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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 9

Energy Flow & Chemical Change

The First Law of Thermodynamics (Law of

Conservation of Energy): The total energy of the universe is constant. Energy can be

neither destroyed nor created.

∆Euniverse = 0

Kinetic Energy: Energy due to motion.

E = ½ mv2

Potential Energy: Stored energy (energy due to position)

Chemical Energy: Energy related to bonds breaking and bonds forming during

chemical reactions

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Examples of Energy Relationships

Units of Energy

• Joule (J)

• calorie (cal)

• Calorie (Cal)

Conversions to know

1,000 J = 1 kJ

1 cal = 4.184 J

1 Cal = 1,000 cal = 1 kcal

Problem Solving

During a snow day, Ron eats 75% of a bag

of Tostitos chips. The bag holds 13.5 oz. One serving is considered to be 6 chips.

The 6 chips is 1 oz. and contains 120. Cal.

How much energy, in kJ, will Ron have to go snowboarding?

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More about Thermodynamics

The system and the surroundings.

The system: The part of the universe whose

change is being observed.

The surroundings: Everything around the system relevant to the change being

observed.

Energy Changes

∆E = Efinal - Einitial or,

∆E = Eproducts – Ereactants

Flow of Energy

Energy can flow into the system from the surroundings

Endothermic process

∆Esystem > 0 ; positive

∆Esurroundings < 0 ; negative

Income Expense

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Or …,

Energy can flow out of the system and into the surroundings

Exothermic process

∆Esystem < 0 ; negative

∆Esurroundings > 0 ; positive

Income Expense

Delta E

The magnitude of ∆E is the same for the

system and the surroundings. However, the sign is opposite.

∆Esystem = -∆Esurroundings or,

-∆Esystem = ∆Esurroundings

Initialstate

Finalstate

Finalstate

Initialstate

Efinal Einitial

En

erg

y, E

En

erg

y, E

Einitial Efinal

A E of system decreases B E of system increases

Efinal < Einitial

∆E < 0Efinal > Einitial

∆E > 0Energy lost tosurroundings

Energy gainedfrom surroundings

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∆E > 0

Einitial Efinal

Tsys > Tsurr

Tsurr

Tsys = Tsurr

∆E < 0

Tsurr

Efinal

Tsys = Tsurr

Tsurr

Tsys < Tsurr

Tsurr

Einitial

En

erg

y, E

En

erg

y, E

Hot H2OTsys

Room tempH2OTsys

B E gained as heatA E lost as heat

Ice H2OTsys

Room tempH2OTsys

Heat (q) lostto surroundings(q < 0)

Heat (q) gainedfrom surroundings(q > 0)

Energy Transfer

Energy can be transferred as heat and/or work.

∆E = heat + work

∆E = q + w

However, in THIS chemistry class: ∆E = q

because systems are open and therefore no work is done.

Enthalpy (H)

Enthalpy (H): Heats of reactions in chemical change.

H = E + PV

However, PV goes to 0 in open systems.

Thus,

∆H = ∆E or,

∆H = qp also,

∆H = ∆Hf - ∆Hi

∆H = ∆Hproducts - ∆Hreactants

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Delta H for Reactions

Assume 0 energy

Example

2 SO2(g) + O2(g) � 2 SO3(g) ∆Hrxn = -198.4 kJ

What does the sign of delta H indicate?

What does this mean in terms of heat

transfer with respect to the system and the surroundings?

Initialstate

Finalstate

Finalstate

Initialstate

Efinal Einitial

En

erg

y, E

En

erg

y, E

Einitial Efinal

A E of system decreases B E of system increases

Efinal < Einitial

∆E < 0Efinal > Einitial

∆E > 0Energy lost tosurroundings

Energy gainedfrom surroundings

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Specific Heat Capacity (c)Specific heat capacity – The quantity of heat

required to change the temperature of 1.00 gram of substance by 1 Kelvin.

