Slide 1 / 118
Slide 2 / 118 Thermochemistry
Slide 3 / 118 Table of Contents The Nature of Energy State Functions** Click on the topic to go to that section Enthalpy Measuring Enthalpy Changes: Calorimetry Energy Associated with Changes 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 Contents
Pr es en at Slide 5 / 118 Thermochemistry 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!
Slide 6 / 118 A Review of Energy from Physics Potential Energy is the energy that objects have energy due to their position. Gravitational Potential Energy GPE = mgh Elastic Potential Energy EPE = 1/2 kx 2 Electric Potential Energy U E = kq 1 Q 2 r 2
Slide 7 / 118 A Review of Energy from Physics 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 = Fd parallel
Slide 8 / 118 A Review of Energy from Physics The total energy of an isolated system is constant. An outside force can change the energy of a system by doing work on it. work Algebraically, these two statements combine to become: E 0 + W = E f Since E f - E o = E, this can also be written as E = W
Slide 9 / 118 Units of Energy The SI unit of energy is the Joule (J). Another unit of energy is the calorie (cal). 1 cal = 4.184 J The energy of food is measured in Calories (C). [note the capital "C"] 1 Calorie = 1000 calories = 4184 Joules
Slide 10 / 118 1 A reaction produces 3.8 cal of energy. How many joules of energy is produced?
Slide 11 / 118 2 A reaction uses 235 J of energy. How many calories have been burned?
Slide 12 / 118 3 A 20 ounce coke contains 240 Calories. How many kilojoules of energy are present in a 20 ounce Coke?
Slide 13 / 118 Energy & Heat From last year, we know that E = W. 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. A T = 20 T = 10 B 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. Consider the following reaction occurring within a metal cylinder. 2H 2 (g) + O 2 (g) --> 2H 2 O(g) 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).
Slide 16 / 118 Changes in Internal Energy Internal energy, E H 2 (g) + O 2 (g) E < 0 (negative) H 2 O(l) E >0 (positive) If E > 0, E final > E initial The system absorbed energy from the surroundings. If E < 0, E final < E initial The system released energy from the surroundings.
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? A NaCl B water C Na+ D beaker E A and B F A, B, and C G A, B, and D
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... A the reaction performed work on the flask B the system absorbed heat from the surroundings C the system released heat to the surroundings D the surroundings released heat to the system
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? A E sys is positive and E sur is negative B E sys is positive and E sur is positive C E sys is negative and E sur is positive D E sys is negative and E sur is negative
Slide 20 / 118 Changes in Internal Energy 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
Slide 21 / 118 q, w, E, and Their Signs Sign Conventions 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
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. A -74 B -26 C 0 D +26 E +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. A -80 B -160 C 0 D +80 E +160
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. A -240 B -120 C 0 D +120 E +240
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. A -24000 B -12000 C 0 D +12000 E +24000
Slide 26 / 118 Exchange of Heat between System and Surroundings 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.
Slide 27 / 118 11 The reaction that occurs inside the foot warmer packet is endothermic? True False
Slide 28 / 118 12 What will happen when a hot rock is put into cold water? A B C D 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
Slide 29 / 118 13 If you put a hot rock in cold water, and your system is the rock, the process is. A B C D exothermic endothermic neither, there is no net change of energy 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. A B C D exothermic endothermic neither, there is no net change of energy 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. A B C D exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures
Slide 32 / 118 16 If you put an ice cube in water, and your system is the water, the process is. A B C D exothermic endothermic neither, there is no net change of energy it depends on the exact temperatures
Slide 33 / 118 17 When NaOH dissolves in water, the temperature of solution increases. This reaction is. A B exothermic endothermic
Slide 34 / 118 18 When CaCl 2 dissolves in water the temperature of water drops. This reaction is. A B endothermic exothermic
Slide 35 / 118 19 Water droplets evaporating from the skin surface will make you feel cold. This process is. A B C D E exothermic for water endothermic for skin exothermic for skin endothermic for water A and C F C and D
Slide 36 / 118 State Functions** Return to Table of Contents
Slide 37 / 118 ## State Functions The internal energy of a system is independent of the path by which the system achieved that state. Below, 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 100C 50g 25C 50g 0C Cooling Heating
Slide 38 / 118 ## 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. D E depends only on E initial and E final
Slide 39 / 118 ## State Functions These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate. They move between energy states while performing some task.
