Lecture Presentation Chapter 6 Thermochemistry Sherril Soman Grand

Lecture Presentation Chapter 6 Thermochemistry Sherril Soman Grand Valley State University © 2014 Pearson Education, Inc.

Chemical Hand Warmers • Most hand warmers work by using the heat released from the slow oxidation of iron 4 Fe(s) + 3 O 2(g) → 2 Fe 2 O 3(s) • The amount your hand temperature rises depends on several factors: – The size of the hand warmer – The size of your glove, etc. Mainly, the amount of heat released by the reaction. © 2014 Pearson Education, Inc.

Nature of Energy • Even though chemistry is the study of matter, energy affects matter. • Energy is anything that has the capacity to do work. • Work is a force acting over a distance. – Energy = work = force × distance • Heat is the flow of energy caused by a difference in temperature. • Energy can be exchanged between objects through contact. – For example, through collisions © 2014 Pearson Education, Inc.

Energy, Heat, and Work • You can think of energy as a quantity an object can possess or as a collection of objects. • You can think of heat and work as the two different ways that an object can exchange energy with other objects. – Either out of it, or into it © 2014 Pearson Education, Inc.

Classification of Energy • Kinetic energy is energy of motion or energy that is being transferred. • Thermal energy is the energy associated with temperature. – Thermal energy is a form of kinetic energy. © 2014 Pearson Education, Inc.

Manifestations of Energy © 2014 Pearson Education, Inc.

Classification of Energy • Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object. – Energy stored in the structure of a compound is potential energy. © 2014 Pearson Education, Inc.

Some Forms of Energy • Electrical – Kinetic energy associated with the flow of electrical charge • Heat or thermal energy – Kinetic energy associated with molecular motion • Light or radiant energy – Kinetic energy associated with energy transitions in an atom • Nuclear – Potential energy in the nucleus of atoms • Chemical – Potential energy due to the structure of the atoms, the attachment between atoms, the atoms’ positions relative to each other in the molecule, or the molecules’ relative positions in the structure © 2014 Pearson Education, Inc.

Conservation of Energy • The law of conservation of energy states that energy cannot be created nor destroyed. • When energy is transferred between objects, or converted from one form to another, the total amount of energy present at the beginning must be present at the end. © 2014 Pearson Education, Inc.

System and Surroundings • We define the system as the material or process within which we are studying the energy changes within. • We define the surroundings as everything else with which the system can exchange energy. • What we study is the exchange of energy between the system and the surroundings. © 2014 Pearson Education, Inc.

Comparing the Amount of Energy in the System and Surroundings during Transfer • Conservation of energy means that the amount of energy gained or lost by the system has to be equal to the amount of energy lost or gained by the surroundings. © 2014 Pearson Education, Inc.

Units of Energy • The amount of kinetic energy an object has is directly proportional to its mass and velocity. ü KE = ½mv 2 • When the mass is in kg and velocity is in m/s, the unit for kinetic energy is. • 1 joule of energy is the amount of energy needed to move a 1 kg mass at a speed of 1 m/s. © 2014 Pearson Education, Inc.

Units of Energy • A joule (J) is the amount of energy needed to move a 1 kg mass a distance of 1 meter. – 1 J = 1 N ∙ m = 1 kg ∙ m 2/s 2 • A calorie (cal) is the amount of energy needed to raise the temperature of one gram of water 1 °C. – kcal = energy needed to raise 1000 g of water 1 °C – food Calories = kcals © 2014 Pearson Education, Inc.

Energy Use © 2014 Pearson Education, Inc.

The First Law of Thermodynamics: Law of Conservation of Energy • Thermodynamics is the study of energy and its interconversions. • The first law of thermodynamics is the law of conservation of energy. – This means that the total amount of energy in the universe is constant. • Therefore, you can never design a system that will continue to produce energy without some source of energy. © 2014 Pearson Education, Inc.

Energy Flow and Conservation of Energy • Conservation of energy requires that the sum of the energy changes in the system and the surroundings must be zero. DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings D is the symbol that is used to mean change. • Final amount–initial amount © 2014 Pearson Education, Inc.

