Based on Prentice Hall Chemistry, Chapter 17 HEAT HYPERINDEX BUTTON Jump to a place! Daniel R. Barnes Init 3/12/2014 Based on Prentice Hall Chemistry, Chapter 17 WARNING: This presentation includes images and other intellectual property taken from the Internet without the permission of the owners. It is meant to be viewed only by Mr. Barnes’ current chemistry students on a non-profit basis. Do not download, copy, post, or otherwise distribute this presentation. Do not store this presentation anywhere, including on any hard drive, flash drive, or other storage device. Do not upload or store any copy of it on the cloud or any other remote storage facility.

17.1 The Flow of Energy -- Heat and Work

WORD OF ADVICE You will not have time in class to copy down every single word in this presentation. * Use abbreviations & symbols so you can write quickly. * Write only the most important ideas, but leave lots of space in your notes so you can write down other stuff you remember when you study your notes at home tonight. * Click through this ppt online on hhscougars.org – the media center has computers even if you don’t, and it’s usually open late.

Relative # of molecules 221oC Look at Figure 13.3 on page 388. 100oC water 21oC water 50oC water 10oC water 1oC water Relative # of molecules Molecular speed

Relative # of molecules 221oC Look at Figure 13.3 on page 388. 100oC water 21oC water 50oC water 10oC water 1oC water Relative # of molecules Molecular speed

Relative # of molecules 221oC Look at Figure 13.3 on page 388. 21oC water 21oC water, larger amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

Relative # of molecules 221oC Look at Figure 13.3 on page 388. 21oC water >21oC water, larger amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

Relative # of molecules 221oC Look at Figure 13.3 on page 388. 21oC water <21oC water, smaller amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

SWBAT . . . . . . explain the difference between temperature and heat. Temperature = average molecular kinetic energy T = KEave Kinetic energy = the energy of motion KE = ½ mv2 m = mass v = speed (In physics, you learn that velocity is speed in a certain direction.) The numerical value of temperature is proportional to average molecular kinetic energy when expressed in Kelvins, but not when expressed in oC or oF. If all you care about is how hot something is, use any units you want. If the energy of each particle matters, use only Kelvins.

SWBAT . . . . . . explain the difference between temperature and heat. T = KEave KE = ½ mv2

“Thermal equilibrium” SWBAT . . . . . . explain the difference between temperature and heat. Temperature = average molecular kinetic energy Heat = q = energy transferred between two systems as a result of a temperature difference. Heat flows because of differences in temperature. “Thermal equilibrium” Heat flows from the hotter system to the colder system.

T = KEave H = KEtotal SWBAT . . . . . . explain the difference between temperature and heat. T = KEave . WARNING: OVERSIMPLIFICATION ALERT!. H = KEtotal [$/person and total $ for group analogy]

SWBAT . . . . . . explain the difference between endothermic and exothermic Endothermic = absorbing heat  q = positive Exothermic = releasing heat  q = negative Do some Q-squats! H = enthalpy = heat energy content q = DH = change in heat energy content = energy absorbed q & DH, being energy, are measured in calories or joules

Cha-CHING! The bank has $3.7 million in deposits. That’s like . . . h KE q T ? I have $20,000 in my little bank account. That’s like . . . h KE q T ? The average bank account has $13,237 in it. That’s like . . . h KE q T ? On Sunday, all the customers combined made a total of $40,821 in deposits That’s like . . . h KE q T ?

SWBAT . . . DT = Tf - Ti (Solving for Cp) Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems DT = Tf - Ti DT = change in temperature Tf = final temperature = temperature at the end of the story Ti = initial temperature = temperature at the beginning of the story Do #1 from Mr. Barnes’ Heat Math Worksheet (First Exposure) Do #2 from Mr. Barnes’ Heat Math Worksheet (First Exposure)

SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems C = specific heat capacity = “specific heat” C = how hard it is to change the temperature of a particular material Look at Table 17.1 on page 508 in your textbook Which material has the highest C? Which material has the lowest in the table? Notice anything about the trend in the metals in the table? The bigger the metal’s atomic mass is . . . the easier it is to DT it.

C = how hard it is to DT a material Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems C = how hard it is to DT a material C = how much energy a material abs/rel when it DT’s

SWBAT . . . q = m DT Cp (Solving for Cp) . . . solve q = m DT Cp problems q = m DT Cp (Cp = C at constant pressure = when the material changes temperature in an open container – don’t stress about the “p”) q = m DT Cp q = energy absorbed m = mass DT = change in temperature = Tf - Ti * The more massive something is, the more energy it takes to heat it up. * Bigger temperature increases require more energy to happen. * The higher the specific heat of a material, the more energy it takes to heat it up.

