Week 8.1 - Thermodynamic and Equilibria

(1)

Prepared by:

Mrs Faraziehan Senusi

PA-A11-7C

Physical Transform

Physical Transformation of ation of Pure SubstancesPure Substances

Chemical Equilibrium Chemical Equilibrium

Chapter 4

Chapter 4 Thermodynamic and Equilibria

Thermodynamic and Equilibria

First Law of Thermodynamics First Law of Thermodynamics

Reference: Chemistry: the Molecular Nature of Matter and

Reference: Chemistry: the Molecular Nature of Matter and Change,Change,

Second Law of Thermodynamics Second Law of Thermodynamics

Simple Mixtures Simple Mixtures

(2)

••

Thermo

dynamics ~

dynamics ~ energy changes that accompany

energy changes that accompany

physical

physical and

and chemical

chemical processes.

processes. Usually

Usually these

these

energy changes involve

energy changes involve heat.

heat.

•• Thermochemistry ~

Thermochemistry ~ is the study of heat change in

is the study of heat change in

chemical reactions.



concerned with how we

concerned with how we observe, measure, and

observe, measure, and

predict energy

predict energy changes for

changes for both physical

both physical changes and

changes and

chemical reactions



use energy changes to tell whether or not a given

process

process can

can occur

occur under

under specified

specified conditions

conditions to

to give

give

predominan

predominantly

tly products

products (or

(or reactants)

reactants) and

and how

how to

to

make a process more (or less) favorable.

Thermochemistry

(3)

•• System :

System :



specific part of the universe that is of interest to us.



The substances involved in the chemical and

physical changes that we are studying

•• Surroundings: the rest of the universe

Surroundings: the rest of the universe

surrounding

boundary

system

Basic concepts

(4)

••

Open system

: can exchange

mass and energy (heat) with

surrounding

••

system where mass and energy

can cross the boundary

••

Closed

system

:

allows

transfer of heat but not mass

••

consists of a fixed amount of

mass and no mass can cross its

boundary

••

Energy in the form of heat or

work can cross the boundary

••

Isolated system

: does not

allow transfer either mass or

energy

••

A system where no mass, heat

system system surroundings surroundings matter matter energy energy system system surroundings surroundings matter matter energy energy system system surroundings surroundings matter matter energy energy

(5)

THE FIRST LAW OF THERMODYNAMICS

•

The first law of thermodynamics (the conservation of energy

principle)

principle) provides a sound basis for studying the relationships among the

provides a sound basis for studying the relationships among the

various forms of energy and energy interactions.

The first law states that

e

en

ne

err gy c

gy can

an be

be n

ne

eiith

the

er

r

cr

cre

eate

ated nor

d nor de

des

str

tr oy

oye

ed;

d; iit can

t can onl

onl y c

y ch

hange

ange ffor

orms

ms..

When a rock falls, the

decrease in potential

energy is equals to the

increase in kinetic

energy.

The increase in the energy of a potato

in an oven is equal to the amount of

heat transferred to it.

(6)

INTERNAL ENERGY,

E

Each particle in a system has po

Each particle in a system has potential and kinetic energy;

tential and kinetic energy;

the sum of these energies for all particles in a system is the

internal energy,

internal energy, E

E ..

In a

In a chemical

chemical reaction:

reaction: when

when reactants

reactants are

are converted

converted to

to

products,

products, E

E changes (

changes (

=

= E

E

_final_final

-- E

E

_initial_initial

=

= E

E

_products_products

-- E

E

_reactants_reactants

E

E ).

).

E

(7)

Energy diagrams for the transfer of internal energy

Energy diagrams for the transfer of internal energy ((

E

))

between a system and its surroundings

(8)

q

q +

+ w

w

wherewhere q q = heat and= heat and w w = work = work

•

Energy transfer outward from the system or inward from

the surroundings can appear in two forms,

the surroundings can appear in two forms, heat and work

heat and work

..