Internal Energy • The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system. • The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end. – A state function is a mathematical function whose result only depends on the initial and final conditions, not on the process used. DE = Efinal – Einitial DEreaction = Eproducts − Ereactants © 2014 Pearson Education, Inc.

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. The distance from the base to the peak of the mountain is a state function. It depends only on the difference in elevation between the base and the peak, not on how you arrive there! © 2014 Pearson Education, Inc.

Energy Diagrams • Energy diagrams are a “graphical” way of showing the direction of energy flow during a process. • If the reactants have a lower internal energy than the products, the change in energy will be positive. © 2014 Pearson Education, Inc.

Energy Diagrams • Energy diagrams are a “graphical” way of showing the direction of energy flow during a process. • If the reactants have a higher internal energy than the products, the change in energy will be negative. © 2014 Pearson Education, Inc.

• The total amount of internal energy in 1 mole of C(s) and 1 mole of O 2(g) is greater than the internal energy in 1 mole of CO 2(g). – At the same temperature and pressure • In the reaction C(s) + O 2(g) → CO 2(g), there will be a net release of energy into the surroundings. – −DEreaction = DEsurroundings • In the reaction CO 2(g) → C(s) + O 2(g), there will be an absorption of energy from the surroundings into the reaction. DEreaction = − DEsurroundings © 2014 Pearson Education, Inc. Internal Energy Flow in a Chemical Reaction C(s), O 2(g) CO 2(g) energy released DErxn = ─ Surroundings System C + O 2 → CO 2

• The total amount of internal energy in 1 mole of C(s) and 1 mole of O 2(g) is greater than the internal energy in 1 mole of CO 2(g) – At the same temperature and pressure • In the reaction C(s) + O 2(g) → CO 2(g), there will be a net release of energy into the surroundings. – −DEreaction = DEsurroundings • In the reaction CO 2(g) → C(s) + O 2(g), there will be an absorption of energy from the surroundings into the reaction. DEreaction = − DEsurroundings © 2014 Pearson Education, Inc. Internal Energy Flow in a Chemical Reaction C(s), O 2(g) CO 2(g) energy absorbed DErxn = + Surroundings System C + O 2 → CO 2

Energy Flow • When energy flows out of a system, it must all flow into the surroundings. • When energy flows out of a system, DEsystem is negative. • When energy flows into the surroundings, DEsurroundings is positive. • Therefore, ─ DEsystem= DEsurroundings © 2014 Pearson Education, Inc. Surroundings DE + System DE ─

Energy Flow • When energy flows into a system, it must all come from the surroundings. • When energy flows into a system, DEsystem is positive. • When energy flows out of the surroundings, DEsurroundings is negative. • Therefore, DEsystem= ─ DEsurroundings © 2014 Pearson Education, Inc. Surroundings DE ─ System DE +

Energy Exchange • Energy is exchanged between the system and surroundings through heat and work. – q = heat (thermal) energy – w = work energy – q and w are NOT state functions; their value depends on the process. DE = q + w © 2014 Pearson Education, Inc.

Energy Exchange • Energy is exchanged between the system and surroundings through either heat exchange or work being done. © 2014 Pearson Education, Inc.

Heat and Work • The white ball has an initial amount of 5. 0 J of kinetic energy. • As it rolls on the table, some of the energy is converted to heat by friction. • The rest of the kinetic energy is transferred to the purple ball by collision. © 2014 Pearson Education, Inc.

Heat and Work On a smooth table, most of the kinetic energy is transferred from the white ball to the purple ball, with a small amount lost through friction. Energy change for the white ball is as follows: DE = KEfinal − KEinitial = 0 J − 5. 0 J = − 5. 0 J Kinetic energy transferred to purple ball is w = − 4. 5 J. Kinetic energy lost as heat is q = − 0. 5 J. q + w = (− 0. 5 J) + (− 4. 5 J) = − 5. 0 J = DE © 2014 Pearson Education, Inc.