SWBAT . . . q = m DT Cp (Solving for Cp) . . . solve q = m DT Cp problems q = m DT Cp This is the form of the formula on your CST reference sheet. As written, it’s really set up to solve for . . . q What if you want to solve for Cp? q = m DT Cp q Cp = m DT m DT m DT Remember that DT = Tf – Ti Other than that you can just plug in the #’s & solve.

SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems Let’s try one as a class! This is #7 from Mr. Barnes’ “Heat Math Worksheet (First Exposure)”. 7. What is the specific heat of an oily liquid if it takes 2000 joules to heat up 7 grams of the liquid from 20oC to 720oC? Cp = ? J/(g oC) q = 2000 J m = 7 g Ti = 20oC Tf = 720oC DT = Tf - Ti = 720oC – 20oC = 700oC q = m DT Cp q q = m DT Cp Cp = m DT m DT m DT 2000 J 2000 J/goC Cp = = = 0.408 J/goC = Cp (7 g)(700oC) 4900

C = ? Q = ? T = ? m = ?

SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems Take about two minutes to get started on #8 . . . TPS . . . SKIP TO ANSWER

SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems C = ?

SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems Let’s try one where q is the unknown. This is easier in a way because the formula on the reference sheet is already set up to solve for q. Take about two minutes to get started on #3 . . . TPS . . . 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius?

SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius? Cp = 1 cal/goC q = ? calories m = 2 g Ti = 23oC Tf = 93oC DT = Tf - Ti = 93oC – 23oC = 70oC q = m DT Cp = (2 g)(70oC)(1 cal/goC) = 140 cal = q SKIP TO ANSWER

SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius? Cp = 1 cal/goC q = ? calories m = 2 g Ti = 23oC Tf = 93oC DTf = Tf - Ti = 93oC – 23oC = 70oC q = m DT Cp = (2 g)(70oC)(1 cal/goC) = 140 cal = q

SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems If you’re an honors section, do #4 for homework. If you’re a normal section, do it now. You have two minutes.

SWBAT . . . YOU MUST BECOME A PROBLEM-SOLVER! (Solving for T) Q = ? T = ? m = ? SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems And now, to solve for temperature . . . Unfortunately, there isn’t one on the worksheet, so we’ll have to pull one out of my question bank . . . Let’s make it #17 on the ws. YOU MUST BECOME A PROBLEM-SOLVER!

SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems 17. A corn cob is heated up in boiling water to a temperature of 100 oC. The corn cob has a mass of 400 g and absorbed 16,000 cal of heat from the boiling water. What was the original temperature of the corn cob? Assume that the specific heat of the corn cob is 1 cal/goC. Tf = 100 oC m = 400 g q = 16,000 cal Ti = ? oC Cp = 1 cal/goC DT = Tf – Ti = ? Ti = Tf – DT q = m DT Cp q = m DT Cp 16,000 cal DT = = 40 oC = DT m Cp m Cp (400 g) (1 cal/goC) SKIP TO ANSWER Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems 17. A corn cob is heated up in boiling water to a temperature of 100 oC. The corn cob has a mass of 400 g and absorbed 16,000 cal of heat from the boiling water. What was the original temperature of the corn cob? Assume that the specific heat of the corn cob is 1 cal/goC. Tf = 100 oC m = 400 g q = 16,000 cal Ti = ? oC Cp = 1 cal/goC DT = Tf – Ti = ? Ti = Tf – DT q = m DT Cp q = m DT Cp 16,000 cal DT = = 40 oC = DT m Cp m Cp (400 g) (1 cal/goC) Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

SWBAT . . . (Solving for m) . . . solve q = m DT Cp problems And now, let’s do a q = m DT Cp problem where m = ? . . . 6. How much unobtainium can be heated from 40 oC to 440 oC if only 40,000 joules of energy are available to heat it up? Assume that the specific heat of unobtanium is 0.4 J/goC. m = ? g Ti = 40 oC Tf = 440 oC q = 40,000 J Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC = DT q = m DT Cp q = m DT Cp 40,000 J m = DT Cp DT Cp (400 oC) (0.4 J/goC) 40,000 g 40,000 g m = = = 250 g = m (400)(0.4) 160 SKIP TO ANSWER

SWBAT . . . (Solving for m) . . . solve q = m DT Cp problems And now, let’s do a q = m DT Cp problem where m = ? . . . 6. How much unobtainium can be heated from 40 oC to 440 oC if only 40,000 joules of energy are available to heat it up? Assume that the specific heat of unobtanium is 0.4 J/goC. m = ? g Ti = 40 oC Tf = 440 oC q = 40,000 J Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC = DT q = m DT Cp q = m DT Cp 40,000 J m = DT Cp DT Cp (400 oC) (0.4 J/goC) 40,000 g 40,000 g m = = = 250 g = m (400)(0.4) 160

Q = m DT Cp DT = m Cp Q m = DT Cp Q Cp = m DT Q

Q = m DT Cp DT = m Cp Q DT = Tf - Ti Tf - Ti = m Cp Q Tf = m Cp Q + Ti Tf = DT + Ti Ti = Tf - m Cp Q Ti = Tf - DT

SWBAT . . . . . . solve q = m DT Cp problems Try to do problems 1, 2, 4, & 5 on the Heat Math Worksheet (First Exposure).