Heat and Work

•

Heat (or thermal energy, symbol q) is the energy transferred

between

between a

a system

system and

and its

its surroundings

surroundings as

as a

a result

result of

of aa

difference in their

difference in their temperatures only.

temperatures only.

•

All other forms of energy transfer (mechanical, electrical, and

so on) involve some type of work (w), the energy transferred

when an object is moved by a force.

E

(9)

Sign Conventions for

q

,

w

and

q

₊₊

w

₌₌

+

-

+

-

depends on

depends on magnitudesmagnitudes of of q q

and

and w w

depends on

depends on magnitudesmagnitudes of of q q

and

and w w

For

qq

::

(+) means system gains heat, (-) means system loses heat.(+) means system gains heat, (-) means system loses heat.

E

•

The numerical values of

The numerical values of q

q and

and w

w can be either

can be either positive

positive

or

negative

, depending on the change the

, depending on the change the system

system undergoes.

undergoes.

•

Energy coming

Energy coming into

into the system is

the system is positive

positive

; energy

; energy going

going out

out

fro

(10)

D

E

_universe

=

D

E

_system

+

D

E

_surroundings

=

= 0

0 Units of Energy

Units of Energy

joule (J)

calorie (cal)

British Thermal Unit

1 1 cal

cal =

= 4.184

4.184 JJ

1 1 J

J =

= 1

1 kg

kg m

m

22

_/s

22

1 1 Btu

Btu =

= 1055

1055 JJ

Law of Conservation of Energy

(First Law of

(First Law of Thermodynamic

Thermodynamics)

s)

The energy of the

The energy of the system plus the energy of the

system plus the energy of the surroundings remains constant:

surroundings remains constant:

energy is conserved.

(11)

Thermodynamic state of a system

•

The properties of a system

The properties of a system —

— such as P, V, T

such as P, V, T —

— are

are

called

called state functions

state functions

•

The value of a state function

The value of a state function depends only on the

depends only on the

state of the system

and not on the way in which the

system came to be in that state.

•

A change in a state function describes a difference

between

between the

the two

two states.

states. It

It is

is independent

independent of

of the

the

process or

(12)

•

The most important use of state functions in

thermodynamics is to describe

thermodynamics is to describe changes

changes

..

Δ

X =

Δ

X

_final_final

–

Δ

X

_initial_initial

•

When X increases, the final value is greater than

the initial value, so

Δ

X is positive; a decrease in X

makes

(13)

Calorimetry

•• We can

We can determine the energy change

determine the energy change

associated with a

chemical or physical process by using an experimental

technique called

technique called calorimetry

calorimetry

..

•• This technique is based on observing the

This technique is based on observing the temperature

temperature

change

when a system

when a system absorbs or releases energy

absorbs or releases energy

in the

form of heat.

•• The experiment is carried out in a device called a

The experiment is carried out in a device called a

calorimeter

, in which the temperature change

, in which the

temperature change

of a known

amount of substance (often water) of known specific heat

is measured

..

•• The temperature change is caused by the absorption or

The temperature change is caused by the absorption or

release of heat by the chemical or physical process under

study.

(14)

•

Specific heat, c

: amount required to raise temperature of one

gram of the substance by one degree Celsius (J/g.

oo

C)

••

Heat capacity

or calorimeter constant,

or calorimeter constant, C

C

: amount of heat

required to raise the temperature of a given quantity of the

substance by one degree Celsius (J/

oo

_C)

C = mc

(where m is the mass)

•• If specific heat and amount of substance is known, change in

If specific heat and amount of substance is known, change in

sample temperature

sample temperature ∆T

∆T

will tell us the

will tell us the amount of heat, q

amount of heat, q

, that

has been released/ absorbed in particular process.

q =

mc∆Tmc∆T

•• q is positive (endothermic) and negative for exothermic

q is positive (endothermic) and negative for exothermic

process

(15)

•• Calorimeter can be used to

Calorimeter can be used to measure the amount of

measure the amount of

heat absorbed or released

when a reaction takes place

in aqueous solution.

