Daniel R. Barnes Init 3/12/2014 Based on Prentice Hall Chemistry, Chapter 17 WARNING: This...

Daniel R. BarnesInit 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.

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17.1The Flow of Energy --

Heat and Work

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.

Rel

ativ

e #

of m

olec

ules

Molecular speed

221oC

Look at Figure 13.3 on page 388.

21oC water

1oC water

10oC water

100oC water

50oC water

Rel

ativ

e #

of m

olec

ules

Molecular speed

221oC


21oC water21oC water, larger amount

THINK-PAIR-SHARE!

Rel

ativ

e #

of m

olec

ules

Molecular speed

221oC


21oC water>21oC water, larger amount

THINK-PAIR-SHARE!

Rel

ativ

e #

of m

olec

ules

Molecular speed

221oC


21oC water<21oC water, smaller amount

THINK-PAIR-SHARE!

. . . explain the difference between temperature and heat.

Temperature = average molecular kinetic energy T = KEave

KE = ½ mv2Kinetic energy = the energy of motion

The numerical value of temperature is proportional to average molecular kinetic energy when expressed in Kelvins, but not when expressed in oC or oF.

m = mass v = speed

(In physics, you learn that velocity is speed in a certain direction.)

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.


T = KEave

KE = ½ mv2


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.

Heat flows from the hotter system to the colder system.

“Thermal equilibrium”

.

WARNING: OVERSIMPLIFICATION ALERT!.


T = KEave

H = KEtotal

[$/person and total $ for group analogy]

. . . explain the difference between endothermic and exothermic

Endothermic = absorbing heat

Exothermic = releasing heat

H = enthalpy = heat energy content

q = DH = change in heat energy content = energy absorbed

q = positive

q = negative

q & DH, being energy, are measured in calories or joules

CHA-CHING!

I have $20,000 in my little bank account. That’s like . . . h KE q T ?

The bank has $3.7 million in deposits. 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 ?

. . . solve q = m DT Cp problems

(Solving for Cp)

DT = Tf - Ti

DT = change in temperature

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

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)


(Solving for Cp)

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 = ? Q = ? T = ? m = ?


(Solving for Cp)

C = how hard it is to DT a material

C = how much energy a material abs/rel when it DT’s

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


q = energy absorbed m = mass

DT = change in temperature = Tf - Ti

(Solving for 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

* 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.

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

q = m DT Cp


(Solving for 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

m DT

m DT

Cp =q

m DT

Remember that DT = Tf – Ti

Other than that you can just plug in the #’s & solve.

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

q = m DT Cp


(Solving for Cp)

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 = m DT Cp

m DT

m DT

Cp =q

m DT

Cp =2000 J

(7 g)(700oC)=

2000 J/goC

4900 = 0.408 J/goC = Cp

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

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


(Solving for Cp)

Take about two minutes to get started on #8 . . . TPS . . .

SKIP TO ANSWER

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


(Solving for Cp)C = ? Q = ? T = ? m = ?


(Solving for q)

Take about two minutes to get started on #3 . . . TPS . . .

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.

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?

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



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

(Solving for q)

SKIP TO ANSWER

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



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

(Solving for q)C = ? Q = ? T = ? m = ?


(Solving for q)C = ? Q = ? T = ? m = ?

If you’re an honors section, do #4 for homework. If you’re a normal section, do it now. You have two minutes.


(Solving for T)

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.

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


(Solving for T)

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

SKIP TO ANSWER

m = 400 g q = 16,000 cal Ti = ? oC

Cp = 1 cal/goC DT = Tf – Ti = ? q = m DT Cp

q = m DT Cpm Cp m Cp

DT = 16,000 cal

(400 g)(1 cal/goC)= 40 oC = DT

Ti = Tf – DT

Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

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


(Solving for T)

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 = ? q = m DT Cp

q = m DT Cpm Cp m Cp

DT = 16,000 cal

(400 g)(1 cal/goC)= 40 oC = DT

Ti = Tf – DT

Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

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


(Solving for m)

And now, let’s do a q = m DT Cp problem where m = ? . . .

m = ? g

SKIP TO ANSWER

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.

Ti = 40 oC Tf = 440 oC q = 40,000 J

Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC

q = m DT CpDT

CpDT Cp

q = m DT Cp m =

40,000 J

(400 oC)(0.4 J/goC)

m = 40,000 g

(400)(0.4)=

40,000 g

160= 250 g = m

= DT

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


(Solving for m)

And now, let’s do a q = m DT Cp problem where m = ? . . .

m = ? g

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.

Ti = 40 oC Tf = 440 oC q = 40,000 J

Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC

q = m DT CpDT

CpDT Cp

q = m DT Cp m =

40,000 J

(400 oC)(0.4 J/goC)

m = 40,000 g

(400)(0.4)=

40,000 g

160= 250 g = m

= DT

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

m =DT Cp

QCp =

m DT

Q

Q = m DT CpDT =

m Cp

Q

Q = m DT Cp

DT = Tf - Ti

Tf - Ti =m Cp

Q

Tf =m Cp

Q+ Ti

Ti = Tf -m Cp

Q

DT =m Cp

Q

Tf = DT + Ti

Ti = Tf - DT


Try to do problems 1, 2, 4, & 5 on the Heat Math Worksheet (First Exposure).

Okay, it’s time for the

17.1 Post-Quiz

This might or might not count toward your grade. Barnes?

17.2Measuring and Expressing

Enthalpy Changes

. . . 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

system

The hot egg is surrounded by cool water.

A freshly-boiled egg is a single object, but it’s made of parts (yolk, white, shell)


system

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.

You would have to measure the temperature of the water before the egg was dropped into it . . .

Ti = 20 oC

As the egg cools off, it gives energy to the water, causing the water to heat up.

Eventually, the water and the egg reach the same temperature.

surroundings

Tf = 28 oC

Ti = 100 oC

Tf = 28 oC


system

The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated.

styrofoam

surroundings

Ti = 20 oC

Tf = 28 oC


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.

polystyrene

air

airair

air. . . Compare and contrast constant-pressure calorimeters and bomb calorimeters

air

air

air

air

airA bunch of these bubbles stuck together makes a “foam”.

air

air

airair

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

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

airair

air


air

air

air

air

air

air

air

air

airair

air

Dead air spaces are almost as good as vacuums at preventing heat flow.

Dead air spaces are excellent thermal insulators.

Each bubble in a “closed cell” foam is a tiny dead air space.

Each bubble in a “closed cell” foam is a tiny dead air space.


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.


metal styrofoam

100 oC 100 oC

0 oC 0 oC

50 oC

50 oC

99 oC

1 oC


metal styrofoam

50 oC

50 oC

99 oC

1 oC

Metals are excellent thermal conductors

Styrofoam is an excellent thermal insulator


system

The egg is exothermic and the water is endothermic.

The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated.

styrofoam

surroundings

Ti = 20 oC

Tf = 28 oC q egg = - q water

q water = m DT Cp

q = (1000 g)(8 oC)(1 cal/goC)

q water = 8000 cal

q egg = - 8000 cal


system

The egg may have given off more heat than that, but some heat may have leaked into the air.

styrofoam

surroundings

This calorimeter would be more accurate if it had a styrofoam lid.

The lid provides even more insulation.

styrofoam A styrofoam lid on a styrofoam cup is not typically an airtight seal.

Therefore, the air pressure inside the calorimeter is always equal to the air pressure outside.


system

styrofoam

surroundings

styrofoam

This apparatus is a “CONSTANT PRESSURE CALORIMETER”

Therefore, the air pressure inside the calorimeter is always equal to the air pressure outside.

These are pretty leaky. They leak gas, and they leak heat.

At least they’re cheap.


This apparatus is a “CONSTANT PRESSURE CALORIMETER”

Q: WHY is this 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.BUT the gas expands, leaking out. Its pressure drops as it expands, til its pressure is equal to the atmosphere again.

101.3 kPa

101.3 kPa

Leakiness pressure equilibrium


A more dangerous, expensive, and accurate calorimeter is the “bomb calorimeter”.

The inner container, the airtight “bomb”, has to be very strong. Why?

When the green stuff, the fuel, reacts with oxygen, the exothermic reaction raises the pressure threat of explosion.

O2

See page 512 for better a picture!

H2O CO2

food


Q: The bomb calorimeter is a “constant V” device. Is it also a “constant n” device? Why/why not?

H2O CO2

Hmm? (TPS, 30 sec!)

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.

n = moles of gas

= # of gas molecules


system

system

Q: What is the “system” in each calorimeter below?

A: The “system” in the P = k calorimeter is the egg.

Hmmm? TPS 15 seconds!

A: The “system” in the V = k calorimeter is the “bomb”.


system

system

Q: What do you call the water in both calorimeters?

A: the “surroundings”


surroundingssurroundings

. . . apply the law of conservation of energy to calorimetry

system

surroundings

system

Q: DHsystem is positive, which way did thermal energy flow?

A: From the surroundings (the water) into the system.


surroundings


system

Q: If the egg absorbs 2000 J, what is DHwater?


surroundings

A: DHwater = -2000 J


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


system

Q: Which of the following is the change in enthalpy of the egg?

surroundings

DHwater

- DHwater

DHegg

-DHegg

- Qegg

- Qwater

DTwater

DTegg

Cwater

Cegg

Qegg

Qwater

mwater

megg

-DTwater

-DTegg

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.

. . . decode thermochemical equations

2H2(g) + O2(g) 2H2O(g) DH = -483.6 kJ

2H2(g) + O2(g) 2H2O(g) + 483.6 kJ

ENDO / EXO?

EXOTHERMIC!

. . . decode thermochemical equations

2H2O(l) 2H2(g) + O2(g) DH = 571.6 kJ

2H2O(l) + 571.6 kJ 2H2(g) + O2(g)

ENDO / EXO?

ENDOTHERMIC!

This is how much energy is released when four

moles of iron is oxidized.

. . . 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 . . .

12 moles of iron oxidizes?

12 mol Fe4 mol Fe

-1644.2 kJx

DH = - 4932.6 kJ12 moles is three times as much as 4, so 12 moles should release three times as much as 1644.2 kJ. 12/4 = 3


4Fe(s) + 3O2(g) 2Fe2O3(s) DH = -1644.2 kJ

What is DH if . . .

8 moles of iron oxide forms?

8 mol Fe2O3 2 mol Fe2O3

-1644.2 kJx

DH = - 6576.8 kJ

TPS – 1 min – GO! . . .


4Fe(s) + 3O2(g) 2Fe2O3(s) DH = -1644.2 kJ

What is DH if . . .

96 grams of oxygen oxidize some iron?

96 g3 mol O2

-1644.2 kJx

DH = - 1644.2 kJ

32 g

1 molx

(The molar mass of O2 is 32 g/mol, and molar mass is needed

here to convert grams into moles.)


4Fe(s) + 3O2(g) 2Fe2O3(s) DH = -1644.2 kJ

What is DH if . . .

112 grams of iron oxidizes?

112 g4 mol Fe

-1644.2 kJx

DH = - 822.1 kJ

56 g

1 molx

TPS – 1 min – GO! . . .

. . . 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.


What is DH when five moles of propane burns?

DH for propane = -2220 kJ/mol

-2220 kJ/mol x 5 mol = -11,100 kJ

TPS – 30 seconds – GO!


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.


What is DH when a 17 gram marshmallow burns? Assume the marshmallow is made entirely of sucrose.

DHcomb = -5645 kJ/mol . . . for sucrose

-5645 kJ/mol x 17 g x 1 mol/ 342 g

DH = - 280 kJC12H22O11:C = 12 x 12 = 144H = 22 x 1 = 22O = 11 x 16 = 176

342 g/mol

TPS – 1 min – GO!

17.3Heat in Changes of State

. . . interpret the “heating curve” for water

T / oC

Q / J

Ice melting

Water getting hotter

Water boiling

Steam getting hotter

100 oC

0 oC

Ice getting hotter

A “heating curve” is a graph that tells a story.


T / oC

Q / J

S L

liquid

liquid gas

gas

100 oC

0 oC

solid

What states of matter are involved in each segment?

S L

liquid

liquid gas

gas

solid

gas

liquid

solid

STATES of MATTER

more entropy

more order

condensation

freezing

vaporization

“fusion” = melting


T / oC

Q / J

melting point

boiling point

100 oC

0 oC

There are two critical temperatures in a heating curve.

100 oC

0 oC


T / oC

Q / J

The horizontal parts of the graph represent phase changes.

100 oC

0 oC

During a phase change on this graph, the temperature remains constant.


T / oC

Q / J

Q: Which stage of this story involved the most energy? How do you know?

100 oC

0 oC

TPS: 30 sec – GO!

I

II

III

IV

V


T / oC

Q / J

Q: Which stage of this story involved the most energy? How do you know?

100 oC

0 oC

A: Segment IV (boiling). It has the largest Dx.

I

II

III

IV

V

. . . 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.


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 the opposite of melting, so DHfreeze = -DHfus

Freezing is exothermic.


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.


Now do question # 14 on Mr. Barnes’ Heat Math Worksheet


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.

. . . solve math problems involving solution formation.

What is DH if 20 moles of ammonium nitrate dissolves in water?

NH4NO3(s) NH4+(aq) + NO3

-(aq) DHsoln = 25.7 kJ/mol

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.

Translation: Ammonium nitrate dissolves in water to form ammonium cations and nitrate anions.


Calcium chloride dissolving in water is exothermic.

CaCl2(s) Ca2+(aq) + 2Cl-(aq) DHsoln = -82.8 kJ/mol

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.


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.


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.

17.4Calculating Heats of Reaction

. . . 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

y = 2x + 12+

2x + 7y = 2x + 28

If you do this right, this allows you to solve for one of the variables by eliminating the other one.

7y = 28

y = 4

Once you know of the variables, you can plug it in and solve for the other variable.

4 = 2x + 12 4 - 12 = 2x + 12 - 12 -8 = 2x

x = -4

Sometimes, you need to flip one of the equations before adding them to get one of the terms to disappear.

9y = 3x - 57

2y + 22 = 3x +If we flip the second equation, and THEN add the two equations, the “x” terms will disappear.

9y = 3x - 57

3x = 2y + 22

3x + 9y = 3x + 2y - 35

9y = 2y - 35

7y = -35

y = -5

2(-5) + 22 = 3x

-10 + 22 = 3x

12 = 3x

x = 4


4x + 23y = 4x + 69

Other times, you need to multiply an equation by an integer to get what you want.

4y = x + 5

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 . . .

16y = 4x + 20

4x + 7y = 49

23y = 69

y = 34(3) = x + 5

12 = x + 5

x = 7


Sometimes, you can use these old algebra tricks with chemical equations to solve for unknowns . . .


. . . 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.


Hess’s law allows us to figure out DHrxn for a reaction without

ever having to make the reaction happen in real life.

Sn(s) + Cl2(g) SnCl2(s) DH = -325.1 kJ

SnCl2(s) + Cl2(g) SnCl4(l) DH = -186.2 kJ

HLEx1: Imagine that experiments tell you the following:

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 . . .

Sn(s) + 2Cl2(g) SnCl4(l) DH = ? kJ

yielding . . .


Sn(s) + Cl2(g) SnCl2(s) DH = -325.1 kJ

SnCl2(s) + Cl2(g) SnCl4(l) DH = -186.2 kJ

Sn(s) + 2Cl2(g) SnCl4(l) DH = ? kJ

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.

(-325.11 kJ) + (-186.2 kJ) = -511.3 kJ

DH = -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.


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.


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 . . .


HLEx2:

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.

Os(cr) + 2O2(g) OsO4(g) DH = -335 kJ



OsO4(g) OsO4(cr) DH = -56.4 kJ


HLEx2: Os(cr) + 2O2(g) OsO4(g) DH = -335 kJ



Os(cr) + 2O2(g) OsO4(g) DH = -335 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.






Os(cr) + 2O2(g) OsO4(g) DH = -335 kJ

Now we can add the equations . . .Os(cr) + 2O2(g) + OsO4(g) OsO4(g) + OsO4(cr)

and simpify . . .

Now we have the equation we were looking for. Now what?






Os(cr) + 2O2(g) OsO4(g) DH = -335 kJ

Add the DH’s to get the DH for the final equation.Os(cr) + 2O2(g) + OsO4(g) OsO4(g) + OsO4(cr)

DH = -335 kJ + (-56.4 kJ) = -391.4 kJ = DH


At this time, please attempt the problems on Mr. Barnes’ “Hess’s Law Worksheet”


[Example needed where multiplication is necessary]


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 kJNOTE: The target equation was given first, so it’s equation I this time. Try not to let that throw you off the horse.

I

IIIII



2CH4(g) + 2O2(g) CH2CO(g) + 3H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -802.3 kJNotice 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.



2CH4(g) + 2O2(g) CH2CO(g) + 3H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -802.3 kJWe’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.



2CH4(g) + 2O2(g) CH2CO(g) + 3H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -802.3 kJWe’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.



2CH4(g) + 2O2(g) CH2CO(g) + 3H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -802.3 kJLet’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).

-II + 2III = 2CO2(g) + H2O(g)

+ 2CH4(g) + 4O2(g)

CH2CO(g) + 2O2(g)

+2CO2(g) + 4H2O(g)Let’s cross out stuff.



2CH4(g) + 2O2(g) CH2CO(g) + 3H2O(g)



CH4(g) + 2O2(g) CO2(g) + 2H2O(g) DH = -802.3 kJ

DHI = -DHII + 2DHIII

[UNDER CONSTRUCTION!]

. . . use standard heats of formation to determine DHrxn

DH0rxn = SDHf

0(products) – SDHf0(reactants)

DT = Tf - Ti

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.

This old formula has something in common with this new formula:


Let’s try an example.


SHFEx1: Use standard heats of formation to find DHrxn for the

thermite reaction:

2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

DH0rxn = SDHf


The book puts it a little more concisely:

DH0 = DHf0(products) – DHf

0(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.

2(0 kj/mol) -822.1 kj/mol 2(0 kj/mol) -1676.0 kj/mol



thermite reaction:


DH0rxn = SDHf


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.



thermite reaction:


DH0rxn = SDHf


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.



thermite reaction:


DH0rxn = SDHf



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 . . .

17.4 Practice Problems #’s 32a, 32b, 32c, 33

17.4 Section Assessment #37

Ch 17 Assessment #’s 67a, 67b


SHFEx2: What is DH for the following chemical equation?

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

DH = Products - Reactants

Reactants: -74.86 kJ/mol + 2(0 kJ/mol)

-74.86 kJ/mol 0 kJ/mol -393.5 kJ/mol -285.8 kJ/mol

Products: -393.5 kJ/mol + 2(-285.8 kJ/mol)

= -74.86 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]

Press a button! Go to a place!™

17.1

17.1Q = m DT Cp

Problems

17.2

17.2 Calorimetry17.1 Temperature &

Heat

17.1 Endothermic & Exothermic

17.2Thermochemical

Equations

17.2Heat of

Combustion

17.3

17.3 Heating Curve for Water

17.3 State Change Math Problems

17.3 Heat of Solution Math

Problems

17.4

17.4 Alegbra Review

17.4 Hess’ Law

17.4 Standard Heat of Formation

Daniel R. Barnes Init 3/12/2014 Based on Prentice Hall Chemistry, Chapter 17 WARNING: This...

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