This might or might not count toward your grade. Barnes? Okay, it’s time for the 17.1 Post-Quiz This might or might not count toward your grade. Barnes?

Measuring and Expressing Enthalpy Changes 17.2 Measuring and Expressing Enthalpy Changes

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Calorimetry = “the precise measurement of the heat flow into or out of a system for chemical and physical processes” (Prentice Hall Chemistry, 2005) System = “a part of the universe upon which you focus your attention” (Prentice Hall Chemistry, 2005) surroundings A freshly-boiled egg is a single object, but it’s made of parts (yolk, white, shell) system The hot egg is surrounded by cool water.

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters You could use this apparatus to measure the energy given off by a boiling-hot (100 oC) egg as it cools to room temperature in one liter of cool water. Tf = 28 oC You would have to measure the temperature of the water before the egg was dropped into it . . . Ti = 20 oC surroundings As the egg cools off, it gives energy to the water, causing the water to heat up. system Eventually, the water and the egg reach the same temperature. Ti = 100 oC Tf = 28 oC

SWBAT . . . CONSERVATION of ENERGY . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated. Tf = 28 oC CONSERVATION of ENERGY Ti = 20 oC surroundings system styrofoam

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters This is a close-up of a tiny bubble, but it’s not a normal bubble. polystyrene Instead of the skin being made of water or soapy water, the skin of this bubble is made of polystyrene plastic.

The inside of the bubble is filled with air. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The inside of the bubble is filled with air. polystyrene air

A bunch of these bubbles stuck together makes a “foam”. SWBAT . . . air . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters air air air A bunch of these bubbles stuck together makes a “foam”. air air The air in each bubble is a “dead air space” because the air in one bubble can’t mix with air in other bubbles. air air air air air air air

SWBAT . . . air . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters air air air The air in each bubble is a “dead air space” because the air in one bubble can’t mix with air in other bubbles. air air air air air air air air air

SWBAT . . . air . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters air air Dead air spaces are almost as good as vacuums at preventing heat flow. Each bubble in a “closed cell” foam is a tiny dead air space. air air air air Dead air spaces are excellent thermal insulators. Each bubble in a “closed cell” foam is a tiny dead air space. air air air air air air

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters It’s not the polystyrene plastic in the styrofoam that makes styrofoam such a good insulator . . . . . . it’s the air trapped in the bubbles that makes it such a good insulator. Dead air spaces, trapped air masses that don’t mix with other air, are excellent insulators.

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q 100 oC 50 oC 100 oC 99 oC Q metal styrofoam 50 oC 0 oC 0 oC 1 oC

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Metals are excellent thermal conductors Styrofoam is an excellent thermal insulator Q 50 oC 99 oC Q metal styrofoam 1 oC 50 oC

SWBAT . . . CONSERVATION of ENERGY . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated. The egg is exothermic and the water is endothermic. Tf = 28 oC CONSERVATION of ENERGY q egg = - q water Ti = 20 oC q water = m DT Cp surroundings q = (1000 g)(8 oC)(1 cal/goC) system q water = 8000 cal q egg = - 8000 cal styrofoam

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The egg may have given off more heat than that, but some heat may have leaked into the air. This calorimeter would be more accurate if it had a styrofoam lid. The lid provides even more insulation. A styrofoam lid on a styrofoam cup is not typically an airtight seal. styrofoam surroundings Therefore, the air pressure inside the calorimeter is always equal to the air pressure outside. system styrofoam

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters These are pretty leaky. They leak gas, and they leak heat. This apparatus is a “CONSTANT PRESSURE CALORIMETER” At least they’re cheap. styrofoam surroundings Therefore, the air pressure inside the calorimeter is always equal to the air pressure outside. system styrofoam

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q: WHY is this a “constant pressure” calorimeter? This apparatus is a “CONSTANT PRESSURE CALORIMETER” Hmm? (TPS, 30 sec!) A: If it gets hot inside, gas molecules move faster, colliding more often and more violently, causing an increase in pressure. 101.3 kPa 101.3 kPa BUT the gas expands, leaking out. Its pressure drops as it expands, til its pressure is equal to the atmosphere again. Leakiness  pressure equilibrium

SWBAT . . . See page 512 for better a picture! . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters A more dangerous, expensive, and accurate calorimeter is the “bomb calorimeter”. The inner container, the airtight “bomb”, has to be very strong. Why? H2O CO2 O2 food When the green stuff, the fuel, reacts with oxygen, the exothermic reaction raises the pressure  threat of explosion.

