Slide 1 / 118 Slide 2 / 118 Thermochemistry Table of ontents Slide 3 / 118 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
Slide 4 / 118 The Nature of Energy Return to Table of ontents Pr es en at Thermochemistry Slide 5 / 118 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! Review of Energy from Physics Potential Energy is the energy that objects have energy due to their position. Gravitational Potential Energy GPE = mgh Slide 6 / 118 Elastic Potential Energy EPE = 1/2 kx 2 Electric Potential Energy UE = kq1q2 r 2
Review of Energy from Physics Slide 7 / 118 Kinetic Energy is the energy that an object has by virtue of its motion: KE = 1/2 mv 2 Work is defined by the formula W = Fdparallel Review of Energy from Physics Slide 8 / 118 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 lgebraically, these two statements combine to become: E0 + W = Ef Since E f - E o = E, this can also be written as E = W Units of Energy Slide 9 / 118 The SI unit of energy is the Joule (J). 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
1 reaction produces 3.8 cal of energy. How many joules of energy is produced? Slide 10 / 118 2 reaction uses 235 J of energy. How many calories have been burned? Slide 11 / 118 3 20 ounce coke contains 240 alories. How many kilojoules of energy are present in a 20 ounce oke? Slide 12 / 118
Energy & Heat From last year, we know that E = W. Slide 13 / 118 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. The First Law of Thermodynamics E = w + q Energy is neither created nor destroyed. Slide 14 / 118 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 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. 2H 2(g) + O 2(g) --> 2H 2O(g) Slide 15 / 118 Surroundings system The system includes the reactants and products (here, the hydrogen, oxygen and water molecules). The surroundings are everything else (here, the cylinder and piston).
hanges in Internal Energy Slide 16 / 118 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. 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? Slide 17 / 118 Nal water Na+ beaker E and F,, and G,, and 5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because... Slide 18 / 118 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
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 Slide 19 / 118 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 hanges in Internal Energy Slide 20 / 118 System E>0 Heat q > 0 Surroundings Work w > 0 When energy is exchanged between the system and the surroundings, it is either exchanged as either heat (q) or work (w). E = q + w q, w, E, and Their Signs Slide 21 / 118 Sign onventions for q, w and E q + system gains heat - system loses heat w + work done on system - work done by system E + net gain of energy by system - net loss of energy by system
7 The E of a system that gains 50 kj of heat and performs 24 kj of work on the surroundings is kj. Slide 22 / 118-74 -26 0 +26 E +74 8 The E of a system that releases 120 J of heat and does 40 J of work on the surroundings is J. Slide 23 / 118-80 -160 0 +80 E +160 9 The E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is J. Slide 24 / 118-240 -120 0 +120 E +240
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. Slide 25 / 118-24000 -12000 0 +12000 E +24000 Exchange of Heat between System and Surroundings Slide 26 / 118 Recall, when heat is absorbed by the system from the surroundings, the process is endothermic. Surroundings System Surroundings System Heat -q +q Heat When heat is released by the system into the surroundings, the process is exothermic. 11 The reaction that occurs inside the foot warmer packet is endothermic? Slide 27 / 118 True False
12 What will happen when a hot rock is put into cold water? Slide 28 / 118 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 13 If you put a hot rock in cold water, and your system is the rock, the process is. Slide 29 / 118 exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures 14 If you put a hot rock in cold water, and your system is the water, the process is. Slide 30 / 118 exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures
15 If you put an ice cube in water, and your system is the ice, the process is. Slide 31 / 118 exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures 16 If you put an ice cube in water, and your system is the water, the process is. Slide 32 / 118 exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures 17 When NaOH dissolves in water, the temperature of solution increases. This reaction is. Slide 33 / 118 exothermic endothermic
18 When al 2 dissolves in water the temperature of water drops. This reaction is. Slide 34 / 118 endothermic exothermic 19 Water droplets evaporating from the skin surface will make you feel cold. This process is. E exothermic for water endothermic for skin exothermic for skin endothermic for water and F and Slide 35 / 118 Slide 36 / 118 State Functions** Return to Table of ontents
## State Functions The internal energy of a system is independent of the path by which the system achieved that state. Slide 37 / 118 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. # 50g 100 50g 25 50g 0 ooling Heating ## State Functions Slide 38 / 118 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. E depends only on E initial and E final ## State Functions Slide 39 / 118 These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate. They move between energy states while performing some task.
