Chemistry Form 6 Sem 2 01

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CHAPTER 1:

THERMOCHEMISTRY

Pre – U Chemistry

Semester 2

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1.1 Introduction

Energy is one of the most fundamental parts of our universe. We use energy to do work. Energy lights our cities. Energy powers our vehicles, trains, planes and rockets. Energy warms our homes, cooks our food, plays our music, gives us pictures on television. Energy powers

machinery in factories and tractors on a farm.

According to the conservation of energy law, energy can be neither created nor destroyed; it can only be converted from one form into another

From the angle of chemistry, when a chemical reaction occur, energy From the angle of chemistry, when a chemical reaction occur, energy

changes occur generally in 2 ways, where it can be explained in terms of kinetic energy and energetic energy. In this chapter, we focus more on the study of energy changes, in the form of heat, which take place during a chemical reaction occur, which is well known as

thermochemistry.

In order to understand thermochemistry, we must first understand what is the difference between system and surrounding. System is the

specific part of substances that involved in chemical and physical change, while surrounding is defined as the rest of the universe outside the system.

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There are generally 3 types of systems.

Open system Closed system Isolated system

An open system can exchange mass and energy, usually in the form of heat with its surroundings

closed system, which allows the transfer of energy (heat) but not mass.

isolated system, which does not allow the

transfer of either mass or energy.

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We shall focus more on a closed system throughout our lesson, with the assumption that energy lost by system in a chemical reaction is the same with the energy gained by surrounding. In thermochemistry energy that were gained / lost by system were measured by heat energy. In the laboratory, heat changes in physical and chemical processes are

measured with a calorimeter, a closed container designed specifically for this purpose. Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity,

The specific heat capacity (c) of a substance is the amount of heat The specific heat capacity (c) of a substance is the amount of heat

required to raise the temperature of one gram of the substance by one degree Celsius. It has the units J g-1_°C-1_.

The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. Its units are J °C-1_.

Specific heat is an intensive property whereas heat capacity is an extensive property.

The relationship between the heat capacity and specific heat capacity of a substance is C = c x m (mass)

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1.2 Enthalpy and Enthalpy Change

Measurement of energy transferred during chemical reaction is made under control conditions. However, in a closed system, we assume that there’s no changes in the volume of a system, hence no work is done toward the heat change occur within the system. By that, we shall

deduce the energy transferred in a system is corresponding to the heat transfer towards the surrounding. Heat transfer in this case is described as enthalpy, H.

In a chemical reaction, where reactants products

The difference of energy changes occur on a chemical reaction is known as enthalpy change, ∆H, as the difference between the enthalpies of the products and the enthalpies of the reactants

∆H = [ΣΣΣΣ ∆H_product – ΣΣΣΣ ∆H_reactant].

Such enthalpy is also known as enthalpy change of reaction

Since the enthalpy changes is a quantitative value use to measure the difference by the heat given off before and after a reaction, so it may be a positive value or negative value

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Enthalpy – heat content of the system

Enthalpy changes ; ∆H ~

heat changes occur

during a chemical reaction

.

∆H = [Σ

Σ

∆H

_product

– Σ

Σ

∆H

_reactant

]

Unit = kJ mol

-1

.

Σ

_∆H

product

> Σ ∆H

reactant

Σ

∆H

product

< Σ ∆H

reactant

∆H = positive (+ve)

∆H = negative (–ve)

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Process

Endothermic

Exothermic

Definition

Process of heat

absorbed by system

Process of heat

released by system

ΔH

Positive

Negative

Energy

profile

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Process

Endothermic

Exothermic

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3 STEPS ON CALCULATING

ENTHALPY CHANGE

1 q = m c

θ

MV

mass

2 mol =

3 ∆H =

1000

MV

or

M

mass

R

mol

q

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Question 1 : Solution

Equation : Zn (s) + H

₂

SO

₄

_{(aq) ZnSO}

₄

(aq) + H

₂

(g)

Step 1 : q = m c θ

θ

_{@ q = (25.0) (4.18) (31.5 – 27.0)}

q = 470.25 J

Step 2 : determine limitant

mol of Zn = mass / mol mol of H

₂

SO

₄

= MV /1000

mol of Zn = mass / mol mol of H

₂

SO

₄

= MV /1000

= 6.00 / 65.3

= (0.100) (25.0) / 1000

= 0.0919 mol

= 0.00250 mol (lim)

Step 3 :

∆H = q / mol @ ∆H = 470.25 / 0.00250

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Question 2 : Solution

Equation : Na

₂

SO

₄

+ Ba(NO

₃

)

₂

_{2 NaNO}

₃

+ BaSO

₄

Step 1 : q = m c θ

θ

@ q = (20.0 + 30.0) (4.18) (34.0 – 30.0)

q = 836 J

Step 2 : determine limitant

mol Na

₂

SO

₄

= MV /1000 mol Ba(NO

₃

)

₂

= MV /1000

mol Na

₂

SO

₄

= MV /1000 mol Ba(NO

₃

)

₂

= MV /1000

= (0.500) (20.0) / 1000

= (0.300) (30.0) / 1000

= 0.010 mol

= 0.0090 mol (lim)

Step 3 : ∆H = q / mol @

∆H = 836 / 0.0090

∆H = 92889 J / mol – 92.9 kJ / mol

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Question 3 : Solution

Equation : 2 KI + Pb(NO

₃

)

₂

_{2 KNO}

₃

+ PbI

₂

Step 1 : q = m c θ

θ

@ q = (20 + 30) (4.18) (34 – 29)

q = 1045 J

Step 2 : determine limitant

mol KI = MV /1000

mol Pb(NO ) = MV /1000

mol KI = MV /1000

mol Pb(NO

₃

)

₂

= MV /1000

= (0.18) (30) / 1000

= (0.15) (20) / 1000

= 0.0054 mol

= 0.0030 mol

*Since 2 mol of KI ≡ 1 mol of Pb(NO

₃

)

₂

;

KI is limitant

mol of reaction = 0.0027 mol

Step 3 : ∆H = q / mol @

∆H = 1045 / 0.0027 mol

= (0.800)(15.0) / 1000

= 0.010 mol (lim)

= 0.012 mol

= 0.010 mol (lim)

= 0.012 mol

so

q = ∆H x mol @

q = (– 63400) (0.010)

q = 634 J

Step 1 : q = m c θ

θ

_{@ θ}

θ

_{= q / mc}

θ

_{= 634 / [(4.18) (10.0 + 15.0)}

θ

_{= 6.07}

o

_C

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1.2.2 Standard condition for calculating enthalpy changes

The standard conditions of temperature and pressure for

thermochemical measurement are 298 K and 1 atm. Any

enthalpy changes measured under these conditions is

described as standard enthalpy of reaction, and the symbol

is written as ∆H

∅∅∅∅

In this Chapter, there are a total of 9 standard enthalpy

change of reaction that we shall learned through.

There are 3 basic rules applied when thermochemical

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The total amount of energy released or absorbed is directly

proportional to the number of moles of the reactant used. For example, in the combustion of methane :

CH₄ (g) + 2 O₂ _{(g) CO}₂ (g) + 2 H₂O (l) ∆H∅ _{= – 890 kJ mol}_-1

If there’s 2 mole of methane, CH₄ are combusted

2 CH₄ (g) + 4 O₂ _{(g) 2 CO}₂ (g) + 4 H₂O (l) ∆H =

The enthalpy change for the reverse reaction is equal in

magnitude but opposite in sign to the enthalpy change for the magnitude but opposite in sign to the enthalpy change for the forward reaction.

Na+ _{(g) + Cl}– _(g) _{NaCl (s)} _∆H∅

= – 770 kJ mol-1

If the reaction is reversed

NaCl (s) Na+ _{(g) + Cl}– _(g) _∆H∅

=

The value of ∆H∅∅∅∅ for a reaction is the same whether it occurs in

one step or a series of steps. This shall be further discussed on the coming sub-topic about Hess' Law

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Enthalpy change of formation, ∆H

∅ f

Energy changes occur when

1 mol of substance is

formed

from

its individual elements

under standard

condition.

∆H

∅

f

and Stability of Compound

Product formed via

exothermic process

are

more

stable

than product formed via

endothermic process

Example : compare ∆H

∅_f

of sodium halide

NaI < NaBr < NaCl < NaF

∆H

∅

f

more exothermic

Reaction between Na and X

₂

more vigorous

Stability of compound formed increase

∆H

_rxn

of a reaction can be calculated using ∆H

∅_f

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Example 5

CO (g) + ½ O

₂

_{(g) CO}

₂

(g)

∆H

_rxn

= Σ ∆H

∅ f

(products) – Σ ∆H

∅ f

(reactants)

= (H

_f

(s)

∆H

_rxn

= Σ ∆H

∅ f

(products) – Σ ∆H

∅ f

(reactants)

= (2 x H

_f

FeCl

₃

) – H

_f

(2 FeCl

₂

+ Cl

₂

)

=(2 x –405 kJ / mol) – ( 2 x –341 kJ / mol + 0)

= – 128 kJ

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*Extra Note – Cyclohexene, C₆H₁₀ contain one carbon – carbon

double bond, C=C. When cyclohexene undergoes hydrogenation, the enthalpy change is –120 kJ / mol.

If a benzene ring (which has 3 C=C), react with hydrogen :

supposedly the enthalpy of hydrogenation must be 3 x -120 = -360 kJ supposedly the enthalpy of hydrogenation must be 3 x -120 = -360 kJ

/ mol. However, when experiment involving hydrogenation is carried out, the ∆H of benzene is – 208 kJ / mol, indicating that benzene molecule does not contain three double bonds in its structure.

1 mol of benzene is 152 kJ / mol more stable than 1 mole of

cyclohexene.

The more stable the structure, the less heat given out during a

reaction. The real structure of benzene is a resonance hybrid between the structure above

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1.4 Enthalpy change of combustion, ∆H

∅ c

Energy liberated

occur when

1 mol of substance is

burned

with excess air (oxygen) under standard

condition.

E.g. : CH

₄

(g) + 2 O

₂

_{(g) CO}

₂

(g) + 2 H

₂

O (l)

Note a few things in the thermochemical equation

above :

The standard combustion of substance must be 1

mole of the reactant burned. The mole of

oxygen used must be balanced accordingly.

O]

₂

(g) + 3 H

₂

O (l)

C

₂

H

₅

OH : C

₂

H

₅

OH (l) + 3 O

₂

O

₁₀

(s)

Al

: Al (s) + 3/4 O

₂

_{(g) ½ Al}

₂

O

₃

₂

O (l)

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Calorimeter ~ instrument used to measure the heat

transferred during a chemical reaction.

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Advantages :

Simple to be prepared and set-up

Disadvantages :

The experimental value is always lesser than the actual

∆H

∅

c

because of the following reason

Heat is easily lost to surrounding

Combustion of the sample is incomplete

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Example 7

Step 1 : q = m c θ

θ

@ q = (150) (4.18) (71.0 – 27.8)

q = 27.1 kJ

Step 2 : calculate the mol of pentane burned

mol = mass / RMM = 4.30 / 72

= 0.0597 mol

Step 3 : ∆H = q / mol @

∆H = 27.1 kJ / 0.0597 mol

∆H = – 454 kJ / mol

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Bomb Calorimeter

Bomb calorimeter consist of a

thick stainless steel pressure vessel called “bomb”

“Bomb” is then sealed after

weighted sample is placed. A volume of water is added

to ensure the surface is covered

Pure oxygen is pumped into the

valve until 25 atm. initial temperature is recorded. Temperature of water is

taken from time until it reached maximum temperature. The difference of temperature is taken as θ.

Then, benzoic acid (C₆H₅COOH) is used to calibrate the instrument to

determine the heat capacity of the instrument.

Heat capacity ~ heat required to raise the temperature of the whole

apparatus by 1 K ) ( ) ( ,

θ

change e temperatur q change enthalpy C capacity Heat =

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Steps of calculating ∆H

∅_c

using bomb calorimeter

Calibration

Sample

(using benzoic acid)

(burned sample)

∆H = q

mol

mol = mass

RMM

mol = mass

RMM

C = q /

θ

mol = mass

RMM

q = ∆H x mol

q = C

θ

q = ∆H x mol

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Example 10 :

mol = 0.625

122 = 5.12 x 10

-3

_mol

q = -3230 x 5.12 x 10

-3

ΔH = 16.2 kJ

0.0123

=-1310 kJ/ mol

mol = 0.712

q = 10.5 x 1.54

= 16.2 kJ

q = -3230 x 5.12 x 10

-3

= 16.5 kJ

C =16.5 / 1.58

= 10.5 kJ / K

mol = 0.712

58 = 0.0123 mol

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1.5– Hess Law

~ stated that the heat absorbed or liberated during a

chemical reaction, is independent of route by which the

chemical changes occur.

Consider the following equation : A + B C + D required 2

steps

_{A + B Z}

_{Z C + D}

A + B

∆H

_M

Z

∆H

_N

C + D

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Example : In the reaction of formation of SO₃, it is a 2 steps reaction. Step 1 : 1/8 S₈ (s) + O₂ _{(g) SO}₂ (g) ∆H₁ = – 297 kJ / mol Step 2 : SO₂ (g) + ½ O₂ _{(g) SO}₃ (g) ∆H₂ = – 99 kJ Overall : 1/8 S₈ (s) + 3/2 O₂ _{(g) SO}₃ (g) ∆H_f∅ ₌ Energy / kJ 1/8 S₈ (s) + 3/2 O₂ (g) – 396 kJ / mol 1/8 S₈ (s) + 3/2 O₂ (g) SO₂ (g) + ½ O₂ (g) SO₃ (g)

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Using Hess’s Law, the energy required to form intermediate can

also be determined.

Example : In the reaction of processing ammonia, the equation is

N₂ (g) + 3 H₂ _{(g) 2 NH}₃ (g) ∆H = – 92.2 kJ

The 2 steps involve in the process of forming ammonia

Step 1 : N₂ (g) + 2 H₂ _{(g) N}₂H₄ (g) ∆H₁ = x kJ/mol Step 2 : N₂H₄ (g) + H₂ _{(g) 2 NH}₃ (g) ∆H₂ = – 187 kJ / mol Since ∆H _{required is N + 2 H N H} Since ∆H_rxn required is N₂ + 2 H₂ _N₂H₄ While : N₂ (g) + 3 H₂ _{(g) 2 NH}₃ (g) ∆H = – 92.2 kJ Eq.2 is reversed 2 NH₃ _{(g) N}₂H₄ (g) + H₂ (g) ∆H₂ = + 187 kJ N₂ (g) + 2 H₂ _{(g) N}₂H₄ ∆_H f = + 94.8 kJ / mol

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Energy / kJ

N₂H₄ (g) + H₂ (g)

N₂ (g) + 3 H₂ (g) N₂ (g) + 3 H₂ (g)

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Example 9 :

Find H₂ (g) + O₂ _{(g) H}₂O₂ (l)

H₂ (g) + ½ O₂ _{(g) H}₂O (l) ∆H_f∅

= - 286 kJ/mol (1) H₂O₂ _{(g) H}₂O (l) + ½ O₂ (g) ∆H = - 188 kJ/mol (2)

Reverse equation (2) => equation (3)

H₂O (l) + ½ O₂ _{(g) H}₂O₂ (g) ∆H = +188 kJ (3)

H₂ (g) + ½ O₂ _{(g) H}₂O (l) ∆H_f∅ _{= - 286 kJ/mol} ₍₁₎

So, when equation (1) + (3)

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Energy / kJ

H₂ (s) + O₂ (g)

H₂O₂ (l) H₂O₂ (l)

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Example 10 :

= - 296.36 kJ/mol

Calculate the enthalpy change for the transformation

S(rhombic) → S(monoclinic)

(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)

Since the equation required is S(rhombic) → S(monoclinic)

Make sure S(rhombic) is at the left while S(monoclinic) is at the right By reversing eq (2) and compare to eq (1)

S(rhombic) + O₂(g) → SO₂(g) ∆H_rxn = - 296.06 kJ/mol SO₂(g) → S(monoclinic) + O₂(g) ∆H_rxn = + 296.36 kJ/mol ---S(rhombic) → S(monoclinic) ∆H_rxn = + 0.30 kJ / mol

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Energy / kJ

sulphur (rhombic) + O₂ (g)

sulphur (monoclinic) + O₂ (g)

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1.5.3 Relationship between ∆H_c∅

and ∆H_f∅

using Hess Law

For example, in determining the ∆H_f∅ of butane, C₄H₁₀. Given the

∆H_c∅

for C₄H₁₀, C and H₂ are – 2 877 kJ / mol ; – 393 kJ / mol and -296 kJ / mol respectively.

Solution : C (s) + O₂ (g) CO₂ (g) ∆H_c∅ = - 393 kJ / mol .. (1) H₂ (g) + ½ O₂ _{(g) H}₂O (l) ∆H_c∅ = - 286 kJ / mol .. (2) C₄H₁₀ (l) + 13/2 O₂ _{(g) 4 CO}₂ (g) + 5 H₂O (l) ∆H_c∅ _{= - 2877 kJ /mol .. (3)} 4 C (s) + 5 H₂ _{(g) C}₄H₁₀ (l) ∆H_f∅ = ? kJ / mol \. (4)

Since the equation of formation require 4 C (s) and 5 H (g), so Since the equation of formation require 4 C (s) and 5 H₂ (g), so

the overall equation for (1) and (2) are multiply by 4 and 5 respectively, where as in equation (3) are reversed.

4 C (s) + 4 O₂ _{(g) 4 CO}₂ (g) ∆H_c∅ _{= – 1572 kJ} 5 H₂ (g) + 5/2 O₂ _{(g) 5 H}₂O (l) ∆H_c∅ = – 1430 kJ 4 CO₂ (g) + 5 H₂_{O (l) C}₄H₁₀ (l) + 13/2 O₂ (g) ∆H_c∅ = + 2877 kJ 4 C (s) + 5 H₂ _{(g) C}₄H₁₀ (l) ∆H_f∅ _{= – 125 kJ / mol}

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Energy / kJ

4 C (s) + 5 H

₂

(g)

+ 13/2 O

₂

(g)

C

₄

H

₁₀

(l)

+ 13/2 O

₂

(g)

4 CO

₂

= –1300 (1)

_{= –3270 (2)}

Multiply equation (1) by 3

Reverse equation (2)

= –3900

= +3270

3 C

₂

H

₂

_C

₆

H

₆

(l)

∆H

∅

_{= – 630 kJ}

(40)

Energy / kJ

3 C₂H₂ (g) + 15 / 2 O₂ (g)

C₆H₆ (l) + 15/2 O₂ (g)

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Example 12 :

C

₃

H

₆

(g) + H

₂

_{(g) C}

₃

H

₈

(g)

∆H

∅

= –124 kJ/mol

= –2222 kJ/mol

H

₂

(g) + ½ O

₂

_{(g) H}

₂

O (l)

∆H

_c∅

= – 286 kJ / mol

(g)

∆H

_c∅

_{= + 286 kJ / mol}

H

₂

_{O (l) H}

₂

(g) + ½ O

₂

(g)

∆H

_c

= + 286 kJ / mol

C

₃

H

₆

(g) + H

₂

_{(g) C}

₃

H

₈

(g)

∆H

∅

= –124 kJ/mol

= –2222 kJ/mol

C

₃

H

₆

(g) + 9/2 O

₂

_{(g)3 CO}

₂

(g) + 3 H

₂

O(l) ∆H

_c∅

_{= –2060 kJ/mol}

(42)

Energy / kJ

C

(g)

+

H

₂

O (l)

C

₃

H

₆

(g)

+ 9/2 O

₂

(g)

+

H

₂

O (l)

3 CO

₂

(g)

+ 4 H

₂

O (l)

(43)

1.6 Enthalpy change of Neutralisation ∆H

∅

neut

~ amount of energy

liberated

when 1 mol of hydrogen ion from

acid react with 1 mol of hydroxide ion from alkali to form 1

mole of water under standard condition.

Equation : H

+

_{(aq) + OH}

–

_{(aq) H}

2

O (l)

∆H

∅_neut

for strong acid and strong base under standard

condition is – 57.3 kJ / mol.

The value of ∆H

is still the same as we

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Example

Equation : HCl (aq) + NH

₃

(aq) NH

₄

Cl (aq)

Step 1 : q = m c θ

θ

_{@ q = (25.0 + 30.0) (4.18) (31.3 – 27.6)}

q = 850.63 J

Step 2 : determine limitant

mol of HCl = MV / 1000

mol of NH

₃

= MV /1000

= (1.00)(25.0)/1000

= (0.800)(30.0) / 1000

= (1.00)(25.0)/1000

= (0.800)(30.0) / 1000

= 0.025 mol

= 0.024 mol (lim)

Step 3 : ∆H = q / mol @ ∆H = 850.63 / 0.0240

∆H = – 35443 J / mol @ – 35.4 kJ / mol

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∆H∅

neut for weak acid or weak alkali reaction.

If ∆H∅_neut is ≠ 57.3 kJ / mol depend on :

the example above, it can be tell that, the ∆H

∅_neut

for

weak acid and strong alkali is ≠ - 57.3 kJ / mol. This is

due to, some heat is absorbed by CH

₃

COO-H to break the

O-H to form hydrogen ion. Therefore, it is less exothermic

than the expected value.

Basicity of an acid : HCl H+ + Cl– [monoproctic acid]

H₂SO₄ _{2 H}+ _{+ SO}

42- [diproctic acid]

H₃PO₄ _{3 H}+ _{+ PO}

43- [triproctic acid]

H₃PO₄ _{3 H + PO}₄ [triproctic acid] For example, when NaOH (aq) react with H₂SO₄ (aq)

Stage 1 :H₂SO₄ _{(aq) + NaOH (aq) NaHSO}₄ (aq) + H₂O (l) ∆H∅

neut = –61.95 kJ / mol

Stage 2 : NaHSO₄ _{(aq) + NaOH (aq) Na}₂SO₄ (aq) + H₂O (l) ∆H∅

neut = –70.90 kJ / mol

Overall : 2 NaOH (aq) + H₂SO₄ _{(aq) Na}₂SO₄ (aq) + 2 H₂O (l) ∆H∅

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Reaction involving HF :

HF (aq) + NaOH (aq) NaF (aq) + H

2

O (l)

∆H

∅

neut

= –102.4 kJ / mol

The reaction become more exothermic than expected

despite that HF is consider as a weak acid. When HF is

dissolve in water, H-F dissociate in water to form H

+

_{and F}

-

_.

The enthalpy of hydration, ∆H

∅

hyd

of the fluoride ion

is very

exothermic

, making the overall process to be much

exothermic

F

-

_{(g) + water F}

-

_(aq)

_∆H

∅

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1.6 Standard Enthalpy Change of Atomisation, ∆H∅ atom

~ energy absorbed when 1 mole of gaseous atoms are formed

from its element under standard condition.

Eq : A (s) A (g) ∆H∅ atom = + ve kJ/mol Example : Mg (s) Mg (g) ¼ P₄ _{(s) P (g)} ½ Cl₂ _{(g) Cl (g)} _{1/8 S (s) S (g)} CH₄ _{(g) C (g) + 4 H (g)} PBr₃_{(s) P (g) + 3 Br (g)}

Since the reaction required the substance involve to become

gaseous atom, so the process involved an endothermic process.

For a solid, the ∆H∅_atom involves 2 processes. For example, in

sodium, Na, to become a gaseous sodium, the solid metal undergoes melting process before vapourising to gas.

Energy required to change 1 mol of solid to liquid is named as

enthalpy change of fusion, while the energy required to change 1 mol of liquid to gas to called as enthalpy change of vapourisation, according to the following equation

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Na (s) → Na (l) ∆H_fusion

Na (l) → Na (g) ∆H_{vapourisation}

Since noble gas exist naturally as monoatom gas the Enthalpy

Change of Atomisation for noble gas 0

As for the Bond enthalpy, it is the energy required to break the bond

between 2 covalently bond atoms.

For example, the bonding enthalpy of chlorine gas

Na (s) Na (g)

∆H

∅

atom

For example, the bonding enthalpy of chlorine gas

Cl – Cl (g) 2 Cl (g) ∆H∅

BE = + 242 kJ / mol

Compare to the , ∆H∅_atom of chlorine atom ;

½ Cl₂ _{(g) Cl (g)} ∆H∅

(49)

1.7 Ionisation energy, ∆H

∅ IE

~

energy absorbed when 1 mole of electron is removed from a

gaseous atom under standard condition.

Eq : A (g) A

+

_{(g) + e-}

_∆H

∅

IE

= + ve kJ/mol

The process is always endothermic as heat is absorbed to

free one mole of electron from an atom (to overcome the

electrostatic forces of attraction between the nucleus and

outermost electron)

Generally, when goes down to Group, ionisation energy

decrease, while across the Period, ionisation energy

increase. These trend shall be further discussed in Chapter 3

It is believed that, when enormous amount of energies is

supplied, electrons in an atom can be removed completely

from an atom. The energies required to consecutively

= + 2740 kJ / mol

Overall : Al (g) Al

3+

_{(g) + 3e}

–

_∆H

IE

= + 5137 kJ

The information of the 1

st

until the 4

th

ionisation energy of

elements can be obtained through Data Booklet supplied

during examination

(51)

1.8 Electron Affinity ∆H∅ EA

~ energy liberated when 1 mole of electron is received from gaseous atom under standard condition.

Eq : O (g) + e- O– _(g) _∆H∅

EA = – X kJ / mol

For 1st Electron Affinity, the process is always exothermic, since

upon receive an electron, the energy carries by the electron is released upon combining with the gaseous atom.

The trend of of 1st electron affinity is the same as in Ionisation

energy, where 1st electron affinity decrease when going down energy, where 1st electron affinity decrease when going down to group, whereas the 1st electron affinity increase when going across Period.

However, unlike 2nd ionisation energy, after an atom received an

electron an form negative charged ion, upon receiving the second electron, a repulsion forces is felt between the anion and electron receive, due to the mutual charge between both substance.

Hence, for second electron affinity, heat is absorbed

(endothermic) by the anion to overcome the repulsion forces between the anion and electron.

(52)

When forming O

2–

from O

–

(2

nd

EA), electron is received by

negative ion.

repulsion forces formed between anion

and electron received. Heat is absorbed

to overcome the forces of repulsion.

1

st

_{EA : O (g) + e}

–

_O

–

_{(g) ∆H}

∅ EA

Overall : O (g) + 2e

–

_O

2–

_{(g) ∆H}

EA

= + 702 kJ

(53)

1.9 Lattice Energy, ∆H

∅ LE

~ energy liberated when 1 mole of solid crystal lattice is

formed from oppositely charged gaseous ions under

standard condition.

Eq : M

+

_{(g) + X}

–

_{(g) MX (s)}

_∆H

∅

LE

= –X kJ/mol

LE – always negative (exothermic) : heat is released when

ionic bond is formed.

Examples of writing thermochemical equation :

NaF : Na

+

_{(g) + F}

-

_{(g) NaF (s)}

MgO : Mg

2+

_{(g) + O}

2-

_{(g) MgO (s)}

CaCl

₂

: Ca

2+

_{(g) + 2 Cl}

-

_{(g) CaCl}

2

(s)

K

₂

O : 2 K

+

(g) + O

2-

_{(g) K2}

O (s)

Al

₂

O

₃

: 2 Al

3+

_{(g) + 3 O}

2-

_{(g) Al2}

_O

3

(s)

AlN : Al

3+

_{(g) + N}

3-

_{(g) AlN (s)}

(54)

Factors influencing Lattice Energy –

i)

charge of ion

ii)

inter-ionic distance

Charge of ions (Z

n+

_{. Z}

n–

₎

_{Inter-ionic distance (r}

+

_{+ r}

–

₎

Greater the charge ; greater

the forces of attraction ;

greater the value of Lattice

Energy (more exothermic)

Smaller the distance, greater

the attraction forces

between ions, greater the

lattice energy

− + − +

+

• ∝

r

Z

energy

Lattice

n n

(55)

Compound

Total

charge

∑ Ionic

radius

Compound

Total

charge

Ionic

radius

NaF

1

0.231 NaCl

1

0.276 KBr

1

0.328 KCl

1

0.314 CaO

4

0.239 MgO

4

0.205 Al O

6

0.190 K O

2

0.273 The trend of lattice energy of these 8 compounds are

KBr < KCl < NaCl < NaF < K

₂

O < CaO < MgO < Al

₂

O

₃

Lattice energy increase

(56)

1.10 Born Haber Cycle

Lattice energy cannot be determined experimentally. They can

only be obtained by applying Hess’s Law in an energy cycle called Born-Haber Cycle, which is a cycle of reactions used for

calculating the lattice energies of ionic crystalline solids.

There are basically 5 types of Born Haber Cycle which is mostly

tested all times. i) A+_B- _ii) _A2+_B

2- iii) A2+B2- iv) A2+B2- v) A23+B3

2- To build the Born Haber cycle, students must be able to write To build the Born Haber cycle, students must be able to write

∆H_f∅∅∅∅

of the compound and ∆H∅∅∅∅ LE.

Here, we are going to build the Born Haber cycle using the 5

examples aboveSodium chloride, NaCl

Calcium chloride, CaCl₂ Potassium oxide, K₂O Magnesium oxide, MgO

(57)

∆_H_atom _{of Na} Na (g) + ½ Cl₂ (g) ∆_H IE of Na Na+ _{(g) + ½ Cl} 2 (g) + e -∆_H atom of Cl Na+ _{(g) + Cl (g) + e} -∆_H EA of Cl Na+ _{(g) + Cl}- _(g) ∆_H LEof NaCl Na (s) + ½ Cl₂ (g) NaCl (s) ∆_H_f _{of NaCl} ∆_H_atom _{of Na} _∆_H LEof NaCl ∆H∅ f = ∆H ∅ LE + [∆H ∅ atom Na + ∆H ∅ atom Cl + ∆H ∅ 1st IE Na + ∆H ∅ 1st EA Cl] ∆H∅ LE = (-411) – [(+108) + (+121) + (+494) + (-364)] = – 770 kJ/mol

(58)

∆_H _{of Ca} Ca (g) + Cl₂ (g) ∆_H 1st IE of Ca + ∆_H 2nd IE of Ca Ca2+ _{(g) + Cl} 2 (g) + 2 e -2 x ∆H_atom of Cl Ca2+ _{(g) + 2 Cl (g) + 2 e} -2 x ∆H_EA of Cl Ca2+ _{(g) + 2 Cl}- _(g) ∆_{H of CaCl} Ca (s) + Cl₂ (g) CaCl₂ (s) ∆_H_f _{of CaCl}₂ ∆_H atom of Ca ∆H_LEof CaCl₂ ∆H∅ f= ∆H ∅ LE + [∆H ∅ atomCa + 2∆H ∅ atom Cl + ∆H ∅ 1st IE Ca + ∆H∅ 2nd IE Ca + 2∆H ∅ 1st EA Cl] ∆H∅ LE = (-795) – [(+132) + 2(+121) +(+590) +(1150) + 2(-364)] = – 2181 kJ/mol

(59)

2 X ∆H of K 2 K (g) + ½ O₂ (g) 2 X ∆H_{1st IE} of K 2 K+ _{(g) + ½ O} 2 (g) + 2 e -∆_H atom of O 2 K+ _{(g) + O (g) + 2 e} -∆_H LEof K2O ∆_H_{1st EA}₊ ∆_H 2nd EA 2 K+ _{(g) + O}2- _(g) 2 K (s) + ½ O₂ (g) K₂O (s) ∆_H f of K2O 2 X ∆H_atom of K ∆H∅ f= ∆H ∅ LE + [2∆H ∅ atom K + ½∆H ∅ BEO + 2∆H ∅ 1st IE K + ∆H∅ 1st EA O + ∆H ∅ 2nd EA O] ∆H∅ LE = (-362) – [2(+129) + ½(+498) + 2(418) + (-141)+(+844)] = – 2408 kJ/mol

(60)

Mg (g) + ½ O₂ (g) ∆_H 1st IE of Mg + ∆_H_{2nd IE} _{of Mg} Mg2+ _{(g) + ½ O} 2 (g) + 2 e -∆_H atom of O Mg2+ _{(g) + O (g) + 2 e} -∆_H LEof MgO ∆_H_{1st EA}₊ ∆_H 2nd EA of O Mg2+ _{(g) + O}2- _(g) Mg (s) + ½ O₂ (g) MgO (s) ∆_H f of MgO ∆_H atom of Mg ∆H∅ f= ∆H ∅ LE + [∆H ∅ atomMg + ½∆H ∅ BEO +∆H ∅ 1st IE Mg + ∆H∅ 2nd IEMg + ∆H ∅ 1st EA O + ∆H ∅ 2nd EA O] ∆H∅ LE = (-612) – [(+146) + ½(+498)+(736) + (1450) + (-141)+(+844)] = – 3896 kJ/mol

(61)

2 Cr (g) + 3/2 O₂ (g) 2 x (∆H_{1st IE} of Cr + ∆_H_{2nd IE} _{of Cr +} ∆_H 3rd IE of Cr) 2 Cr3+ _{(g) + 3/2 O} 2 (g) + 6 e -3 x ∆H_atom of O 2 Cr3+ _{(g) + 3 O (g) + 6 e} -∆_H LEof Cr2O3 3 x (∆H_{1st EA}O + ∆_H 2nd EA of O) 2 Cr3+ _{(g) + 3 O}2- _(g) 2 Cr (s) + 3/2 O₂ (g) Cr₂O₃ (s) ∆_H f of Cr2O3 2 x ∆H_atom of Cr 2 Cr (g) + 3/2 O₂ (g) ∆H∅ LE = – 16408 kJ/mol

(62)

1.12 Enthalpy Change of Hydration, ∆h_hyd

In terms of Thermochemistry, the solubility of ionic compound in

water depend on 2 factors

The enthalpy change of hydration Lattice energy of the salt involved

Standard enthalpy change of hydration, ∆H∅_hyd is

\\\\\\\\\\\\\\\\\..\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\...under standard condition.

~ energy liberated when one mole of gaseous ion is hydrated by water.

\\\\\\\\\\\\\\\\\...under standard condition. Equation :

Intermolecular forces occur during hydration of ions are ion-dipole

forces, which were stronger than hydrogen bonding. Diagram below shows the ion-dipole forces between a positively and negatively charged ion with water respectively.

water.

Mn+ _{(g) + water M}n+ _(aq) _∆H

hyd = – x kJ/mol

Qn- _{(g) + water Q}n- _(aq) _∆H

(63)

(64)

Since the intermolecular forces between ion and water is

strong, the ∆H

∅

hyd

is always exothermic. Similar to lattice

energy, the magnitude of ∆H

∅

hyd

depends on 2 factors :

Charge of ion - Greater the charge of ion, stronger the

attraction between the water and ion, the more exothermic it

is enthalpy change of hydration of ions

Size of ion - Smaller the size of ion, stronger the attraction

between the ions and water, the more exothermic it is the

enthalpy change of hydration of ions

(65)

Q

n-

δ-Mn+

δ+

∆H

_hyd

for cation increase

∆H

_hyd

for anion increase

(66)

1.12 Enthalpy change of solution, ∆H

_soln

Energy change when 1 mole of solute is dissolved in a large

O

₆

(aq)

∆H

_soln

-

_(aq)

∆H

_hyd

: M

+

_{(g) + X}

-

_{(g) + water M}

+

_{(aq) + X}

-

_(aq)

– ∆H

_LE

:

_{MX (s) M}

+

_{(g) + X}

-

_(g)

MX (s) + water M

+

_{(aq) + X}

-

_(aq)

As a conclusion, ∆H

_soln

= ∆H

_hyd

+ (– ∆H

_LE

)

If ∆H

_soln

= - ve, then the salt is soluble in water

If ∆H

_soln

= + ve, then the salt is insoluble in water

(67)

In the solubility of Group 2 sulphate

both lattice energy and enthalpy change of hydration are

proportional to of the ions. Hence, when going down to Group 2 sulphate, both of these energies \\\\\\. As the size of metal ion \\\\\\\\..

However, the rate decrease in lattice energy is \\\\\\\\

than the rate of decrease in ∆H∅ hyd

This is because the size of sulphate ion is much larger than the

size of metal ions, so even though the size of cation increases,

decrease

increase

Less significant

size of metal ions, so even though the size of cation increases, the increase of (r+ _{+ r}-_{) is very small. This makes the lattice energy}

changes become less significant when goes down to Group 2.

While in ∆H∅_hyd it depend on both cation and anion. Since the

∆H∅

hyd for anion is constant, so the ∆H ∅

hyd is mainly depend on

the size of cation. When goes down to Group 2, the metal ion size \\\\\\\\\. , making ∆H∅

hyd become \\\\\

exothermic. So, the \\\\\. of the heat become more significant thus causing the rate of ∆H∅

hyd is greater than lattice

energy.

increase

less

(68)

Group 2 sulphate Be SO₄ Mg SO₄ Ca SO₄ Sr SO₄ Ba SO₄ ∆H_solution (kJ / mol) -95.3 -91.2 + 17.8 + 18.70 +19.4 Solubility (g / 100mL) 41.0 36.4 0.21 0.010 0.00025 ∆H_hydration ∆H_{lattice energy}

(69)

Example : Solubility of Group 2 sulphate :

Sr2+ Ba2+ SO42-Be2+ Mg2+ Ca2+