SWBAT . . . n = moles of gas = # of gas molecules . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q: The bomb calorimeter is a “constant V” device. Is it also a “constant n” device? Why/why not? Hmm? (TPS, 30 sec!) H2O CO2 A: The “bomb” is airtight, so gas molecules can’t get in or out . . . n = k, right? A: BUT food molecules are liquid or solid, and they turn into gas when they burn, so that might increase the moles of gas = n. So, NO.

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q: What is the “system” in each calorimeter below? Hmmm? TPS 15 seconds! A: The “system” in the P = k calorimeter is the egg. A: The “system” in the V = k calorimeter is the “bomb”. system system

SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q: What do you call the water in both calorimeters? Hmmm? TPS 15 seconds! A: the “surroundings” system system surroundings surroundings

SWBAT . . . . . . apply the law of conservation of energy to calorimetry Q: DHsystem is positive, which way did thermal energy flow? Hmmm? TPS 30 seconds! A: From the surroundings (the water) into the system. system system surroundings surroundings

SWBAT . . . . . . apply the law of conservation of energy to calorimetry Q: If the egg absorbs 2000 J, what is DHwater? A: DHwater = -2000 J Hmmm? TPS 30 seconds! system surroundings

SWBAT . . . . . . apply the law of conservation of energy to calorimetry Q: If there are 1000 grams of water in the outer container, and it goes from 20 oC to 50 oC when a small piece of candy is burned inside the bomb, what is the Calorie content of the piece of candy? Hmmm? TPS 2 minutes! m = 1000 g Ti = 20 oC Tf = 50 oC Q = ? Calories Q = m DT Cp DT = Tf - Ti = 50 oC – 20 oC = 30 oC Cp = 1 cal/goC Q = (1000 g)(30 oC)(1 cal/goC) Q = 30,000 cal x 1 Calorie / 1000 calories = 30 Calories

SWBAT . . . . . . apply the law of conservation of energy to calorimetry Q: Which of the following is the change in enthalpy of the egg? Cwater - Qegg Cegg Qwater DHwater Qegg - Qwater - DHwater DTegg DHegg system DTwater -DHegg -DTegg surroundings mwater -DTwater megg

Hot Nail Lab Barnes, hand out the instruction sheet. If you haven’t done it already, it’s time for the Hot Nail Lab Barnes, hand out the instruction sheet. Kis, dress for lab tomorrow.

SWBAT . . . . . . decode thermochemical equations ENDO / EXO? EXOTHERMIC! 2H2(g) + O2(g)  2H2O(g) DH = -483.6 kJ 2H2(g) + O2(g)  2H2O(g) + 483.6 kJ

SWBAT . . . . . . decode thermochemical equations ENDO / EXO? ENDOTHERMIC! 2H2O(l)  2H2(g) + O2(g) DH = 571.6 kJ 2H2O(l) + 571.6 kJ  2H2(g) + O2(g)

SWBAT . . . . . . solve for DH using thermochemical equations when given g or mol of reactant or product 4Fe(s) + 3O2(g)  2Fe2O3(s) DH = -1644.2 kJ This is how much energy is released when four moles of iron is oxidized. What is DH if . . . 12 moles of iron oxidizes? -1644.2 kJ 12 mol Fe x 4 mol Fe DH = - 4932.6 kJ 12 moles is three times as much as 4, so 12 moles should release three times as much as 1644.2 kJ. 12/4 = 3

SWBAT . . . . . . solve for DH using thermochemical equations when given g or mol of reactant or product 4Fe(s) + 3O2(g)  2Fe2O3(s) DH = -1644.2 kJ What is DH if . . . 8 moles of iron oxide forms? TPS – 1 min – GO! . . . -1644.2 kJ 8 mol Fe2O3 x 2 mol Fe2O3 DH = - 6576.8 kJ

SWBAT . . . . . . solve for DH using thermochemical equations when given g or mol of reactant or product 4Fe(s) + 3O2(g)  2Fe2O3(s) DH = -1644.2 kJ What is DH if . . . 96 grams of oxygen oxidize some iron? 1 mol -1644.2 kJ 96 g x x 32 g 3 mol O2 DH = - 1644.2 kJ (The molar mass of O2 is 32 g/mol, and molar mass is needed here to convert grams into moles.)

