Chapter 5 Thermochemistry John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation © 2012 Pearson Education, Inc.

Thermochemistry © 2012 Pearson Education, Inc. Energy Energy is the ability to do work or transfer heat. –Energy used to cause an object that has mass to move is called work. –Energy used to cause the temperature of an object to rise is called heat.

Copyright © 2011 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 x Distance Heat is the flow of energy caused by a difference in temperature Energy can be exchanged between objects through contact collisions 3

Copyright © 2011 Pearson Education, Inc. Energy, Heat, and Work You can think of energy as a quantity an object can possess or 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 4

Copyright © 2011 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 5

Copyright © 2011 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 6

Thermochemistry © 2012 Pearson Education, Inc. Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 2 E k =  mv 2

Thermochemistry © 2012 Pearson Education, Inc. Potential Energy Potential energy is energy an object possesses by virtue of its position or chemical composition. The most important form of potential energy in molecules is electrostatic potential energy, E el : KQ1Q2KQ1Q2 d E el =

Copyright © 2011 Pearson Education, Inc. Manifestations of Energy 9

Copyright © 2011 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 10

Thermochemistry © 2012 Pearson Education, Inc. Units of Energy The SI unit of energy is the joule (J): An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = J 1 J = 1  kg m 2 s2s2

Copyright © 2011 Pearson Education, Inc. The amount of kinetic energy an object has is directly proportional to its mass and velocity KE = ½mv 2 Units of Energy 12

Copyright © 2011 Pearson Education, Inc. Units of Energy 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 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 Energy Conversion Factors 1 calorie (cal)=4.184 joules (J) (exact) 1 Calorie (Cal)=1000 cal = 1 kcal = 4184 J 1 kilowatt-hour (kWh)=3.60 x 10 6 J 13

Copyright © 2011 Pearson Education, Inc. Energy Use 14

Copyright © 2011 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 with What we study is the exchange of energy between the system and the surroundings Surroundings System Surroundings System 15

Thermochemistry © 2012 Pearson Education, Inc. Definitions: System and Surroundings The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). The surroundings are everything else (here, the cylinder and piston).

Copyright © 2011 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 17

Thermochemistry © 2012 Pearson Education, Inc. Definitions: Work Energy used to move an object over some distance is work: w = F  d where w is work, F is the force, and d is the distance over which the force is exerted.

Thermochemistry © 2012 Pearson Education, Inc. Heat Energy can also be transferred as heat. Heat flows from warmer objects to cooler objects.

Thermochemistry © 2012 Pearson Education, Inc. Conversion of Energy Energy can be converted from one type to another. For example, the cyclist in Figure 5.2 has potential energy as she sits on top of the hill.

Thermochemistry © 2012 Pearson Education, Inc. Conversion of Energy As she coasts down the hill, her potential energy is converted to kinetic energy. At the bottom, all the potential energy she had at the top of the hill is now kinetic energy.

Thermochemistry © 2012 Pearson Education, Inc. First Law of Thermodynamics Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Copyright © 2011 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 You can therefore never design a system that will continue to produce energy without some source of energy 23

Copyright © 2011 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  Energy universe = 0 =  Energy system +  Energy surroundings  Is the symbol that is used to mean change  final amount – initial amount 24

Thermochemistry © 2012 Pearson Education, Inc. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Thermochemistry © 2012 Pearson Education, Inc. Internal Energy By definition, the change in internal energy,  E, is the final energy of the system minus the initial energy of the system:  E = E final − E initial

Copyright © 2011 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  E = E final – E initial  E reaction = E products − E reactants 27

Copyright © 2011 Pearson Education, Inc. 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 steep. Regardless of which trail you take, 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! 28

Copyright © 2011 Pearson Education, Inc. Energy Diagrams 29 Energy diagrams are a “graphical” way of showing the direction of energy flow during a process Internal Energy initial final energy added  E = + Internal Energy initial final energy removed  E = ─ If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be + If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be ─