Heat and Work On a rough table, most of the kinetic energy of the white ball is lost through friction—less than half is transferred to the purple ball. Energy change for the white ball is as follows: DE = KEfinal − KEinitial = 0 J − 5. 0 J = − 5. 0 J Kinetic energy transferred to purple ball is w = − 3. 0 J. Kinetic energy lost as heat is q = − 2. 0 J. q + w = (− 2. 0 J) + (− 3. 0 J) = − 5. 0 J = DE © 2014 Pearson Education, Inc.

Heat, Work, and Internal Energy • In the previous billiard ball example, the DE of the white ball is the same for both cases, but q and w are not. • On the rougher table, the heat loss, q, is greater. – q is a more negative number. • But on the rougher table, less kinetic energy is transferred to the purple ball, so the work done by the white ball, w, is less. – w is a less negative number. • The DE is a state function and depends only on the velocity of the white ball before and after the collision. – In both cases it started with 5. 0 k. J of kinetic energy and ended with 0 k. J because it stopped. – q + w is the same for both tables, even though the values of q and w are different. © 2014 Pearson Education, Inc.

Heat Exchange • Heat is the exchange of thermal energy between a system and surroundings. • Heat exchange occurs when system and surroundings have a difference in temperature. • Temperature is the measure of thermal energy within a sample of matter. • Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature. – Thermal equilibrium © 2014 Pearson Education, Inc.

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. – Units of C are J/°C or J/K. q = C × DT • The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat. © 2014 Pearson Education, Inc.

Factors Affecting Heat Capacity • The heat capacity of an object depends on its amount of matter. – It is usually measured by its mass. – 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. – 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. © 2014 Pearson Education, Inc.

Quantifying Heat Energy • The heat capacity of an object is proportional to the following: – Its mass – 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. © 2014 Pearson Education, Inc.

Specific Heat Capacity • Measure of a substance’s intrinsic ability to absorb heat. • The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1 °C. – Cs – Units J/(g ∙ °C) • The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1 °C. © 2014 Pearson Education, Inc.

Specific Heat of Water • Water can absorb a lot of heat energy without a large increase in its temperature due to its high specific heat capacity. • The large amount of water absorbing heat from the air keeps beaches cool in the summer. – Without water, Earth’s temperature would be about the same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F). • Water is commonly used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating. – Water can even prevent melting. – It can also be used to transfer the heat to something else because it is a fluid. © 2014 Pearson Education, Inc.

Heat Transfer and Final Temperature • When two objects at different temperatures are placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature. • Heat flows until both materials reach the same final temperature. • The amount of heat energy lost by the hot material equals the amount of heat gained by the cold material. © 2014 Pearson Education, Inc.

Thermal Energy Transfer • A block of metal at 55 °C is added to water at 25 °C. • Thermal energy transfers heat from the metal to the water. • The exact temperature change depends on the following: – The mass of the metal – The mass of water – Specific heat capacities of the metal and of water © 2014 Pearson Education, Inc.

Pressure –Volume Work • PV work is work caused by a volume change against an external pressure. • When gases expand, DV is positive, but the system is doing work on the surroundings, so wgas is negative. • As long as the external pressure is kept constant, ─ Workgas = External Pressure × Change in Volumegas w = ─PDV. – To convert the units to joules use 101. 3 J = 1 atm ∙ L. © 2014 Pearson Education, Inc.

Exchanging Energy between System and Surroundings • Exchange of heat energy q = mass × specific heat × DTemperature • Exchange of work w = −Pressure × DVolume © 2014 Pearson Education, Inc.

Measuring DE: Calorimetry at Constant Volume • Because DE = q + w, we can determine DE 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, so w = 0. – At constant volume, DEsystem = qsystem. • In practice, we cannot observe the temperature changes of the individual chemicals involved in a reaction, so instead we measure the temperature change in the surroundings. – Use insulated, controlled surroundings – qsystem = −qsurroundings • The surrounding area is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water. qsurroundings = qcalorimeter = ─qsystem © 2014 Pearson Education, Inc.

Bomb Calorimeter • It is used to measure DE because it is a constant volume system. • 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, k. J/ºC © 2014 Pearson Education, Inc.

Enthalpy • The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume. – H is a state function H = E + PV • The enthalpy change, DH, of a reaction is the heat evolved in a reaction at constant pressure. DHreaction = qreaction at constant pressure • Usually DH and DE are similar in value; the difference is largest for reactions that produce or use large quantities of gas. © 2014 Pearson Education, Inc.