SWBAT . . . . . . solve for DH using thermochemical equations when given g or mol of reactant or product 4Fe(s) + 3O2(g)  2Fe2O3(s) DH = -1644.2 kJ What is DH if . . . 112 grams of iron oxidizes? TPS – 1 min – GO! . . . 1 mol -1644.2 kJ 112 g x x 56 g 4 mol Fe DH = - 822.1 kJ

SWBAT . . . . . . calculate DH when given heat of combustion and g or mol of fuel What is DH when three moles of methane burns? Look at Table 17.2 on page 517. What is the heat of combustion of methane? -890 kJ/mol -890 kJ/mol x 3 mol = -2670 kJ The negative sign means that . . . . . . the combustion of methane is exothermic. Notice also that since no chemical equation is given, DH is just a certain number of kJ per mole of fuel. You don’t have to divide by a coefficient or anything.

SWBAT . . . . . . calculate DH when given heat of combustion and g or mol of fuel What is DH when five moles of propane burns? TPS – 30 seconds – GO! DH for propane = -2220 kJ/mol -2220 kJ/mol x 5 mol = -11,100 kJ

SWBAT . . . . . . calculate DH when given heat of combustion and g or mol of fuel What is DH when 240 grams of carbon burns? DHcomb = -394 kJ/mol -394 kJ/mol x 240 g x 1 mol/ 12 g = - 7880 kJ Notice here that the amount of carbon was given in grams, but heat of combustion was given in kJ per mole. Therefore, it was necessary to use molar mass to turn grams into moles.

SWBAT . . . . . . calculate DH when given heat of combustion and g or mol of fuel What is DH when a 17 gram marshmallow burns? Assume the marshmallow is made entirely of sucrose. TPS – 1 min – GO! DHcomb = -5645 kJ/mol . . . for sucrose -5645 kJ/mol x 17 g x 1 mol/ 342 g C12H22O11: DH = - 280 kJ C = 12 x 12 = 144 H = 22 x 1 = 22 O = 11 x 16 = 176 342 g/mol

Heat in Changes of State 17.3 Heat in Changes of State

SWBAT . . . . . . interpret the “heating curve” for water A “heating curve” is a graph that tells a story. Steam getting hotter Water boiling 100 oC T / oC Water getting hotter Ice melting 0 oC Ice getting hotter Q / J

SWBAT . . . . . . interpret the “heating curve” for water What states of matter are involved in each segment? gas liquid  gas 100 oC T / oC liquid S  L 0 oC solid Q / J

S  L liquid liquid  gas gas solid

S  L liquid liquid  gas gas solid

S  L liquid liquid  gas gas solid

S  L liquid liquid  gas gas solid

S  L liquid liquid  gas gas solid

STATES of MATTER gas liquid solid more entropy condensation vaporization liquid freezing “fusion” = melting solid more order

fuses

SWBAT . . . . . . interpret the “heating curve” for water There are two critical temperatures in a heating curve. boiling point 100 oC 100 oC T / oC 0 oC 0 oC melting point Q / J

SWBAT . . . . . . interpret the “heating curve” for water The horizontal parts of the graph represent phase changes. 100 oC During a phase change on this graph, the temperature remains constant. T / oC 0 oC Q / J

SWBAT . . . . . . interpret the “heating curve” for water Q: Which stage of this story involved the most energy? How do you know? V IV 100 oC T / oC III TPS: 30 sec – GO! II 0 oC I Q / J

SWBAT . . . . . . interpret the “heating curve” for water Q: Which stage of this story involved the most energy? How do you know? V IV 100 oC A: Segment IV (boiling). It has the largest Dx. T / oC III II 0 oC I Q / J

SWBAT . . . . . . solve math problems involving phase changes. QA: How much energy is required to melt a 10 gram ball of wax? The latent heat of fusion of the wax is 150 J/g.

SWBAT . . . . . . solve math problems involving phase changes. QB: What is Q when 4 grams of water freezes at its freezing point? The latent heat of fusion of water is 80 cal/g. Freezing is exothermic. Freezing is the opposite of melting, so DHfreeze = -DHfus

SWBAT . . . . . . solve math problems involving phase changes. QC: How much energy must be removed from 6400 g of oxygen gas in order to make it liquefy at its condensation point? The molar heat of vaporization of O2(g) is 7 kJ/mol.