Slide 40 / 118 ## State Functions Understanding thermodynamics enables us to harness energy flow for useful purposes. For instance, 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. 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. A B Charged battery A Heat Heat Work B #E Energy lost by battery Discharged battery
## Slide 41 / 118 Work Done by a Gas When a gas expands it does work on its surroundings: Initial state Final state W = Fd. P= F/A P= F/A In this case: F = PA and d = h. V h W = Fd W = (PA) h h W = P V A = cross sectional area
Slide 42 / 118 ## Work Done By a Gas 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/A P= F/A (-) work is done by system W = - P V V h h A = cross sectional area
Slide 43 / 118 ## Work Done By a Gas 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/A h Final state P= F/A V h A = cross sectional area
Slide 44 / 118 ## Work Done by a Gas We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston. HCl Solution HCl Solution + Zinc
Slide 45 / 118 Enthalpy Return to Table of Contents
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. H = E + W
Slide 47 / 118 Enthalpy At 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)
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) At 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. Combustion of 16 grams of CH 4 H = -891 kj Combustion of 32 grams of CH 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. CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(g) H = -891 kj 2H 2 O(g) + CO 2 (g) --> CH 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. CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(l) H = -891 kj H = -979 kj
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?
Slide 52 / 118 Endothermic and Exothermic Processes A process is endothermic when H is positive. Surroundings System A process is exothermic when H is negative. Surroundings System +q Heat Heat -q H>0 Endothermic H<0 Exothermic
Slide 53 / 118 21 The reaction A + B --> C is endothermic. The DH for this reaction is +50 J. What is the DH for the reaction C --> A + B? A B C D Cannot be determined. 0.02 J - 50 J 100 J
Slide 54 / 118 22 The reaction A + B --> C is exothermic. The H for this reaction is -150 J. What is the H for the reaction C --> A + B? A +150 B zero C -150 D this reaction will not happen
Slide 55 / 118 23 Dissolving NaOH in water will increase the temperature of the solution. This reaction is A B C D exothermic endothermic adiabatic isothermal
Slide 56 / 118 24 NH 3 NO 3 + H 2 O --> NH 4 + + NO 3 - + OH - H = +25.69 kj/mol. This reaction is exothermic. Yes No
Slide 57 / 118 Measuring Enthalpy Changes: Calorimetry Return to Table of Contents
Slide 58 / 118 Measuring Enthalpy Changes: Calorimetry 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.
Slide 59 / 118 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. The amount of energy required to raise the temperature of one gram of a substance by 1 K (1 C) is its specific heat (c).
Slide 60 / 118 25 Heat capacity is an example of an A Intensive property B Extensive property
Slide 61 / 118 26 Specific heat is an example of an A Intensive property B Extensive property
Slide 62 / 118 Heat Capacity and Specific Heat Specific heat = heat transferred mass x temperature change 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 C to 45 C? q = mc T
Slide 64 / 118 28 An 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 C to C. q = mc T
Slide 65 / 118 29 A 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 C to 80 C. What is the mass of the sample? q = mc T
Slide 66 / 118 Constant Pressure Calorimetry 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
Slide 67 / 118 Constant Pressure Calorimetry stirrer insulated cover thermometer Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: D H = q = mc T (at constant pressure) styrofoam cup sparknotes.com
Slide 68 / 118 Constant Pressure Calorimetry Example: A student wishes to determine the enthalpy change when ammonium chloride (NH 4 Cl) 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 Celsius. The student observes the temperature to decrease to 13.2 Celsius. 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 C 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 4 Cl = +6334 J
Slide 69 / 118 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. Water Thermometer Ignition wires Oxygen atmosphere Sample
Slide 70 / 118 Bomb Calorimetry (Constant Volume) 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. 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 Celsius to start and the heat capacity of the calorimeter was 34 J/C, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 Celsius? 4Fe(s) + 3O 2 (g) --> 2Fe 2 O 3 (s)
Slide 72 / 118 31 A 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 C and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10 C. After the metal is added, the temperature rises to 11.3 C. Identify the metal. A Metal Cu 0.386 Specific Heat (J/gC) B Au 0.126 C Al 0.900
Slide 73 / 118 Energy Changes Associated with Changes of State Return to Table of Contents
Slide 74 / 118 Energy Changes Associated with Changes of State Chemical 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.
Slide 75 / 118 Energy Changes Associated with Changes of State Fill in the blanks Endothermic processes Exothermic processes Energy is taken into the system from the surroundings ( H > 0) Energy is released from the system to surroundings ( H < 0) Answer boiling or evaporating a liquid condensing a gas deposition of a gas freezing a liquid melting a solid sublimation of a solid
Slide 76 / 118 Phase Changes Gas Vaporization Condensation Energy of system Sublimation Liquid Melting Freezing Solid Deposition
Slide 77 / 118 32 Which of the following is/are exothermic? I. boiling II. melting III. freezing A B C D I only I and II only III only I, II and III
Slide 78 / 118 Energy Changes Associated with Changes of State The heat of fusion ( H fus ) is the energy required to change a solid at its melting point to a liquid. Heat of fusion(h 2 O) = 6.0 kj/mol The heat of vaporization ( H vap ) is defined as the energy required to change a liquid at its boiling point to a gas. Heat of vaporization(h 2 O) = 41.0 kj/mol Class 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.
Heat of fusion and vaporization kj/mol 80 40 10 24 5 Heat of fusion Heat of vaporization 29 41 Slide 79 / 118 Energy Changes Associated with Changes of State Butane 7 Diethyl 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 80 / 118 33 The heat of vaporization for butane is 24 kj/mol. How much energy is required to vaporize 2 mol of butane? A 2kJ B C D 12 kj 22 kj 48 kj
Slide 81 / 118 34 The heat of vaporization for butane is 24 kj/ mol. How much energy is required to vaporize 0.33 mol of butane? A B C D 8kJ 12 kj 16 kj 72 kj
Slide 82 / 118 35 How much energy is required to melt 0.5 mol of water? A B C D 2kJ 3 kj 6 kj 12 kj Heat of fusion for H 2 O (s) Heat of vaporization for H 2 O (l) 6 kj/mol 41 kj/mol Answer
Slide 83 / 118 36 How much energy is released when 3.0 mol of water freezes? A B C D 2kJ 3 kj 18 kj 123 kj Heat of fusion for H 2 O (s) Heat of vaporization for H 2 O (l) 6 kj/mol 41 kj/mol Answer
Slide 84 / 118 37 How much energy is needed to melt 10.0 mol solid Hg? A B C D 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 Answer
Slide 85 / 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 Answer
Slide 86 / 118 Energy Changes Associated with Changes of State 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 C) 125 100 75 50 25 0-25 B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) A Heat added (each division corresponds to 4 kj)
Slide 87 / 118 Energy Changes Associated with Changes of State From A to B, ice is heating up from -25 o C to 0 o C. Since Q = mc T T = Q (mc) So the slope is 1 (mc) Temperature ( 0 C) 125 100 75 50 25 0-25 B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) A Heat added (each division corresponds to 4 kj)
Slide 88 / 118 Energy Changes Associated with Changes of State From B to C, 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 C) 125 100 75 50 25 0-25 B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) A Heat added (each division corresponds to 4 kj)
From C to D, liquid water is heating up from 0 C to 100 C. Once again, the slope is 1/(mc). But "c" is different for all the phases of a substance, so the slope is different for solid, liquid and gaseous H 2 O. Slide 89 / 118 Energy Changes Associated with Changes of State Temperature ( 0 C) 125 100 75 50 25 0-25 A B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) Heat added (each division corresponds to 4 kj)
Slide 90 / 118 Energy Changes Associated with Changes of State From D 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 C) 125 100 75 50 25 0-25 B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liqui d water (melting) A Heat added (each division corresponds to 4 kj)
Slide 91 / 118 Energy Changes Associated with Changes of State From E to F water vapor, steam, is heating up from 100 C to 125 C. Once again, the slope is 1/(mc). But "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 C) 125 100 75 50 25 0-25 A B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) Heat added (each division corresponds to 4 kj)
Recall that slope = Slide 92 / 118 Energy Changes Associated with Changes of State 1 mc where m = mass and C = specific heat of the substance This graph shows heat transfer versus change in temperature for 1 gram of 4 different substances. (DT) Temperature (C) A B C D (q) Energy Added (Joules)
Slide 93 / 118 39 Which substance has the lowest specific heat? A B C D A B C D (DT) Temperature (C) A B C D (q) Energy Added (Joules)
Slide 94 / 118 40 Which substance requires the highest amount of heat added to raise the temperature? A B C D A B C D (DT) Temperature (C) A B C D (q) Energy Added (Joules)
Slide 95 / 118 41 Which segment(s) contain solid sodium chloride? A AB only C AB, BC and CD B AB and BC D BC, CD and DE
Slide 96 / 118 42 Which segment(s) contain liquid sodium chloride? A AB only C AB, BC and CD B AB and BC D BC, CD and DE
Slide 97 / 118 43 What is the melting point (in o C) of sodium chloride?