Endothermic and Exothermic Reactions • When DH is negative, heat is being released by the system. – This is called an exothermic reaction. • When DH is positive, heat is being absorbed by the system. – This is called an endothermic reaction. © 2014 Pearson Education, Inc.

Endothermic and Exothermic 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 transferred to your hands. • Chemical cold packs contain NH 4 NO 3 that dissolves in water in an endothermic process—your hands get cold because the pack is absorbing your heat. © 2014 Pearson Education, Inc.

Molecular View of Exothermic Reactions • For an exothermic reaction, the surrounding’s temperature rises due to a release of thermal energy by the reaction. • 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, existing bonds are broken and new bonds are made. • The products of the reaction have less chemical potential energy than the reactants. • The difference in energy is released as heat. © 2014 Pearson Education, Inc.

Molecular View of Endothermic Reactions • In an endothermic reaction, the surrounding’s temperature drops due to absorption of some of its thermal energy by the reaction. • During the course of a reaction, existing bonds are broken and new bonds are made. • The products of the reaction have more chemical potential energy than the reactants. • To acquire this extra energy, some of thermal energy of the surroundings is converted into chemical potential energy stored in the products. © 2014 Pearson Education, Inc.

Enthalpy of Reaction • The enthalpy change in a chemical reaction is an extensive property. – 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. C 3 H 8(g) + 5 O 2(g) → 3 CO 2(g) + 4 H 2 O(g) DH = − 2044 k. J 1 mol C 3 H 8(g) = – 2044 k. J or 5 mol O 2(g) = – 2044 k. J © 2014 Pearson Education, Inc.

Measuring DH Calorimetry at Constant Pressure • Reactions done in aqueous solution are at constant pressure. – Open to the atmosphere • The calorimeter is often nested foam cups containing the solution. qreaction = ─ qsolution = ─(masssolution × Cs, solution × DT) DHreaction = qconstant pressure = qreaction – To get DHreaction per mol, divide by the number of moles. © 2014 Pearson Education, Inc.

Relationships Involving DHrxn • When reaction is multiplied by a factor, DHrxn is multiplied by that factor. – Because DHrxn is extensive, C(s) + O 2(g) → CO 2(g) DH = − 393. 5 k. J 2 C(s) + 2 O 2(g) → 2 CO 2(g) DH = 2(− 393. 5 k. J) = − 787. 0 k. J. • If a reaction is reversed, then the sign of DH is changed. CO 2(g) → C(s) + O 2(g) © 2014 Pearson Education, Inc. DH = +393. 5 k. J

Relationships Involving DHrxn Hess’s Law • If a reaction can be expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step. © 2014 Pearson Education, Inc.

Standard Conditions • The standard state is the state of a material at a defined set of conditions. – Pure gas at exactly 1 atm pressure – Pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest • Usually 25 °C – Substance in a solution with concentration 1 M • The standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states. • The standard enthalpy of formation, DHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements. – The elements must be in their standard states. – The DHf° for a pure element in its standard state = 0 k. J/mol. © 2014 Pearson Education, Inc.

Standard Enthalpies of Formation © 2014 Pearson Education, Inc.

Formation Reactions • Reactions of elements in their standard state to form 1 mole of a pure compound. – if you are not sure what the standard state of an element is, find the form in Appendix IIB that has DHf° = 0. – Because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions. © 2014 Pearson Education, Inc.

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. – There are several forms of solid C, but the one with DHf° = 0 is graphite. – Oxygen’s standard state is the diatomic gas. C(s, graphite) + O 2(g) → CO(g) • The equation must be balanced, but the coefficient of the product compound must be 1. – 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 O 2(g) → CO(g) © 2014 Pearson Education, Inc.

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 DH° for the reaction is then the sum of the DHf° for the component reactions. DH°reaction = S n DHf°(products) − S n DHf°(reactants) S means sum. n is the coefficient of the reaction. © 2014 Pearson Education, Inc.