SWBAT . . . . . . solve math problems involving phase changes. Now do question # 14 on Mr. Barnes’ Heat Math Worksheet

SWBAT . . . . . . solve math problems involving phase changes. COMPOUND QUESTION: How much energy does 100 grams of steam give off when it condenses onto someone’s skin? How much energy does 1 grams of just-condensed water give off as it cools from 100oC to 37oC (human body temperature)? Use calories instead of joules to keep the math easy. Qcond = m DHcond = (1 g)(-540 cal/g) = -540 cal Q100 to 37 Celsius = m DT Cp = (1g )(37 oC – 100 oC)(1 cal/g/oC) = (1g )(– 63 oC)(1 cal/g/oC) = – 63 cal The condensation of 100 oC vapor gives off 8.57 times as much energy as it gives off as it cools down to body temperature after condensing. Steam is much more dangerous than boiling water.

SWBAT . . . . . . solve math problems involving solution formation. NH4NO3(s)  NH4+(aq) + NO3-(aq) DHsoln = 25.7 kJ/mol Translation: Ammonium nitrate dissolves in water to form ammonium cations and nitrate anions. What is DH if 20 moles of ammonium nitrate dissolves in water? DH = (20 mol)(25.7 kJ/mol) = 514 kJ Ammonium nitrate dissolving in water is endothermic. When you dissolve ammonium nitrate in water, the water gets colder.

SWBAT . . . . . . solve math problems involving solution formation. CaCl2(s)  Ca2+(aq) + 2Cl-(aq) DHsoln = -82.8 kJ/mol Calcium chloride dissolving in water is exothermic. You can tell because DHsoln is a negative number. If you sprinkle CaCl2(s) on an icy road, what will happen and why? The ice will probably melt because (1) calcium chloride dissolving in water releases heat, and (2) dissolving anything in water lowers its freezing point, making it harder to be solid.

SWBAT . . . . . . solve math problems involving solution formation. I think these trucks in NJ spread rock salt (NaCl, not CaCl2). It damages concrete more . . . so they say . . . CaCl2 is regarded as a better de-icer than NaCl.

SWBAT . . . . . . solve math problems involving solution formation. You can probably find some interesting icy road accident videos on youtube. It will give you some appreciation of why roads need to be de-iced in places that get freezing cold in the winter.

Calculating Heats of Reaction 17.4 Calculating Heats of Reaction

SWBAT . . . . . . manipulate and combine algebraic equations. In algebra I, you learned that you can combine the like terms from two equations to make a third equation. 2x + 6y = 16 If you do this right, this allows you to solve for one of the variables by eliminating the other one. y = 2x + 12 + 2x + 7y = 2x + 28 Once you know of the variables, you can plug it in and solve for the other variable. 7y = 28 y = 4 4 = 2x + 12 4 - 12 = 2x + 12 - 12 -8 = 2x x = -4

SWBAT . . . . . . manipulate and combine algebraic equations. Sometimes, you need to flip one of the equations before adding them to get one of the terms to disappear. 9y = 3x - 57 9y = 3x - 57 2y + 22 = 3x 3x = 2y + 22 + If we flip the second equation, and THEN add the two equations, the “x” terms will disappear. 3x + 9y = 3x + 2y - 35 9y = 2y - 35 2(-5) + 22 = 3x 7y = -35 -10 + 22 = 3x y = -5 12 = 3x x = 4

SWBAT . . . . . . manipulate and combine algebraic equations. Other times, you need to multiply an equation by an integer to get what you want. 4y = x + 5 16y = 4x + 20 4x + 7y = 49 + 4x + 7y = 49 Even if we flip one of the equations, nobody disappears after adding the equations, but if we multiply the first equation by four first . . . 4x + 23y = 4x + 69 23y = 69 y = 3 4(3) = x + 5 12 = x + 5 x = 7

SWBAT . . . . . . manipulate and combine algebraic equations. Sometimes, you can use these old algebra tricks with chemical equations to solve for unknowns . . .

SWBAT . . . DEFINITION: . . . use Hess’ law to determine DHrxn Hess’s law of heat summation: If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

SWBAT . . . . . . use Hess’ law to determine DHrxn Hess’s law allows us to figure out DHrxn for a reaction without ever having to make the reaction happen in real life. HLEx1: Imagine that experiments tell you the following: Sn(s) + Cl2(g)  SnCl2(s) DH = -325.1 kJ SnCl2(s) + Cl2(g)  SnCl4(l) DH = -186.2 kJ If you add the two equations above, you get the following: Sn(s) + SnCl2(s) + 2Cl2(g)  SnCl2(s) + SnCl4(l) Just as with normal, algebraic equations, when the same term appears on the left and right, it can be crossed out . . . yielding . . . Sn(s) + 2Cl2(g)  SnCl4(l) DH = ? kJ

