9.0 THERMOCHEMISTRY

Concept of Enthalpy

Important Terms Heat is energy transferred between two bodies of

different temperatures

System is any specific part of the universe

Surroundings is everything that lies outside the system

Open system is a system that can exchange mass and energy with its surroundings

Closed system is a system that allows the exchange of energy with its surroundings

Isolated system is a system that does not allow the exchange of either mass or energy with its surroundings

Energy is the ability to do work SI unit of energy is kg m2 s-2 or Joule (J) Non SI unit of energy is calorie (Cal) 1 Cal = 4.184 J

Thermochemistry A study of heat change in chemical reactions. Two types of chemical reactions:

Exothermic Endothermic

Exothermic reactions Enthalpy of products < Enthalpy of reactants, ΔH is

negative.

Energy is released from the system to the surroundings.

Consider the following reaction: A (g) + B (g) → C (g) ΔH = ve (reactants) (product)

reactantsenthalpy

reaction pathway

= -veH

products

Energy profile diagram for exothermic reaction

Enthalpy of products > enthalpy of reactants, ΔH is positive

Energy is absorbed by the system from the surrounding

Endothermic Reactions

Consider the following reactionA (g) + B (g) → C(g) ΔH = + ve (reactants) (product)

Energy profile of diagram endothermic reactions

Enthalpy, H The heat content of a system or total energy in the

system

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

Example: system undergoes combustion or ionisation.

Enthalpy of Reaction, ∆H and Standard Condition

Enthalpy of reaction: The enthalpy change associated with a chemical

reaction.

Standard enthalpy, ∆Hº The enthalpy change for a particular reaction that

occurs at 298K and 1 atm (standard state)

Thermochemical Equation The thermochemical equation shows the enthalpy changes.

Example : H2O(s) → H2O(l) ΔH = +6.01 kJ

1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C, ΔH = +6.01 kJ

However, when 1 mole of H2O(s) is formed from 1 mole of H2O(l), the magnitude of ΔH remains the same with the opposite sign of it.

H2O(l) → H2O(s) ΔH = 6.01 kJ

Types of Enthalpies There are many kind of enthalpies such as:

Enthalpy of formation Enthalpy of combustion Enthalpy of atomisation Enthalpy neutralisation Enthalpy hydration Enthalpy solution

Enthalpy of Formation, ∆Hf

The change of heat when 1 mole of a compound is formed from its elements at their standard states.

H2 (g) + ½ O2(g) → H2O (l) ∆Hf = 286 kJ mol1

The standard enthalpy of formation of any element in its most stable state form is ZERO.∆H (O2 ) = 0 ∆H (Cl2) = 0

Enthalpy of Combustion, ∆Hc

The heat released when 1 mole of substance is burned completely in excess oxygen.

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

Enthalpy of Atomisation, Ha The heat change when 1 mole of gaseous atoms is formed

from its element

Ha is always positive because it involves only breaking of bonds

e.g:

Na(s) Na(g) Ha = +109 kJ mol-1

½Cl2(g) Cl(g) Ha = +123 kJ mol-1

Enthalpy of Neutralization, ∆Hn The heat change when 1 mole of water, H2O is formed from the

neutralization of acid and base .

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

Enthalpy of Hydration, Hhyd

The heat change when 1 mole of gaseous ions is hydrated in water.

e.g:Na+

(g) Na+(aq) Hhyd = 406 kJ mol-1

Cl-(g) Cl-

(aq) Hhyd = 363 kJ mol-1

Enthalpy of Solution, Hsoln The heat change when 1 mole of a substance is

dissolves in water. e.g: KCl(s) K+

(aq) + Cl(aq) Hsoln = +690 kJ mol-1

Enthalpy of Sublimation, Hsubl

The heat change when one mole of a substance sublimes (solid into gas).

