkanes (CnH2n+2), cycloalkanes and alkenes (CnH2n), etc. Bond energies are measured spectroscopically. Works best when all reactants and products in gas state. In this way, the energies of the intermolecular interactions do not have to be accounted for. © 2017 Pearson Education, Inc. Using Bond Energies to Calculate DH°rxn For HF Formation Figure 9.17 ©McGraw-Hill Education. ∆𝐻𝑓° 𝐻𝐹 = −546 𝑘𝐽 𝑚𝑜𝑙 Using Bond Energies to Calculate DH°rxn for the Combustion of Methane Figure 9.18 ©McGraw-Hill Education. ° ∆𝐻𝑟𝑥𝑛 = −891 𝑘𝐽 𝑚𝑜𝑙 Sample Problem 9.4 – Problem and Plan Using Bond Energies to Calculate DH°rxn PROBLEM: Calculate DH°rxn for the chlorination of methane to form chloroform. PLAN: All the reactant bonds break, and all the product bonds form. Find the bond energies in Table 9.2 and substitute the two sums, with correct signs, into Equation 9.2. ©McGraw-Hill Education. Sample Problem 9.4 - Solution SOLUTION: For bonds broken: 3 x C-H = (3 mol)(413 kJ/mol) = 1239 kJ 3 x Cl-Cl = (3 mol)(243 kJ/mol) = 729 kJ SDHºbonds broken = 1968 kJ For bonds formed: 3 x C-Cl = (3 mol)(–339 kJ/mol) = –1017 kJ 3 x H-Cl = (3 mol)(–427 kJ/mol) = –1281 kJ SDHºbonds formed = –2298 kJ DHºreaction = SDHºbonds broken + SDH bonds formed = 1968 kJ + (–2298 kJ) = –330 kJ ©McGraw-Hill Education.