CH 4(g)+ 2 O 2(g)→ CO 2(g) + H 2 O(g) C(s, graphite) + 2 H 2(g) → CH 4(g) DHf°= − 74. 6 k. J/mol CH 4 C(s, graphite) + O 2(g) → CO 2(g) DHf°= − 393. 5 k. J/mol CO 2 H 2(g) + ½ O 2(g) → H 2 O(g) © 2014 Pearson Education, Inc. DHf°= − 241. 8 k. J/mol H 2 O

CH 4(g)+ 2 O 2(g)→ CO 2(g) + H 2 O(g) CH 4(g) → C(s, graphite) + 2 H 2(g) C(s, graphite) + O 2(g) → CO 2(g) 2 H 2(g) + O 2(g) → 2 H 2 O(g) D H° = + 74. 6 k. J DHf°= − 393. 5 k. J/mol CO 2 DH° = − 483. 6 k. J CH 4(g) + 2 O 2(g) → CO 2(g) + 2 H 2 O(g) DH° = − 802. 5 k. J © 2014 Pearson Education, Inc.

CH 4(g)+ 2 O 2(g)→ CO 2(g) + H 2 O(g) DH° = [((− 393. 5 k. J)+ 2(− 241. 8 k. J)− ((− 74. 6 k. J)+ 2(0 k. J))] = − 802. 5 k. J DH° = [(DHf° CO 2(g) + 2 ∙ DHf° H 2 O(g)) − (DHf° CH 4(g) + 2 ∙ DHf° O 2(g))] CH 4(g) + 2 O 2(g) → CO 2(g) + 2 H 2 O(g) DH° = − 802. 5 k. J © 2014 Pearson Education, Inc.

Energy Use and the Environment • In the United States, each person uses over 105 k. Wh of energy per year. • Most comes from the combustion of fossil fuels. – Combustible materials that originate from ancient life. C(s) + O 2(g) → CO 2(g) DH°rxn = − 393. 5 k. J CH 4(g) +2 O 2(g) → CO 2(g) + 2 H 2 O(g) DH°rxn = − 802. 3 k. J C 8 H 18(g) +12. 5 O 2(g) → 8 CO 2(g) + 9 H 2 O(g) DH°rxn = − 5074. 1 k. J © 2014 Pearson Education, Inc.

Energy Use and the Environment • Fossil fuels cannot be replenished. • At current rates of consumption, oil and natural gas supplies will be depleted in 50– 100 years. © 2014 Pearson Education, Inc.

Energy Consumption © 2014 Pearson Education, Inc.

The Effect of Combustion Products on Our Environment • Because of additives and impurities in the fossil fuel, incomplete combustion, and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy. • Therefore, fossil fuel emissions contribute to air pollution, acid rain, and global warming. © 2014 Pearson Education, Inc.

Global Warming • CO 2 is a greenhouse gas. – It allows light from the sun to reach Earth, but does not allow the heat (infrared light) reflected off Earth to escape into outer space. • It acts like a blanket. • CO 2 levels in the atmosphere have been steadily increasing. • Current observations suggest that the average global air temperature has risen 0. 6 °C in the past 100 years. • Atmospheric models suggest that the warming effect could worsen if CO 2 levels are not curbed. • Some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats. © 2014 Pearson Education, Inc.

Global Warming • CO 2 is a greenhouse gas. – It allows light from the sun to reach Earth, but does not allow the heat (infrared light) reflected off Earth to escape into outer space. • It acts like a blanket. • CO 2 levels in the atmosphere have been steadily increasing. • Current observations suggest that the average global air temperature has risen 0. 6 °C in the past 100 years. © 2014 Pearson Education, Inc.

Renewable Energy • Our greatest unlimited supply of energy is the sun. • New technologies are being developed to capture the energy of sunlight. – Parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity. – Solar energy is used to decompose water into H 2(g) and O 2(g); the H 2 can then be used by fuel cells to generate electricity. H 2(g) + ½ O 2(g) → H 2 O(l) DH°rxn = − 285. 8 k. J © 2014 Pearson Education, Inc.

Renewable Energy • Hydroelectric power • Wind power © 2014 Pearson Education, Inc.