chapter 5 principles of chemical reactivity: energy and ... · pdf fileprinciples of chemical...

Jeffrey Mack

California State University,

Sacramento

Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

Questions that need to be addressed:

• How do we measure and calculate the energy

changes that are associated with physical changes

and chemical reactions?

• What is the relationship between energy changes,

heat, and work?

• How can we determine whether a chemical reaction is

product-favored or reactant-favored at equilibrium?

• How can we determine whether a chemical reaction

or physical change will occur spontaneously, that is,

without outside intervention?

Energy & Chemistry

1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C.

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food ―calorie‖)

SI units for energy: joule (J)

1 cal = exactly 4.184 J

James Joule

1818-1889

Units of Energy

Some Basic Principles

• Energy as the capacity to do work or transfer heat (q).

• Heat (q) is NOT temperature, temperature is a

measure of kinetic energy.

• Energy is divided into two basic categories: – Kinetic energy (the energy associated with motion)

– Potential energy (energy that results from an object’s position).

• The law of conservation of energy requires that

energy can neither be created nor destroyed.

• However, energy can be converted from one type into

another.

Energy & Chemistry

Energy can be divided into two forms:

Kinetic: Energy of motion:

Thermal

Mechanical

Electrical

Potential: Stored energy:

Gravitational

Electrostatic

Chemical & Nuclear

Energy & Chemistry

Kinetic energy — energy of motion

• Translation

Potential & Kinetic Energy

Potential energy — energy a motionless body has by virtue of its composition and position.

Potential & Kinetic Energy

• Burning peanuts supply sufficient energy to boil a cup of water.

• Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

Energy & Chemistry

• Energy transfer as heat will occur spontaneously from an

object at a higher temperature to an object at a lower

temperature.

• Transfer of energy as heat continues until both objects are at

the same temperature and thermal equilibrium is achieved.

• At thermal equilibrium, the object with a temperature

increase has gained thermal energy, the object with a

temperature decrease has lost thermal energy.

Thermal Equilibrium

• SYSTEM – The object under study

• SURROUNDINGS – Everything outside the

system

• Energy flows between the two

System & Surroundings

11

• Open: energy and

matter can be

exchanged with the

surroundings.

• Closed: energy can

be exchanged with

the surroundings,

matter cannot.

• Isolated: neither

energy nor matter

can be exchanged

with the

surroundings.

Types of Systems

A closed system;

energy (not

matter) can be

exchanged.

After the lid of the jar

is unscrewed, which

kind of system is it?

• When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMIC.

• In the case of thermal energy, the temperature of the system decreases. (qsystem < 0)

• Tsystem = (Tfinal – Tinitial) < 0

Directionality of Energy Transfer

• When energy enters the system and from the surroundings, the process is said to be ENDOTHERMIC.

• In the case of thermal energy, the temperature of the system increases. (qsystem > 0)

• Tsystem > 0

Directionality of Energy Transfer

• Heat flows between the system and

surroundings.

• U is defined as the internal energy of the

system.

• q = the heat absorbed or lost

Heat & Changes in Internal Energy

∆U = Ufinal – Uinitial = ± q

If heat enters the system:

U > 0 therefore q is positive (+)

If heat leaves the system:

U < 0 therefore q is negative (–)

The sign of q is a ―convention‖, it designates the

direction of heat flow between the system and

surroundings.



Surroundings

System

heat in

qsystem > 0

(+)

Usystem > 0

heat out

qsystem < 0

(–)

Usystem < 0

increases decreases

q = the heat absorbed or lost by the system.

Usyste

m

Usyste

m



17

Thermodynamics is the scientific study of the

interconversion of heat and other kinds of energy.

State functions are properties that are determined by the state of the system,

regardless of how that condition was achieved.