The relationship between specific heat, mass, and heat energy is:

q = mc∆Τ

m = mass

c = specific heat capacity

∆T = Tfinal - Tinitial

Note: If a substance absorbs heat ∆T is positive

If a substance releases heat ∆T is negative

Examples of Specific Heats

Al(s) 0.900 J/g-K

Au(s) 0.129 J/g-K

H2O(l) 4.184 J/g-K * Know this one!

CCl4(l) 0.862 J/g-K

Molar Heat Capacity (C)

Molar heat capacity (C) – Amount of heat

required to change the temperature of 1.00 mole of a substance by 1 K.

Problem Solving: What is the molar heat capacity of gold?

Gold's atomic weight = 196.97 g/mol.

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More Examples

1. How much heat is required to increase

the temperature of 3.00 g of gold from 25oC to 55oC?

2. If 9.05 grams of aluminum at 30. oC lost 407.25 J of heat, what is the final

temperature?

Two More Examples

Calcium chloride dissolving in water.

Ammonium nitrate dissolves in water.

Calorimetry

Calorimetry: Lab measurements of ∆Hrxn

- Coffee Cup Calorimeter

- Bomb Calorimeter

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Enthalpy (H)

Enthalpy (H): Heats of reactions in chemical change.

H = E + PV

However, PV goes to 0 in open systems.

Thus,

∆H = ∆E or,

∆H = qp also,

∆H = ∆Hf - ∆Hi

∆H = ∆Hproducts - ∆Hreactants

Coffee Cup Calorimeter

Bomb Calorimeter

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Calorimeters

Recall that

∆Euniverse = ∆Esystem + ∆Esurroundings

And 0 = qsystem + qsurroundings

Then, -qsystem = q surroundings

Reactions in Aqueous SolutionsWhen reactions occur we consider them to be the

system.

Let’s consider the acid/base reaction below in an aqueous solution in a coffee cup calorimeter.

HCl(aq) + NaOH(aq) � H2O(l) + NaCl(aq)

Net Ionic: H+ + OH- � H2O(l)

What is the system? The surroundings? The heat exchange? Explain.

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Measuring Heats of Rxns

Using a coffee cup calorimeter:

System = Rxn

Surroundings = Solution in which the rxn occurs

qsystem = - qsurroundings

= - mc∆T

Bomb Calorimetry

There are TWO ways of looking at bomb

calorimetry.

#1: -qrxn = qcal where qcal = qbomb + qwater

#2: -qrxn = qbomb where the water is already included in the heat capacity of the

bomb calorimeter.

Bomb Calorimeter

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Bomb Calorimetry Problem Solving

Suppose a 0.562 g sample of graphite is

placed in a bomb calorimeter with an excess of oxygen gas at 25.00oC and 1.00

atm. The calorimeter temperature rises to

25.89oC. The heat capacity of the calorimeter is 20.7 kJ/oC. What is the

∆Hrxn per mole of Cgraphite?

Cgraphite + O2(g) � CO2(g)

Chemical Rxns and Bonds

During chemical reactions bonds break and

bonds form.

Bonds breaking is an endothermic process.

Bonds forming is an exothermic process.

Enthalpy and Rxn Stoichiometry

When 2 moles of H2(g) react with 1 mole of

O2(g), water is produced along with 572 kJ of heat. Write a “thermochemical

equation.”

What unit factors can be derived from this

thermochemical equation?

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Problem Solving

How much heat evolves if 52.568 grams of

hydrogen gas react with oxygen gas to produce gaseous water vapor?

Problem Solving

Consider the rxn of solid calcium oxide reacting with solid carbon to form calcium carbide and carbon monoxide. For this reaction ∆H = -464.8 kJ.

(a) Is heat released or absorbed?

(b) Is the reaction exothermic or endothermic?

(c) If 15.00 grams of carbon react with excess calcium oxide, how much energy is produced?

(d) If 986.21 kJ of heat are produced, how many grams of carbon monoxide are produced?

Standard Heat of Reaction (∆Horxn)

For standard heats of reaction solutions are

1 Molar, gases are at 1 atm, temperature is 25oC. The pure substances (elements

and compounds) must be in their most

stable form.