SWBAT . . . . . . use Hess’ law to determine DHrxn Now here’s the real magic. Hess’s law says that you can also add the DH’s for the reactions to get the DH for the final reaction. Sn(s) + Cl2(g)  SnCl2(s) DH = -325.1 kJ SnCl2(s) + Cl2(g)  SnCl4(l) DH = -186.2 kJ (-325.11 kJ) + (-186.2 kJ) = -511.3 kJ The sneaky miracle here is that we figured this out without ever having to make tin metal and chlorine gas react to form tin (IV) chloride in real life. Sn(s) + 2Cl2(g)  SnCl4(l) DH = ? kJ DH = -511.3 kJ

SWBAT . . . . . . use Hess’ law to determine DHrxn That was a very simple usage of Hess’ law. We didn’t have to manipulate any equations before adding them. Let’s try a harder problem.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx2: Let’s say experiments have told us the following: Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(cr)  OsO4(g) DH = 56.4 kJ Use Hess’s Law to figure out DH for the following reaction: Os(cr) + 2O2(g)  OsO4(cr) DH = ? kJ (The final equation looks a lot like the first equation, but notice that OsO4 is a gas in the first equation and a crystal in the third.) What would you have to do to figure this one out? Think a moment . . .

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx2: Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(cr)  OsO4(g) DH = 56.4 kJ Os(cr) + 2O2(g)  OsO4(cr) DH = ? kJ OsO4(g) is in both of the initial equations, but doesn’t appear in the final equation, so it needs to be eliminated somewhow. OsO4(g) is on the right on both equations, so they won’t cancel each other out if you add the equations as they are. One filthy little trick you can do is to flip the second equation to put OsO4(g) on the left, and THEN add the two equations together.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx2: Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(cr)  OsO4(g) DH = 56.4 kJ Os(cr) + 2O2(g)  OsO4(cr) DH = ? kJ Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(g)  OsO4(cr) DH = -56.4 kJ Notice how the sign of DH changed on the equation that we flipped. Remember that DHsolid = -DHfus and DHcond = -DHvap Freezing is the opposite of melting and condensation is the opposite of vaporization, so their DH’s have opposite signs.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx2: Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(cr)  OsO4(g) DH = 56.4 kJ Os(cr) + 2O2(g)  OsO4(cr) DH = ? kJ Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(g)  OsO4(cr) DH = -56.4 kJ Os(cr) + 2O2(g) + OsO4(g)  OsO4(g) + OsO4(cr) Now we can add the equations . . . and simpify . . . Now we have the equation we were looking for. Now what?

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx2: Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(cr)  OsO4(g) DH = 56.4 kJ Os(cr) + 2O2(g)  OsO4(cr) DH = ? kJ Os(cr) + 2O2(g)  OsO4(g) DH = -335 kJ OsO4(g)  OsO4(cr) DH = -56.4 kJ Os(cr) + 2O2(g) + OsO4(g)  OsO4(g) + OsO4(cr) Add the DH’s to get the DH for the final equation. DH = -335 kJ + (-56.4 kJ) = -391.4 kJ = DH

SWBAT . . . . . . use Hess’ law to determine DHrxn At this time, please attempt the problems on Mr. Barnes’ “Hess’s Law Worksheet”

[Example needed where multiplication is necessary] SWBAT . . . . . . use Hess’ law to determine DHrxn [Example needed where multiplication is necessary]

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ I II III NOTE: The target equation was given first, so it’s equation I this time. Try not to let that throw you off the horse.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ Notice that there is no CO2(g) in the target equation. This means we’re going to need to . . . . . . cancel out the CO2(g)’s by making sure that . . . . . . they end up on opposite sides of our final equation. But there’s a problem. CO2(g) is on the right in both II & III.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ We’re going to need to . . . . . . flip equation II or III. We need to make sure we pick the right one to flip, though. CH4(g) is on the left in both the target equation and in III, so we can’t flip III. CH2CO(g) is on the right of the target equation but on the left of equation I, so I is the one we have to flip.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ We’ve got another problem, though. There are two methane molecules in the target equation, but only one methane molecule is equation III. What shall we do? We’ll have to multiply everything in equation III by two, including all the coefficients AND the DH value.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ Let’s add the flipped version of equation II to the doubled version of equation III and see if it gives us the target equation (I). 2CO2(g) + H2O(g) + 2CH4(g) + 4O2(g) CH2CO(g) + 2O2(g) +2CO2(g) + 4H2O(g) -II + 2III = Let’s cross out stuff.

SWBAT . . . . . . use Hess’ law to determine DHrxn HLEx3: What is DH for the following equation? 2CH4(g) + 2O2(g)  CH2CO(g) + 3H2O(g) The following equations have known DH values: CH2CO(g) + 2O2(g)  2CO2(g) + H2O(g) DH = -981.1 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802.3 kJ DHI = -DHII + 2DHIII [UNDER CONSTRUCTION!]

DH0rxn = SDHf0(products) – SDHf0(reactants) SWBAT . . . . . . use standard heats of formation to determine DHrxn DT = Tf - Ti This old formula has something in common with this new formula: DH0rxn = SDHf0(products) – SDHf0(reactants) THE BIG IDEA: You can calculate DH for a reaction by adding up the standard heats of formation of all the reactants and subtracting that from the total of the standard heats of formation of all the products.

SWBAT . . . . . . use standard heats of formation to determine DHrxn Let’s try an example.

SWBAT . . . . . . use standard heats of formation to determine DHrxn SHFEx1: Use standard heats of formation to find DHrxn for the thermite reaction: 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) 2(0 kj/mol) -822.1 kj/mol 2(0 kj/mol) -1676.0 kj/mol DH0rxn = SDHf0(products) – SDHf0(reactants) The book puts it a little more concisely: DH0 = DHf0(products) – DHf0(reactants) So, we need to look up DHf0 for all the reactants and products. There’s a table on page 530. During a test the DHf0’s will probably be provided in the question itself.

SWBAT . . . . . . use standard heats of formation to determine DHrxn SHFEx1: Use standard heats of formation to find DHrxn for the thermite reaction: 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) 2(0 kj/mol) -822.1 kj/mol 2(0 kj/mol) -1676.0 kj/mol DH0rxn = SDHf0(products) – SDHf0(reactants) DHf0 for solid aluminum and solid iron is zero. That’s true for any element that is in the state it’s normally in at 25oC and 1 atm. DHf0 for iron oxide and aluminum oxide can be found on page 530 of your textbook, but, on a test, their numbers would probably be provided in the question itself.

SWBAT . . . . . . use standard heats of formation to determine DHrxn SHFEx1: Use standard heats of formation to find DHrxn for the thermite reaction: 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) 2(0 kj/mol) -822.1 kj/mol 2(0 kj/mol) -1676.0 kj/mol DH0rxn = SDHf0(products) – SDHf0(reactants) Notice the two being multiplied by zero in these two terms. Whenever a coefficient appears in a chemical equation, you have to multiply that coefficient times the DHf0 for that substance. In this equation, the only substances with coefficients were elements, so the coefficients get multiplied by zero, which gives zero, which is boring, but I went through the motions anyway.

SWBAT . . . . . . use standard heats of formation to determine DHrxn SHFEx1: Use standard heats of formation to find DHrxn for the thermite reaction: 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) 2(0 kj/mol) -822.1 kj/mol 2(0 kj/mol) -1676.0 kj/mol DH0rxn = SDHf0(products) – SDHf0(reactants) Let’s get REALLY concise and just say . . . DH = Products - Reactants Products = 2(0 kj/mol) + (-1676.0 kj/mol) = -1676.0 kj/mol) Reactants = 2(0 kj/mol) + (-822.1 kj/mol) = -822.1 kj/mol) DH = -1676.0 kJ/mol – (-822.1 kJ/mol) = -853.9 kJ/mol = DH

For some practice and reinforcement, you can try . . . SWBAT . . . . . . use standard heats of formation to determine DHrxn For some practice and reinforcement, you can try . . . 17.4 Practice Problems #’s 32a, 32b, 32c, 33 17.4 Section Assessment #37 Ch 17 Assessment #’s 67a, 67b

SWBAT . . . . . . use standard heats of formation to determine DHrxn SHFEx2: What is DH for the following chemical equation? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) -74.86 kJ/mol -285.8 kJ/mol 0 kJ/mol -393.5 kJ/mol DH = Products - Reactants Reactants: -74.86 kJ/mol + 2(0 kJ/mol) = -74.86 kJ/mol Products: -393.5 kJ/mol + 2(-285.8 kJ/mol) = -965.1 kJ/mol DH = P – R = (-965.1 kJ/mol) – (-74.86 kJ/mol) DH = -965.1 kJ/mol + 74.86 kJ/mol DH = -890.24 kJ/mol

[presentation under construction] SWBAT . . . . . . use standard heats of formation to determine DHrxn [presentation under construction]

HYPERINDEX Press a button! Go to a place!™ 17.1 17.2 17.3 17.4 17.1 Temperature & Heat 17.2 Calorimetry 17.3 Heating Curve for Water 17.4 Alegbra Review 17.1 Endothermic & Exothermic 17.2 Thermochemical Equations 17.3 State Change Math Problems 17.4 Hess’ Law 17.1 Q = m DT Cp Problems 17.2 Heat of Combustion 17.3 Heat of Solution Math Problems 17.4 Standard Heat of Formation