## State Functions Understanding thermodynamics enables us to harness energy flow for useful purposes. Slide 40 / 118 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 ischarged battery #E Energy lost by battery ## Slide 41 / 118 Work one by a Gas When a gas expands it does work on its surroundings: Initial state Final state W = Fd. P= F/ P= F/ In this case: F = P and d = h. V h W = Fd W = (P) h h W = P V = cross sectional area ## Work one y a Gas Slide 42 / 118 Since the gas expands, and does work on its surroundings, the energy of the gas decreases, this is considered negative work. Initial state Final state P= F/ P= F/ (-) work is done by system W = - P V V h h = cross sectional area
## Work one y a Gas Slide 43 / 118 If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work. (+) work is done on system W = + P V Initial state P= F/ h Final state P= F/ V h = cross sectional area ## 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. Slide 44 / 118 Hl Solution Hl Solution + Zinc Slide 45 / 118 Enthalpy Return to Table of ontents
Enthalpy Slide 46 / 118 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. H = E + W Enthalpy Slide 47 / 118 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) ** Enthalpy Slide 48 / 118 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
** Enthalpy Slide 49 / 118 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. Enthalpy Slide 50 / 118 Enthalpy is an extensive property; its value depends on the quantity of the substance present. 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 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 51 / 118
Endothermic and Exothermic Processes Slide 52 / 118 process is endothermic when H is positive. Surroundings System process is exothermic when H is negative. Surroundings System +q Heat Heat -q H>0 Endothermic H<0 Exothermic 21 The reaction + --> is endothermic. The H for this reaction is +50 J. What is the H for the reaction --> +? Slide 53 / 118 annot be determined. 0.02 J - 50 J 100 J 22 The reaction + --> is exothermic. 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 / 118
23 issolving NaOH in water will increase the temperature of the solution. This reaction is Slide 55 / 118 exothermic endothermic adiabatic isothermal 24 NH 3NO 3 + H 2O --> NH 4 + + NO 3 - + OH - H = +25.69 kj/mol. This reaction is exothermic. Slide 56 / 118 Yes No Slide 57 / 118 Measuring Enthalpy hanges: alorimetry Return to Table of ontents
Measuring Enthalpy hanges: alorimetry Slide 58 / 118 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 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. Heat apacity and Specific Heat Slide 59 / 118 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). 25 Heat capacity is an example of an Slide 60 / 118 Intensive property Extensive property
26 Specific heat is an example of an Slide 61 / 118 Intensive property Extensive property Heat apacity and Specific Heat Slide 62 / 118 Specific heat = heat transferred mass x temperature change c = q m T 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 / 118 q = mc T
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 / 118 q = mc T 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 / 118 q = mc T onstant Pressure alorimetry Slide 66 / 118 stirrer thermometer Many chemical reactions happen in aqueous solutions. insulated cover The apparatus to the left is a calorimeter. styrofoam cup How could you use it to measure the heat change for a chemical reaction in an aqueous solution? sparknotes.com
onstant Pressure alorimetry Slide 67 / 118 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) styrofoam cup sparknotes.com onstant Pressure alorimetry Slide 68 / 118 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 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 omb alorimetry Slide 69 / 118 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
omb alorimetry (onstant Volume) Slide 70 / 118 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 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 / 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 / 118 Metal u 0.386 Specific Heat (J/g) u 0.126 l 0.900
Slide 73 / 118 Energy hanges ssociated with hanges of State Return to Table of ontents Energy hanges ssociated with hanges of State Slide 74 / 118 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. When energy is released by the system, the process is called. Energy hanges ssociated with hanges of State Fill in the blanks Endothermic processes Exothermic processes Slide 75 / 118 Energy is taken into the system from the surroundings ( H > 0) 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
Phase hanges Slide 76 / 118 Gas Vaporization ondensation Energy of system Sublimation Liquid Melting Freezing Solid eposition 32 Which of the following is/are exothermic? I. boiling II. melting III. freezing Slide 77 / 118 I only I and II only III only I, II and III 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 78 / 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. Slide 79 / 118 33 The heat of vaporization for butane is 24 kj/mol. How much energy is required to vaporize 2 mol of butane? 2kJ Slide 80 / 118 12 kj 22 kj 48 kj 34 The heat of vaporization for butane is 24 kj/ mol. How much energy is required to vaporize 0.33 mol of butane? Slide 81 / 118 8kJ 12 kj 16 kj 72 kj