Copyright © 2011 Pearson Education, Inc. Energy Flow When energy flows out of a system, it must all flow into the surroundings When energy flows out of a system,  E system is ─ When energy flows into the surroundings,  E surroundings is + Therefore: ─  E system =  E surroundings Surroundings  E + System  E ─ 30

Copyright © 2011 Pearson Education, Inc. Energy Flow When energy flows into a system, it must all come from the surroundings When energy flows into a system,  E system is + When energy flows out of the surroundings,  E surroundings is ─ Therefore:  E system = ─  E surroundings Surroundings  E ─ System  E + 31

Copyright © 2011 Pearson Education, Inc. energy released  E rxn = ─ energy absorbed  E rxn = + Energy Flow in a Chemical Reaction The total amount of internal energy in 1mol 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 −  E reaction =  E surroundings In the reaction CO 2 (g) → C(s) + O 2 (g), there will be an absorption of energy from the surroundings into the reaction  E reaction = −  E surroundings Internal Energy CO 2 (g) C(s), O 2 (g) Surroundings System C + O 2 → CO 2 System CO 2 →C + O 2 32

Copyright © 2011 Pearson Education, Inc. 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  E = q + w 33

Copyright © 2011 Pearson Education, Inc. Energy Exchange Energy is exchanged between the system and surroundings through either heat exchange or work being done 34

Copyright © 2011 Pearson Education, Inc. 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 Heat & Work 35

Copyright © 2011 Pearson Education, Inc. energy change for the white ball,  E = KE final − KE initial = 0 J − 5.0 J = −5.0 J kinetic energy transferred to purple ball, w = −4.5 J kinetic energy lost as heat, q = −0.5 J q + w = (−0.5 J) + (−4.5 J) = −5.0 J =  E On a smooth table, most of the kinetic energy is transferred from the white ball to the purple – with a small amount lost through friction energy change for the white ball,  E = KE final − KE initial = 0 J − 5.0 J = −5.0 J kinetic energy transferred to purple ball, w = −3.0 J kinetic energy lost as heat, q = −2.0 J q + w = (−2.0 J) + (−3.0 J) = −5.0 J =  E 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 36 Heat & Work

Copyright © 2011 Pearson Education, Inc. Heat, Work, and Internal Energy In the previous billiard ball example, the  E 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  E 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 kJ of kinetic energy and ended with 0 kJ because it stopped q + w is the same for both tables, even though the values of q and w are different 37

Copyright © 2011 Pearson Education, Inc. Heat Exchange Heat is the exchange of thermal energy between the system and surroundings Heat exchange occurs when system and surroundings have a difference in temperature Temperature is the measure of the amount 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 38

Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy If  E > 0, E final > E initial –Therefore, the system absorbed energy from the surroundings. –This energy change is called endergonic.

Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy If  E < 0, E final < E initial –Therefore, the system released energy to the surroundings. –This energy change is called exergonic.

Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). That is,  E = q + w.

Thermochemistry © 2012 Pearson Education, Inc.  E, q, w, and Their Signs

Thermochemistry © 2012 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.

Thermochemistry © 2012 Pearson Education, Inc. Exchange of Heat between System and Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic. When heat is released by the system into the surroundings, the process is exothermic.

Thermochemistry © 2012 Pearson Education, Inc. State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry © 2012 Pearson Education, Inc. State Functions However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. –In the system depicted in Figure below, the water could have reached room temperature from either direction.

Thermochemistry © 2012 Pearson Education, Inc. State Functions Therefore, internal energy is a state function. It depends only on the present state of the system, not on the path by which the system arrived at that state. And so,  E depends only on E initial and E final.

Thermochemistry © 2012 Pearson Education, Inc. State Functions However, q and w are not state functions. Whether the battery is shorted out or is discharged by running the fan, its  E is the same. –But q and w are different in the two cases.

