1

Chapter Six:

THERMOCHEMISTRY

2

Contentsp228

6-1 Nature of Energyp229

The Law of Conservation of Energy

Kinetic Energy Heat Work

Pathway State Function or State Property

Chemical Energy

4

Initial Position

In the initial position, ball A has a higherpotential energy than ball B.

Figure 6.1

5

After A has rolled down the hill, the potential

energy lost by A has been converted to

random heading) and to the increase motions

of the components of the hill (frictional in the

potential energy of B.

p230

Final Position

6

Final Position

After A has rolled down the hill, thepotential energy lost by A has beenconverted to random motions of thecomponents of the hill (frictional heading)and to the increase in the potential energyof B.

7

System

Surrounding

Exothermic: exo- is meaning “out off”; energy flows out ofthe system.

Endothermic: the heat flow is into a system is endothermic.Reactions that absorb energy fromsurroundings are said to be endothermic.

p231

8

Chemical Energy p231

Figure 6.2

The combustion of methane releases the quantity ofenergy Δ (PE) to surrounding via heat flow, This anexothermic process.

9

The study of energy and its interconversions is called thermodynamics.The conservation of energy is often called the firstlaw of thermodynamics. The energy of the universeis constant.

The internal energy of E of a system can be definedmost precisely as the sum of the kinetic and potentialenergies of all the “particles”in the system.

Surroundings Surroundings

p232

ΔE = q + w

10

Ex 6.1 Internal EnergyCalculate ΔE for a system undergoing an

endothermic process in which 15.6 kJ of heat flows

and where 1.4 kJ of work is done on the system.Solution:

P233

ΔE = q + w

Where q = + 15.6 kJ, since the process isendothermic, and w = + 1.4 kJ, since work isdone on the system. Thus

ΔE = 1.56 kJ + 1.4 kJ = 17.0 kJ

11

p233

Work = force x distance = F x △h

Since P = F/A or F = P x A

Work = F x △h = P x A x △h

△V = final volume –initial volume

= A x △h

Work = P x A x △h = P △V

W = - P △V

12

Calculate the work associated with the expansion ofa gas from 46 L to 64 L at a constant externalpressure of 15 atm.Solution:

P234Ex 6.2 PV Work

For a gas at constant pressure,

w = - PΔV

In this case P = 15 atm , andΔV = 64 - 46 = 18 L. Hence

w = -15 atm x 18 L = - 270 atm ∙L

13

Ex 6.3 Internal Energy, Heat, and WorkA balloon is being inflated to its full extent by heating theair inside it. In the final stages of this process, the volumeof the balloon changes from 4.00 × 106 L to 4.50 × 106 L bythe addition of 1.0 atm, calculate ΔE for the process. (To

convert between L˙atm = 101.3 J.)Solution:

P234

V = 4.50 x 106 L - 4.00 x 106 L = 0.50 x 106 L

Thus w = - (1.0 atm) x (5.0 x 105 L) x (101.3 J/(L· atm)

= - 5.1 x 107 J

Then E = q + w = (+1.3 x 108 J) + ( -5.1 x 107 J)

= 8 x 107 J

14

Ex 6.4 EnthalpyWhen 1 mole of methane (CH4) is burned atconstant pressure, 890 kJ of energy is released asheat. Calculate ΔH for a process in which a 5.8-gsample of methane is burned at constant pressure.Solution:

P236

5.8 g ÷(16 g/mole) = 0.36 mol CH4, and

0.36 mol x ( - 890 kJ/mol) = - 320 kJ

Thus, when 5.8-g Sample of CH4 is burned atconstant pressure,

ΔH = heat flow = -320 kJ

15

Coffee Creamer Flammability

16

Sugar and Potassium Chlorate

17

Thermite

18

Is the freezing of water an endothermic or

exothermic process? Explain.

React 1

19

Classify each process as exothermic or endothermic.

Explain.

a) Your hand gets cold when you touch ice.

b) The ice gets warmer when you touch it.

c) Water boils in a kettle being heated on a

stove.

d) Water vapor condenses on a cold pipe.

e) Ice cream melts.

React 2

20

Work vs. Energy Flow

21

6-2 Enthalpy and Calorimetryp235

Enthalpy H = E + PV

Change in H = (Change in E) + (Change in PV)

△E = qp + W; △E = qp- P△V; qp = △E + P△V

△H = △E + △ (PV); △(PV) = P △V

△H = △E + P△V

qp =△E + P△V

△H = qp

22

p236

ΔH = ΔE + Δ (PV), Δ(PV) = P ΔV

ΔH = ΔE + PΔV

qP = ΔE + PΔV

ΔH = qP

ΔH = Hproducts - Hreactants

23

Calorimetry p237

Specific heat capacity:its unit L/K ‧ g

Molar heat capacity:it has the units J/K‧ mol

Constant-pressure calorimetryA Coffee Cup CalorimeterMade of Two StyrofoamCups

Figure 6-5

H+(aq) + OH-(aq) → H2O(l)

Heat capacity

24

p238Energy released by the neutralization reaction

p242

26

p242Ex 6.6 Constant-Volume CalorimetryIt has been suggested that hydrogen gas obtainedby the decomposition of water might be a substitutefor natural gas (principally methane). To comparethe energies of combustion of these fuels, thefollowing experiment was carried out using a bombcalorimeter with a heat capacity of 11.3 kJ/℃.When a 1.50-g sample of methane gas burned withexcess oxygen in the calorimeter, the temperatureincreased by 7.3℃. When a 1.15-g sample ofhydrogen gas was burned with excess oxygen, thetemperature increase was 14.3℃. Calculate theenergy of combustion (per gram) for hydrogen andoxygen.

27

Solution

Energy released in the combustion of 1.5 g CH4

= 11.3 kJ/℃)(7.3℃) = 8.3 kJ

Energy released in the combustion of 1 g CH4

= 83 kJ/(1.5 g) = 55 kJ/g

Similarly, for hydrogen

Energy released in the combustion of 1.15 g H2

= (11.3 kJ/℃)(14.3℃) = 162 kJ

Energy released in the combustion of 1 g H2

= 162 kJ/(1.15 g) = 141 kJ/g

28

You have a Styrofoam cup with 50.0 g of water at

10C. You add a 50.0 g iron ball at 90C to the

water. The final temperature of the water is:

a) Between 50°C and 90°C.

b) 50°C

c) Between 10°C and 50°C.

Calculate the final temperature of the water.

React 3

29

p242

30

6-3 Hess’s Law p242

Figure 6-7 The principle of Hess’slaw. The same change in

enthalpy occurs when nitrogen and oxygen react to form

nitrogen dioxide, regardless of whether the reaction occurs

in one (red) or two (blue) steps.

31

p243

N2(g) +2O2(g) → 2NO2 (g), △H1 = 68 kJ

N2(g) + O2(g) → 2NO(g), ΔH2 = 180 kJ

2NO(g) + O2(g) → 2NO2(g), ΔH3 = -112 kJ

Net reaction: N2(g) + 2O2(g) → 2NO2(g), ΔH2 + ΔH3 = 68 kJ

ΔH1 = ΔH2 + ΔH3 = 68 kJ

32

Characteristic of Enthalpy Changes p243

1. If a reaction is reversed, the sign of ΔH is also reversed.

2.The magnitude of ΔH is directly proportional to thequantities of reactants and products in a reaction. If thecoefficients in a balanced reaction are multiplied by integer,the value of ΔH is multiplied by the same integer.

Xe(g) + 2F2(g) → XeF4(s), ΔH = -251 kJ

XeF4(s) → Xe(g) + 2F2(g), ΔH = ?

Crystals of xenon tetrafluoride, the first reportedbinary compound containing a noble gas element.

33

Hess’s Law

34

Ex 6.7 Hess’s Law

Two forms of carbon are graphite, the soft, black,

slippery material used in “lead”pencils and as a lubricant

for locks, and diamond, the brilliant, hard gemstone.

Using the enthalpies of combustion for graphite (-394

kJ/mol) and diamond (-396 kJ/mol), calculate ΔH for the

conversion of graphite to diamond:

P244

)()( sCsC diamondgraphite Solution:

35

Hints for Using Hess’s Law p246

1. Work backward from the required reaction, using the

reactants and products to know to manipulate the

o t h e r g i v e n r e a c t a n t s a t y o u r d i s p o s a l .

2. Reverse any reactions as needed to give the required

reactants and products.

3. Multiply reactions to give the correct numbers of

reactants and products.

36

6-4 Standard Enthalpies of Formationp246

Cgraphite(s) → Cdiamond(s)

Standard enthalpy of formation is defined as the change

in enthalpy that accompanied the formation of one mole

of a compound from its elements with all substances in

their standard states.

A degree symbol on a thermodynamic function, for

example, ΔH0, indicates that the corresponding process

has been carried out under standard conditions.

37

Conventional Definitions of Standard states p246

For a compound

1. The standard state of a gaseous substance is a pressure ofexactly 1 atmosphere.

2. For a pure substance in a condensed state (liquid or solid),the standard state is the pure liquid or solid.

3. For a substance present in a solution, the standard is aconcentration of exactly 1 M.

For an Element

The standard state of an element is the which the element existsunder conditions of 1 atmosphere and 25 C.( The standard state foroxygen is O2(g) at a pressure of 1 atmosphere; the standard state forsodium is Ma(s); the standard for mercury is Hg(l); and so on. )

p247

)(2H)(C)(CH 24 gsg

kJ/mol75ΔΗ)(CH)(2H)(C 0f42 ggs

kJ/mol286)(OH)(O)(H 0f222

2

1 Hlgg

?)(OH)(CO)(O2)(CH oreaction2224 Hlggg

kJ/mol394)(CO)(O)(C of22 Hggs

39

A Schematic Diagram of the Energy Changesfor the Reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

40

p247

H°reaction = npHf(products) - nrHf(reactants) (6.1)

Change in enthalpy can be calculated from enthalpiesof formation of reactants and products.

41

Standard States

Compound

For a gas, pressure is exactly 1 atmosphere.

For a solution, concentration is exactly 1 molar.

Pure substance (liquid or solid)

Element

The form [N2(g), K(s)] in which it exists at 1 atmand 25°C.

42

Using the standard enthalpies of formation listed in

Table 6.2, calculate the standard enthalpy change for

the overall reaction that occurs when ammonia is

burned in air to form nitrogen dioxide and water. This

is first step in the manufacture of nitric acid.

P249Ex 6.9 Enthalpies from Standard Enthalpiesof Formation

)(6)(4)(7)(4 2223 lOHgNOgOgNH Solution:

ΔH0reaction = { - 4 mol [-(46 kJ/mol)]} + [ - 7 mol ( 0

kJ/mol )] + [ 4 mol ( + 34 kJ/mol)] +

[ 6 mol ( - 286 kJ)]

= - 1396 kJ

43

Ex 6.11 Enthalpies from Standard of Formation III

Methanol (CH3OH) is often used as fuel in high-performance

engines in race cars. Using the data in Table 6.2, compare the

standard enthalpy of combustion per gram of methanol with

that per gram of gasoline. Gasoline is actually a mixture of

compounds, but assume for this problem that gasoline is

liquid octane (C8H18).

Solution )(O4H)(CO2)(O3)(OHCH2 2223 lggl

p252

p252For Methanol

For Octane

6-5 Present Sources of Energy p252

Petroleum andNatural Gas

p257

Figure 6.14

Hydrogen as a Fuel

HgOHgCH 3)()( 24

kJ286ΔHO(l)H(g)O(g)H 0222

2

1

p257

CO(g)(g)3HO(g)H(g)CH 224

Wind turbines to create electricity.

p259

49

Ex 6.12 Enthalpies of CombustionCompare the energy available from the combustion of given

volume of methane and the same volume of hydrogen at the

same temperature and pressure.

P260

Solution:

50

Ex 6.13 Comparing Enthalpies of CombustionAssuming that the combustion of hydrogen gas

provides three times as much energy per gram as

gasoline, calculate the volume of liquid H2 (density

= 0.0710 g/mL) required to furnish the energy

contained in 80.0 L (about 20 gal) of gasoline

(density = 0.740 g/mL). Calculate also the volume

that this hydrogen would occupy as a gas at 1.00

atm and 25℃.

P261

Solution: p261

52

Energy Sources Used in the United States

53

The Earth’sAtmosphere