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 Slide 82 / 118 36 How much energy is released when 3.0 mol of water freezes? Slide 83 / 118 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 37 How much energy is needed to melt 10.0 mol solid Hg? Slide 84 / 118 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
How much energy is needed to vaporize 38 2.0 mol Hg (l)? Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kj/mol 58 kj/mol nswer Slide 85 / 118 Energy hanges ssociated with hanges of State Slide 86 / 118 This graph is called a heating curve. 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 Energy hanges ssociated with hanges of State Slide 87 / 118 From to, ice is heating up from -25 o to 0 o. Since Q = mc T T = Q (mc) So the slope is 1 (mc) 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
Energy hanges ssociated with hanges of State Slide 88 / 118 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. 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, 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. 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) F Slide 89 / 118 Energy hanges ssociated with hanges of State Slide 90 / 118 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 liqui d water (melting) Ice Heat added (each division corresponds to 4 kj) F
Energy hanges ssociated with hanges of State Slide 91 / 118 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. 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 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 () Slide 92 / 118 (q) Energy dded (Joules) 39 Which substance has the lowest specific heat? Slide 93 / 118 (T) Temperature () (q) Energy dded (Joules)
40 Which substance requires the highest amount of heat added to raise the temperature? (T) Temperature () Slide 94 / 118 (q) Energy dded (Joules) 41 Which segment(s) contain solid sodium chloride? only, and and, and E Slide 95 / 118 42 Which segment(s) contain liquid sodium chloride? only, and and, and E Slide 96 / 118
43 What is the melting point (in o ) of sodium chloride? Slide 97 / 118 44 What is the freezing point (in o ) of sodium chloride? Slide 98 / 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 / 118
Features of a Heating urve Slide 100 / 118 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 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 Energy hanges ssociated with hanges of State Slide 101 / 118 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. Specifics about Water Slide 102 / 118 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 Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g
alculating Energy hanges from a Heating urve Slide 103 / 118 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. 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 104 / 118 Segment 1 Warming the ice from -15.0 up to the substance's MP which is 0. Temperature () 20 10 0-4 6 12 24-14 Time (s) 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 Segment 2 Melting the ice at 0. Temperature () 20 10 0-4 6 12 24-14 Time (s)
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 106 / 118 Segment 3 Warming the liquid from 0 to 25. 20 Temperature () 10 0-4 6 12 24-14 Time (s) alculating Energy hanges from a Heating urve Slide 107 / 118 Sample Problem: alculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 to water at 25.0. 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 alculating Energy hanges from a Heating urve Slide 108 / 118 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 6.01/18 J/g x 10g = 3.34 kj = 3340J
alculating Energy hanges from a Heating urve Slide 109 / 118 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 alculating Energy hanges from a Heating urve Slide 110 / 118 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 46 alculate the enthalpy change in J of converting 75 g of ice at -11 to liquid water at 22. Slide 111 / 118 98,654 33,371 J 8,621 26,474 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 nswer E 35,096 J Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g
47 alculate the enthalpy change of converting 25 g of water vapor at 110 to liquid water at 50. Slide 112 / 118-69,765 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 E -13,935 Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g 48 alculate the enthalpy change in kj of converting 31.8 g of ice at 0 to water vapor at 140. Slide 113 / 118 109 kj Specific heat of ice 2.09 J/g- 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 -109 kj Heat of vaporization ( H vap) of water at 100 40.7 kj/mol or 2600 J/g Slide 114 / 118 Enthalpies of Reaction Return to Table of ontents
Enthalpy of Reaction Slide 115 / 118 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) Enthalpy H 1 = -890 kj H 2 = 890 kj O 2(g) + 2H 2O(l) Enthalpy of Reaction Slide 116 / 118 This quantity, H, is called the enthalpy of reaction or the heat of reaction. The combustion of hydrogen in the balloon below is an exothermic reaction and the energy released is equal to the heat of reaction. 2H 2(g) + O 2(g) Enthalpy H<0 exothermic 2H 2O(g) Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of Hr is formed in this reaction? Slide 117 / 118 H2(g) + r2(g) --> 2Hr(g) H = -72 kj X -72kJ 4.0 mol Hr(g) = 2.0 mol Hr(g) X = -72kJ (4/2) X = -144kJ 144kJ of energy are released
Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of Hr is formed in this reaction? Slide 118 / 118 H2(g) + r2(g) --> 2Hr(g) H = -72 kj 4.0 mol Hr x -72 kj 2 mol Hr = -144kJ or 144 kj released