1. THERMOCHEMISTRY

2. Concept of Enthalpy

3. 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

4. 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

5. 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 = J

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

7. Exothermic reactions
Enthalpy of products < Enthalpy of reactants, ΔH is negative.
Energy is released from the system to the surroundings.

8. Consider the following reaction:
A (g) B (g) → C (g) ΔH = ve
(reactants) (product)
Energy profile diagram for exothermic reaction

9. Endothermic Reactions
Enthalpy of products > enthalpy of reactants, ΔH is positive
Energy is absorbed by the system from the surrounding

10. Consider the following reaction
A (g) B (g) → C(g) ΔH = + ve
(reactants) (product)
Energy profile of diagram endothermic reactions

11. 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.

12. Enthalpy of Reaction, ∆H and Standard Condition
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)

13. Thermochemical Equation
The thermochemical equation shows the enthalpy changes.
Example : H2O(s) → H2O(l) ΔH = kJ
1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C, ΔH = 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

14. Types of Enthalpies There are many kind of enthalpies such as:
Enthalpy of formation
Enthalpy of combustion
Enthalpy of atomisation
Enthalpy of neutralisation
Enthalpy of hydration
Enthalpy of solution
Enthalphy of sublimation

15. 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 mol1
The standard enthalpy of formation of any element in its most stable state form is ZERO.
∆H (O2 ) = ∆H (Cl2) = 0

16. 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 mol1

17. 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

18. 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 mol1

19. 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

20. 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

21. Enthalpy of Sublimation, Hsubl
The heat change when one mole of a substance sublimes (solid into gas).
I2 (s) → I2(g)
Hsubl = kJ mol1

22. 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

23. Simple 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

24. A bomb calorimeter

25. 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 1C1).
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 (JC1)

26. q = mc∆T Heat absorbed by calorimeter Heat released by substance =
q = heat released by substance
m= mass of substance
C= specific heat capacity
∆T = temperature change

27. Basic Principle in Calorimeter
Heat released
by a reaction
Heat absorbed
by surroundings =
Surroundings may refer to the:
Calorimeter itself or;
The water and calorimeter
qreaction= mcΔT or CΔT

28. Example 1
In an experiment, 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 J0C1. The initial temperature of the calorimeter was C and finally it increased to C. Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.

29. Solution
Heat 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
= mol

30. moles of H2O = mole of H2
mol of H2O released 14.1 kJ energy
1 mol H2O released = kJ
= 284 kJ
∆Heat of reaction, ∆H = kJ mol1

31. Example 2
1. Calculate the amount of heat released in a reaction in an aluminum calorimeter with a mass of g and contains mL of water. The initial temperature of the calorimeter is 25.0°C and it increased to 27.8°C.

32. Heat absorbed by aluminium calorimeter
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-1
ΔT = ( )°C = 2.8°C
Solution
q = mwcwΔT + mcccΔT
= ( g)(4.18 Jg-1 °C-1)(2.8 °C) +
( g)(0.553 Jg-1 °C-1)(2.8°C)
= J
= kJ
Heat absorbed by
water
Heat absorbed by aluminium calorimeter
Heat released = +

33. HESS’S LAW

34. 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. A B C

35. Algebraic Method
Step 1
i. List all the thermochemical equations involved

36. Algebraic Method
Step 1
i. List all the thermochemical equations involved

37. ii. Write the enthalpy of formation reaction for C2H6

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

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

41. Example 1
The 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.

42. Example 2
Calculate the standard enthalpy of formation of methane if the enthalpy of combustion of carbon, hydrogen and methane are as follows:
∆H [C(s)] = -393 kJ mol-1
∆H [H2(s)] = -293 kJ mol-1
∆H [CH4(s)] = -753 kJ mol-1

43. Example 3
Standard enthalpy of formation of ammonia, hydrogen chloride and ammonium chloride is kJ mol-1, kJ mol-1, 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)

44. Exercise
1.Calculate the enthalpy of formation of benzene
if :
∆H (CO2(g) ) = kJ mol-1
∆H (H2O(l) ) = kJ mol-1
∆H (C6H6(l) ) = kJ mol-1

45. Born-Haber Cycle

46. 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

47. 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)

48. Hydration Process of Ionic Solid
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

49. Hydrated Na+ and Cl- ion
Na+ and Cl- ion in
the gaseous state
Lattice Energy
Heat of Hydration
Heat of Solution
Na+ and Cl- ion in
the solid state
Hydrated Na+ and Cl- ion

50. 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.

51. Example : Given; Enthalpy of formation NaCl = -411 kJmol-1
Enthalpy of sublimation of Na = +108 kJmol-1
First ionization energy of Na = +500 kJmol-1
Enthalpy of atomization of Cl = +122 kJmol-1
Electron affinity of Cl = -364 kJmol-1
Lattice energy of NaCl = ?

52. Example: A Born-Haber cycle for NaCl
energy
Na+(g) + e
+ Cl(g)
Ionisation Energy of Na
Electron Affinity of Cl
Na+(g)
+ Cl- (g)
Na(g)
+ Cl(g)
HaCl
Na(g)
+ ½ Cl2(g)
+ve
Lattice energy
HaNa
From Hess’s Law:
Hf NaCl = HaNa + HaCl +IENa +
EACl + Lattice Energy
Na(s) + ½ Cl2(g)
E=0
Hf NaCl
-ve
NaCl(s)

53. Calculation:

54. Exercise:
Construct a Born-Haber cycle to explain why ionic compound NaCl2 cannot form under standard conditions. Use the data below:
Enthalpy of sublimation of sodium = +108 kJmol-1
First ionization energy of sodium = +500 kJmol-1
Second ionization energy of sodium = kJmol-1
Enthalpy of atomization of chlorine = +121kJmol-1
Electron affinity of chlorine = -364 kJmol-1
Lattice energy of NaCl2 = kJmol-1