Potential energy of hiker 1 and

hiker 2 is the same even though

they took different paths.

energy, pressure, volume, temperature

E = Efinal - Einitial

P = Pfinal - Pinitial

V = Vfinal - Vinitial

T = Tfinal - Tinitial

The total internal energy of an isolated system is constant.

The ―system‖ is that which we are

interested in.

The ―surroundings‖ are

everything in contact with the

system.

Together: System + Surroundings = Universe

First Law of Thermodynamics

∆Usystem = Ufinal – Uinitial

Any energy lost by the

system is transferred to the

surroundings and vice versa.

Any change in energy is

related to the final and initial

states of the system.

The same holds for the surrounding!

Energy

Energy

First Law of Thermodynamics

Energy cannot be created or destroyed.

Energy of the universes (system + surroundings) is constant.

Any energy transferred from a system must be transferred to

the surroundings (and vice versa).

From the first law of thermodynamics: When a system undergoes

a physical or chemical change, the change in internal energy is

given by the heat added to or absorbed by the system plus the

work done on or by the system:

U q wchange in

system

energy

=

heat

lost or gained

by the system

+ work done by or

on the system

Relating U to Heat and Work

energy transfer out

(exothermic), -q

energy transfer in

(endothermic), +q

SYSTEM

∆U = q + w

w transfer in

(+w) w transfer out

(-w)

U initial

U Final

en

erg

y

q in

work in

Uf > Ui

Usystem >

0 (+)

q out

work out

Uf < Ui

Usystem <

0 (–)

U Final

U initial

Energy in Energy out

Heat

Work

Usystem = 0

work and

heat can

balance!

Energy is the capacity to do work.

Work equals a force applied through a distance.

When you do work, you expend energy.

When you push down on a bike

pump, you do work.

=force

Parea

=2

force

x

change in volume V = x3

3

2

FP V = x = F x

x

P × V = work

Therefore work is equal to a change

of volume at constant pressure.

24

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant

temperature. What is the work done in joules if the gas expands (a) against a

vacuum and (b) against a constant pressure of 3.7 atm?

w = -P V

(a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J

1L•atm = -1430 J

Try one more!

A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy?

Answer: + 21J

= ´Dq C T

The amount of heat (q) transfer related to an object

and temperature is given by:

When heat is absorbed or lost by a body, the

temperature must change as long as the phase

(s, g or l) remains constant.

q = heat lost or gained (J)

C = Heat Capacity of an object

J

C or K

T = Tfinal Tinitial is the temperature change (°C or K)

Heat Capacity

q = m S ∆T

The amount of heat (q) transfer per unit mass of a

substance is related to the mass and temperature by:

When heat is absorbed or lost by a body, the

temperature must change as long as the phase (s, g or l)

remains constant.

q = heat lost or gained (J) m = mass of substance (g)

S= the Specific Heat Capacity of a compound J

g C or K

T = Tfinal Tinitial is the temperature change (°C or K)

Heat & Specific Heat Capacity

Recall that T = Tf –Tin

50.0 °C = 50.0 °C + 273.2 = 323.2 K

25.0 °C = (25.0 °C + 273.2) = 298.2 K

= 25.0 °C = 25.0 K

The are the same!

T T

Does it matter if we calculate a temperature

change in Kelvin or degrees C?

Tf =

–Tin =

let Tin = 25.0 °C and Tf = 50.0 °C

Determine the final temperature of a 25.0 g block of

metal that absorbs 255 cal of energy. The initial

temperature of the block was 17.0 C

The specific heat capacity of the metal is:

J2.72

g C

Data Information: Mass,

initial temp, heat capacity

of metal, heat (q) absorbed.

Solve q = mS T for Tfinal,

plug in data.

Final temperature of the

metal is determined.

Find Tfinal of the metal after

it absorbs the energy

Strategy Map:

Determine the final temperature of a 25.0 g block of

metal that absorbs 255 cal of energy. The initial

temperature of the block was 17.0 C

The specific heat capacity of the metal is:

J2.72

g C

rearranging:

Tf =

+255 cal 4.184 J

1 cal

25.0 g J

2.72 g C

+ 17.0 °C = 32.7 °C

Tf > Tin

as

expected

q = m S ∆T q = m S (TFinal – Tinitial)

´final initial

qT = +T

m C

Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially

at 21.0 °C.

• At thermal equilibrium, the water and iron are both at 23.1 °C

• What is the specific heat capacity of the metal?

Solution:

Heat is transferred from the hot metal to the colder

water.

Energy is conserved so:

+ =

= -

Fe water

Fe water

q q 0

q q


at 21.0 °C.



Solution:


at 21.0 °C.



Heat lost by metal = heat gained by water

-qFe = qw

mFe SFe ∆TFe = -mw Sw ∆Tw

Solution:


at 21.0 °C.



SFe = (-mw Sw ∆Tw)/ [mFe ∆TFe]]

Solution:


at 21.0 °C.



SFe

• When matter absorbs heat, its temperature will rise until

it undergoes a Phase Change.

• The matter will continue to absorb energy, however

during the phase change its temperature remains

constant: Phase changes are ―Isothermal‖ processes.

Energy & Changes of State

• Some changes of state (phase

changes) are endothermic:

• When you perspire, water on your

skin evaporates.

• This requires energy.

• Heat from your body is absorbed by

the water as it goes from the liquid

state to the vapor state, as a result

you cool down.

+ energy

+ ®2 2H O(l) Heat H O(g)

Energy Transfer & Changes of State

• Some changes of state (phase changes) are exothermic:

• When it is muggy outside, water condenses on your skin.

• This releases energy.

• Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state.

• As a result you feel hot.

+ energy

® +2 2H O(g) H O(l) Heat

Energy Transfer & Changes of State

Heating/Cooling Curve for Water

Note the isotherms

at each phase

transition.

• The energy associated with a change of state is given by the

Enthalpy or Heat of the phase change.

• Since there is no temperature change associated with the

process, the units are most often in J/g or J/mol.

Sublimation: subH > 0 (endothermic)

Vaporization: vapH > 0 (endothermic)

Melting or Fusion: fusH > 0 (endothermic)

Deposition: depH < 0 (exothermic)

Condensation: conH < 0 (exothermic)

Freezing: freH < 0 (exothermic)

Where H refers to the ―Heat‖ of a phase change

Heat & Changes of State

q(heating or cooling) = m S ∆T

q(phase change) = (phase change)H n

Heating & Cooling:

Heat absorbed or lost in a Phase change:

(n = moles or grams)

Energy Change Calculations

Problem:

What quantity of heat is required to melt 500. g

of ice at 0 °C then heat the resulting water to

steam at 100 °C?

+333 J/g +2260 J/g

melt the ice form liquid water at 0 °C heat the water to 100 °C boil water

fusH

Swater

VapH

×

Constants Needed:

JHeat of fusion of ice = 333

g

JSpecific heat of water = 4.18

g K

JHeat of vaporization = 2260

g

Equations:

qphase change =

m H

qheat = m S T

Problem:



steam at 100 °C?

Ice

H2O(s)

(0 °C)

fusH

water

H2O(l)

(0 °C)

water

H2O(l)

(100 °C)

Cwater vapH

steam

H2O(g)

(100 °C)

Problem:



steam at 100 °C?

= ´D1 ice fusq n H + = ´ ´D2 water water q m C T + = ´D3 water vap q n Hqtotal =

melt ice heat water boil water

Problem:



steam at 100 °C?

Ice

H2O(s)

(0 °C)

fusH

water

H2O(l)

(0 °C)

water

H2O(l)

(100 °C)

Cwater vapH

steam

H2O(g)

(100 °C)

total

6

Jq 500. g 333

g

J 500. g 4.18 (100.0 C 0.0 C)

g C

J 500. g 2260 1.51 10 J

g

Problem:



steam at 100 °C?

Ice

H2O(s)

(0 °C)

fusH

water

H2O(l)

(0 °C)

water

H2O(l)

(100 °C)

Cwater vapH

steam

H2O(g)

(100 °C)

Enthalpy, ―H‖ is the heat transferred between the system and

surroundings under conditions of constant pressure.

H = qp the subscript “p”

indicates constant

pressure = +H U PV

( )D = D +

= D + D

H U PV

U P Vif no “PV” work is

done by the system,

V = 0

0

H = Up

the change in

enthalpy is the

change in internal

energy at constant

pressure

Enthalpy

For a ―system‖ the overall change in Enthalpy is path

independent.

No matter which path is taken

(A B) the results are the

same:

Final – Initial

Since individual Enthalpies

cannot be directly measured,

we only deal with enthalpy

changes

( H = Hf – Hi)

A

B

Enthalpy is a “State Function”

• Since Enthalpies are state functions, one

must specify the conditions at which they are

measured.

• H(T,P): Enthalpy is a function of temperature

and pressure.

• ―H°‖ indicates that the Enthalpy is taken at

Standard State conditions.

• Standard State Conditions are defined as:

• 1 atm = 760 mm Hg or 760 torr & 298.15 K or

25 °C

Enthalpy Conditions

• Values of H are measured experimentally.

• Negative values indicate exothermic reactions.

• Positive values indicate endothermic reactions.

Enthalpy Diagrams

A decrease in

enthalpy during the

reaction; H is

negative.

An increase in

enthalpy during the

reaction; H is

positive.

Reactants Products

Energ

y

H (Products)

H (Reactants) H (Products)

H (Reactants)

Hr > 0 Hr < 0

Endothermic Exothermic

H = Hfinal Hinitial

Enthalpy of reaction = Hr = Hproducts Hreactants

Enthalpies & Chemical Reactions:

Hr

Just like a regular chemical equation, with an energy term.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) rHo = 802 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ

What does this imply? CONVERSION FACTORS!!!

From the equation:

+

4

802 kJ of Energy Released

1 mol CH (g) consumed2

+802 kJ of Energy Released

2 mol H O (g) produced

energy out…

Exothermic Energy is a product just like CO2 or H2O!

Thermochemical Equations

Problem: How many kJ of energy are released when 128.5 g of

methane, CH4(g) is combusted?

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

4128.5g CH´ 41 mol CH

16.04 g´ =

+802 kJ

1mol rxn6.43 103 kJ

g mols J

molar

mass

Reaction

enthalpy

rHo = 802 kJ

´4

1 mol rxn

1 mol CH

When the reaction is reversed , the sign of H reverses:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) Hro = +802 kJ

Endothermic

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Hro = 802 kJ

Exothermic


[ ] [ ]

H scales with the reaction:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Hro = 802 kJ

1

2´

1

2´

®4 2 2 2

1 1CH (g) + O (g) CO (g) + H O(g)

2 2Hr

o = 401 kJ

Yes you can write the reaction with fractions, so long as

you are writing it on a mole basis…


Change in enthalpy depends on state:

H2O(g) H2O(l)

This means that water’s liquid state lies 44

kJ/mol lower than the gas state

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

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Hr° = 802 kJ

Hr° = 890 kJ

H = 44 kJ/mol

= 88kJ

From your text:

[ ] 2 = 88kJ The difference in the rxn H

is due to the change in state!!


CH4(g) + 2O2(g)

CO2(g) + 2H2O(g)

CO2(g) + 2H2O(l)

Hro = 802 kJ

Hro = 890 kJ

2 44 kJ

The difference is the

88 kJ released when 2

mols of water go from

gas to liquid.

Comparing the reaction with water as a gas or

liquid: E

nth

alp

y (

H)

products

products

reactants

either pathway

gives the same

results!

• A constant pressure

calorimeter can be used

to measure the amount of

energy transferred as

heat under constant

pressure conditions, that

is, the enthalpy change

for a chemical reaction.

Constant Pressure Calorimetry, Measuring H

• The constant pressure

calorimeter used in general

chemistry laboratories is

often a ―coffee-cup

calorimeter.‖ This

inexpensive device

consists of two nested

Styrofoam coffee cups with

a loose-fitting lid and a

temperature-measuring

device such as a

thermometer or

thermocouple.


• Because the ―coffee-cup

calorimeter.‖ is an isolated

system, ―What happens in

the coffee cup, stays in

the coffee cup!‖

• No mass loss to the

surroundings.

• No heat loss to the

surroundings.


Problem: • 5.44 g of ammonium nitrate was added to 150.0mL of water in a coffee-

cup calorimeter.

• This resulted in a decrease in temperature from 18.6 °C to 16.2 °C.

• Calculate the enthalpy change for dissolving NH4NO3(s) in water in

kJ/mol.

• Assume the solution has a specific heat capacity of 4.18 J/g ? K.


cup calorimeter.



kJ/mol.


Data Information: Mass of

reactant, C, water & temperature

change.

Cals. Moles of

NH4NO3

Step 1:

Eq. gives mole ratios

(stoichiometry)

Determine qsolution =

mC T

Step 2:

qsolution

qsolution + qrxn = 0 Step 3:

qrxn = q(NH4NO3)

rH =qrxn/mol NH4NO3 Step 4:

Enthalpy per mole of

reactant


cup calorimeter.



kJ/mol.


4 34 3 4 3

solution solution

solution

3

solution

1 mol NH NO5.44 g NH NO 0.0680 moles NH NO

80.04 g

q m C T

Jq 154.4 g 4.18 (16.2 C 18.6 C)

g C

q 1.55 10 J


cup calorimeter.



kJ/mol.


= - ´

+ =

= - = + ´

+ ´D = = ´ = +

3

solution

rxn solution

3

rxn solution

3

rxnr 3

4 3

q 4.55 10 J

q q 0

q q 1.55 10 J

q 1.55 10 J 1kJ kJH 22.8

moles of reaction 0.0680 moles NH NO 10 J mol

The sign is positive indicating an endothermic process.

Under conditions of

constant volume, any

heat transferred is equal

to a change of internal

energy Ur.

qV = Ur

Constant Volume Calorimetry, Measuring U

Heats of combustion

( rUo

combustion ) are measured

using a device called a Bomb

Calorimeter.

A combustible sample is

reacted with excess O2

rxn cal

cal water bomb

q C T

C C C

= - ´D

= +

The heat capacity of the

bomb is constant.

The heat of reaction is found

by:

Constant Volume Calorimetry, Measuring U

Octane, the primary component of gasoline combusts by the

reaction: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9H2O(l)

A 1.00 g sample of octane is burned in a bomb calorimeter that

contains 1.20 kg of water surrounding the bomb.

The temperature of the water rises to 33.20 °C from 25.00 °C

when the octane is reacted.

If the heat capacity of the bomb is 837 J/°C, calculate the heat of

reaction per mole of octane.

–qrxn = qwater + qbomb

Since the temperature of the water rose, the reaction must

have been exothermic:

Therefore one can write:

Calculating Heat in an Exothermic Reaction

–qrxn = mwater Swater Twater + qbomb Twater

- =RXNq4.184J

g C´ ´ 31.20 10 g 33.20 25.00 C

J837

C33.20 25.00 Cqwater

qbomb

qRXN = – 4803 J or – 48.0 kJ

Heat transferred per mole qV: 3 48.0 kJ kJ

5.48 10mol mol

1.00g114.2g

-= - ´

´

Calculating Heat in an Exothermic Reaction

The overall enthalpy change for a reaction is equal to the sum

of the enthalpy changes for the individual steps in the reaction.

Reactants Products

rH = ?

unknown!

Intermediate

Reaction

rH1 + rH2 = rH

The sum of the H’s in one direction must equal the sum

in the other direction.

What if the enthalpy

changes through

another path are know?

Why?

Because enthalpy is a state function… Path independent!

so we can write…

Hess’s Law

Forming CO2 can

occur in a single

step or in a two

steps.

∆rHtotal is the same

no matter which

path is followed.

Hess’s law & Energy Level Diagrams

Notice that the path from reactants to products in the

desired reaction goes through an “intermediate

compound” in the given reactions.

This means that the path for hydrogen and nitrogen to

produce ammonia goes through hydrazine (N2H4).

Therefore, the path to the enthalpy of the reaction

must be a sum of the two given reactions!

Hess’s Law Problem:

Example: Determine the rH for the reaction:

3H2(g) + N2(g) 2NH3(g) Hro

= ???

Given: (1) 2 H2(g) + N2(g) N2H4(g) ∆H°1 = +95.4 kJ

(2) N2H4(g) + H2(g) 2NH3(g) ∆H°2 = –187.6 kJ

adding equations (1) & (2) yields:

2 H2(g) + N2(g) + N2H4(g) + H2(g) N2H4(g) + 2NH3(g)



3H2(g) + N2(g) 2NH3(g) Hro

= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆H°2 = –187.6 kJ



3H2(g) + N2(g) 2NH3(g) Hro

= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆H°2 = –187.6 kJ



Look what happens…

/ /

3H2(g) + N2(g) 2NH3(g)



3H2(g) + N2(g) 2NH3(g) Hro

= ???


(2) N2H4(g) + H2(g) 2NH3(g) ∆H°2 = –187.6 kJ



Look what happens…

/ /

3H2(g) + N2(g) 2NH3(g)

Hr = ∆H°1 + ∆H°2 = +95.4 kJ + (–187.6 kJ) = –92.2 kJ

therefore…

• Reaction (1) is endothermic by 95.4 kJ

• Reaction (2) is exothermic by 187.6 kJ

• Since the exothermicity has a greater magnitude

than the endothermicity, the overall process is

exothermic ( rH < 0).

intermediates

H2(g) +

N2H4(g)

reactants

3H2(g) +

N2(g)

products

2N3 (g)

(Step 1)

Ho = 95.4kJ (Step 2)

Ho = 187.6kJ

(Overall)

Ho = 92.2kJ

When 1 mole of compound is formed from its elements, the

enthalpy change for the reaction is called the enthalpy of

formation, fHo (kJ/mol).

These enthalpies are always reported at Standard

conditions:

1 atm and 25 °C (298 K).

The standard enthalpies of formation of the most stable form

of any element is zero:

Hf (element) = 0

Hf° O2(g) = 0 Hf° O(g) ≠ 0

elemental form NOT the elemental form

Standard Enthalpies of Formation

H2(g) + ½ O2(g) H2O(g) ∆H˚ = -242 kJ

H2(g) + ½ O2(g) H2O(liq) ∆H˚ = -286 kJ

Same reaction, different phases, different enthalpies.

Since Enthalpy is a state function, enthalpy values

depend on the reaction conditions in terms of the

phases of reactants and products.

Enthalpy Values

The formation of water is given by the reaction:

H2(g) + ½ O2(g) H2O(l)

Each element and the compound are represented by

the physical state they take on at 25.0 °C and 1 atm

pressure. (Standard State conditions)

A chemical reaction that describes the formation of

one mole of a compound from its elements at

standard state conditions is known as a “formation

reaction”.

Formation Reactions

Question:

What is the formation reaction for potassium

permanganate?

KMnO4

• salts are solids at standard state conditions.

• metals are solids at standard state conditions.

• oxygen is a gas at standard state conditions.

• balance for one mole of the product

(s) K + Mn + O2 (s) (s) (g) 2

standard state conditions = 25oC and 1 atm

compound elements

• All components of a reaction can be related back

to their original elements.

• Each compound has an enthalpy of formation

associated with it.

• Reactants require energy to return to component

elements.

• Products release energy when formed form

component elements.

• Since enthalpy is a state function, the sum of the

above must relate somehow to the overall

enthalpy of a reaction.

Enthalpy Changes for a Reaction: Using Standard Enthalpy Values

C3H8(g) + 5O2(g) 3CO2(g) + 4 H2O(l)

• In order to make CO2(g) and H2O(l) one must break

the propane up into its elements.

• This takes energy.

• The elements carbon, hydrogen and oxygen then

combine to make the new compounds, CO2 and

H2O.

• This process releases energy.

3C(s) + 8H2(g) + 5O2(g)

Consider the Combustion of Propane

( ) ( )° °D ° = D - Då år f fH n H products m H reactants

The sum of the fH° for the products multiplied by

the respective coefficients, subtracted by the sum of

the fH° for the reactants multiplied by the

respective coefficients, yields the Hf° for the

reaction.

n and m are the stoichiometric balancing coefficients.

In other words: Energy gained – Energy spent = Net Energy

Enthalpy Changes for a Reaction: Using Standard Enthalpy Values

Hf° [CaO(s)] = – 635.5 kJ/mol

CaO(s) + CO2(g) CaCO3(s)

Hf° [CO2(g)] = – 393.5 kJ/mol

Hf° [CaCO3(s)] = –1207 kJ/mol

Problem:

Calculate the Hr° for CaCO3(s) formed from CaO and CO2

given the following Hf°

Hf° [CaO(s)] = – 635.5 kJ/mol


Hf° [CO2(g)] = – 393.5 kJ/mol

Hf° [CaCO3(s)] = –1207 kJ/mol

( )

( )

°

°

D ° = D

D

å

å

r f

f

H n H products

m H reactants

Problem:

Calculate the rH° for CaCO3(s) given the following Hf°

Problem:

Calculate the rH° for CaCO3(s) given the following fH°

fH° [CaO(s)] = – 635.5 kJ/mol


fH° [CO2(g)] = – 393.5 kJ/mol

fH° [CaCO3(s)] = –1207 kJ/mol

( )

( )

°

°

D ° = D

D

å

å

r f

f

H n H products

m H reactants

rH° = –1207 kJ/mol

1mol CaCO3

– (–635.5 kJ/mol

1mol CaO

–393.5 kJ/mol)

1mol CO2

rH° = – 178 kJ/mol

Problem:

The combustionH for naphthalene, C10H8(l) is – 5156 kJ/mol.

Calculate the Hf° for naphthalene given the following

enthalpies of formation:

Hf° [CO2(g)] = –393.5 kJ/mol

Hf° [H2O(l)] = –285.7 kJ/mol

C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)

Step 1: Write the balanced equation for the reaction…

Problem:


Calculate the fH° for naphthalene given the following


fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– { fH° [C8H10(l)] + 12 fH° [O2(g)]}

combustionH° =

( ) ( )° °D ° = D - Då år f fH n H products m H reactants

Next recall that:

C10H8(l) + 12 O2(g) 10CO2(g) + 4H2O(l)

From the problem, all quantities are know but f 10 8H C H

Problem:




fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

fH° [C10H8(l)] = 10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– { combH° + 12 fH° [O2(g)]}

elements = 0

Problem:




fH° [CO2(g)] = –393.5 kJ/mol

fH° [H2O(l)] = –285.7 kJ/mol

fH° [C10H8(l)] = 10 fH° [CO2(g)] + 4 fH° [H2O(l)]

– { combH° + 12 fH° [O2(g)]}

elements = 0 H° f [C10H8(l)] =

10 (–393.5 kJ/mol + 4 (–285.7 kJ/mol) – (– 5156 kJ/mol)

= + 79 kJ/mol

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