26771935 Matriculation Chemistry Thermochemistry

7/30/2019 26771935 Matriculation Chemistry Thermochemistry

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9.0 THERMOCHEMISTRY


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Concept of Enthalpy


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Important Terms

Heat is energy transferred between two bodies ofdifferent temperatures

System is any specific part of the universe

Surroundings is everything that lies outside the

system


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Open system is a system that can exchangemass and energy with its surroundings

Closed system is a system that allows theexchange of energy with its surroundings

Isolated system is a system that does not allowthe exchange of either mass or energy with itssurroundings


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Energy is the ability to do work

SI unit of energy is kg m2

s-2

orJoule (J) Non SI unit of energy is calorie (Cal)

1 Cal = 4.184 J


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Thermochemistry

A study of heat change in chemical reactions.

Two types of chemical reactions:

Exothermic Endothermic


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Exothermic reactions

Enthalpy of products < Enthalpy of reactants, H is

negative.

Energy is released from the system to the

surroundings.


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Consider the following reaction:

A (g) + B (g) C (g) H = ve

(reactants) (product)

reactantsenthalpy

reaction pathway

= -veH

products

Energy profile diagram for exothermic reaction


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Enthalpy of products > enthalpy of reactants, H ispositive

Energy is absorbed by the system from the surrounding

Endothermic Reactions


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Consider the following reaction

A (g) + B (g) C(g) H = + ve

(reactants) (product)

Energy profile of diagram endothermic reactions


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Enthalpy, H

The heat content of a system or total energy in thesystem

Enthalpy, H of a system cannot be measured whenthere is a change in the system.

Example: system undergoes combustion or ionisation.


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Enthalpy of Reaction, H andStandard Condition

Enthalpy of reaction:

The enthalpy change associated with a chemicalreaction.

Standard enthalpy, H

The enthalpy change for a particular reaction thatoccurs at 298K and 1 atm(standard state)


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Thermochemical Equation The thermochemical equation shows the enthalpy changes.

Example : H2O(s) H2O(l) H = +6.01 kJ

1 mole ofH2O(l) is formed from 1 mole ofH2O(s) at 0C,H = +6.01 kJ

However, when 1 mole ofH2O(s)is formed from 1 mole ofH2O(l), the magnitude ofH remains the same with theoppositesignof it.

H2O(l) H2O(s) H = 6.01 kJ


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Types of Enthalpies

There are many kind of enthalpies such as:

Enthalpy of formation

Enthalpy of combustion

Enthalpy of atomisation

Enthalpy neutralisation

Enthalpy hydration

Enthalpy solution


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Enthalpy of Formation, Hf

The change of heat when 1 mole of a compound is

formed from its elements at their standard states.

H2(g) + O2(g) H2O (l) Hf= 286 kJ mol1

The standard enthalpy of formation of any element in

its most stable state form is ZERO.

H (O2 ) = 0 H (Cl2) = 0


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Enthalpy of Combustion, Hc

The heat released when 1 mole of substance is

burned completely in excess oxygen.

C(s) + O2(g) CO2(g) Hc = 393 kJ mol1


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Enthalpy of Atomisation, Ha

The heat change when 1 mole ofgaseous atoms is formedfrom its element

Ha is always positive because it involves only breaking ofbonds

e.g:

Na(s)

Na(g)

Ha = +109 kJ mol-1

Cl2(g) Cl(g) Ha = +123 kJ mol-1


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Enthalpy of Neutralization, Hn

The heat change when 1 mole of water, H2O is formed from the

neutralization of acid and base .

HCl(aq)+ NaOH(aq) NaCl(aq) +H2O(aq) Hn = 58 kJ mol1


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Enthalpy of Hydration, Hhyd

The heat change when 1 mole ofgaseous ionsis

hydratedin water.

e.g:Na+(g) Na

+(aq) Hhyd = 406 kJ mol

-1

Cl-(g) Cl-(aq) Hhyd = 363 kJ mol

-1


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Enthalpy of Solution, Hsoln

The heat change when 1 mole of a substance is

dissolves in water.

e.g:KCl(s) K

+(aq) + Cl

(aq) Hsoln = +690 kJ mol

-1


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Enthalpy of Sublimation, Hsubl

The heat change when one mole of a substance

sublimes (solid into gas).

I2 (s) I2(g)

sublH

Hsubl= +106 kJ mol1


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Calorimetry

A method used in the laboratory to measure the

heat change of a reaction.

Apparatus used is known as the calorimeter Examples of calorimeter

Simple calorimeter

Bomb calorimeter


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The outer Styrofoam cup

insulate the reaction

mixture from the

surroundings (it is

assumed that no heat is

lost to the surroundings)

Heat release by the

reaction is absorbed by

solution and the

calorimeter

Simple calorimeter


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Important Terms in Calorimeter

Specific heat capacity, c

Specific heat capacity, cof a substance is the amount of

heatrequired to raise the temperature ofone gramof the

substance by one degree Celsius(Jg 1 C 1). Heat capacity, C

Heat capacity,Cis the amount of heat required to raise

the temperature ofagiven quantity of the substance by

one degree Celsius (J C 1)


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q = mcT

Heat released by

substance=

Heat absorbed

by calorimeter

q = heat released by substance

m= mass of substance

C= specific heat capacityT = temperature change


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Basic Principle in Calorimeter

Heat released

by a reaction=

Heat absorbed

by surroundings

Surroundings may refer tothe:i. Calorimeter itself or;

ii. The water and calorimeter qreaction= mcT or CT


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Example 1

In an experiment, 0.100 g of H2 and excess of O2 were compressed

into a 1.00 L bomb and placed into a calorimeter with heat capacity

of9.08 x 104 J0C 1. The initial temperature of the calorimeter was

25.0000C and finally it increased to 25.155 0C. Calculate the

amount of heat released in the reaction to form H2O, expressed in kJ

per mole.


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Solution

Heat released = Heat absorbed by the

calorimeter

q = CT

= (9.08 X 104 J0C-1) X (0.1550C)

= 1.41 X 104 J

= 14.1 kJ

H2(g) + O2(g) H2O(c)

mole of H2 = 0.1002.016

= 0.0496 mol


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moles of H2O = mole of H2

0.0496 mol of H2O released 14.1 kJ energy

1 mol H2O released = kJ

= 284 kJ

Heat of reaction, H = - 284 kJ mol 1

0496.0

1.14


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Example 2

1. Calculate the amount ofheat released in a reaction inan aluminum calorimeter with a mass of3087.0g andcontains 1700.0 mL of water. The initial temperature ofthe calorimeter is 25.0C and it increased to 27.8C.


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Given:

Specific heat capacity of aluminum = 0.553Jg-1C-1

Specific heat capacity of water = 4.18 Jg-1C-1

Water density = 1.0 g mL-1T = (27.8 -25.0 )C = 2.8C

Solution

q = mwcwT + mcccT

= (1700.0 g)(4.18 Jg-1C-1)(2.8 C) +

(3087.0 g)(0.553 Jg-1 C-1)(2.8C)

= 24676.71 J

= 24.7 kJ

Heat absorbed bywater

Heat absorbed byaluminium

calorimeter

Heat released = +


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HESSS LAW


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Hess Law

Hesss Law states that when reactants are converted to

products, the change in enthalpy is the same whether the

reaction takes place in one step or in the series of steps.

The enthalpy change depends only on the nature of thereactants and products and is independent of the route

taken.

1HA B

C

3H

2H

321HHH


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i. List all the thermochemical equations involved

Algebraic Method

1-

-1560kJmolH)(2

3)(2

2)(22

7)(62

.

1-

-286kJmolH)(2)(22

1)(2

.

1-393kJmol-H

)(2)(2)(.

gOH

gCO

gO

gHCiii

gOH

gO

gHii

gCO

gO

SCi

Step 1


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i. List all the thermochemical equations involved

Algebraic Method

1-

-1560kJmolH)(2

3)(2

2)(22

7)(62

.

1-

-286kJmolH)(2)(22

1)(2

.

1-393kJmol-H

)(2)(2)(.

gOH

gCO

gO

gHCiii

gOH

gO

gHii

gCO

gO

SCi

Step 1


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ii. Write the enthalpy of formation reaction for C2H6

)(62

?

)(23

)( gHC

fH

gH

sC


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iii. Add the given reactions so that the result is the desiredreaction.

-84kJ

3H

2H

1H

fH

84kJ-)(62

?

)(23

)(2

_______________________________________________________________

1560kJ3H)(227)(62)(232(g)2CO(iii)

-286kJ32

H)(2

3)(22

3)(2

33)(

kJ-39321

H)(2

2)(2

2)(

22)(

gHC

fH

gH

sC

gOgHCgOHreverse

gOH

gO

gHii

gCO

gO

SCi


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Energy Cycle Method

Draw the energy cycle and apply Hesss Law to calculate theunknown value.

2C(s) + 3H2(g) C2H6(g)

2CO2(g) + 3H2O(g)

2O2(g) 32O2(g)HH

H

H72O2(g)

OO

O

Of

12

3=2(-393)

=3(-286)

=-(-1560)


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1-kJmol-84

1560858--786

3

H+)2

H3(+)1

H2(=f

H


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Example 1

The thermochemical equation of combustion of carbonmonoxide is shown as below.

C(s) + O2(g) CO(g) = ?

given :

C(s) + O2(g) CO2(g) H= -394 kJ mol-1

CO(s)

+ O2(g)

CO2(g)

H= -283 kJ mol-1

Calculate the enthalpy change of the combustion of carbon tocarbon monoxide.

H

H

H


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Example 2

Calculate the standard enthalpy of formation of methane if the

enthalpy of combustion of carbon, hydrogen and methane are

as follows:H [C(s)] = -393 kJ mol

-1

H [H2(s)] = -293 kJ mol-1

H [CH4(s)] = -753 kJ mol-1

cH

cH

cH


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Example 3

Standard enthalpy of formation of ammonia, hydrogenchloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJmol-1, 314.4 kJ mol-1 respectively. Write the thermochemicalequation for the formation of each substance and calculatethe enthalpy change for the following reaction.

NH3(g) + HCl (g) NH4Cl(s)


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Exercise

1.Calculate the enthalpy of formation of benzene

if :

H (CO2(g) ) = -393.3 kJ mol-1

H (H2O(l) ) = -285.5 kJ mol-1

H (C6H6(l) ) = -3265.3 kJ mol-1

fH

fH

fH


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Born-Haber

Cycle


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Lattice Energy, Hlattice

is the energy required to completely separate one mole of a

solid (ionic compound) into gaseous ions

e.g:

NaCl(s) Na+(g) + Cl-(g) Hlattice = +771 kJ mol-1(lattice dissociation)

Na+(g) + Cl-(g) NaCl(s) Hlattice = -771 kJ mol-1

(lattice formation)

The magnitude of lattice energy increases as the ionic charges increase

the ionic radii decrease


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There is a strong attraction between small ions and

highly charged ions so the

H is more negative.

H for MgO is more negative than H for Na2O

because Mg2+ is smaller in size and has bigger charge

than Na+, therefore

Hlattice (MgO) > Hlattice (Na2O)


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Na+ and Cl- ions in the solid crystal are separated fromeach other and converted to the gaseous state (Hlattice)

The electrostatic forces between gaseous ions and polarwater molecules cause the ions to be surrounded by watermolecules (Hhydr)

Hsoln = Hlattice + Hhdyr

Hydration Process of Ionic Solid

Na+ and Cl- ion in


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Heat of Solution

Na+ and Cl- ion in

the gaseous state

Na+ and Cl- ion inthe solid state

Hydrated Na+ and Cl- ion


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Born-Haber Cycle

The process of ionic bond formation occurs in a few stages. Ateach stage the enthalpy changes are considered.

The Born Haber cycle is often used to calculate the lattice

energy of an ionic compound. In the Born-Haber cycle energy diagram, by convention,

positive values are denoted as going upwards, negative valuesas going downwards.

Consider the enthalpy changes in the formation of sodium

chloride.


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Example :

Given;

i. Enthalpy of formation NaCl = -411 kJmol-1

ii. Enthalpy of sublimation of Na = +108 kJmol-1

iii. First ionization energy of Na = +500 kJmol-1

iv. Enthalpy of atomization of Cl = +122 kJmol-1

v. Electron affinity of Cl = -364 kJmol-1

vi. Lattice energy of NaCl = ?

(s)2(g)21

(s) NaClClNa


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Example: A Born-Haber cycle for NaCl

Na(s) + Cl2(g)

Na(g) + Cl2(g)

NaCl(s)

energy

E=0

Na(g) + Cl(g)

Na+(g) + e + Cl(g)

Na+(g) + Cl- (g)

HaNa

HaCl

IonisationEnergy of Na

Electron Affinity of Cl

Lattice energy

Hf NaCl

From Hesss Law:

HfNaCl = HaNa + HaCl +IENa +

EACl + Lattice Energy-ve

+ve


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Calculation:

kJ777H

kJ364kJ122kJ500kJ108kJ411H

EAHIEHHH

HEAHIEHH

lattice

lattice

)Cl(aS

0

flattice

lattice)Cl(aS

0

f


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Exercise:

Construct a Born-Haber cycle to explain why ionic compoundNaCl2 cannot form under standard conditions. Use the databelow:

i. Enthalpy of sublimation of sodium = +108 kJmol-1

ii.

First ionization energy of sodium = +500 kJmol

-1

iii. Second ionization energy of sodium = +4562 kJmol-1

iv. Enthalpy of atomization of chlorine = +121kJmol-1

v. Electron affinity of chlorine = -364 kJmol-1

vi. Lattice energy of NaCl2 = -2489 kJmol-1

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26771935 Matriculation Chemistry Thermochemistry