(16)

Example 1

We add

We add 3.358 kJ of heat

3.358 kJ of heat

to a calorimeter that contains

to a calorimeter that contains 50.00 g of

50.00 g of

water

. The

. The temperature

temperature

of the water and the calorimeter, originally

at

at 22.34°C, increases to 36.74°C

22.34°C, increases to 36.74°C

. Calculate the heat capacity of the

calorimeter in J/°C. The

calorimeter in J/°C. The specific heat of water

specific heat of water

is

is 4.184 J/g.°C

4.184 J/g.°C

..



Calculate the amount of heat gained by the water in the

calorimeter.



The rest of the heat must have been gained by the

calorimeter.



(17)

Example 1

amount of heat gained by the water amount of heat gained by the water

amount of heat gained by the calorimeter amount of heat gained by the calorimeter

(18)

A

A 50.0 mL sample of 0.400 M copper(II) sulfate

50.0 mL sample of 0.400 M copper(II) sulfate

solution at

solution at 23.35°C

23.35°C

is

mixed with

mixed with 50.0 mL of 0.600M sodium hydroxide solution

50.0 mL of 0.600M sodium hydroxide solution

, also at

23.35°C

, in the coffee-cup calorimeter.

, in the coffee-cup calorimeter. Heat capacity of calorimeter

Heat capacity of calorimeter

is

24.0 J/

°°

C

..

After the reaction occurs, the temperature of the resulting

mixture is measured to be

mixture is measured to be 25.23°C

25.23°C

. The

. The density of the final solution

density of the final solution

is

1.02 g/mL

. Calculate the

. Calculate the amount of heat evolved

amount of heat evolved

. Assume that the

specific heat of the solution

is the same as that of pure water,

is the same as that of pure water, 4.184

4.184 J/g.°C.

J/g.°C.

Example 2



The amount of heat released by the reaction is absorbed by the

calorimeter and by the solution.



To find the amount of heat absorbed by the solution, we must

know the mass of solution; to find that, we assume that the

volume of the reaction mixture is the sum of volumes of the

original solutions.

(19)

Example 2

(20)

•• Two common types are :

Two common types are :



constant-pressure calorimeters

A "coffee-cup" calorimeter is often used to

measure the heat transferred (q

_p_p

) in processes

open to the atmosphere.



constant-volume calorimeters

One type of constantvolume apparatus is the

bomb

bomb calorimeter,

calorimeter, designed

designed to

to measure

measure very

very

precisely

precisely the

the heat

heat released

released in

in a

a combustion

combustion

reaction.

The practice of calorimetry

(21)

Constant-pressure Calorimetry

•

One common use is

One common use is to find the specific heat

to find the specific heat

capacity of a solid

that

that does not react with or

does not react with or

dissolve in water

..

•

The solid (system) is weighed, heated to some

known temperature, and added to a sample of

water (surroundings) of known temperature and

mass in the calorimeter.

•

With stirring, the final water temperature, which

is also the final temperature of the solid, is

measured.

•

The heat lost by the system (-q

_sys_sys

, or -q

_solid_solid

)

) is

is

equal in magnitude but opposite in sign to the

heat gained by the surroundings (+q

_surn_surn

or +q

_H2O_H2O

):

- q

_solid_solid

= q

_H2O_H2O

Or Or,,

- (c

(22)

Determining the Specific Heat Capacity of a Solid

PROBLEM:

PROBLEM: AA 25.64 g sample of a solid25.64 g sample of a solid waswas heatedheated in a test tubein a test tube to 100.00to 100.00 ooCC in boilingin boiling water and carefully added to a coffee-cup calorimeter containing

water and carefully added to a coffee-cup calorimeter containing 50.00 g of 50.00 g of

water

water . The. The water temperature increased from 25.10water temperature increased from 25.10 ooC to 28.49C to 28.49 ooCC.. What is the

What is the specific heat capacity of the solidspecific heat capacity of the solid? (Assume all the heat is? (Assume all the heat is gained by the water)

gained by the water)

SOLUTION:

PLAN:

PLAN: It is helpful to use a table to summarIt is helpful to use a table to summarize the data given. ize the data given. Then work the pThen work the problemroblem realizing that heat lost by the system must be equal to that gained by the realizing that heat lost by the system must be equal to that gained by the surroundings.

surroundings.

mass (g)

mass (g) c c (J/g(J/g..K)K) _T_T_initial_initial _T_T_final_final 25.64 25.64 ?? 100.00100.00 28.4928.49 -71.51-71.51 50.00 50.00 4.1844.184 25.1025.10 28.4928.49 3.393.39 solid solid H H₂₂OO cc xx 25.64 g 25.64 g xx -71.51 -71.51 K K = = - - 50.00 g50.00 g xx 4.184 4.184 J/gJ/g..K K xx 3.39 K 3.39 K cc_solid_solid == -- 50.00 g50.00 g xx 4.184 J/g4.184 J/g..K K xx 3.39 K 3.39 K 25.64 g 25.64 g xx -71.51 K -71.51 K = = 0.387 0.387 J/gJ/g..K K T T

Example 3

(23)

Constant-volum

Constant-volume

e Calorimetry

Calorimetry

•

Figure 6.8 depicts the preweighed

combustible sample in a metal-walled

chamber (the bomb), which is filled

with oxygen gas and immersed in an

insulated water bath fitted with

motorized stirrer and thermometer.

•

A heating coil connected to an

electrical source ignites the sample, and

the heat evolved raises the temperature

of the bomb, water, and other

calorimeter parts.

•

Because we know the mass of the

sample and the heat capacity of the

entire calorimeter, we can use the

measured

ΔΔ

T to calculate the heat

released.

(24)

Calculating the Heat of Combustion

PROBLEM:

PROBLEM: A manufacturer claims that its new diet dessert hasA manufacturer claims that its new diet dessert has ―fewer ―fewer than 10than 10 Calories (10 kcal) per

Calories (10 kcal) per serving‖serving‖. . TTo o test the test the claim, a claim, a chemist chemist at theat the Department of Consumer Affairs places one serving in a bomb Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O

calorimeter and burns it in O₂₂ (the heat capacity of the calorimeter =(the heat capacity of the calorimeter = 8.151 kJ/K). The temperature increases by 4.937

8.151 kJ/K). The temperature increases by 4.937 ooC. Is theC. Is the manufacturer’s

manufacturer’s claim correct?claim correct?

SOLUTION:

PLAN:

PLAN: -- q q _sample_sample == q q _calorimeter_calorimeter q

q _calorimeter_calorimeter = heat capacity x= heat capacity x DDTT = 8.151 = 8.151 kJ/K kJ/K x 4.93x 4.937 K 7 K = 40.24 kJ = 40.24 kJ 40.24 40.24 kJ kJ x x kcalkcal 4.184 kJ 4.184 kJ

= 9.62 kcal < 10 Calories = 10 kcal = 9.62 kcal < 10 Calories = 10 kcal

The manufacturer’s claim is correct.

Example 4

(25)

Enthalpy

•• The

The quantity of heat transferred into or out of a

quantity of heat transferred into or out of a

system

as it undergoes a chemical or physical

change at constant pressure.

•• Extensive property

Extensive property

: magnitude

: magnitude depends on amount

depends on amount

of substance present

•• Impossible to determine enthalpy of substance

Impossible to determine enthalpy of substance

(26)

•• Enthalpy of reaction,

Enthalpy of reaction, ∆H

∆H

∆H = H

= H

_(product)_(product)

–

– H

H

_(reactant)_(reactant)

•

Exothermic : negative

•• Endothermic: positive

Endothermic: positive

•• EXOTHERMIC PROCESS

EXOTHERMIC PROCESS –

– a process that

a process that

releases energy in the form of heat into its

surroundings. (Ex: combustion reaction)

•• ENDOTHERMIC PROCESS

ENDOTHERMIC PROCESS –

– a process that

a process that

absorbs energy from its surroundings

(27)

Enthalpy diagr

Enthalpy diagrams for exothermic

ams for exothermic and endothermic processes

and endothermic processes

CH CH₄₄((g g ) + 2O) + 2O₂₂((g g ) ) COCO₂₂((g g ) + 2H) + 2H₂₂O(O(g g ) +) + heatheat heat heat + H+ H₂₂O(O(s s ) ) HH₂₂O(O(l l ))

Heat is released;

enthalpy

enthalpy decreases

decreases..

Heat is absorbed;

enthalpy

(28)

Some Important Types of Enthalpy Change

heat of combustion (

heat of formation (

heat of fusion (

heat of vaporization (

1 1CC₄₄HH₁₀₁₀((l l ) ) ++ 13/213/2OO₂₂((g g ) ) 4CO4CO₂₂((g g ) + 5H) + 5H₂₂O(O(g g )) K( K(s s ) +) + 1/21/2Br Br ₂₂((l l )) 11KBr(KBr(s s )) 1

1 NaCl( NaCl(s s ) ) NaCl(NaCl(l l ))

1 1CC₆₆HH₆₆((l l ) ) CC₆₆HH₆₆((g g )) Standard quantity Standard quantity of either reactant of either reactant or

or product: product: 1 1 molmol

H

_comb_comb

))

H

_f_f

))

H

_fus_fus

))

H

_vap_vap

))

(29)

Thermochemical equations

•• A balanced chemical equation, together with its

A balanced chemical equation, together with its

value of

Δ

H

•• The

The

Δ

H

_rxn_rxn

value shown refers to the amounts (moles)

of substances and their states of matter in that

specific equation.

Combustion of methane :

CH

₄₄

(( g

g ) + 2O

) + 2O

₂₂

(( g

g ))

_

CO

₂₂

(( g

g ) + 2H

) + 2H

₂₂

O(

O(l

l ))

∆

∆H= -890.4kJ

H= -890.4kJ

(30)

••

1367 kJ of heat is released when one mole of C

₂₂

H

₅₅

OH(l) reacts

with three moles of O

₂₂

(g) to give two moles of CO

₂₂

(g) and three

moles of H

₂₂

O(l).

•• We can refer to this amount of reaction as one mole of reaction,

We can refer to this amount of reaction as one mole of reaction,

which we abbreviate

which we abbreviate ―mol

―mol

rxn.

rxn.‖‖

•• We can also write the thermochemical equation as

We can also write the thermochemical equation as

•

We always interpret

ΔΔ

H as the enthalpy change for the reaction

as written; as (enthalpy change)/(mole of reaction), where the

denominator means

denominator means ―for

―for the number of moles of each substance

the number of moles of each substance

shown in the balanced equation.

(31)

•• Guidelines

Guidelines

for

writing

and

interpreting

thermochemical equations

–

– Stoichiometric coefficient refer to number of moles of

Stoichiometric coefficient refer to number of moles of

substance

–

– Reverse equation, magnitude of

Reverse equation, magnitude of ∆H

∆H same but sign

same but sign

changes

–

– Multiply equation by factor of n then

Multiply equation by factor of n then ∆H

∆H must also

must also

change by the same factor

–

(32)

AMOUNT (mol)

of compound A

AMOUNT

(mol)(mol) of compound B of compound B

HEAT (kJ)

gained or lost

molar ratio from molar ratio from balanced equation balanced equation

(kJ/mol) (kJ/mol)

Summary of the relationship between

amount (mol) of substance and

the heat

(kJ) transferred during a reaction

H H _rxn_rxn

(33)

Using the Heat of Reaction (

SOLUTION:

PLAN:

PROBLEM:

PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminumThe major source of aluminum in the world is bauxite (mostly aluminum oxide).

oxide). Its thermal Its thermal decomposition decomposition can be can be represented by:represented by:

If aluminum is produced this way, how many grams of aluminum can form If aluminum is produced this way, how many grams of aluminum can form when 1.000 x 10

when 1.000 x 1033kJ of heat is transferred?kJ of heat is transferred? Al

Al₂₂OO₃₃(( s s) ) 2Al(2Al( s s) + 3/2O) + 3/2O₂₂(( g g )) DD H H _rxn_rxn = 1676 kJ= 1676 kJ

heat (kJ) heat (kJ) mol of Al mol of Al g of Al g of Al 1676 kJ = 1676 kJ = 2 mol Al2 mol Al x x M M (g/mol)(g/mol) 1.000 x 10 1.000 x 1033kJ kJ xx 2 mol Al2 mol Al 1676 kJ 1676 kJ 26.98 g Al 26.98 g Al 1 mol Al 1 mol Al = 32.20 g Al = 32.20 g Al x x

Example

(34)

Specifying Standard States

For a gas, the standard state is 1 atm; ideal gas behavior is For a gas, the standard state is 1 atm; ideal gas behavior is assumed.

assumed.

For a substance in aqueous solution, the standard state is For a substance in aqueous solution, the standard state is 1

1 M M concentration (1 mol/liter solution).concentration (1 mol/liter solution).

For a pure substance (element or compound), the standard state For a pure substance (element or compound), the standard state is usually the most stable form of the substance at 1 atm and the is usually the most stable form of the substance at 1 atm and the temperature of interest (usually 25

temperature of interest (usually 25 ooC (298 K).C (298 K).

= standard heat of reaction

(enthalpy change determined with all

substances in their standard states)

Standard heat of reaction

H

Hoo

rxn

(35)

Standard enthalpy of formation

•• Standard enthalpy of formation (

Standard enthalpy of formation (∆H

∆H

oo_f_f

) : heat

change that results when one mole of a compound

is formed from its elements at a pressure of 1 atm

(standard state).

• • ∆H

∆H

oo_f_f

of any element in its most stable form is

zero

∆H

(36)

Formation Equations

In a formation equation, 1 mol of a compound forms

from

from its

its elements.

elements. The

The standard

standard heat

heat of

of formation

formation

((

DD

H

oo

f

) is the enthalpy change for the formation

equation when all substances are in their standard

states.

C(graphite) + 2H

C(graphite) + 2H₂₂(( g g ) ) CHCH₄₄(( g g )) DD H H oo_f_f= -74.9 kJ= -74.9 kJ

An element in its standard state is assigned a

DD

H

oo f

f

of 0.

Direct method of measuring

∆H

o

f

C

_(graphite)_(graphite)

+ O

_{2 (g)}_{2 (g)}

CO

_{2 (g)}_{2 (g)}

∆

∆H

H

oo rxn rxn

= - 393.5kJ

(37)