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Formation Reactions (∆Hof)

A formation reaction is one that forms 1 mole of a compound from the most stable form of its elements.

Note: ∆Hof = 0 for the most stable form of an

element. See appendix

Examples: Write the formation reaction for ammonium nitrate.

Write the formation reaction for sodium sulfite.

Delta H for Reactions

Assume 0 energy

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Using ∆Hof to Determine ∆Ho

rxn

∆Horxn = Σn ∆Ho

f(products) - Σn ∆Hof(reactants)

Example: Determine the heat of reaction for the following unbalanced equation:

NH3(g) + O2(g) � NO(g) + H2O(g)

Hess’s Law

The enthalpy change, ∆H, of an overall process is the sum of the individual enthalpy changes of

each step.

Example: Calculate ∆Hrxn for . . .

2H2(g) + O2(g) � 2 H2O(l)

given the following:

H2(g) + ½ O2(g) � H2O(g) ∆H = -241.8 kJ

H2O(g) � H2O(l) ∆H = -44.0 kJ

Example: Calculate ∆Hrxn for . . .

2H2(g) + O2(g) � 2 H2O(l)

given the following:

H2(g) + ½ O2(g) � H2O(g) ∆H = -241.8 kJ

H2O(g) � H2O(l) ∆H = -44.0 kJ

2H2(g) +O2(g) � 2H2O(g) ∆Hrxn = -482.6 kJ

2H2O(g) � 2H2O(l) ∆Hrxn = -88.0 kJ

2H2(g) +O2(g) + 2H2O(g) � 2H2O(g) + 2H2O(l) ∆Hrxn = -562.6 kJ

2H2(g) +O2(g) � 2H2O(l) ∆Hrxn = -562.6 kJ

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Hess’s Law

E

H2O(l)

H2O(g)

H2, O2

ΔHrxn

ΔHrxn

ΔHrxn

ΔHrxn = ΔHrxn + ΔHrxn

Algebra Review

2X + 1 = 3Y + 4

6Y + 8 = 9

4X + 2 = 6Y + 8

6Y + 8 = 9

4X + 2 + 6Y + 8 = 6Y + 8 + 94X + 2 = 9

Another Example

Determine ∆Ho for the rxn . . .

C2H6(g) � C2H4(g) + H2(g)

Given,2 C2H6(g) + 7O2(g) ���� 4 CO2(g) + 6 H2O(l) ∆∆∆∆Ho = -3119.4 kJ

C2H4(g) + 3 O2(g) ���� 2 CO2(g) + 2 H2O(l) ∆∆∆∆Ho = -1410.9 kJ

2 H2(g) + O2(g) ���� 2 H2O(l) ∆∆∆∆Ho = -571.66 kJ

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Review: Phase Changes - Changes in Physical States

Solid � Liquid Melting (Fusion)

Liquid � Solid Freezing

Liquid � Gas Vaporization

Gas � Liquid Condensation

Solid � Gas Sublimation

Gas � Solid Deposition

Examples of Physical Properties of

Three States of Matter

Which of the physical changes are endothermic

or exothermic?

Endothermic Exothermic

Melting Freezing

Sublimation Deposition

Vaporization Condensation

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∆H of a Phase Change: The amount of

energy required to produce a phase change

for one mole of a substance (kJ/mol).

Note: The VALUE of ∆Hvap. = ∆Hcond.

However, the sign is different

depending on exothermic or

endothermic processes.

Phase Changes of pure substances require a

specific amount of energy per mole (∆H)

Examples

H2O(l) � H2O(g) ∆Hvap = 40.7 kJ/mol

H2O(g) � H2O(l) ∆Hvap = - 40.7 kJ/mol

H2O(s) � H2O(l) ∆Hfus = 6.02 kJ/mol

H2O(l) � H2O(s) ∆Hfus = - 6.02 kJ/mol

Enthalpy of Vaporization & Fusion for Various Substances

HgMP = -38oC

BP = 357oC

ArMP = -189oC

BP = -186oC

C6H6

MP = 6 oC

BP = 80 oC

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Quantitative Aspects of Phase Changes

Calculating the amount of energy for a substance to undergo phase changes.

If you had 25.00 grams of water at 130.0oC, how much energy would be released when the water is cooled to

-40.0oC.

Some important facts to know:

Specific Heat of H2O(g): 33.1 J/mol-oCSpecific Heat of H2O(l) : 75.4 J/mol-oC (4.184 J/go C)

Specific Heat of H2O(s): 37.6 J/mol-oC

Stage 1 Stage 5

GAS – LIQUID

LIQUID

∆H 0vap

LIQUID–SOLID

Stage 4Stage 3Stage 2

Tem

pera

ture

(ºC)

SOLID

GAS

100

0

130

– 40

∆H 0fus

Heat removed

Tem

pera

ture

(ºC)

GAS

100

0

130

– 40Heat removed

Stage 1

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Calculating the amount of heat absorbed or released when a substance undergoes a temperature change:

q = mc∆T or q = nc∆T

n = (25.00gH2O)(1mol H2O)

(18.02 g H2O)

n = 1.387 mol H2O

q = (1.387 mol)(33.1J/moloC)(100.0oC -130.0oC)

q = -1377.291 J

GAS – LIQUID

∆H 0vap

Tem

pera

ture

(ºC)

GAS

100

0

130

– 40Heat removed

Stage 1 Stage 2

Calculating the amount of heat gained or

lost during a phase change.

q = n∆H?

q = 1.387 mol (- 40.7 kJ/mol)

q = -56.4509 kJ

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GAS – LIQUID

LIQUID

∆H 0vap

Tem

pera

ture

(ºC)

GAS

100

0

130

– 40Heat removed

Stage 1 Stage 2 Stage 3

q = nc∆T

q = 1.387mol (75.4 J/moloC)(0.0oC – 100.0oC)

q = -10457.98 J

GAS – LIQUID

LIQUID

∆H 0vap

LIQUID–SOLID

Tem

pera

ture

(ºC)

GAS

∆H 0fus

100

0

130

– 40Heat removed

Stage 1 Stage 4Stage 3Stage 2

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q = n∆Hfus

q = 1.387mol(-6.02 kJ/mol)

q = -8.34974 kJ

GAS – LIQUID

LIQUID

∆H 0vap

LIQUID–SOLID

Heat removed

Tem

pera

ture

(ºC)

SOLID

GAS

∆H 0fus

100

0

130

– 40

Stage 1 Stage 5Stage 4Stage 3Stage 2

q = nc∆T

q = 1.387mol (37.6 J/moloC)(-40.0oC – 0.0oC)

q = -2086.048 J

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Total Heat Change

q = -1377.291 J + -56.4509 kJ +

-10457.98 J + -8.34974 kJ + -2086.048 J

However, you can’t add kJ with J!

1,000 J = 1 kJ

-1.377291 kJ

-56.4509 kJ

-10.45798 kJ

-8.34974 kJ

-2.086048 kJ

-78.721959 kJ

Final Answer: - 78.7 kJ of heat are released

GAS – LIQUID

LIQUID

∆H 0vap

LIQUID–SOLID

Heat removed

Tem

pera

ture

(ºC)

SOLID

GAS

∆H 0fus

100

0

130

– 40

Stage 1 Stage 5Stage 4Stage 3Stage 2

-40°C to 130°C:

130°C to -40°C: - 78.7 kJ

+78.7 kJ

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Possible Exam Question(s)

You should be able to calculate the amount

of heat released or absorbed for any substance given ∆H’s, C’s, and temps.

ΔEuniverse = 0

ΔHrxn ≈ ΔE ≈ qsys

qsys = –qsurr

q = mcΔT or ncΔT

q = nΔHvap or fus

ΔH°rxnA = ΔH°rxnB + ΔH°rxnC (Hess’s Law)

ΔH°rxn = ΣmΔHf°prdt – ΣnΔHf°react. (Hess’s Law)