Slide 98 / 118 44 What is the freezing point (in o C) of sodium chloride?
Slide 99 / 118 45 Which is greater: the specific heat of solid NaCl, or the specific heat of molten (liquid) NaCl? A B solid liquid C D Cannot be determined. They are equal.
Slide 100 / 118 Features of a Heating Curve Segments AB, CD, EF slope = nonzero T increasing KE increasing PE constant apply q = mc T Segments BC & DE slope = 0 T = 0 KE constant PE increasing apply H fus or H vap Temperature ( 0 C) 125 100 75 50 25 0-25 A B Ice C D Water vapor E Liquid water and vapor (vaporization) Liquid water F Ice and liquid water (melting) Heat added (each division corresponds to 4 kj)
Slide 101 / 118 Energy Changes Associated with Changes of State Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas. Additionally, 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 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
Slide 103 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. It is a good idea to sketch out the segments first on a graph.
Slide 104 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. Segment 1 Warming the ice from -15.0 C up to the substance's MP which is 0 C. Temperature (C) 20 10 0-4 6 12 24-14 Time (s)
Slide 105 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. Segment 2 Melting the ice at 0 C. Temperature (C) 20 10 0-4 6 12 24-14 Time (s)
Slide 106 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. Segment 3 Warming the liquid from 0 C to 25 C. 20 Temperature (C) 10 0-4 6 12 24-14 Time (s)
Slide 107 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. We are now ready to apply either the calorimetry equation or H fus or H vap. Segment 1 Warming the ice from -15.0 C up to the substance's MP which is 0 C. q 1 = mc T q 1 = (10.0g) (2.09 J/g- o C) (15.0 o C) q 1 = 313.5 J q 1 = 0.314 kj or 313.5J
Slide 108 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. Segment 2 Melting the ice at 0 C. 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
Slide 109 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. Segment 3 Warming the water from 0 C to 25.0 C. q 3 = mc T q 3 = (10.0g) (4.18 J/g- o C) (25.0 o C) q 3 = 1045 J q 3 = 1.05 kj
Slide 110 / 118 Calculating Energy Changes from a Heating Curve Sample Problem: Calculate the enthalpy change (in kj) for converting 10.0 g of ice at -15.0 C to water at 25.0 C. 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
Slide 111 / 118 46 Calculate the enthalpy change in J of converting 75 g of ice at -11 to liquid water at 22. A 98,654 B 33,371 J C 8,621 D 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 Answer E 35,096 J Heat of vaporization ( H vap ) of water at 100 40.7 kj/mol or 2600 J/g
Slide 112 / 118 47 Calculate the enthalpy change of converting 25 g of water vapor at 110 to liquid water at 50. A -69,765 Specific heat of ice 2.09 J/g- B 69,765 C -59,315 D -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 Answer E -13,935 Heat of vaporization ( H vap ) of water at 100 40.7 kj/mol or 2600 J/g
Slide 113 / 118 48 Calculate the enthalpy change in kj of converting 31.8 g of ice at 0 to water vapor at 140. A 109 kj Specific heat of ice 2.09 J/g- B 96 kj C 104 kj Specific heat of water Specific heat of steam 4.18 J/g- 1.84 J/g- Answer D - 88 kj Heat of fusion ( H fus ) of water at 0 6.01 kj/mol or 330 J/g 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 Contents
Slide 115 / 118 Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: D H = H products H reactants CH 4 (g) + 2 O 2 (g) Enthalpy H 1 = -890 kj H 2 = 890 kj CO 2 (g) + 2H 2 O(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 exothermic reaction and the energy released is equal to the heat of reaction. 2H 2 (g) + O 2 (g) Enthalpy H<0 exothermic 2H 2 O(g)
Slide 117 / 118 Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of HBr is formed in this reaction? H 2 (g) + Br 2 (g) --> 2HBr(g) H = -72 kj X -72kJ 4.0 mol HBr(g) = 2.0 mol HBr(g) X = -72kJ (4/2) X = -144kJ 144kJ of energy are released
Slide 118 / 118 Enthalpy of Reaction Example Problem #1 How much energy is released when 4.0 mol of HBr is formed in this reaction? H 2 (g) + Br 2 (g) --> 2HBr(g) H = -72 kj 4.0 mol HBr x -72 kj 2 mol HBr = -144kJ or 144 kj released