T

Table able 6.5 6.5 Selected Selected Standard Standard Heats Heats of of Formation Formation at at 2525 ooC (298 K)C (298 K) Formula Formula calcium calcium Ca( Ca(s s )) CaO( CaO(s s )) CaCO CaCO₃₃((s s )) carbon carbon C(graphite) C(graphite) C(diamond) C(diamond) CO( CO(g g )) CO CO₂₂((g g )) CH CH₄₄((g g )) CH CH₃₃OH(OH(l l )) HCN( HCN(g g )) CS CS₂₂((l l )) chlorine chlorine Cl( Cl(g g )) 0 0 -635.1 -635.1 -1206.9 -1206.9 0 0 1.9 1.9 -110.5 -110.5 -393.5 -393.5 -74.9 -74.9 -238.6 -238.6 135 135 87.9 87.9 121.0 121.0 hydrogen hydrogen nitrogen nitrogen oxygen oxygen Formula Formula H( H(g g )) H H₂₂((g g )) N N₂₂((g g )) NH NH₃₃((g g )) NO( NO(g g )) O O₂₂((g g )) O O₃₃((g g )) H H₂₂O(O(g g )) H H₂₂O(O(l l )) Cl Cl₂₂((g g )) HCl( HCl(g g )) 0 0 0 0 0 0 -92.3 -92.3 0 0 218 218 -45.9 -45.9 90.3 90.3 143 143 -241.8 -241.8 -285.8 -285.8 107.8 107.8 Formula Formula silver silver Ag( Ag(s s )) AgCl( AgCl(s s )) sodium sodium Na( Na(s s )) Na( Na(g g )) NaCl( NaCl(s s )) sulfur sulfur S S₈₈(rhombic)(rhombic) S S₈₈(monoclinic)(monoclinic) SO SO₂₂((g g )) SO SO₃₃((g g )) 0 0 0 0 0 0 -127.0 -127.0 -411.1 -411.1 2 2 -296.8 -296.8 -396.0 -396.0 H

(38)

Indirect method measuring

∆H

o

f

•• Many compounds cannot be directly synthesized from

Many compounds cannot be directly synthesized from

their elements (proceed too slowly/ side reactions produce

other substance than desired compound)

• • H

Heessss’’ss law: when reactants are converted to products, the

law: when reactants are converted to products, the

change in enthalpy is the same whether the reaction takes

place in one step or in a series

place in one step or in a series of steps

of steps

•• General rule of applying:

General rule of applying:

–

– Arrange series of chemical equation in a way that when

Arrange series of chemical equation in a way that when

added together all species will cancel except reactant

and product that appear in the

and product that appear in the overall reaction

overall reaction

–

(39)

Hess’s Law of Heat Summation

The

The enthalpy change of an overall process

enthalpy change of an overall process

is the

is the sum of

sum of

the enthalpy changes of its individual steps

..

Used to predict the enthalpy change (a) of an overall reaction

that cannot be studied directly, and/or (b) of an overall reaction

that can be separated into distinct reactions whose enthalpy

changes can

(40)

Using

Using Hess’sHess’s _{Law to Calculate an Unknown}_{Law to Calculate an Unknown}

SOLUTION:

PLAN:

PROBLEM:

PROBLEM: TTwo gaseouwo gaseous pollutants s pollutants that form that form auto exhaust auto exhaust are CO are CO and NO. and NO. AnAn environmental chemist is studying ways to

environmental chemist is studying ways to convert them into less harmfulconvert them into less harmful gases through the following equation:

gases through the following equation: CO(

CO(g g ) + NO() + NO(g g ) ) COCO₂₂((g g ) ) + + 1/2N1/2N₂₂((g g )) DD H H = ?= ? Given the following information, calculate the

Given the following information, calculate the unknownunknown DD H H .. Equation

Equation A: A: CO(CO(g g ) +) + 1/21/2OO₂₂((g g ) ) COCO₂₂((g g )) DD H H _A_A = -283.0 kJ= -283.0 kJ

Equation

Equation B: B: NN₂₂((g g ) + O) + O₂₂((g g ) ) 2NO(2NO(g g )) DD H H _B_B = +180.6 kJ= +180.6 kJ

Equations

Equations A and B have to be manipulated by reversal A and B have to be manipulated by reversal and/or multiplied byand/or multiplied by factors in order to sum to the target equation.

factors in order to sum to the target equation. Multiply Equation B by 1/2 and reverse it. Multiply Equation B by 1/2 and reverse it.

D

D H H _B_B = = -90.3 -90.3 kJkJ

CO(

CO(g g ) +) + 1/21/2OO₂₂((g g ) ) COCO₂₂((g g )) DD H H _A_A = -283.0 kJ= -283.0 kJ NO(

NO(g g ) ) 1/2N1/2N₂₂((g g ) + 1/2O) + 1/2O₂₂((g g ))

H

D

D H H _rxn_rxn = -373.3 = -373.3 kJkJ CO(

CO(g g ) + NO() + NO(g g ) ) COCO₂₂((g g ) ) + + 1/2N1/2N₂₂((g g ))

Example

(41)

•• From

From ∆H

∆H

oo f

f

values, the standard enthalpy of

reaction

reaction ∆

∆H

H

oo_rxn_rxn

can be calculated

aA + bB



cC + dD

∆

∆H

H

oo_rxn_rxn

= [c∆H

= [

c∆H

oo_f_f

(C) +

(C) + d∆H

d∆H

oo_f_f

(D)]

(D)] –

– [[a∆H

a∆H

oo_f_f

(A) +

(A) + b∆H

b∆H

oo_f_f

((

B)]

∆

∆H

H

oo rxn

(42)

The general process for determining

H

oo

from

values

rxn

(43)

Calculating the Standard Heat of Reaction from

Standard Heats of Formation

SOLUTION:

PROBLEM:

PROBLEM: Nitric acid, Nitric acid, whose whose worldwide worldwide annual annual production production is is about about 8 8 billion billion kg, kg, isis used to make many products, including fertilizers, dyes and explosives. used to make many products, including fertilizers, dyes and explosives. The first step in the industrial production process is the oxidation of The first step in the industrial production process is the oxidation of ammonia:

ammonia:

Calculate

Calculate DD H H oo_rxn_rxn fromfrom DD H H oo_f_fvalues.values. 4NH

4NH₃₃((g g ) + 5O) + 5O₂₂((g g ) ) 4NO(4NO(g g ) + 6H) + 6H₂₂O(O(g g ))

D

D H H oo_rxn_rxn == SS nnDD H H oo_f_f(products) -(products) - SS mmDD H H oo_f_f(reactants)(reactants)

D

D H H oo_rxn_rxn = [4= [4DD H H oo_f_fNO(NO(g g ) + 6) + 6DD H H oo_f_fHH₂₂O(O(g g )] )] - - [4[4DD H H oo_f_fNHNH₃₃((g g ) + 5) + 5DD H H oo_f_fOO₂₂((g g )])] = [(4 mol)(90.3 kJ/mol) + (6 mol)(241.8 kJ/mol)]

= [(4 mol)(90.3 kJ/mol) + (6 mol)(241.8 kJ/mol)]

-[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]

D

D H H oo_rxn_rxn = -906 kJ= -906 kJ

Example

(44)

Bond Energies

•

Chemical reactions involve the breaking and making of chemical

bonds.

•

Energy is always required to break a chemical bond.

•

The

The bond energy (B.E.) is the amount of energy necessary to

bond energy (B.E.) is the amount of energy necessary to

break

one molone mole e of of

bonds in

bonds

in a

a gaseous

gaseous covalent

covalent substance

substance to

to form

form

products in

products in the gaseous s

the gaseous state at constant

tate at constant temperature

temperature and pressure.

and pressure.

•

The greater the bond energy, the more stable (stronger) the bond is,

and the harder it is to break. Thus bond energy is a measure of bond

strengths.

•

A special case of

A special case of Hess’s

Hess’s Law involves the use of bond energies to

Law involves the use of bond energies to

estimate heats of

(45)

A schematic representation of the relationship between bond energies A schematic representation of the relationship between bond energies and

and ΔΔ H H _rxn_rxn for gas phase for gas phase reactions.reactions.

(a)

(46)

(47)

Example 1

Use the bond energies listed in Table 15-2 to estimate the heat

of reaction at 298 K for the following reaction:

(48)

Example 2 Example 2

Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K Use the bond energies listed in Table 15-2 to estimate the heat of reaction at 298 K for the following reaction: