14 Thermochemistry

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Chapter 14

Thermochemistry



Introduction

• Every chemical change is accompanied byenergy change, principally in the form of

HEAT ENERGY .

• i.e: The evolution or the absorption of heat

• Exothermic reaction: Eheat is liberated

Temperature of rxn mixture increases • Endothermic reaction: Eheat is absorbed

Temperature of rxn mixture decreases



GENERALLY (For stable compounds)

• Bond breaking is an endothermicprocess.

• Bond making/ formation is an

exothermic process.



•ENTHALPHY CHANGES:

–enthalpy of reaction, formation,

combustion, solution,neutralisation, atomisation, andlattice energy

•Molar heat capacity

•Specific heat capacity

TERMS & DEFINITION:



14.1 Entalphy changes

of reactions



Enthalpy Changes

• Entalphy = Total energy content of

reacting materials. Symbol: H

• We cannot measure entalphy, we can

only measure the entalphy change.



• Enthalpy change for a reaction, Δ H

= The total heat energy absorbed or

liberated in a chemical reaction= The energy exchange with the

surroundings where a reaction takes place

• Enthalpy change is measured in kJ mol –1



Let’s start with something you

know… •Symbol: Δ H

HΔ HΔ = HΔ tstanacReoductsPr



Endothermic Rxn - Energy Diagram

• Heat is absorbed from the surroundings,temp. decreases

• ∆H has a positive value

E n e r g y

Reactants

Products

∆H= +ve



If ΔHproducts > Δ Hreactants

then ∆H is positive Process is ENDOTHERMIC

ENTHALPY CHANGE, ΔH

∆Hrxn = ∆HProducts - ∆HReactants



11

• Heat is given out to the surroundings, temp.increases

• ∆H has a negative value

E n e r g y

Reactants

Products

∆H= -ve

Exothermic Rxn – Energy Diagram



If ΔHproducts < Δ Hreactants

then ∆H is negative

Process is EXOTHERMIC

ENTHALPY CHANGE, ΔH

∆Hrxn = ∆HProducts - ∆HReactants



13

THERMOCHEMICAL EQUATION:

E.g.:

CH4

(g) + 2O2

(g) CO2

(g) + 2H2

O(l)

ΔHθ = -890 kJ mol–1

Standard conditions:• 25 °C/ 298 K

• 1 atm

• [aq] = 1.0 mol dm –3



14

2 Basic Rules in usingThermochemical

equations in calculations



15

Rule 1:

Enthalpy change/

Total amount ofenergy releasedor absorbed

The number

of moles ofthe reactantsused

Δ H is directly proportional to thequantity of reactants and products.



16

Reactants Products ……ΔH

2(Reactants) 2(Products)…(∆H x 2)



17

E.g:

• H2(g) + 0.5 O2(g) H2O(l)

Δ H = -285.8 kJ

• 2H2(g) + O2(g) 2H2O(l) Δ H = -571.6 kJ



18

A B ΔH

B A -(ΔH)

Rule 2:

Δ H for a reaction is equal in magnitude but opposite in sign to Δ H for the

reverse reaction.



19

E.g.:• HgO(s) Hg(l) + ½O2(g)

Δ H =+90.7 kJ

• Hg(l) + ½O2(g)

HgO(s) Δ H =-90.7 kJ



20

Example 1:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔHθ = -890 kJ mol–1

(a) Is the rxn endothermic orexothermic?

(b) Is the total entalphy for the

reactants larger or smaller than thetotal entalphy for the products?



21

Example 1:

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

ΔHθ = -890 kJ mol–1 (c) What is the value of ΔHθ for the

reaction:i. 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l)

ii. ½CH4(g) + O2(g) ½CO2(g) + H2O(l)

iii. CO2(g) + 2H2O(l) CH4(g) + 2O2(g)iv. 3CO2(g) + 6H2O(l) 3CH4(g) + 6O2(g)



Five factors that affect the

value of the entalphy change:1. temp. of the experiment being carried

out2. Physical states of the reactants

3. Allotropic forms of the reactants

4. Pressure of the gaseous reactants

5. Concentration of the reactants



The Standard Conditions ForCalculating Enthalpy Changes

- T = 25°C or 298 K

- P = 101 kPa or 1 atm- [Reactant] = 1.0 mol dm-3

- For Allotropic reactants, the most stableallotrope at 298 K and 1 atm is used



Entalphy change measured

under standard conditionsis described as StandardEntalphy change of

reaction, ∆Hθ

.



25

14.2 Hess’s Law

• Hess’s Law states that the heat

liberated or absorbed during a chemicalreaction is independent of the route by

which the chemical change occurs.



26

• The Hess’s Law can be illustrated byenergy diagram/ entalphy diagram.

E n e r g y

Reactants

Products

∆H1

Intermediate

∆H2

∆H3

∆H2 + ∆H3 = ∆H1



27

• Another way to draw the energy diagram/

entalphy diagram.

Reactants

Products

∆H1

Intermediate

∆H2

∆H3

∆H2 + ∆H3 = ∆H1



14.3 Standard Entalphy changeof Formation, ∆Hf

θ • Definition: the enthalpy change when one

mole of compound is formed from its

elements under standard condition.

H2(g) + ½O2(g) H2O(l)

∆Hf Φ = -241 kJ mol-1

2H2(g) + O2(g) 2H2O(l)

∆Hf Φ =2 x -241 kJ

X

E le of the o he i l



Examples of thermochemicalequations for some compounds:

• The formation of Carbon dioxide:C(s) + O2(g) CO2(g) ∆Hf

θ= -394 kJ mol –1

• The formation of Urea:C(s) + 2H2(g) + N2(g) + ½O2(g) CO(NH2) 2 (s)

∆Hf θ= -333.5 kJ mol –1

• The formation of Potassium nitrate:

K(s) + N2(g) + O2(g) KNO3(s)

∆Hf

θ= -x kJ mol –1

12

32



30

• ∆Hf Φ of elements = 0

–Eg: H2 (g) H2 (g)

Na (s) Na (s) Ne (g) Ne (g)

He (g)

He (g)



31

Some typical Standard Entalphy change

of formation, ∆Hfθ (kJ mol –1)

Substance/ Compound ∆Hfθ (kJ mol –1) * 298 KC (s), Graphite 0

C (s), Diamond +1.90

CH4 (g), methane – 74.81C3H6 (g), propene + 20.42

CO (g), carbon monoxide – 110.92

CO2 (g), carbon dioxide – 395.01

HCOOH (l), methanoic acid – 424.72

CH3COOH (l), ethanoic acid – 484.5

C6H12O6 (s), glucose – 1268



• Most compounds have negative

standard enthalpies of formation

• They are called exothermic compounds

– VERY EXOTHERMIC compounds are stable

–Formation is spontaneous

• The more negative the ∆Hf Φ, the morestable is the product.



33

• Few compounds have positive standardenthalpies of formation

• They are called endothermic compounds – VERY ENDOTHERMIC Compounds are

unstable



34

ΔHfθ and Stability of

a compound



35

ΔHfθ and Stability of a compound

Example 1 (At room conditions):C (diamond) → C (graphite) ∆Hf

θ = -2 kJ mol –1

• The –ve ∆Hf θ shows that graphite is more

stable than diamond. I.e. graphite is more

stable energetically.

• Diamond tends to change to graphite.

• But this reaction (diamond → graphite) is

very slow. Hence, diamond is energetically

unstable, but kinetically stable.

Example 2 (At room conditions):



36

Example 2 (At room conditions):

H2 (g) + O2 (g) → H2O2 (l) ∆Hf θ = -188

1. Is H2O2 (l) stable energetically?

2. Does the above reaction occur spontaneously?

3. Given:

H2(g) + ½O2(g) → H2O(l) … ∆Hf θ = –286 kJ mol –

1

a) Which substance is more stable energetically,H2O2 (l) or H2O (l)? Explain.

b) Find the value of the ∆Hf θ of:

H2O2 (l) → H2O (l) + ½O2 (g)kJ mol –1

ΔH θ and the energetic stability of



37

ΔHfθ and the energetic stability of

hydrogen halides (Page 462):

Hydrogenhalides

HF (g) HCl (g) HBr(g) HI (g)

∆Hfθ

(kJ mol –1

)

-271 -92 -36 +26.5

H F H Cl Br H

I

H

The size of halogen atoms increasesThe bond length of H – X increasesThe strength of H – X bonds decreases

HF is a very exothermic compound.It is energetically stable. It does not

decompose easily upon heating.



38

ΔHfθ and Reactivity

of halogens(Page 461)



39

Using ΔHfθ to

calculate ΔHθreaction



40

)tstanac(ReHΔmΣ )oducts(PrHΔnΣ=

HΔ reaction, of changeentalphyStandard

θ

f

θ

f

θrxn

• Example 14.6 (Pg 462)

• Example 14.7 (Pg 463)



41



42



43



44

E l



45

Example:

Given, 1

1

22

394

)()()(

kJmol H

gCOgOsC

1

2

222

286

)()(2

1

)(

kJmol H

lO H gOg H

1

3

22222

1300

)()(2)(2

5)(

kJmol H

lO H gCOgOg H C



46

Calculate ∆H for the following reaction:

2 methods:• Algebraic method

• Entalphy diagram

)()()(2 222 g H C g H sC



47

Answer :

∆H = +226kJ

M th d 1 Al b i th d



48

Method 1: Algebraic method

2C + 2O2 2CO2 -394 kJ x 2

H2 + O2 H2O -286 kJ

2C + H2 + 2 O2 2CO2 + H2O -1074kJ

Given,

C2H2 + 2 O2 2CO2 + H2O -

1300 kJ

2CO2 + H2O C2H2 + 2 O2 +1300 kJ

12

12

+

12

12

1

2



49

2C + H2 + 2 O2 2CO2 + H2O -1074 kJ

2CO2 + H2O C2H2 + 2 O2 +1300 kJ

+ :

2C + H2 C2H2 +226 kJ

Hence, the entalphy change for the reaction:2C + H2 C2H2 is +226 kJ.

12

12

1

22

Method 2: Entalphy cycle



50

Δ H= y

-1300 kJ

Method 2: Entalphy cycle

C2H2 + 2 O22C + H2 + 2 O2

2CO2+H2O

-(394 x 2) +(-286) kJ

1

2

1

2

Using Hess’s Law, y = [-(394 x 2) + (-286)] – (-1300)

y = + 266 kJ

+1300 kJ

et o : s ng ormu a



51

et o : s ng ormu aGiven, ∆Hf

θ (CO2) = – 394 kJ mol – 1

∆Hf θ (H2O) = – 286 kJ mol – 1

Let: ∆Hf θ (C2H2) = x kJ mol – 1

Thermochemical equation:

C2H2(g) + O2(g) 2CO2(g) + H2O (l) …∆Hθ = –1300 kJ mol –1

Solution:

C2H2(g) + O2(g) 2CO2(g) + H2O (l)

∆Hf θ(kJ mol – 1): x 0 2( – 394) ( – 286)

∆Hθ = Σ∆Hf θ(products) – Σ∆Hf

θ(reactants)

[2( – 394) + ( – 286)] – [ x + 0] = – 1300

x

= +266 kJ mol

– 1

52

52



52

•Quick check 14.2 (Pg 463)

14 4 St d d M l E th l



14.4 Standard Molar EnthalpyChange of Combustion, ∆Hc

θ

• Definition: The enthalpy change when onemole of substance is completely burned in

excess oxygen under standard conditions\

• E.g. 1:

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

∆Hc

Φ

= -890 kJmol

-1

● When one mole of methane, CH4 (g) is completely

burnt in excess oxygen under standard conditions,

890 kJ of heat energy is released.

b f b



54

• E.g. 2: Combustion of carbon:

C (s) + O2(g) CO2(g) ∆HcΦ = -394 kJmol-1

● When one mole of carbon, C (s) is completely burnt

in excess oxygen under standard conditions, 394 kJ

of heat energy is released.

• E.g. 3: Combustion of carbon:

H2(g) + ½O2(g) H2O(l) ∆HcΦ = -242 kJmol-1

● When one mole of hydrogen gas, H2 (g) is

completely burnt in excess oxygen under standard

conditions, 242 kJ of heat energy is released.

●Sometimes, ∆HcΦ = ∆Hf

Φ



55



● ∆HcΦ is always exothermic

● ∆HcΦ can be determined by using a

bomb calorimeter



57

Which of the following are equations for

ΔHcθ?

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

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

3. 3CH4 (g) + 6O2(g) 3CO2(g) + 6H2O(l)

4. Mg (s) + ½O2 (g) MgO (s)

5. C2H5OH (l) + 2[O] CH3COOH (l) + H2O(l)

6. C2H5OH (l) + 3O2(g) 2CO2(g) + 3H2O(l)



58

7. 2Fe (s) + O2(g) Fe2O3(s)

8. Fe (s) + ¾O2(g) ½Fe2O3(s)

9. 2H2 (g) + O2 (g) 2H2O (l)

10. H2 (g) + ½O2 (g) H2O (g)

11. H2 (g) + ½O2 (g) H2O (l)

12. ½N2 (g) + O2 (g) NO2 (g)

13. N2 (g) + 2O2 (g) 2NO2 (g)

32



59

∆Hcθ and molecular structure

• Isomers usually have similar ∆HcΦ.

• Chemical reactions occurs in two steps:

1. Energy is absorbed to break bonds

between atoms in reactants.

2. Energy is liberated when the atomsform new bonds to form the products.

• E.g.: For the combustion of methane



60

gCH4(g) + 2O2(g) CO2(g) + 2H2O(l)

∆HcΦ = -890 kJmol-1

O O

H HC

H

H

O O

Energy

CH4 + 2O2

C + 4H + 4O

Heat is absorbed

• E.g.: For the combustion of methane



61

E.g.: For the combustion of methane

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

∆Hc

Φ

= -890 kJmol-1

O

HC

H

O

O

H

O

H

CO2 + 2H2O

C + 4H + 4O

EnergyHeat is released



62

• Hence, ∆HΦreaction

= Total of heat change

= +(Heat absorbed) – (Heat released)

CO2 + 2H2O

C + 4H + 4O

Energy

CH4 + 2O2



63

• If the quantity of absorbed heat >

released heat, the reaction becomes

endothermic.

• If the quantity of absorbed heat <

released heat, the reaction becomes

exothermic.



64

∆HcΦ and bond strength:

• Alkanes

• Butane VS methyl propane

• 1-butene VS cyclobutane

Standard entalphies of combustion of the first ten



65

∆HcΦ

(kJmol-1)

1 2 3 4 5 6 7 8 9 10Number of carbon atoms

(straight chain) alkanes

CH4C2H6

C3H8

C4H10

C5H12

C6H14

C7H16

C8H18

C9H20C10H22

Alkane ∆HcΦ (kJmol-1) Difference in ∆Hc

Φ

CH4

C2H6

C3H8

C4H10

– 890 – 1560

– 2220

– 2877

670

660

657

CH2

CH2

CH2

• Each alkane is different from the next alkane



66

• Each alkane is different from the next alkaneby the addition of a –CH2 – group.

• The difference in the ∆HcΦ

between eachsuccessive member of the series is about 650kJ mol –1.

• This 650 kJ mol –1

corresponds to thebreaking and forming of bonds when onemole of –CH2 – group undergoes combustionto form CO2 and H2O.

–CH2 – + O2 CO2(g) + H2O(l)32

–650 kJ mol –1

E.g. 1: ∆HcΦ of butane & methyl propane



67

E.g. 1: ∆Hc of butane & methyl propane

Hydrocarbon Butane, C4H10 Methylpropane,

C4H10

Structural

formula

∆HcΦ

(kJmol-1)

– 2877 – 2870

Type of

bonds

Three C –C bonds and

ten C –H bonds

Three C –C bonds

& ten C –H bonds

H H H H

H H H H

H C C C C H

H H H

H H

H C C C H

H C H

H

e.g. 2: ∆HcΦ of 1-butene & cyclobutane



68

e.g. 2: ∆Hc of 1 butene & cyclobutane

Hydrocarbon 1-butene, C4H8 Cyclobutane, C4H8

Structural

formula

∆HcΦ

(kJmol-1) – 2717 – 2800

Type of

bonds

Two C –C bonds, oneC=C bond & eight C –

H bonds

Four C –C bonds & eight C –H bonds

H H H H

H H

H C C C C H C

C

C

C

H

H

H

H

H

H

H

H

E.g. 1: ∆HcΦ of butane & methyl propane



69

E.g. 1: ∆Hc of butane & methyl propane

Hydrocarbon Butane, C4H10 Methylpropane,

C4H10

Structural

formula

∆HcΦ

(kJmol-1)

– 2877 – 2870

Type of

bonds

Three C –C bonds and

ten C –H bonds

Three C –C bonds

& ten C –H bonds

H H H H

H H H H

H C C C C H

H H H

H H

H C C C H

H C H

H



70

• Both butane and methylpropane contain

the same number of atoms and the same

type of chemical bonds.

• Their ∆Hc

Φ

are similar.• This suggests that when each type of

bond is broken or formed, a fixed value

of energy is absorbed/ liberated.

e.g. 2: ∆HcΦ of 1-butene & cyclobutane



71

e.g. 2: ∆Hc of 1 butene & cyclobutane

Hydrocarbon 1-butene, C4H8 Cyclobutane, C4H8

Structural

formula

∆HcΦ

(kJmol-1) – 2717 – 2800

Type of

bonds

Two C –C bonds, oneC=C bond & eight C –

H bonds

Four C –C bonds & eight C –H bonds

H H H H

H H

H C C C C H C

C

C

C

H

H

H

H

H

H

H

H



72

• The ∆Hc

Φ for 1-butene and cyclobutane

are different.

• This is because different type of bonds inreactants (1-butene and cyclobutane) are

broken to form 4 moles of CO2

and 4

moles of H2O.



73

• C=C bond is shorter and stronger than C –Cbond.

• More heat is absorbed to break/ dissociate

a C=C bond.

• Hence, the combustion of but-1-ene

releases less heat than cyclobutane.

∆H Φ of 1-butene & cyclobutane



74

∆Hc of 1 butene & cyclobutane

4CO2 + 4H2O

4C + 8H + 12O

Energy

1-butene +6O2

4CO2 + 4H2O

4C + 8H + 12O

Energy

cyclobutane+ 6O2

∆HcΦ ∆Hc

Φ

Your turn…



75

• Which bond is stronger, the C –C bond or

C=C bond?

• The formation of a stronger bond releases

energy. Hence, the formation of

bond is more exothermic.

• The breaking of a stronger bond also

requires energy. Hence, thedissociation of a bond is more

endothermic than a bond.

more

C=C

moreC=C

C –C

Your turn…

Example:



pWhen 0.540 g of benzoic acid is completely burnt in

a bomb calorimeter, it causes the temperature of

water in the calorimeter to increase by 1.38 K. Thecombustion of 0.148g of butan-1-ol in the same

bomb calorimeter causes an increase of 0.51 K.

a. Calculate the calorimeter constant (heatcapacity of the calorimeter), in kJ K–1.

b. Calculate the standard enthalpy change of

combustion of butan-1-ol.

[The standard heat combustion of benzoic acid is

-3230 kJmol-1. The Mr of benzoic acid and butan-1-

ol are 122 and 74 respectively]

Solution:



a. Heat liberated from the combustion of 0.540 gbenzoic acid

= (0.540/122) x 3230= 14.30 kJ

Calorimeter constant = 14.30/1.38

= 10.36 kJ K -1

b. Number of moles of butan-1-ol

= 0.148/74 = 0.002 mole

Heat liberated from the combustion of butan-1-ol= 10.36 x 0.51 = 5.2836 kJ

Standard enthalpy change of combustion

= 5.2836/0.002 = 2641.8 kJ mol-1



Exercise :

Quick check 14.3



14.5 Standard Molar

Enthalpy Change of Neutralisation, ∆Hneut

Φ

∆H Φ



∆Hneut

•The enthalpy change when 1 mole of H+(aq)ions from an acid reacts with 1 mole of OH-

(aq) ions from an alkali to form one mole of water molecules under standard conditions.

• OR: The entalphy change when 1 mole of water is formed from the reaction betweenan acid and a base under standard

conditions.• H+(aq) + OH-(aq) H2O(l)… ∆Hneut

θ • ∆Hneut

Φ is always negative

∆HneutΦ for some acid-base rxns:



81

neut

Acid Base∆Hneut

Φ

(kJmol –1

)

Type of

neutralisation

HCl NaOH – 57.3Strong acid - Strong

base

HNO3 KOH – 57.3

Strong acid - Strong

base

CH3COOH NaOH – 55.6Weak acid - Strong

base

CH3COOH NH3 – 50.4 Weak acid - Weakbase

Write the thermochemical equations for the

neutralisation reactions.

Strong acid + Strong base



Strong acid + Strong base

• Strong acids and alkalis dissociate completely inaqueous solution.

• ∆HneutΦ is constant, i.e. -57.3 kJ mol-1

• E.g.:

NaOH(aq)+ HCl(aq) NaCl(aq) + H2O(l)

∆HneutΦ = -57.3 kJ mol-1

• What are “Spectator ions”?

Wh HCl ( ) [St id] d N OH ( )



83

• When HCl (aq) [Strong acid] and NaOH (aq)

[Strong alkali] react,

Na+(aq) + Cl-(aq) + Na+(aq) + Cl-(aq) + H2O(l)

H+(aq) + OH-(aq) ∆HneutΦ = -57.3 kJ mol-1

• Na+(aq) & Cl-(aq) do not take part in

neutralisation.

• Hence, for strong acid-strong base reaction:

H+(aq) + OH-(aq) → H2O(l) ∆HneutΦ = -57.3 kJmol-1

Example:



84

Example:

Find the standard entalphy for the followingreaction:

1. KOH(aq) + HNO3(aq) KNO3(aq) +

H2O (l) ∆H1Φ

2. Ca(OH)2(aq) + 2HNO3(aq) Ca(NO3)2

(aq) + 2H2O (l) ∆H2Φ

Refer to example 14.14 on page 470.

Neutralisation involving Weak acids/



• ∆HneutΦ

for a weak acid and a strong alkali orvice versa is lower (less negative) than

-57.3 kJ mol-1

● Less heat energy is liberated in this reaction.● E.g.:

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) +

H2O(l) ∆HneutΦ = -55.6 kJ mol-1

gWeak bases



●Weak acids and alkalis do not dissociatecompletely

●Part of the heat energy liberated during

neutralisation is reabsorbed to cause

dissociation of the weak acid/ base.

●CH3COOH H+

+ CH3COO –

Endothermic

Neutralisation involving HF



• Neutralisation between HF (weak acid) &

NaOH, ∆HneutΦ is more negative than -57.3 kJ

mol-1 (ie. -68.6 kJ mol-1)

• When HF dissociates, the process is highly

exothermic.

• The small F- (high charge density) are

hydrated, releasing high amount of energy.• I.e.: HF (aq) H+ (aq) + F- (aq)

highly exothermic

g

Hydration energy?



Hydration energy?

• Hydration of X Formation of bondingbetween particle X with H2O.

• Formation of strong bonding liberates high

hydration energy.

• Formation of weak bonding liberates low

hydration energy.• When a substance is hydrated, it becomes

wet.

Hydration of anion



-δ-

δ+

δ-

δ+

δ+

δ-

Water

Anion

Hydration of anion

Hydration of cation



90

+δ+

δ-

δ+

δ-

δ-

δ+

Water

Cation

Hydration of cation

Hydration of fluoride ion



F–

HO

H

Hydration of fluoride ion



Hydration energy?

• The size of F – is very small.

• It has an extremely high charge

density.• Therefore, strong F – – H2O bond isformed.

• Formation of strong bonding liberateshigh hydration energy.

∆HneutΦ of Strong Polyprotic Acid



neut g yp

• More exothermic.

• Eg:NaOH(aq) + H2SO4(aq) NaHSO4(aq) + H2O(l)

∆H1 = -61.95 kJ mol-1

NaOH(aq) + NaHSO4(aq) Na2SO4(aq) + H2O(l)

∆H2 = -70.90 kJ mol-1

∆Hneut

Φ

= (-61.95 + -70.90)/ 2 kJ mol-1

= -66.4 kJ mol-1 (per mole H2O formed)> -57.2 kJ mol-1



• Heat energy is also liberated whenpolyprotic acid is diluted as the alkali is

added to the strong diprotic acid.

Example



Example

200 cm3 of 0.50 mol dm-3 hydrochloric acid is

mixed with the same volume of 0.50 mol dm-3 sodium hydroxide at the same temperature.

The temperature of the mixture is raised to

3.4 K. Calculate the enthalpy of neutralisation.[Specific heat capacity of solution

= 4.2 J g-1 K -1 ]

Answer: -57.12 kJ mol-1



96

•Quick check 14.4

14 10 S d d B d



14.10 Standard Bond

Dissociation Enthalpy

• Endothermic

• The energy absorbed to separate the

two atoms in a bond under standard

conditions

Energy



A B

A–B bond is

broken ALL reactants &products are ingaseous state.

Energy

• E.g.:



g

• (g) C (g) + 4H (g)

∆HΦ=+1662 kJ mol-1

• Four C–H bonds are broken.• i.e.: 1662 kJ is needed to break 4 (moles) ofC–H bonds.

• Then (1662 ÷ 4) = 416 kJ is needed to break1 (mole) of C–H bonds.

• The average bond entalphy of the C–H

bond = +416 kJ mol-1

H

H C H

H

State the bonds that have to be



State the bonds that have to bebroken during a reaction:





• Example 12 & 13.

14.11 Entalphy Change of Fusion



14.11 Entalphy Change of Fusion& Vaporisation:

• The standard entalphy change of fusion,

Δ Hθfusion, is the entalphy change when one

mole of a solid changes to one mole of liquidat its melting point at standard pressure (1.0

atm).

• E.g.: H2O (s) H2O (l)

Δ Hθfusion= +x kJmol –1



• The standard entalphy change of

vaporisation, Δ Hθvaporisation, is the entalphy

change when one mole of a liquid changesto one mole of gas at its boiling point at

standard pressure (1.0 atm).

• E.g.: H2O (l) H2O (g)

Δ Hθvaporisation= +y kJmol –1

14.12 Standard Molar Enthalpy



Change of Atomisation, ∆HatΦ

• Endothermic

• The heat energy absorbed when one mole of

gaseous atoms (Product) are formed from itselement (Reactant) under standard conditions

Element Atom(At 298 K, 1.0 atm) (g)

Atomisation



106

AtomisationFor diatomic gaseous:

Gaseous atomsFor metals (solid):

Gaseous atomsElements at std

conditions

Gaseous atoms

• E.g.:



• Na (s) Na (g) ∆HatΦ=+107 kJ mol-1

• ½ H2 (g) H (g) ∆HatΦ=+218 kJ mol-1

• ∆HatΦ for monoatomic gases (Noble gases) iszero because no chemical occurred.

• E.g.:

• He (g) He (g) ∆HatΦ= 0 kJ mol-1

• Ne (g) Ne (g) ∆Hat

Φ= 0 kJ mol-1

Φ



• ∆HatΦ for liquids include the enthalpy

change of vapourisation

• ∆Hat

Φ

for solids include the enthalpychange of fusion & the enthalpy change of

vapourisation.

• ∆HatΦ for noble gases (monoatomic gas) is

always zero.

About ∆HatΦ

Which of the following equations



represent atomization?

1. Mg (s) Mg (g)

2. Na (s) Na (l)

3. ½Cl2 (g) Cl (g)

4. Cl2 (g) 2Cl (g)

5. C (s) C (g)

6. H2O(l) H2O(g)

7. NaCl (s) Na (g)

+ Cl (g)

8. ½N2 (g) N (g)

9. O2 (g) 2O (g)

10.H2O(l) H2(g) +

½O2 (g)

The First Electron Affinity



• 1st EA = Enthalpy change when a moleof gaseous atom accepts a mole of electron to form a mole of gaseous

uninegative ions• i.e.

M(g) + e M-(g)1st EA



• E.g:

• H (g) + e H- (g) ∆HΦ = -72.8 kJ mol-1

• Cl (g) + e Cl- (g) ∆HΦ = -364.0 kJ mol-1

• The more exothermic the EA, the more

stable the anion formed.

Second Electron Affinity



• 2nd

EA = Enthalpy change when a mole of gaseous uninegative ions accepts a mole

of electron to form a mole of gaseous

dinegative ions.• i.e.

M-(g) + e M2-(g)

2nd EA

The Electron Affinity



• E.g.:

1. O (g) + e O- (g) ∆HΦEA1=-142 kJ mol-1

2. O-

(g) + e

O2-

(g) ∆HΦ

EA2=+844 kJ mol-

1

• 2nd EA is endothermic, 1st EA isexothermic.

2nd EA is endothermic…



• A negatively charged ion will repel theaddition of 2nd electron

• Energy must be absorbed to overcome therepulsive forces between the 2 negativelycharged particles

S – e – Repulsion

14.13 Lattice Energy



Definition:

• The enthalpy change/ energy liberatedwhen one mole of solid ionic crystalline is

formed from its gaseous constituent ions.

Ax+(g) + By-(g) AyBx(s)



• Eg:



• Eg:

•Na+

(g) + Cl-

(g) NaCl (s) ∆Hlat

Φ=-788 kJ mol-1

•K +(g) + Cl-(g) KCl (s) ∆Hlat

Φ=-701 kJ mol-1

• Always -ve because the formation of chemical bond (ionic) is exothermic.

Lattice dissociation Energy



• Opposite reaction, crystal lattice is broken.

• Eg:

• NaCl (s) Na+(g) + Cl-(g)

∆HΦ

lattice dissociation= +788 kJ mol-1

• KCl (s) K +(g) + Cl-(g) ∆HΦ

lattice dissociation= +701 kJ mol-1

• Always +ve because the dissociation of

chemical bond is endothermic

Factors affecting ∆HΦ



lattice

The magnitude of Δ Hlat depends on:• The charge on the ions (attractive forces

between anion & cation), and

• The distance between the ions

∆Hlat α l Z+

x Z-

l(r+ + r-)

• The greater the ionic charges,h h h



the more exothermic the

lattice energy.Compound Z+ Z- ∆Hlat(kJ mol-1)

NaCl +1 -1 -770

Na2O +1 -2 -2478

MgO +2 -2 -3850

Al2O3 +3 -2 -15 920

Exo/ endo-thermic?

• The smaller the distance between



h sma r th stanc tw nions, the more exothermic the

lattice energy.

Ionic

Compound

r+

(nm)

r-

(nm)

∆Hlat

(kJ mol-1)

NaCl 95 181 -770

NaBr 95 196 -731

NaI 95 216 -684

• The lattice energy give us some ideaabout the strength of ionic bonds



about the strength of ionic bonds.

How?

• The more exothermic the latticeenergy, the stronger the ionic bond.

Compound Z+ Z- ∆Hlat(kJ mol-1)

NaCl +1 -1 -770

Na2O +1 -2 -2478MgO +2 -2 -3850

Al2O3 +3 -2 -15 920

14 14 The Born-Haber Cycle



14.14 The Born-Haber Cycle

• A cycle of reactions used for calculating

the lattice energies of ionic crystalline

solids.

• An application of Hess’ s Law.



• It involves;- enthalpy of atomisation

- ionisation energy

- electron affinity

- lattice energy

- enthalpy of formation

Important formula:



Important formula:

ΔH+

ΔH+

ΔH+

ΔH+ΔH

=HΔ

lataffinitye.

ionization

atom(2)atom(1)

f





Na(s) + ½Cl2(g) NaCl(s)



( ) 2(g) ( )

∆Hf

Φ

=-411kJ mol-1

From the Born-Haber Cycle,

-411 = +107+122+496+(-349)+

lat. energy

Lattice energy = -787 kJ mol-1

Example:

Draw a Born-Haber cycle for rubidium



Draw a Born Haber cycle for rubidium

iodide, RbI(s). Hence use the data below to

calculate the enthalpy change of formation

of RbI(s).

Δ Hat of rubidium = +86 kJ mol-1

Δ Hat of iodine = +107 kJ mol-1

Δ Hlattice of rubidium iodide = -609 kJ mol-1

IE of rubidium = +402 kJ mol-1

AE of iodine = -314 kJ mol-1

Rb+(g) + I(g)

Energy (kJ mol-1)



RbI(s)

Rb(s) + ½ I2(g)

Rb(g) + ½ I2(g)

Rb+(g) + ½ I2(g)

Rb+(g) + I-(g)

Hf

+86

+402

+107 -314

-609

Example:

Draw a Born-Haber cycle for calcium oxide. Hence



Draw a Born Haber cycle for calcium oxide. Hence

use the data below to calculate the enthalpy

change of formation of calcium oxide.

Enthalpy change of atomisation of calcium = +177 kJmol-1

Enthalpy change of atomisation of oxygen = +249 kJmol-1

Lattice energy of calcium oxide = -3409 kJ mol-1

1st Ionisation energy of calcium = +592 kJ mol-1

2nd Ionisation energy of calcium = +1100 kJ mol-1

Electron Affinity of oxygen = -140 kJ mol-1

Formation of O2- = +795 kJ mol-1

Answer



Answer

-636 kJ mol-1



•The more exothermic theenthalpy change of

formation, the morestable the compound.

Test Yourself 16



Test Yourself 16

14.15 Hydration Energies, ∆Hhyd



y g hyd

• Enthalpy change of hydration, ∆Hhyd is

the heat liberated when one mole of free

gaseous ion dissolves in water to form

hydrated ions of infinite dilution under

standard conditions .

• An Exothermic process.

• Bonds are formed with water molecules.

∆Hhyd



Mn+(g) + H2O Mn+(aq)

∆HhydΦ of the cation

Xn-

(g) + H2O

Xn-

(aq)∆Hhyd

Φ of the anion

What happens during hydration?



pp g y

+δ+ δ-

δ+

δ-

δ-

δ+

Water

Cation



-δ- δ+

δ-

δ+

δ+

δ-

Water

Anion



F–

HO

H

Water-ion bonds are formed.

Energy is liberated.

The stronger the water-ion bonds, themore exothermic the hydration energy.

Hydration energies depends on:



Hydration energies depends on:

1. Ionic radius, r

2. Ionic Charge, Z

The larger the chargedensity (=Z/r), the

stronger the water-ionbond.

And hence, the more

exothermic the hydrationenergy.Heh… heh……

Try these:M t h th i ΔH ith it i



Match the given ΔHhyd with its ions:

Ion Li+ Na+ K+ Rb+

Radius

(pm) 78 98 133 149ΔHhyd

(kJ mol–1)

-320 -410 -295 -520

Try these:M t h th i ΔH ith it i



Match the given ΔHhyd with its ions:

Ion Na+ Mg2+ Al3+

Radius (pm) 78 98 133

ΔHhyd

(kJ mol–1)

-410 -4680 -1930





Enthalpy Change ofSolution/ Solvation, ∆H l



Solution/ Solvation , ∆Hsol

Enthalpy change when 1 mole of solidionic compound dissolves in water to

form a solution of infinite dilution under standard conditions

Mn+ X n-(s) + H2O Mn+(aq) + X n-(aq) ∆HΦ= Enthalpy change of solution

The process of Solution:



Lattice (solid)

Hydrated ions(aq)

Separated ions(g)

Th ll f l ti

The process of solution



The overall process of solution

can be divided into 2 stages:• The breaking down of the crystal

lattice into gaseous ions Endothermic (= - ∆Hlat)

• The solvation (hydration of separated

gaseous ions by water molecules)

Exothermic (Hydration energy of thecation and of the anion)

Stage 1: Breaking down thecrystal lattice into ions (g)



M+ X -(s) M+(g) + X -(g) ∆H1= Endothermic (- ∆Hlat )

y (g)

+

+ +

+

+ + +

+

– – – –

– –

– Solid lattice structure

+ +

– – Free ions (g)

Stage 2: Hydration of ions(g)by water molecules



Mn+(g) + water Mn+(aq)

∆H2=exothermic

And:

X n-(g) + water X n-(aq)

∆H3=exothermic



+

+

–

–

The enthalpy change of solution, ∆Hsol



= ∆H1 + ∆H2 + ∆H3

= (-lattice energy) + (enthalpy changes of

hydration of the cation and anion)

= (+ve) + (-ve + -ve)

= +ve (insoluble) or -ve (soluble)

Is a compound soluble/insoluble in water?



insoluble in water?

•If ∆Hhyd > ∆Hlat Soluble

•If ∆Hhyd < ∆Hlat INsoluble

Energy Solubility in water



Insoluble

Soluble Latticeenergy

Enthalpy of hydration

Compounds

Solubility of the Group II Sulphates

- The solubility of the sulphates decreases



The solubility of the sulphates decreases

down the group- Beryllium sulphate and magnesium

sulphates are soluble in water

- Calcium sulphate and strontium sulphateare sparingly soluble in water

- Barium sulphate is insoluble in water

- When a group 2 sulphate dissolves in waterMSO4 (s) + water M2+(aq) + SO4

2-(aq)

• The dissolution process take place in 2distinct stages



distinct stages

(i) the breaking of ionic bondsMSO4 (s) M2+(g) + SO4

2-(g)

(ii) hydration of the gaseous ions bywater molecules

M2+

(g) + SO42-

(g) + aq

M2+(aq) + SO4

2-(aq)

• Ionic radius increases

Going down Group 2:



Ionic radius increases

• Δ Hlat and Δ Hhyd decreases (becomes less exo)• The drop off in the Δ Hlat of Group 2 sulphates

is less rapid compared to the decrease in the

enthalpy change of hydration.• Ionic radius of SO4

2- is very large, thechanging size of the much smaller cations

makes very little difference to the value of (r+ + r-) No significant reduction in Δ Hlat downGroup 2 sulphate.

• But, the increasing size of the cations causes

id bl h i ΔH



considerable changes in Δ Hhyd

• In short, decrease in solubility in Group

2 sulphate is due to the decrease of

enthalpy of hydration of the metal ions

• Enthalpy of hydration must be greater

than lattice energy in order for the

compound to dissolve in water

EnergyThe Variation of ∆Hlat and∆Hhyd: Group 2 sulphates



Insoluble

Soluble Latticeenergy

Enthalpy of hydration

BeSO4 MgSO4 CaSO4 SrSO4 BaSO4

hyd

• The solubility of gr. 2 hydroxides and fluoridesincreases on going down the group



•The presence of covalent character causes thecompounds to have higher ∆Hlat than expected

• Solubility of covalent compound is determinedby its polarity~ Like-dissolves-like Rule

• Non-polar compound dissolve in non-polarcovalent solvent, naphthalene dissolves inbenzene (v.d.w forces present in both comp.)

• Cannot dissolves in water because cannotovercome the strong attraction in watermolecules

• Polar compound dissolves in polar solvent

• Hydroxyl group ( OH) carboxylic acid



• Hydroxyl group (-OH), carboxylic acid

group (COOH), amine group (NH2),sulphonic acid group (SO3H) promote thesolubility of the compound in water byforming H-bonding with water

• More hydrophilic groups, more soluble andmore H- bond formed in the water

• Hydrophobic groups, such as alkyl groups

decrease the solubilities of organiccompounds

Exercise: Checkpoint 15.6



p

Examination Ques. 15:Objectives,Structural Questions

1,2,3,4 & 5

14 Thermochemistry

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