CHAPTER 1:

THERMOCHEMISTRY

Pre – U Chemistry

Semester 2

THERMOCHEMISTRY

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.

� 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.

� 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)

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 ∆Hproduct – ΣΣΣΣ ∆Hreactant].

� 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

� Enthalpy – heat content of the system

� Enthalpy changes ; ∆H ~ heat changes occur

during a chemical reaction.

� ∆H = [ΣΣΣΣ ∆Hproduct – ΣΣΣΣ ∆Hreactant]

Unit = kJ mol-1.

Σ ∆Hproduct > Σ ∆Hreactant Σ ∆Hproduct < Σ ∆Hreactant

∆H = positive (+ve) ∆H = negative (–ve)

Endothermic exothermic

Process Endothermic Exothermic

Definition Process of heat

absorbed by system

Process of heat

released by system

ΔH Positive Negative

Energy profile

Process Endothermic Exothermic

θ Temperature decrease Temperature increase

3 STEPS ON CALCULATING

ENTHALPY CHANGE

1 q = m c θθθθ

MVmass2

mol =

3∆H =

1000

MVor

M

mass

R

mol

q

� Question 1 : Solution

Equation : Zn (s) + H2SO4 (aq)� ZnSO4 (aq) + H2 (g)

Step 1 : q = m c θθθθ @ q = (25.0) (4.18) (31.5 – 27.0)

q = 470.25 J

Step 2 : determine limitant

mol of Zn = mass / mol mol of H2SO4 = MV /1000mol of Zn = mass / mol mol of H2SO4 = 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

∆H = 188 100 J / mol @ – 188 kJ / mol

� Question 2 : Solution

Equation : Na2SO4 + Ba(NO3)2 � 2 NaNO3 + BaSO4

Step 1 : q = m c θθθθ

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

q = 836 J

Step 2 : determine limitant

mol Na2SO4 = MV /1000 mol Ba(NO3)2 = MV /1000mol Na2SO4 = MV /1000 mol Ba(NO3)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

� Question 3 : Solution

Equation : 2 KI + Pb(NO3)2 � 2 KNO3 + PbI2

Step 1 : q = m c θθθθ

@ q = (20 + 30) (4.18) (34 – 29)

q = 1045 J

Step 2 : determine limitant

mol KI = MV /1000 mol Pb(NO ) = MV /1000mol KI = MV /1000 mol Pb(NO3)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(NO3)2 ;

KI is limitant mol of reaction = 0.0027 mol

Step 3 : ∆H = q / mol @ ∆H = 1045 / 0.0027 mol

∆H = 387037 @ = – 390 kJ / mol

� Question 4 : Solution

NaCl + AgNO3 � NaNO3 + AgCl

Step 3 : From ∆H and mol ; find q

Step 2 :

mol of AgNO3 = MV / 1000 mol of NaCl = MV / 1000

= (1.00)(10.0) / 1000 = (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 θθθθ @ θθθθ = q / mc

θθθθ = 634 / [(4.18) (10.0 + 15.0)

θθθθ = 6.07 oC

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.change of reaction that we shall learned through.

� There are 3 basic rules applied when thermochemical

equations were used for calculations.

� 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) � CO2 (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 CO2 (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

Enthalpy change of formation, ∆H∅f

� Energy changes occur when 1 mol of substance is

formed from its individual elements under standard

condition.

E.g. : H2 (g) + ½ O2 (g) � H2O (l)

� Note the following important things

�The physical states of the substance involved is �The physical states of the substance involved is

stated accordingly under standard condition

�The ∆Hf∅ of water is not written as H2O (g) as it

is not in gas under standard condition.

�2 H2 (g) + O2 (g) � 2 H2O (l) is not consider as

standard as substance formed is not 1 mole.

� ∆H∅f of pure element is = 0 kJ / mol

� Examples

CO2 : C (s) + O2 (g) � CO2 (g)

MgCO3 : Mg (s) + C (s) + 3/2 O2 (g) � MgCO3 (s)

NH3 : ½ N2 (g) + 3/2 H2 (g) � NH3 (g)

NaCl : Na (s) + ½ Cl2 (g) � NaCl (s)

C H O : 6 C (s) + 6 H (g) + 3 O (g)�C H O (s)C6H12O6 : 6 C (s) + 6 H2 (g) + 3 O2 (g)�C6H12O6 (s)

SO3 : 1/8 S8 (s) + 3/2 O2 (g) � SO3 (g)

CH3COOH :2C(s) + 2H2(g) + O2(g)�CH3COOH (l)

Al2O3 : 2 Al (s) + 3/2 O2 (g) � Al2O3 (s)

∆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 X2 more vigorous

Stability of compound formed increase

� ∆Hrxn of a reaction can be calculated using ∆H∅f

∆Hrxn = Σ ∆H∅f (products) – Σ ∆H∅

f (reactants)

� Example 5

CO (g) + ½ O2 (g) � CO2 (g)

∆Hrxn = Σ ∆H∅f (products) – Σ ∆H∅

f (reactants)

= (Hf CO2) – Hf (CO + O2)

= (– 393 kJ / mol) – ( – 110 kJ / mol + 0)

= – 283 kJ

Example 6� Example 6

2 FeCl2 (s) + Cl2 � 2 FeCl3 (s)

∆Hrxn = Σ ∆H∅f (products) – Σ ∆H∅

f (reactants)

= (2 x Hf FeCl3) – Hf (2 FeCl2 + Cl2)

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

= – 128 kJ

� *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

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. : CH4 (g) + 2 O2 (g) � CO2 (g) + 2 H2O (l)

� Note a few things in the thermochemical equation

above :above :

�The standard combustion of substance must be 1

mole of the reactant burned. The mole of

oxygen used must be balanced accordingly.

Oxygen is always combust in excess

�For ∆H∅c is always exothermic. MAKE SURE

THE ‘ – ‘ MUST BE PLACED.

� Examples

C : C (s) + O2 (g) � CO2 (g) [= ∆H∅fof CO2]

H2 : H2 (g) + ½ O2 (g) � H2O (l) [= ∆H∅fof H2O]

C2H5COOH : C2H5COOH (l) + 7/2 O2 (g) �

3 CO2 (g) + 3 H2O (l)

C2H5OH : C2H5OH (l) + 3 O2 (g) � 2 CO2 (g) + 3 H2O (l)

Mg : Mg (s) + ½ O (g) � MgO (s)Mg : Mg (s) + ½ O2 (g) � MgO (s)

P : P4 (s) + 5 O2 (g) � P4O10 (s)

Al : Al (s) + 3/4 O2 (g) � ½ Al2O3 (s)

C6H12O6 : C6H12O6 (s) + 6 O2 (g) � 6 CO2 (g) + 6 H2O (l)

� Calorimeter ~ instrument used to measure the heat

transferred during a chemical reaction.

� Simple calorimeter :

� 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� Combustion of the sample is incomplete

� Combustion is not done under standard condition

� Example 7

Step 1 : q = m c θθθθ

@ 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 / 72mol = mass / RMM = 4.30 / 72

= 0.0597 mol

Step 3 : ∆H = q / mol @

∆H = 27.1 kJ / 0.0597 mol

∆H = – 454 kJ / mol

� 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

)(

)(,

θchangeetemperatur

qchangeenthalpyCcapacityHeat =

� 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 / θθθθ

mol = mass

RMM

q = ∆H x mol

q = C θθθθ

q = ∆H x mol

Example 10 :

mol = 0.625

122= 5.12 x 10-3mol

q = -3230 x 5.12 x 10-3

ΔH = 16.2 kJ0.0123

=-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

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

∆HM

Z

∆HN

C + D

� Example : In the reaction of formation of SO3, it is a 2 steps reaction.

Step 1 : 1/8 S8 (s) + O2 (g) � SO2 (g) ∆H1 = – 297 kJ / mol

Step 2 : SO2 (g) + ½ O2 (g) � SO3 (g) ∆H2 = – 99 kJ

Overall : 1/8 S8 (s) + 3/2 O2 (g) � SO3 (g) ∆Hf∅ =

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)

� 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 NH3 (g) ∆H = – 92.2 kJ

� The 2 steps involve in the process of forming ammonia

Step 1 : N2 (g) + 2 H2 (g) � N2H4 (g) ∆H1 = x kJ/mol

Step 2 : N2H4 (g) + H2 (g) � 2 NH3 (g) ∆H2 = – 187 kJ / mol

Since ∆H required is N + 2 H � N HSince ∆Hrxn required is N2 + 2 H2 � N2H4

While : N2 (g) + 3 H2 (g) � 2 NH3 (g) ∆H = – 92.2 kJ

Eq.2 is reversed 2 NH3 (g) � N2H4 (g) + H2 (g) ∆H2 = + 187 kJ

N2 (g) + 2 H2 (g) � N2H4∆Hf = + 94.8 kJ / mol

Energy / kJ

N2H4 (g) + H2 (g)

N2 (g) + 3 H2 (g)N2 (g) + 3 H2 (g)

2 NH3 (g)

Example 9 :

Find H2 (g) + O2 (g) � H2O2 (l)

H2 (g) + ½ O2 (g) � H2O (l) ∆Hf∅ = - 286 kJ/mol (1)

H2O2 (g) � H2O (l) + ½ O2 (g) ∆H = - 188 kJ/mol (2)

Reverse equation (2) => equation (3)

H2O (l) + ½ O2 (g) � H2O2 (g) ∆H = +188 kJ (3)

H2 (g) + ½ O2 (g) � H2O (l) ∆Hf∅ = - 286 kJ/mol (1)

So, when equation (1) + (3)

H2 (g) + O2 (g) � H2O2 (l) ∆H = - 98 kJ/mol

Energy / kJ

H2 (s) + O2 (g)

H2O2 (l)H2O2 (l)

H2O (g) + ½ O2 (g)

Example 10 :

� From these data,

S(rhombic) + O2(g) → SO2(g) ∆Hrxn = - 296.06 kJ/mol

S(monoclinic) + O2(g) → SO2(g) ∆Hrxn = - 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) ∆Hrxn = - 296.06 kJ/mol

SO2(g) → S(monoclinic) + O2 (g) ∆Hrxn = + 296.36 kJ/mol

-----------------------------------------------------------------------------------

S(rhombic) → S(monoclinic) ∆Hrxn = + 0.30 kJ / mol

Energy / kJ

sulphur (rhombic) + O2 (g)

sulphur (monoclinic) + O2 (g)

SO2 (g)

1.5.3 Relationship between ∆Hc∅ and ∆Hf

∅ using Hess Law

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

∆Hc∅ 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) ∆Hc∅ = - 393 kJ / mol .. (1)

H2 (g) + ½ O2 (g) � H2O (l) ∆Hc∅ = - 286 kJ / mol .. (2)

C4H10 (l) + 13/2 O2 (g) � 4 CO2 (g) + 5 H2O (l)

∆Hc∅ = - 2877 kJ /mol .. (3)

4 C (s) + 5 H2 (g) � C4H10 (l) ∆Hf∅ = ? 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 CO2 (g) ∆Hc∅ = – 1572 kJ

5 H2 (g) + 5/2 O2 (g) � 5 H2O (l) ∆Hc∅ = – 1430 kJ

4 CO2 (g) + 5 H2O (l) � C4H10 (l) + 13/2 O2 (g)

∆Hc∅ = + 2877 kJ

4 C (s) + 5 H2 (g) � C4H10 (l) ∆Hf∅ = – 125 kJ / mol

Energy / kJ

4 C (s) + 5 H2 (g) + 13/2 O2 (g)

C4H10 (l) + 13/2 O2 (g)

4 CO2 (g) + 5 H2 (g) + 5/2 O2 (g)

4 CO2 (g) + 5 H2O (l)

Example 11 :

Given ∆Hc∅ of C2H2 and C6H6 are – 1300 kJ / mol and – 3270

kJ/mol respectively. Find 3 C2H2 (g) � C6H6 (l)

C2H2 (g) + 5/2 O2 (g) � 2 CO2 (g) + H2O (l) ∆Hc∅ = –1300 (1)

C6H6 (l) + 15/2 O2(g) � 6 CO2(g) + 3 H2O(l) ∆Hc∅ = –3270 (2)

Multiply equation (1) by 3

Reverse equation (2)Reverse equation (2)

3C2H2 (g) + 15/2 O2 (g) � 6 CO2 (g) + 3 H2O (l) ∆Hc∅ = –3900

6 CO2(g) + 3 H2O(l) � C6H6 (l) + 15/2 O2 (g) ∆Hc∅ = +3270

3 C2H2 � C6H6 (l) ∆H∅ = – 630 kJ

Energy / kJ

3 C2H2 (g) + 15 / 2 O2 (g)

C6H6 (l) + 15/2 O2 (g)

3 H2O (g) + 6 CO2 (g)

Example 12 :

C3H6 (g) + H2 (g) � C3H8 (g) ∆H∅ = –124 kJ/mol

C3H8(g) + 5 O2(g) � 3 CO2(g) + 4 H2O(l) ∆Hc∅ = –2222 kJ/mol

H2 (g) + ½ O2(g) � H2O (l) ∆Hc∅ = – 286 kJ / mol

Find for C3H6(g) + 9/2 O2(g) � 3 CO2(g) + 3 H2O(l) ∆Hc∅ = ?

Reverse equation (3)

H2O (l) � H2 (g) + ½ O2(g) ∆Hc∅ = + 286 kJ / molH2O (l) � H2 (g) + ½ O2(g) ∆Hc = + 286 kJ / mol

C3H6 (g) + H2 (g) � C3H8 (g) ∆H∅ = –124 kJ/mol

C3H8(g) + 5 O2(g) � 3 CO2(g) + 4 H2O(l) ∆Hc∅ = –2222 kJ/mol

C3H6(g) + 9/2 O2(g)�3 CO2(g) + 3 H2O(l) ∆Hc∅= –2060 kJ/mol

Energy / kJ

C3H6 (g) + H2 (g) + 5 O2 (g)

C3H8 (g) + 5 O2 (g)

C3H6 (g) + 9/2 O2 (g) + H2O (l) C3H6 (g) + 9/2 O2 (g) + H2O (l)

3 CO2 (g) + 4 H2O (l)

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) � H2O (l)

� ∆H∅neut for strong acid and strong base under standard

condition is – 57.3 kJ / mol.

� The value of ∆H∅neut will be different with weak acid / base is � The value of ∆H∅neut will be different with weak acid / base is

used or if the acid used is a polyproctic acid

� In laboratory, ∆H∅neut can be determine using simple cup

calorimeter (MPM Experiment 6)

� The ways of calculating ∆H∅neut is still the same as we

learned previously.

Example

� Equation : HCl (aq) + NH3 (aq) � NH4Cl (aq)

Step 1 : q = m c θθθθ @ 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 NH3 = 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

∆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 CH3COO-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+ + SO42- [diproctic acid]

H3PO4 � 3 H+ + PO43- [triproctic acid]H3PO4 � 3 H + PO4 [triproctic acid]

For example, when NaOH (aq) react with H2SO4 (aq)

Stage 1 :H2SO4 (aq) + NaOH (aq) � NaHSO4 (aq) + H2O (l)

∆H∅neut = –61.95 kJ / mol

Stage 2 : NaHSO4 (aq) + NaOH (aq) � Na2SO4 (aq) + H2O (l)

∆H∅neut = –70.90 kJ / mol

Overall : 2 NaOH (aq) + H2SO4 (aq) � Na2SO4 (aq) + 2 H2O (l)

∆H∅neut = [-61.95 + (-70.90)]= - 132.85 kJ

� Reaction involving HF :

HF (aq) + NaOH (aq) � NaF (aq) + H2O (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 exothermic

F- (g) + water � F- (aq) ∆H∅hyd = – 63.4 kJ/mol

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

� Na (s) → Na (l) ∆Hfusion

� Na (l) → Na (g) ∆Hvapourisation

� 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∅atom = + 121 kJ / mol

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)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

remove the electrons from an atom is called as successive

ionisation energies

� The total ionisation energy is the sum of all the successive

ionisation of the element involve. Example

1st IE of Al : Al (g) � Al+ (g) + e– ∆H∅IE = + 577 kJ / mol

2nd IE of Al : Al+ (g) � Al2+(g) + e– ∆H∅IE = + 1820 kJ / mol

3rd IE of Al : Al2+ (g) � Al3+(g) + e– ∆H∅IE = + 2740 kJ / mol

Overall : Al (g) � Al3+ (g) + 3e– ∆HIE = + 5137 kJ

� The information of the 1st until the 4th ionisation energy of

elements can be obtained through Data Booklet supplied elements can be obtained through Data Booklet supplied

during examination

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.

� When forming O2– from O– (2nd EA), electron is received by negative ion.

repulsion forces formed between anionrepulsion forces formed between anionrepulsion forces formed between anionrepulsion forces formed between anionand electron received. Heat is absorbedand electron received. Heat is absorbedand electron received. Heat is absorbedand electron received. Heat is absorbed

to overcome the forces of repulsion.to overcome the forces of repulsion.to overcome the forces of repulsion.to overcome the forces of repulsion.

1st EA : O (g) + e– � O– (g) ∆H∅EA= – 142 kJ / mol

2nd EA : O– (g) + e– � O2– (g) ∆H∅EA = + 844 kJ / mol

e-

2nd EA : O– (g) + e– � O2– (g) ∆H∅EA = + 844 kJ / mol

Overall : O (g) + 2e– � O2– (g) ∆HEA = + 702 kJ

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 :� Examples of writing thermochemical equation :

NaF : Na+ (g) + F- (g) � NaF (s)

MgO : Mg2+ (g) + O2- (g) � MgO (s)

CaCl2 : Ca2+ (g) + 2 Cl- (g) � CaCl2 (s)

K2O : 2 K+ (g) + O2- (g) � K2O (s)

Al2O3 : 2 Al3+ (g) + 3 O2- (g) � Al2O3 (s)

AlN : Al3+ (g) + N3- (g) � AlN (s)

� Factors influencing Lattice Energy –

i) charge of ion ii) inter-ionic distance

Charge of ions (Zn+ . Zn–) 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

−+

−+

+

•∝

rr

ZZenergyLattice

nn

CompoundTotal

charge

∑ Ionic

radiusCompound

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 < K2O < CaO < MgO < Al2O3

Lattice energy increase

Al2O3 6 0.190 K2O 2 0.273

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+B2- 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

∆Hf∅∅∅∅ 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

� Chromium (III) oxide, Cr2O3

∆Hatom of Na

Na (g) + ½ Cl2 (g)

∆HIE of Na

Na+ (g) + ½ Cl2 (g) + e-

∆Hatom of Cl

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

∆HEA of Cl

Na+ (g) + Cl- (g)

∆HLEof NaCl

Na (s) + ½ Cl2 (g)

NaCl (s)

∆Hf of NaCl

∆Hatom of Na ∆HLEof NaCl

∆H∅f = ∆H

∅LE + [∆H∅

atomNa + ∆H∅atomCl + ∆H∅

1st IE Na + ∆H∅1st EA Cl]

∆H∅LE = (-411) – [(+108) + (+121) + (+494) + (-364)]

= – 770 kJ/mol

∆H of Ca

Ca (g) + Cl2 (g)

∆H1st IE of Ca +

∆H2nd IE of Ca

Ca2+ (g) + Cl2 (g) + 2 e-

2 x ∆Hatom of Cl

Ca2+ (g) + 2 Cl (g) + 2 e-

2 x ∆HEA of Cl

Ca2+ (g) + 2 Cl- (g)

∆H of CaCl

Ca (s) + Cl2 (g)

CaCl2 (s)

∆Hf of CaCl2

∆Hatom of Ca ∆HLEof 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

2 X ∆H of K

2 K (g) + ½ O2 (g)

2 X ∆H1st IE of K

2 K+ (g) + ½ O2 (g) + 2 e-

∆Hatom of O

2 K+ (g) + O (g) + 2 e-

∆HLEof K2O

∆H1st EA +

∆H2nd EA

2 K+ (g) + O2- (g)

2 K (s) + ½ O2 (g)

K2O (s)

∆Hf of K2O

2 X ∆Hatom 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

Mg (g) + ½ O2 (g)

∆H1st IE of Mg +

∆H2nd IE of Mg

Mg2+ (g) + ½ O2 (g) + 2 e-

∆Hatom of O

Mg2+ (g) + O (g) + 2 e-

∆HLEof MgO

∆H1st EA +

∆H2nd EA of O

Mg2+ (g) + O2- (g)

Mg (s) + ½ O2 (g)

MgO (s)

∆Hf of MgO

∆Hatom 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

2 Cr (g) + 3/2 O2 (g)

2 x (∆H1st IE of Cr +

∆H2nd IE of Cr +

∆H3rd IE of Cr)

2 Cr3+ (g) + 3/2 O2 (g) + 6 e-

3 x ∆Hatom of O

2 Cr3+ (g) + 3 O (g) + 6 e-

∆HLEof Cr2O3

3 x (∆H1st EA O +

∆H2nd EA of O)

2 Cr3+ (g) + 3 O2- (g)

2 Cr (s) + 3/2 O2 (g)

Cr2O3 (s)

∆Hf of Cr2O3

2 x ∆Hatom of Cr

2 Cr (g) + 3/2 O2 (g)

∆H∅LE = – 16408 kJ/mol

1.12 Enthalpy Change of Hydration, ∆hhyd

� 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 � Mn+ (aq) ∆Hhyd = – x kJ/mol

Qn- (g) + water � Qn- (aq) ∆Hhyd = – x kJ/mol

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

enthalpy change of hydration of ions

Qn-δ-

Mn+

δ+

∆Hhyd for cation increase ∆Hhyd for anion increase

Na+ < Mg2+ < Al3+ l– < Br– < Cl–

1.12 Enthalpy change of solution, ∆Hsoln

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

excess water to form an infinite dilute solution.

For ionic substance : MX (s) + water � M+ (aq) + X- (aq)

Some covalent subs : C6H12O6 (s) + water � C6H12O6 (aq)

� ∆Hsoln is determined by ∆Hhyd and ∆HLE

∆HLE : M+ (g) + X- (g) � MX (s) [reverse]

∆H : M+ (g) + X- (g) + water � M+ (aq) + X- (aq)∆Hhyd : M+ (g) + X- (g) + water � M+ (aq) + X- (aq)

– ∆HLE : MX (s) � M+ (g) + X- (g)

MX (s) + water � M+ (aq) + X- (aq)

As a conclusion, ∆Hsoln = ∆Hhyd + (– ∆HLE)

If ∆Hsoln = - ve, then the salt is soluble in water

If ∆Hsoln = + ve, then the salt is insoluble in water

� 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,

decreaseincrease

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

decrease

Group 2 sulphate Be SO4 Mg SO4 Ca SO4 Sr SO4 Ba SO4

∆Hsolution (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

∆Hhydration

∆Hlattice energy

BeSO4 Mg SO4 CaSO4 Sr SO4 BaSO4

� Example : Solubility of Group 2 sulphate :

Sr2+

Ba2+

SO42-Be2+

Mg2+

Ca2+