I2 (s) → I2(g)

sublH

Hsubl = +106 kJ mol1

Calorimetry A method used in the laboratory to measure the

heat change of a reaction. Apparatus used is known as the calorimeter Examples of calorimeter

Simple calorimeter Bomb calorimeter

The outer Styrofoam cup insulate the reaction mixture from the surroundings (it is assumed that no heat is lost to the surroundings)

Heat release by the reaction is absorbed by solution and the calorimeter

Simple calorimeter

A bomb calorimeter

Important Terms in Calorimeter Specific heat capacity, c

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 (Jg 1C1).

Heat capacity, C Heat capacity,C is the amount of heat required to raise

the temperature of a given quantity of the substance by one degree Celsius (JC1)

q = mc∆T

Heat released by substance =

Heat absorbed by calorimeter

q = heat released by substance

m= mass of substance

C= specific heat capacity

∆T = temperature change

Basic Principle in Calorimeter

Heat released by a reaction = Heat absorbed

by surroundings

• Surroundings may refer to the:i. Calorimeter itself or;ii. The water and calorimeter • qreaction= mcΔT or CΔT

Example 1

In an experiment, 0.100 g of H2 and excess of O2 were compressed into a 1.00 L bomb and placed into a calorimeter with heat capacity of 9.08 x 104 J0C1. The initial temperature of the calorimeter was 25.0000C and finally it increased to 25.155 0C. Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.

SolutionHeat released = Heat absorbed by the calorimeter q = C∆T = (9.08 X 104 J0C-1) X (0.1550C) = 1.41 X 104 J = 14.1 kJ

H2(g) + ½O2(g) → H2O(c)

mole of H2 = 0.100 2.016 = 0.0496 mol

moles of H2O = mole of H2

0.0496 mol of H2O released 14.1 kJ energy

1 mol H2O released = kJ

= 284 kJ∆Heat of reaction, ∆H = - 284 kJ mol1

0496.01.14

Example 2

1. Calculate the amount of heat released in a reaction in an aluminum calorimeter with a mass of 3087.0 g and contains 1700.0 mL of water. The initial temperature of the calorimeter is 25.0°C and it increased to 27.8°C.

Given: Specific heat capacity of aluminum = 0.553Jg-1 °C-1

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

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

Solution

q = mwcwΔT + mcccΔT

= (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) + (3087.0 g)(0.553 Jg-1 °C-1)(2.8°C)= 24676.71 J= 24.7 kJ

Heat absorbed by water

Heat absorbed by aluminium calorimeter

Heat released = +

HESS’S LAW

Hess Law Hess’s Law states that when reactants are converted to

products, the change in enthalpy is the same whether the reaction takes place in one step or in the series of steps.

The enthalpy change depends only on the nature of the reactants and products and is independent of the route taken.

1HA B

C

3H

2H

321 H H H

i. List all the thermochemical equations involved

Algebraic Method

1--1560kJmol H )(2

3)(2

2)(22

7)(62

.

1--286kJmol H )(2)(22

1)(2

.

1-393kJmol- H )(2)(2)(

.

gOH

gCO

gO

gHCiii

gOH

gO

gHii

gCO

gO

SCi

Step 1

i. List all the thermochemical equations involved

Algebraic Method

1--1560kJmol H )(2

3)(2

2)(22

7)(62

.

1--286kJmol H )(2)(22

1)(2

.

1-393kJmol- H )(2)(2)(

.

gOH

gCO

gO

gHCiii

gOH

gO

gHii

gCO

gO

SCi

Step 1

ii. Write the enthalpy of formation reaction for C2H6

)(62

?

)(23

)( gHCf

H

gH

sC

iii. Add the given reactions so that the result is the desired reaction.

-84kJ 3H2H1HfH

84kJ- )(62?

)(23)(2

_______________________________________________________________

1560kJ3H )(227

)(62)(232(g)2CO (iii)

-286kJ32H )(23)(223

)(23 3)(

kJ -3932 1H )(22)(22)(2 2)(

gHCfHgHsC

gOgHCgOHreverse

gOHgOgHii

gCOgOSCi

Energy Cycle MethodDraw the energy cycle and apply Hess’s Law to calculate the unknown value.

2 C ( s ) + 3 H2 ( g ) C2 H6 ( g )

2 C O2 ( g ) + 3 H2 O ( g )

2 O2 ( g ) 3 / 2 O2 ( g ) H

H

H

H

7 / 2 O2 ( g ) O

O

O

O f

1 2

3 = 2 ( - 3 9 3 )

= 3 ( - 2 8 6 )

= - ( - 1 5 6 0 )

1-kJmol -84

1560858 - -786

°3ΔH + )°

2H3( + )°1H 2( = °

fΔH

Example 1The thermochemical equation of combustion of carbon monoxide is shown as below.

C(s) + ½ O2(g) CO(g) = ?

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

CO(s) + ½ O2(g) CO2(g) ∆H= -283 kJ mol-1

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

H

H H

Example 2Calculate the standard enthalpy of formation of methane if the enthalpy of combustion of carbon, hydrogen and methane are as follows:

∆H [C(s)] = -393 kJ mol-1

∆H [H2(s)] = -293 kJ mol-1

∆H [CH4(s)] = -753 kJ mol-1

cH cH cH

Example 3 Standard enthalpy of formation of ammonia, hydrogen

chloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJ mol-1, 314.4 kJ mol-1 respectively. Write the thermochemical equation for the formation of each substance and calculate the enthalpy change for the following reaction.

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

Exercise1.Calculate the enthalpy of formation of benzene if :

∆H (CO2(g) ) = -393.3 kJ mol-1

∆H (H2O(l) ) = -285.5 kJ mol-1

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

fH

Born-Haber Cycle

Lattice Energy, Hlattice is the energy required to completely separate one mole of a

solid (ionic compound) into gaseous ions

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

(lattice dissociation)

Na+(g) + Cl-(g) NaCl(s) Hlattice = -771 kJ mol-1 (lattice formation) The magnitude of lattice energy increases as

the ionic charges increase the ionic radii decrease

There is a strong attraction between small ions and highly charged ions so the H is more negative.

H for MgO is more negative than H for Na2O because Mg2+ is smaller in size and has bigger charge than Na+, therefore Hºlattice (MgO) > Hºlattice (Na2O)

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

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

Hsoln = Hlattice + Hhdyr

Hydration Process of Ionic Solid

Lattice Energy Heat of Hydration

Heat of Solution

Na+ and Cl- ion in the gaseous state

Na+ and Cl- ion in the solid state

Hydrated Na+ and Cl- ion

Born-Haber Cycle

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

The Born Haber cycle is often used to calculate the lattice energy of an ionic compound.

In the Born-Haber cycle energy diagram, by convention, positive values are denoted as going upwards, negative values as going downwards.

Consider the enthalpy changes in the formation of sodium chloride.

Example :

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

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

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

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

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

vi. Lattice energy of NaCl = ?

(s)2(g)21

(s) NaClCl Na

Example: A Born-Haber cycle for NaCl

Na(s) + ½ Cl2(g)

Na(g) + ½ Cl2(g)

NaCl(s)

energy

E=0

Na(g) + Cl(g)

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

Na+(g) + Cl- (g)

HaNa

HaCl

Ionisation Energy of Na

Electron Affinity of Cl

Lattice energy

Hf NaCl

From Hess’s Law:Hf NaCl = HaNa + HaCl +IENa + EACl + Lattice Energy -ve

+ve

Calculation:

kJ777HkJ364kJ122kJ500kJ108kJ411H

EAHIEHHH

HEAHIEHH

lattice

lattice

)Cl(aS0flattice

lattice)Cl(aS0f

Exercise: Construct a Born-Haber cycle to explain why ionic compound

NaCl2 cannot form under standard conditions. Use the data below:

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

ii. First ionization energy of sodium = +500 kJmol-1

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

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

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

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