Thermochemistry © 2012 Pearson Education, Inc. Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

Thermochemistry © 2012 Pearson Education, Inc. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −P  V

Thermochemistry © 2012 Pearson Education, Inc. Enthalpy If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

Thermochemistry © 2012 Pearson Education, Inc. Enthalpy When the system changes at constant pressure, the change in enthalpy,  H, is  H =  (E + PV) This can be written  H =  E + P  V

Thermochemistry © 2012 Pearson Education, Inc. Enthalpy Since  E = q + w and w = −P  V, we can substitute these into the enthalpy expression:  H =  E + P  V  H = (q + w) − w  H = q So, at constant pressure, the change in enthalpy is the heat gained or lost.

Copyright © 2011 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,  H, of a reaction is the heat evolved in a reaction at constant pressure  H reaction = q reaction at constant pressure Usually  H and  E are similar in value, the difference is largest for reactions that produce or use large quantities of gas 54

Thermochemistry © 2012 Pearson Education, Inc. Endothermicity and Exothermicity A process is endothermic when  H is positive.

Thermochemistry © 2012 Pearson Education, Inc. Endothermicity and Exothermicity A process is endothermic when  H is positive. A process is exothermic when  H is negative.

Thermochemistry © 2012 Pearson Education, Inc. Enthalpy of Reaction The change in enthalpy,  H, is the enthalpy of the products minus the enthalpy of the reactants:  H = H products − H reactants

Thermochemistry © 2012 Pearson Education, Inc. Enthalpy of Reaction This quantity,  H, is called the enthalpy of reaction, or the heat of reaction.

Thermochemistry © 2012 Pearson Education, Inc. The Truth about Enthalpy 1.Enthalpy is an extensive property. 2.  H for a reaction in the forward direction is equal in size, but opposite in sign, to  H for the reverse reaction. 3.  H for a reaction depends on the state of the products and the state of the reactants.

Thermochemistry © 2012 Pearson Education, Inc. Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure  H through calorimetry, the measurement of heat flow.

Thermochemistry © 2012 Pearson Education, Inc. Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1  C) is its heat capacity.

Thermochemistry © 2012 Pearson Education, Inc. Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1  C).

Thermochemistry © 2012 Pearson Education, Inc. Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass  temperature change s = q m  Tm  T

Copyright © 2011 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 x  T The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat 64 Tro: Chemistry: A Molecular Approach, 2/e

Copyright © 2011 Pearson Education, Inc. Factors Affecting Heat Capacity The heat capacity of an object depends on its amount of matter 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 65 Tro: Chemistry: A Molecular Approach, 2/e

Copyright © 2011 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 C s units are 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 66 Tro: Chemistry: A Molecular Approach, 2/e

Copyright © 2011 Pearson Education, Inc. Specific Heat of Water The rather high specific heat of water allows water to absorb a lot of heat energy without a large increase in its temperature The large amount of water absorbing heat from the air keeps beaches cool in the summer without water, the 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 or even melting it can also be used to transfer the heat to something else because it is a fluid 67

Copyright © 2011 Pearson Education, Inc. 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) x (specific heat) x (temp. change) q = (m) x (C s ) x (  T) 68

Copyright © 2011 Pearson Education, Inc. 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? q = m ∙ C s ∙  T C s = J/gºC (Given) the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise T 1 = −8.0 °C, T 2 = 37.0 °C, m = 3.10 g q, Jq, J Check: Check Solution: Follow the conceptual plan to solve the problem Conceptual Plan: Relationships: Strategize Given: Find: Sort Information C s m,  T q 69

Copyright © 2011 Pearson Education, Inc. Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C (water’s specific heat is 4.18 J/g  ºC) 70

Copyright © 2011 Pearson Education, Inc. Practice – Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C q = m ∙ C s ∙  T C s = 4.18 J/g  C (Table 6.4) T 1 = 49 °C, T 2 = 29 °C, m = 7.40 g q, Jq, J Check: Check Solution: Follow the concept plan to solve the problem Conceptual Plan: Relationships: Strategize Given: Find: Sort Information C s m,  T q 71 the unit is correct, the sign is reasonable as the water must release heat to make its temperature fall

Copyright © 2011 Pearson Education, Inc. Heat Transfer & 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 q hot = −q cold m hot  C s,hot   T hot = −(m cold  C s,cold   T cold ) 72

Copyright © 2011 Pearson Education, Inc. 4. Heat lost by the metal = heat gained by water 1. Heat lost by the metal > heat gained by water 2. Heat gained by water > heat lost by the metal 3. Heat lost by the metal > heat lost by the water 4. Heat lost by the metal = heat gained by water 5. More information is required A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true? 73

Copyright © 2011 Pearson Education, Inc. q = m ∙ C s ∙  T C s, Al = J/gºC, C s, H2O = 4.18 J/gºC(Table 6.4) Example: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures m Al = 32.5 g, T al = 45.8 °C, m H20 = g, T H2O = 15.4 °C T final, °C Check: Solution: Conceptual Plan: Relationships: Given: Find: C s, Al m Al, C s, H2O m H2O  T Al = k   T H2O 74 q Al = −q H2O T final

Copyright © 2011 Pearson Education, Inc. Practice – A hot piece of metal weighing g is heated to °C. It is then placed into a coffee cup calorimeter containing g of water at 22.4 °C. The water warms and the copper cools until the final temperature is 35.2 °C. Calculate the specific heat of the metal and identify the metal. 75

Copyright © 2011 Pearson Education, Inc. 76 Practice – Calculate the specific heat and identify the metal from the data q = m x C s x  T; q metal = −q H2O the units are correct, the number indicates the metal is copper metal: g, T 1 = °C, T 2 = 35.2 °C H 2 O: g, T 1 = 22.4 °C, T 2 = 35.2°C, C s = 4.18 J/g  °C C s, metal, J/g  ºC Check: Solution: Concept Plan: Relationships: Given: Find: m, C s,  T q

Copyright © 2011 Pearson Education, Inc. Pressure –Volume Work PV work is work caused by a volume change against an external pressure When gases expand,  V is +, but the system is doing work on the surroundings, so w gas is ─ As long as the external pressure is kept constant ─ Work gas = External Pressure x Change in Volume gas w = ─P  V to convert the units to joules use J = 1 atm∙L 77

Copyright © 2011 Pearson Education, Inc. Example : If a balloon is inflated from L to 1.85 L against an external pressure of 1.00 atm, how much work is done? the unit is correct; the sign is reasonable because when a gas expands it does work on the surroundings and loses energy V 1 = L, V 2 = 1.85 L, P = 1.00 atm w, Jw, J Check: Solution: Conceptual Plan: Relationships: Given: Find: P,  V w J = 1 atm∙L 78

Copyright © 2011 Pearson Education, Inc. Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? (1 cal = 4.18 J, J = 1.00 atmL) 79

Copyright © 2011 Pearson Education, Inc. Practice – A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of the gas is 10.0 L, what is the final volume of the gas? so both  E and q are −, and  E > q, w must be −; and when w is − the system is expanding, so V 2 should be greater than V 1 ; and it is  E = −68.3 kJ, q = −15.8 kcal, V 1 = 10.0 L, P = 1.00 atm V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find:  E = q + w, w = −P  V, 1 kJ = 1000 J, 1 cal = 4.18 J, J = 1 atm∙L q,  E w V2V2 P, V 1  E = −6.83 x 10 4 J, q = −6.604 x 10 4 J, V 1 = 10.0L, P = 1.00 atm V 2, L 80

Thermochemistry © 2012 Pearson Education, Inc. 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.

Thermochemistry © 2012 Pearson Education, Inc. Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure  H for the reaction with this equation: q = m  s   T

Copyright © 2011 Pearson Education, Inc. Measuring  H 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 q reaction = ─ q solution = ─(mass solution x C s, solution x  T)   H reaction = q constant pressure = q reaction to get  H reaction per mol, divide by the number of moles 83

Copyright © 2011 Pearson Education, Inc. Example : What is  H rxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) if g Mg reacts in mL of solution and changes the temperature from 25.6 °C to 32.8 °C? the sign is correct and the value is reasonable because the reaction is exothermic g Mg, mL, T 1 = 25.6 °C, T 2 = 32.8 °C, assume d = 1.00 g/mL, C s = 4.18 J/g∙°C  H, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol HH C s,  T, m q sol’n q rxn q sol’n = m x C s, sol’n x  T = −q rxn, MM Mg= g/mol x 10 −3 mol Mg, 1.00 x 10 2 g,  T = 7.2 °C,,C s = 4.18 J/g∙°C  H, kJ/mol 84

Thermochemistry © 2012 Pearson Education, Inc. Bomb Calorimetry Reactions can be carried out in a sealed “bomb” such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

Thermochemistry © 2012 Pearson Education, Inc. Bomb Calorimetry Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy,  E, not  H. For most reactions, the difference is very small.

Copyright © 2011 Pearson Education, Inc. Example : When g of sugar is burned in a bomb calorimeter, the temperature rises from °C to °C. If C cal = 4.90 kJ/°C, find  E for burning 1 mole q cal = C cal x  T = −q rxn MM C 12 H 22 O 11 = g/mol the units are correct, the sign is as expected for combustion g C 12 H 22 O 11, T 1 = °C, T 2 = °C, C cal = 4.90 kJ/°C  E rxn, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol EE C cal,  T q cal q rxn x 10 −3 mol C 12 H 22 O 11,  T = 3.41°C, C cal = 4.90 kJ/°C  E rxn, kJ/mol 87

Copyright © 2011 Pearson Education, Inc. Practice – When 1.22 g of HC 7 H 5 O 2 (MM ) is burned in a bomb calorimeter, the temperature rises from °C to °C. If  E for the reaction is −3.23 x 10 3 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? 88

Copyright © 2011 Pearson Education, Inc. Practice – When 1.22 g of HC 7 H 5 O 2 is burned in a bomb calorimeter, the temperature rises from °C to °C. If  E for the reaction is −3.23 x 10 3 kJ/mol, what is the heat capacity of the calorimeter in kJ/°C? q cal = C cal x  T = −q rxn MM HC 7 H 5 O 2 = g/mol the units and sign are correct 1.22 g HC 7 H 5 O 2, T 1 =20.27 °C, T 2 =22.67 °C,  E= −3.23 x 10 3 kJ/mol C cal, kJ/°C Check: Solution: Conceptual Plan: Relationships: Given: Find: q cal,  T C cal mol,  E q rxn q cal x 10 −3 mol HC 7 H 5 O 2,  T = 2.40 °C,  E= −3.23 x 10 3 kJ/mol C cal, kJ/°C 89

Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn When reaction is multiplied by a factor,  H rxn is multiplied by that factor because  H rxn is extensive C(s) + O 2 (g) → CO 2 (g)  H = −393.5 kJ 2 C(s) + 2 O 2 (g) → 2 CO 2 (g)  H = 2(−393.5 kJ) = −787.0 kJ If a reaction is reversed, then the sign of  H is changed CO 2 (g) → C(s) + O 2 (g)  H = kJ 90

Thermochemistry © 2012 Pearson Education, Inc. Hess’s Law  H is well known for many reactions, and it is inconvenient to measure  H for every reaction in which we are interested. However, we can estimate  H using published  H values and the properties of enthalpy.

Thermochemistry © 2012 Pearson Education, Inc. Hess’s Law Hess’s law states that “[i]f a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Thermochemistry © 2012 Pearson Education, Inc. Hess’s Law Because  H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn Hess’s Law If a reaction can be expressed as a series of steps, then the  H rxn for the overall reaction is the sum of the heats of reaction for each step 94

Copyright © 2011 Pearson Education, Inc. [3 NO 2 (g)  3 NO(g) O 2 (g)]  H° = (+174 kJ) [1 N 2 (g) O 2 (g) + 1 H 2 O(l)  2 HNO 3 (aq)]  H° = (−128 kJ) [2 NO(g)  N 2 (g) + O 2 (g)]  H° = (−183 kJ) 3 NO 2 (g) + H 2 O(l)  2 HNO 3 (aq) + NO(g)  H° = − 137 kJ Example: Hess’s Law Given the following information: 2 NO(g) + O 2 (g)  2 NO 2 (g)  H° = −116 kJ 2 N 2 (g) + 5 O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq)  H° = −256 kJ N 2 (g) + O 2 (g)  2 NO(g)  H° = +183 kJ Calculate the  H° for the reaction below: 3 NO 2 (g) + H 2 O(l)  2 HNO 3 (aq) + NO(g)  H° = ? [2 NO 2 (g)  2 NO(g) + O 2 (g)] x 1.5  H° = 1.5(+116 kJ) [2 N 2 (g) + 5 O 2 (g) + 2 H 2 O(l)  4 HNO 3 (aq)] x 0.5  H° = 0.5(−256 kJ) [2 NO(g)  N 2 (g) + O 2 (g)]  H° = −183 kJ 95

Copyright © 2011 Pearson Education, Inc. Practice – Hess’s Law Given the following information: Cu(s) + Cl 2 (g)  CuCl 2 (s)  H° = −206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Calculate the  H° for the reaction below: Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = ? kJ 96

Copyright © 2011 Pearson Education, Inc. Practice – Hess’s Law Given the following information: Cu(s) + Cl 2 (g)  CuCl 2 (s)  H° = −206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Calculate the  H° for the reaction below: Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = ? kJ CuCl 2 (s)  Cu(s) + Cl 2 (g)  H° = +206 kJ 2 Cu(s) + Cl 2 (g)  2 CuCl(s)  H° = − 36 kJ Cu(s) + CuCl 2 (s)  2 CuCl(s)  H° = kJ 97

Thermochemistry © 2012 Pearson Education, Inc. Enthalpies of Formation An enthalpy of formation,  H f, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Copyright © 2011 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,  H°, is the enthalpy change when all reactants and products are in their standard states The standard enthalpy of formation,  H f °, 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  H f ° for a pure element in its standard state = 0 kJ/mol  by definition 99

Thermochemistry © 2012 Pearson Education, Inc. Standard Enthalpies of Formation Standard enthalpies of formation,  H f °, are measured under standard conditions (25 °C and 1.00 atm pressure).

Copyright © 2011 Pearson Education, Inc. Standard Enthalpies of Formation 101

Copyright © 2011 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 that has a  H f ° = 0 because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions 102

Thermochemistry © 2012 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) C 3 H 8 (g)  3C (graphite) + 4H 2 (g)

Thermochemistry © 2012 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: C 3 H 8 (g)  3C (graphite) + 4H 2 (g) 3C (graphite) + 3O 2 (g)  3CO 2 (g) C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)

Thermochemistry © 2012 Pearson Education, Inc. Calculation of  H Imagine this as occurring in three steps: C 3 H 8 (g)  3C (graphite) + 4H 2 (g) 3C (graphite) + 3O 2 (g)  3CO 2 (g) 4H 2 (g) + 2O 2 (g)  4H 2 O(l) C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)

Thermochemistry © 2012 Pearson Education, Inc. C 3 H 8 (g)  3C (graphite) + 4H 2 (g) 3C (graphite) + 3O 2 (g)  3CO 2 (g) 4H 2 (g) + 2O 2 (g)  4H 2 O(l) C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) Calculation of  H The sum of these equations is C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)

Thermochemistry © 2012 Pearson Education, Inc. Calculation of  H We can use Hess’s law in this way:  H =  n  H f,products –  m  H f °,reactants where n and m are the stoichiometric coefficients.

Copyright © 2011 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  H° for the reaction is then the sum of the  H f ° for the component reactions  H° reaction =  n  H f °(products) −  n  H f °(reactants)  means sum n is the coefficient of the reaction 108 Tro: Chemistry: A Molecular Approach, 2/e

Thermochemistry © 2012 Pearson Education, Inc.  H= [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)] = [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)] = (− kJ) – (− kJ) = − kJ Calculation of  H C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l)

Copyright © 2011 Pearson Education, Inc. CH 4 (g) → C(s, graphite) + 2 H 2 (g)  H° = kJ 2 H 2 (g) + O 2 (g) → 2 H 2 O(g)  H° = −483.6 kJ CH 4 (g)+ 2 O 2 (g)→ CO 2 (g) + H 2 O(g) C(s, graphite) + 2 H 2 (g) → CH 4 (g)  H f °= − 74.6 kJ/mol CH 4 C(s, graphite) + O 2 (g) → CO 2 (g)  H f °= −393.5 kJ/mol CO 2 H 2 (g) + ½ O 2 (g) → H 2 O(g)  H f °= −241.8 kJ/mol H 2 O CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(g)  H° = −802.5 kJ  H° = [(  −393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ 110  H° = [(  H f ° CO 2 (g) + 2∙  H f ° H 2 O(g) )− (  H f ° CH 4 (g) + 2∙  H f ° O 2 (g) )]

Copyright © 2011 Pearson Education, Inc. Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l) 1. Write formation reactions for each compound and determine the  H f ° for each 2 C(s, gr) + H 2 (g)  C 2 H 2 (g)  H f ° = kJ/mol C(s, gr) + O 2 (g)  CO 2 (g)  H f ° = −393.5 kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f ° = −285.8 kJ/mol 111

Copyright © 2011 Pearson Education, Inc. 2 C 2 H 2 (g)  4 C(s) + 2 H 2 (g)  H° = 2(−227.4) kJ Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l) 4 C(s) + 4 O 2 (g)  4CO 2 (g)  H° = 4(−393.5) kJ 2 H 2 (g) + O 2 (g)  2 H 2 O(l)  H° = 2(−285.8) kJ 2.Arrange equations so they add up to desired reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l)  H = − kJ 112

Copyright © 2011 Pearson Education, Inc. Example: Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l)  H° reaction =  n  H f °(products) −  n  H f °(reactants)  H rxn = [(4  H CO2 + 2  H H2O ) – (2  H C2H2 + 5  H O2 )]  H rxn = [(4(−393.5) + 2(−285.8)) – (2(+227.4) + 5(0))]  H rxn = − kJ 113

Copyright © 2011 Pearson Education, Inc. 114 Example : How many kg of octane must be combusted to supply 1.0 x kJ of energy? MM octane = g/mol, 1 kg = 1000 g the units and sign are correct, the large value is expected 1.0 x kJ mass octane, kg Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H rxn ° Write the balanced equation per mole of octane kJ mol C 8 H 18 g C 8 H 18 kg C 8 H 18 from above C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g) Look up the  H f ° for each material in Appendix IIB  H°rxn = − kJ Tro: Chemistry: A Molecular Approach, 2/e

Copyright © 2011 Pearson Education, Inc. Practice – Calculate the  H° for decomposing 10.0 g of limestone, CaCO 3, under standard conditions. CaCO 3 (s) → CaO(s) + O 2 (g) 115

Copyright © 2011 Pearson Education, Inc. Practice – Calculate the  H° for decomposing 10.0 g of limestone, CaCO 3 (s) → CaO(s) + O 2 (g) MM limestone = g/mol, the units and sign are correct 10.0 g CaCO 3  H°, kJ Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H° rxn per mol CaCO 3 g CaCO 3 mol CaCO 3 kJ from above CaCO 3 (s) → CaO(s) + O 2 (g)  H° rxn = kJ 116

Thermochemistry © 2012 Pearson Education, Inc. Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats.

Thermochemistry © 2012 Pearson Education, Inc. Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels.