Slide 1 / 118 Slide 2 / 118 Thermochemistry Slide 3 / 118 Slide 4 / 118 Table of ontents The Nature of Energy State Functions** Enthalpy lick on the topic to go to that section Measuring Enthalpy hanges: alorimetry Energy ssociated with hanges of State Enthalpies of Reaction Hess's Law Enthalpies of Formation Energy in Foods and Fuels The Nature of Energy Return to Table of ontents Pr es en at Slide 5 / 118 Thermochemistry Slide 6 / 118 Review of Energy from Physics Potential Energy is the energy that objects have energy due to their position. Gravitational Potential Energy GPE = mgh We know chemical and physical processes release and absorb energy. We use these thermochemical principles to design air conditioners and refrigerators as well as foot warmers that allow us to stay comfortable as we "go big" on the hill! Elastic Potential Energy EPE = 1/2 kx 2 Electric Potential Energy UE = kq1q2 r 2
Slide 7 / 118 Review of Energy from Physics Slide 8 / 118 Review of Energy from Physics Kinetic Energy is the energy that an object has by virtue of its motion: KE = 1/2 mv 2 The total energy of an isolated system is constant. n outside force can change the energy of a system by doing work on it. work Work is defined by the formula lgebraically, these two statements combine to become: W = Fdparallel Since E f - E o = E, this can also be written as E0 + W = Ef E = W Slide 9 / 118 Units of Energy The SI unit of energy is the Joule (J). Slide 10 / 118 1 reaction produces 3.8 cal of energy. How many joules of energy is produced? nother unit of energy is the calorie (cal). 1 cal = 4.184 J The energy of food is measured in alories (). [note the capital ""] 1 alorie = 1000 calories = 4184 Joules Slide 10 (nswer) / 118 1 reaction produces 3.8 cal of energy. How many joules of energy is produced? Slide 11 / 118 2 reaction uses 235 J of energy. How many calories have been burned? nswer 15.9 J
Slide 11 (nswer) / 118 2 reaction uses 235 J of energy. How many calories have been burned? Slide 12 / 118 3 20 ounce coke contains 240 alories. How many kilojoules of energy are present in a 20 ounce oke? nswer 56 cal Slide 12 (nswer) / 118 3 20 ounce coke contains 240 alories. How many kilojoules of energy are present in a 20 ounce oke? Slide 13 / 118 Energy & Heat From last year, we know that E = W. nswer 1.0 x 10 3 kj This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q). When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object. T = 20 T = 10 heat flow E = w + q *Note, we use a lower case "w" in chemistry. Slide 14 / 118 The First Law of Thermodynamics E = w + q 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. Internal energy, E Initial state E < E 0 E < 0 (-) Final state E E of system decreases E 0 Energy lost to surroundings Internal energy, E Final state E > E 0 E > 0 (+) Initial state E E 0 E of system increases Energy gained from surroundings Slide 15 / 118 System and Surroundings When considering energy changes, we need to focus on a welldefined, limited part of the universe. The portion we focus on is called the system and everything else is called the surroundings. onsider the following reaction occurring within a metal cylinder. Surroundings system 2H 2(g) + O 2(g) --> 2H 2O(g) The system includes the reactants and products (here, the hydrogen, oxygen and water molecules). The surroundings are everything else (here, the cylinder and piston).
Slide 16 / 118 hanges in Internal Energy Slide 17 / 118 4 Ten grams of table salt in dissolved in water in a 250 ml beaker. Which of the following is a component of the system? Internal energy, E H 2 (g) + O 2 (g) E < 0 (negative) H 2 O(l) E >0 (positive) If E > 0, Efinal > Einitial The system absorbed energy from the surroundings. If E < 0, E final < E initial The system released energy from the surroundings. Nal water Na+ beaker E and F,, and G,, and Slide 17 (nswer) / 118 4 Ten grams of table salt in dissolved in water in a 250 ml beaker. Which of the following is a component of the system? Nal water Na+ beaker nswer E and F,, and G,, and F Slide 18 / 118 5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because... the reaction performed work on the flask the system absorbed heat from the surroundings the system released heat to the surroundings the surroundings released heat to the system Slide 18 (nswer) / 118 5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because... the reaction performed work on the flask the system absorbed heat from the surroundings the system released heat to the surroundings the surroundings released heat to the system nswer Slide 19 / 118 6 When a strong acid is added to a flask containing water the flask becomes warm to the touch. Which correctly describes the change in energy? E sys is positive and E sur is negative E sys is positive and E sur is positive E sys is negative and E sur is positive E sys is negative and E sur is negative
Slide 19 (nswer) / 118 6 When a strong acid is added to a flask containing water the flask becomes warm to the touch. Which correctly describes the change in energy? E sys is positive and E sur is negative nswer E sys is positive and E sur is positive E sys is negative and E sur is positive System Slide 20 / 118 hanges in Internal Energy Heat q > 0 Surroundings When energy is exchanged between the system and the surroundings, it is either exchanged as either heat (q) or work (w). E sys is negative and E sur is negative Work w > 0 E = q + w E>0 Slide 21 / 118 q, w, E, and Their Signs Sign onventions for q, w and E q + system gains heat - system loses heat w + work done on system - work done by system Slide 22 / 118 7 The E of a system that gains 50 kj of heat and performs 24 kj of work on the surroundings is kj. -74-26 0 +26 E +74 E + net gain of energy by system - net loss of energy by system Slide 22 (nswer) / 118 7 The E of a system that gains 50 kj of heat and performs 24 kj of work on the surroundings is kj. -74 Slide 23 / 118 8 The E of a system that releases 120 J of heat and does 40 J of work on the surroundings is J. -80-26 0 +26 E +74 nswer -160 0 +80 E +160
Slide 23 (nswer) / 118 8 The E of a system that releases 120 J of heat and does 40 J of work on the surroundings is J. Slide 24 / 118 9 The E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is J. -80-240 -160 0 +80 E +160 nswer -120 0 +120 E +240 Slide 24 (nswer) / 118 9 The E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is J. Slide 25 / 118 10 The E of a system that absorbs 12,000 J of heat and the surrounding does 12,000 J of work on the system is J. -240-24000 -120 0 nswer -12000 0 +120 E +240 +12000 E +24000 Slide 25 (nswer) / 118 10 The E of a system that absorbs 12,000 J of heat and the surrounding does 12,000 J of work on the system is J. -24000-12000 0 +12000 E +24000 nswer E Slide 26 / 118 Exchange of Heat between System and Surroundings Recall, when heat is absorbed by the system from the surroundings, the process is. Surroundings System Surroundings System Heat -q +q Heat When heat is released by the system into the surroundings, the process is.
Slide 27 / 118 11 The reaction that occurs inside the foot warmer packet is? Slide 27 (nswer) / 118 11 The reaction that occurs inside the foot warmer packet is? True True False False nswer False Slide 28 / 118 12 What will happen when a hot rock is put into cold water? Slide 28 (nswer) / 118 12 What will happen when a hot rock is put into cold water? the water and rock will both gain energy the water and rock will both lose energy the rock will gain energy and the water will lose energy the rock will lose energy and the water will gain energy the water and rock will both gain energy the water and rock will both lose energy the rock will gain energy and the water will lose energy the rock will lose energy and the [This water object is a pull will gain energy nswer Slide 29 / 118 Slide 29 (nswer) / 118 13 If you put a hot rock in cold water, and your system is the rock, the process is. 13 If you put a hot rock in cold water, and your system is the rock, the process is. neither, there is no net change of energy neither, there is no net change of energy nswer it depends on the exact temperatures it depends on the exact temperatures
Slide 30 / 118 14 If you put a hot rock in cold water, and your system is the water, the process is. Slide 30 (nswer) / 118 14 If you put a hot rock in cold water, and your system is the water, the process is. neither, there is no net change of energy it depends on the exact temperatures neither, there is no net change of energy nswer it depends on the exact temperatures Slide 31 / 118 15 If you put an ice cube in water, and your system is the ice, the process is. Slide 31 (nswer) / 118 15 If you put an ice cube in water, and your system is the ice, the process is. neither, there is no net change of energy it depends on the exact temperatures neither, there is no net change of energy it depends on the exact temperatures nswer Slide 32 / 118 Slide 32 (nswer) / 118 16 If you put an ice cube in water, and your system is the water, the process is. 16 If you put an ice cube in water, and your system is the water, the process is. neither, there is no net change of energy it depends on the exact temperatures neither, there is no net change of energy it depends on the exact temperatures nswer
Slide 33 / 118 17 When NaOH dissolves in water, the temperature of solution increases. This reaction is. Slide 33 (nswer) / 118 17 When NaOH dissolves in water, the temperature of solution increases. This reaction is. nswer Slide 34 / 118 18 When al 2 dissolves in water the temperature of water drops. This reaction is. Slide 34 (nswer) / 118 18 When al 2 dissolves in water the temperature of water drops. This reaction is. nswer Slide 35 / 118 19 Water droplets evaporating from the skin surface will make you feel cold. This process is. E for water for skin for skin for water and F and Slide 35 (nswer) / 118 19 Water droplets evaporating from the skin surface will make you feel cold. This process is. E for water for skin for skin for water and F and nswer F
Slide 36 / 118 State Functions** ## Slide 37 / 118 State Functions The internal energy of a system is independent of the path by which the system achieved that state. elow, the water could have reached room temperature from either direction...it makes no difference to the final energy of the system if it reached its final temperature by heating up or cooling down. # Return to Table of ontents 50g 100 50g 25 50g 0 ooling Heating Slide 38 / 118 Slide 39 / 118 ## State Functions ## State Functions 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. These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate. E depends only on E initial and E final They move between energy states while performing some task. Slide 40 / 118 Slide 41 / 118 ## State Functions ## Work one by a Gas Understanding thermodynamics enables us to harness energy flow for useful purposes. When a gas expands it does work on its surroundings: Initial state Final state For instance, whether the battery is shorted out or is discharged by running the fan, its #E is the same. ut q and w are different in the two cases. If the battery shorts out, all of the energy is lost as heat, whereas if it is used to run the fan, some energy is used to do work. Q and w are NOT state functions. harged battery Heat Heat Work #E Energy lost by battery W = Fd. In this case: F = P and d = h. W = Fd W = (P) h W = P V P= F/ P= F/ V h h ischarged battery = cross sectional area
Slide 42 / 118 Slide 43 / 118 ## Work one y a Gas ## Work one y a Gas Since the gas expands, and does work on its surroundings, the energy of the gas decreases, this is considered negative work. (-) work is done by system Initial state Final state P= F/ P= F/ V h If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work. Initial state P= F/ Final state P= F/ V h W = - P V h (+) work is done on system W = + P V h = cross sectional area = cross sectional area Slide 44 / 118 Slide 45 / 118 ## Work one by a Gas We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston. Enthalpy Return to Table of ontents Hl Solution Hl Solution + Zinc Slide 46 / 118 Enthalpy The word "enthalpy" is derived from the Greek noun "enthalpos" which means heating. The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance to occupy. H = E + W It is impossible to measure enthalpy, H, directly. Only the change ( H) can be measured. Slide 47 / 118 Enthalpy t constant pressure, the change in enthalpy is the heat gained or lost by the system. H = (q+w) - W H = q Note: This is only true at constant pressure. See ** for more information (W = P V) H = E + W
** Slide 48 / 118 Enthalpy The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy. H = E + PV It is impossible to measure enthalpy, H, directly. Only the change ( H) can be measured. H = E + PV ** Slide 49 / 118 Enthalpy If a process takes place at constant pressure (as most processes we study do), we can account for heat flow during the process by measuring the enthalpy of the system. H = E + (PV) H = E + P V H = (q+w) - w H = q p (at constant pressure) t constant pressure, the change in enthalpy is the heat gained or lost by the system. Slide 50 / 118 Enthalpy Enthalpy is an extensive property; its value depends on the quantity of the substance present. Slide 51 / 118 20 When 114 grams of gasoline combust, the H is equal to -5,330 kj. What is the H for combustion of 57 grams of gasoline? ombustion of 16 grams of H 4 H = -891 kj ombustion of 32 grams of H 4 H = -1782 kj H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. H 4(g) + 2O 2(g) --> O 2(g) + 2H 2O(g) H = -891 kj 2H 2O(g) + O 2(g) --> H 4(g) + 2O 2(g) H = +891 kj H for a reaction depends on the state of the products and the state of the reactants. H 4(g) + 2O 2(g) --> O 2(g) + 2H 2O(g) H 4(g) + 2O 2(g) --> O 2(g) + 2H 2O(l) H = -891 kj H = -979 kj Slide 51 (nswer) / 118 20 When 114 grams of gasoline combust, the H is equal to -5,330 kj. What is the H for combustion of 57 grams of gasoline? Slide 52 / 118 Endothermic and Exothermic Processes process is when H is positive. process is when H is negative. Surroundings Surroundings nswer -2665 kj System System +q Heat Heat -q H>0 Endothermic H<0 Exothermic
Slide 53 / 118 21 The reaction + --> is. The H for this reaction is +50 J. What is the H for the reaction --> +? Slide 53 (nswer) / 118 21 The reaction + --> is. The H for this reaction is +50 J. What is the H for the reaction --> +? annot be determined. 0.02 J - 50 J 100 J annot be determined. 0.02 J - 50 J 100 J nswer Slide 54 / 118 22 The reaction + --> is. The H for this reaction is -150 J. What is the H for the reaction --> +? +150 zero -150 this reaction will not happen Slide 54 (nswer) / 118 22 The reaction + --> is. The H for this reaction is -150 J. What is the H for the reaction --> +? +150 zero -150 nswer this reaction will not happen Slide 55 / 118 23 issolving NaOH in water will increase the temperature of the solution. This reaction is Slide 55 (nswer) / 118 23 issolving NaOH in water will increase the temperature of the solution. This reaction is adiabatic isothermal adiabatic isothermal nswer
Slide 56 / 118 Slide 56 (nswer) / 118 24 NH 3NO 3 + H 2O --> NH 4 + + NO 3 - + OH - H = +25.69 kj/mol. This reaction is. Yes No 24 NH 3NO 3 + H 2O --> NH 4 + + NO 3 - + OH - H = +25.69 kj/mol. This reaction is. Yes No nswer No Slide 57 / 118 Slide 58 / 118 Measuring Enthalpy hanges: alorimetry Measuring Enthalpy hanges: alorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow by making use of the expression: H = q p Return to Table of ontents The subscript p on q means the process is occurring at constant pressure. No work is being done only heat is being exchanged. You will not always see the subscript but it is implied when we are speaking of Enthalpy. Slide 59 / 118 Heat apacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1 ) is its heat capacity. The amount of energy required to raise the temperature of one gram of a substance by 1 K (1 ) is its specific heat (c). Slide 60 / 118 25 Heat capacity is an example of an Intensive property Extensive property
Slide 60 (nswer) / 118 25 Heat capacity is an example of an Slide 61 / 118 26 Specific heat is an example of an Intensive property Extensive property Intensive property Extensive property nswer Slide 61 (nswer) / 118 26 Specific heat is an example of an Slide 62 / 118 Heat apacity and Specific Heat Intensive property Extensive property Specific heat = heat transferred mass x temperature change nswer c = q m T Slide 63 / 118 27 The specific heat of marble is 0.858 J/g-K. How much heat (in J) is required to raise the temperature of 20g of marble from 22 to 45? Slide 63 (nswer) / 118 27 The specific heat of marble is 0.858 J/g-K. How much heat (in J) is required to raise the temperature of 20g of marble from 22 to 45? nswer 395J q = mc T q = mc T
Slide 64 / 118 28 n 26 g sample of wood (c = 1.674 J/(g-K)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 to. Slide 64 (nswer) / 118 28 n 26 g sample of wood (c = 1.674 J/(g-K)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 to. nswer 24.6 q = mc T q = mc T Slide 65 / 118 29 sample of silver (c = 0.236 J/g-K) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 to 80. What is the mass of the sample? Slide 65 (nswer) / 118 29 sample of silver (c = 0.236 J/g-K) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 to 80. What is the mass of the sample? nswer 113 g q = mc T q = mc T Slide 66 / 118 onstant Pressure alorimetry Slide 66 (nswer) / 118 onstant Pressure alorimetry stirrer insulated cover thermometer styrofoam cup Many chemical reactions happen in aqueous solutions. The apparatus to the left is a calorimeter. How could you use it to measure the heat change for a chemical reaction in an aqueous solution? thermometer Many chemical reactions happen stirrer y in carrying aqueous out solutions. a reaction in aqueous solution in a simple The apparatus calorimeter to the like left is a insulated cover this calorimeter. one, the heat change for a system can be indirectly How could measured you use by it to measure measuring the heat change the heat for a chemical change reaction for in the an water aqueous in solution? the calorimeter. styrofoam cup nswer sparknotes.com sparknotes.com
Slide 67 / 118 onstant Pressure alorimetry Slide 68 / 118 onstant Pressure alorimetry stirrer insulated cover thermometer ecause the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: H = q = mc T (at constant pressure) Example: student wishes to determine the enthalpy change when ammonium chloride (NH 4l) dissolves in water. The student masses out 20.00 grams of ammonium chloride and adds it to 500 grams of water in a styrofoam cup at a temperature of 16.1 elsius. The student observes the temperature to decrease to 13.2 elsius. What is the enthalpy change for the dissolution of ammonium chloride? 1. Find enthalpy change of solution using q = mc T = 520 g x -2.9 x 4.2 J = -6,334 J g styrofoam cup 2. Since heat released by surroundings (solution) is equal to the heat gained by system (ammonium chloride). H for dissolving of NH 4l = +6334 J sparknotes.com Slide 69 / 118 omb alorimetry Slide 70 / 118 omb alorimetry (onstant Volume) 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. Water Thermometer Ignition wires Oxygen atmosphere Sample ecause 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. Water Thermometer Ignition wires Oxygen atmosphere Sample Slide 71 / 118 30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 elsius to start and the heat capacity of the calorimeter was 34 J/, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 elsius? 4Fe(s) + 3O 2(g) --> 2Fe 2O 3(s) Slide 71 (nswer) / 118 30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 elsius to start and the heat capacity of the calorimeter was 34 J/, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 elsius? 4Fe(s) + 3O 2(g) --> 2Fe 2O 3(s) nswer -115.6 J
Slide 72 / 118 31 mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10. fter the metal is added, the temperature rises to 11.3. Identify the metal. Slide 72 (nswer) / 118 31 mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10. fter the metal is added, the temperature rises to 11.3. Identify the metal. Metal u 0.386 Specific Heat (J/g) Metal u 0.386 Specific Heat (J/g) nswer u 0.126 u 0.126 l 0.900 l 0.900 Slide 73 / 118 Slide 74 / 118 Energy hanges ssociated with hanges of State Energy hanges ssociated with hanges of State hemical and physical changes are usually accompanied by changes in energy. Recall the following terms: When energy is put into the system, the process is called. Return to Table of ontents When energy is released by the system, the process is called. Slide 74 (nswer) / 118 Energy hanges ssociated with hanges of State hemical and physical changes are usually accompanied by changes in energy. Recall the following terms: When energy is put into the system, process is called. When energy is released by the system, the process is called. nswer Endothermic processes Energy is taken into the system from the surroundings ( H > 0) Slide 75 / 118 Energy hanges ssociated with hanges of State Fill in the blanks Exothermic processes Energy is released from the system to surroundings ( H < 0) nswer boiling or evaporating a liquid condensing a gas deposition of a gas freezing a liquid melting a solid sublimation of a solid
Energy of system Slide 76 / 118 Phase hanges Gas Vaporization ondensation Sublimation Liquid Melting Freezing Solid eposition Slide 77 / 118 32 Which of the following is/are? I. boiling II. melting III. freezing I only I and II only III only I, II and III Slide 77 (nswer) / 118 32 Which of the following is/are? I. boiling II. melting III. freezing I only I and II only III only nswer I, II and III Slide 78 / 118 Energy hanges ssociated with hanges of State The heat of fusion ( Hfus ) is the energy required to change a solid at its melting point to a liquid. Heat of fusion(h 2O) = 6.0 kj/mol The heat of vaporization ( Hvap) is defined as the energy required to change a liquid at its boiling point to a gas. Heat of vaporization(h 2O) = 41.0 kj/mol lass Question: Why is the heat of vaporization much higher than the heat of fusion for a substance? In order to change phase from a solid to liquid, the particle attractions need only be strained somewhat. When a material changes from move a for liquid answer to a gas, the particle attractions must be essentially broken. Slide 79 / 118 Slide 80 / 118 Heat of fusion and vaporization kj/mol 80 40 10 Energy hanges ssociated with hanges of State Heat of fusion Heat of vaporization 24 5 utane 29 41 7 iethyl ether 6 Water 58 23 Mercury Note that these quantities are usually per 1.00 mole, whereas q = mc T involves mass in grams. 33 The heat of vaporization for butane is 24 kj/mol. How much energy is required to vaporize 2 mol of butane? 2kJ 12 kj 22 kj 48 kj
Slide 80 (nswer) / 118 Slide 81 / 118 33 The heat of vaporization for butane is 24 kj/mol. How much energy is required to vaporize 2 mol of butane? 2kJ 12 kj 22 kj 48 kj nswer 34 The heat of vaporization for butane is 24 kj/ mol. How much energy is required to vaporize 0.33 mol of butane? 8kJ 12 kj 16 kj 72 kj Slide 81 (nswer) / 118 Slide 82 / 118 34 The heat of vaporization for butane is 24 kj/ mol. How much energy is required to vaporize 0.33 mol of butane? 8kJ 12 kj 16 kj nswer 35 How much energy is required to melt 0.5 mol of water? 2kJ 3 kj 6 kj 12 kj Heat of fusion for H 2O (s) Heat of vaporization for H 2O (l) 6 kj/mol 41 kj/mol nswer 72 kj 36 Slide 83 / 118 How much energy is released when 3.0 mol of water freezes? Slide 84 / 118 37 How much energy is needed to melt 10.0 mol solid Hg? 2kJ 3 kj 18 kj 123 kj Heat of fusion for H 2O (s) Heat of vaporization for H 2O (l) 6 kj/mol 41 kj/mol nswer E 2.3 kj 5.8 kj 23 kj 230 kj 580 kj Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kj/mol 58 kj/mol nswer
Slide 85 / 118 Slide 86 / 118 38 How much energy is needed to vaporize 2.0 mol Hg (l)? Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kj/mol 58 kj/mol nswer This graph is called a heating curve. Energy hanges ssociated with hanges of State It illustrates how temperature changes over time as constant heat is applied to a pure solid substance. Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor E Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) F From to, ice is heating up from -25 o to 0 o. Since Q = mc T T = Q (mc) Slide 87 / 118 Energy hanges ssociated with hanges of State So the slope is 1 (mc) From to, liquid water is heating up from 0 to 100. Once again, the slope is 1/(mc). ut "c" is different for all the phases of a substance, so the slope is different for solid, liquid and gaseous H 2 O. Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor E Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) Slide 89 / 118 Energy hanges ssociated with hanges of State Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) F F E Liquid water and vapor (vaporization) From to, ice is melting. The added heat is breaking the hydrogen bonds of the solid, so the temperature is constant. That's why the slope is zero. Slide 88 / 118 Energy hanges ssociated with hanges of State From to E, liquid water is boiling into vapor. The added heat is breaking the IM forces of the liquid, so the temperature is constant. That's why the slope is zero. Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor E Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) Slide 90 / 118 Energy hanges ssociated with hanges of State Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor Liquid water Ice and liqui d water (melting) Ice Heat added (each division corresponds to 4 kj) F F E Liquid water and vapor (vaporization)
From E to F water vapor, steam, is heating up from 100 to 125. Once again, the slope is 1/(mc). ut "c" is different for all the phases of a substance, so the slope is different for for solid, liquid and gaseous H 2 O. Slide 91 / 118 Energy hanges ssociated with hanges of State Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor E Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) Slide 93 / 118 39 Which substance has the lowest specific heat? F Slide 92 / 118 Energy hanges ssociated with hanges of State 1 Recall that slope = m where m = mass and = specific heat of the substance This graph shows heat transfer versus change in temperature for 1 gram of 4 different substances. (T) Temperature () (q) Energy dded (Joules) Slide 93 (nswer) / 118 39 Which substance has the lowest specific heat? (T) Temperature () (T) Temperature () nswer (q) Energy dded (Joules) (q) Energy dded (Joules) Slide 94 / 118 40 Which substance requires the highest amount of heat added to raise the temperature? (T) Temperature () Slide 94 (nswer) / 118 40 Which substance requires the highest amount of heat added to raise the temperature? (T) Temperature () nswer (q) Energy dded (Joules) (q) Energy dded (Joules)
Slide 95 / 118 41 Which segment(s) contain solid sodium chloride? only, and and, and E Slide 95 (nswer) / 118 41 Which segment(s) contain solid sodium chloride? only, and and, and E nswer Slide 96 / 118 42 Which segment(s) contain liquid sodium chloride? only, and and, and E Slide 96 (nswer) / 118 42 Which segment(s) contain liquid sodium chloride? only, and and, and E nswer Slide 97 / 118 43 What is the melting point (in o ) of sodium chloride? Slide 97 (nswer) / 118 43 What is the melting point (in o ) of sodium chloride? nswer 801
Slide 98 / 118 44 What is the freezing point (in o ) of sodium chloride? Slide 98 (nswer) / 118 44 What is the freezing point (in o ) of sodium chloride? nswer 801 Slide 99 / 118 45 Which is greater: the specific heat of solid Nal, or the specific heat of molten (liquid) Nal? solid liquid annot be determined. They are equal. Slide 99 (nswer) / 118 45 Which is greater: the specific heat of solid Nal, or the specific heat of molten (liquid) Nal? solid liquid annot be determined. They are equal. nswer Segments,, EF slope = nonzero T increasing KE increasing PE constant apply q = mc T Segments & E slope = 0 T = 0 KE constant PE increasing apply H fus or H vap Slide 100 / 118 Features of a Heating urve Temperature ( 0 ) 125 100 75 50 25 0-25 Water vapor E Liquid water and vapor (vaporization) Liquid water Ice and liquid water (melting) Ice Heat added (each division corresponds to 4 kj) F Slide 101 / 118 Energy hanges ssociated with hanges of State Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas. dditionally, melting one mole of a substance ( H fus) and vaporizing one mole of the same substance ( H vap) require different amounts of energy.
Slide 102 / 118 Specifics about Water Slide 103 / 118 alculating Energy hanges from a Heating urve Specific heat of ice Specific heat of water Specific heat of steam Heat of fusion ( H fus) of water at 0 2.09 J/g- 4.18 J/g- 1.84 J/g- 6.01 kj/mol or 330 J/g Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. It is a good idea to sketch out the segments first on a graph. Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g Slide 104 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Slide 105 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Segment 1 Warming the ice from -15.0 up to the substance's MP which is 0. Temperature () 20 10 0-4 6 12 24 Segment 2 Melting the ice at 0. Temperature () 20 10 0-4 6 12 24-14 Time (s) -14 Time (s) Slide 106 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Slide 107 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Segment 3 Warming the liquid from 0 to 25. Temperature () 20 10 0-4 6 12 24 We are now ready to apply either the calorimetry equation or Hfus or Hvap. Segment 1 Warming the ice from -15.0 up to the substance's MP which is 0. q 1 = mc T q 1 = (10.0g) (2.09 J/g- o ) (15.0 o ) q 1 = 313.5 J q 1 = 0.314 kj or 313.5J -14 Time (s)
Slide 108 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Segment 2 Melting the ice at 0. q 2 = ( H fus ) (# moles) q 2 = (6.01 kj/mol) [(10.0 g)/ 18.0 g/mol)] q 2 = 3.34 kj or Slide 109 / 118 alculating Energy hanges from a Heating urve Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Segment 3 Warming the water from 0 to 25.0. q 3 = mc T q 3 = (10.0g) (4.18 J/g- o ) (25.0 o ) q 3 = 1045 J q 3 = 1.05 kj 6.01/18 J/g x 10g = 3.34 kj = 3340J Slide 110 / 118 alculating Energy hanges from a Heating urve Slide 111 / 118 46 alculate the enthalpy change in J of converting 75 g of ice at -11 to liquid water at 22. Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. Now, we add the heat changes for all 3 segments. First, make sure that all quantities are in kilojoules. Total H = (q 1 + q 2 + q 3) kj H = (0.314 + 3.34 + 1.05) kj H = 4.6 kj or 4698.5J 98,654 33,371 J 8,621 26,474 E 35,096 J Specific heat of ice Specific heat of water Specific heat of steam Heat of fusion ( H fus) of water at 0 Heat of vaporization ( H vap) of water at 100 2.09 J/g- 4.18 J/g- 1.84 J/g- 6.01 kj/mol or 330 J/g 40.7 kj/mol or 2600 J/g nswer Slide 112 / 118 47 alculate the enthalpy change of converting 25 g of water vapor at 110 to liquid water at 50. Slide 113 / 118 48 alculate the enthalpy change in kj of converting 31.8 g of ice at 0 to water vapor at 140. -69,765 Specific heat of ice 2.09 J/g- 109 kj Specific heat of ice 2.09 J/g- 69,765-59,315-70,685 J Specific heat of water Specific heat of steam Heat of fusion ( H fus) of water at 0 4.18 J/g- 1.84 J/g- 6.01 kj/mol or 330 J/g nswer 96 kj 104 kj - 88 kj Specific heat of water Specific heat of steam Heat of fusion ( H fus) of water at 0 4.18 J/g- 1.84 J/g- 6.01 kj/mol or 330 J/g nswer E -13,935 E -109 kj Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g
Slide 114 / 118 Slide 115 / 118 Enthalpy of Reaction Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = H products H reactants H 4(g) + 2 O 2(g) Return to Table of ontents Enthalpy H 1 = -890 kj H 2 = 890 kj O 2(g) + 2H 2O(l) Slide 116 / 118 Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction or the heat of reaction. The combustion of hydrogen in the balloon below is an reaction and the energy released is equal to the heat of reaction. Enthalpy 2H 2(g) + O 2(g) H<0 2H 2O(g) Slide 117 / 118 Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of Hr is formed in this reaction? H2(g) + r2(g) --> 2Hr(g) X -72kJ 4.0 mol Hr(g) = 2.0 mol Hr(g) X = -72kJ (4/2) X = -144kJ 144kJ of energy are released H = -72 kj Slide 118 / 118 Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of Hr is formed in this reaction? H2(g) + r2(g) --> 2Hr(g) H = -72 kj 4.0 mol Hr x -72 kj 2 mol Hr = -144kJ or 144 kj released