the ppt is on energy cycles in reactions and deal with Hess's law of constant heat summation where the law of conservation of mass is obeyed. The enthalpy change can be calculated no matter whatever path the reaction has taken place as long as reactants and products are same the enthalpy change can be determined. Thus if enthalpy change of reaction is to be calculated than we can use enthalpy of combustion and formation to calculate it.

1. Ppt 13 Reactivity 1.2 Energy cycles in reactions
SL &HL Reactivity 1.2.1 Bond-breaking absorbs and bond-forming releases energy
SL &HL Reactivity 1.2.2
Hess’s law states that the enthalpy change for a reaction is
independent of the pathway between the initial and final
states.
How does application of the law
of conservation of energy help us
to predict energy changes during
reactions?

4. • DEFINITION
• VALUES ENDOTHERMIC - ENERGY MUST BE PUT IN TO BREAK ANY CHEMICAL BOND
• EXAMPLES CL2(G) ———> 2CL(G) O-H(G) ———> O(G) + H(G)
• NOTES • STRENGTH OF BONDS ALSO DEPENDS ON ENVIRONMENT; MEAN VALUES QUOTED
• • MAKING BONDS IS EXOTHERMIC AS IT IS THE OPPOSITE OF BREAKING A BOND
• • FOR DIATOMIC GASES, BOND ENTHALPY = 2 X ENTHALPY OF ATOMISATION
• • SMALLER BOND ENTHALPY = WEAKER BOND = EASIER TO BREAK

8. •
•
•
• - 90 KJ/MOL
• -2046 KJMOL-1
• NOW 11 GRAMS = 0.25 MOLE OF PROPANE (11 G/44 G MOL-1)
• (0.25 MOL )(-2046 KJ MOL-1) = - 511.5 KJ

12. HESS’ LAW: EXAMPLE 1
N2 (G) +2O2 (G)  + 2 NO2 (G)
THE REQUIRED EQUATION IS REALLY THE SUM OF THE TWO
GIVEN EQUATIONS
SOLUTION:
N2 (G) + O2 (G)  2 NO (G) DH1 = +181 KJ
2 NO (G) + O2 (G)  2 NO2 (G) DH2 = -113 KJ
------------------------------------------------------------
N2 (G) +2O2 (G)+ 2 NO (G)  2 NO (G) + 2 NO2 (G)
N2 (G) +2O2 (G)  + 2 NO2 (G)
DH = DH1 + DH2 = +181 KJ +(-113) = + 68 KJ

13. HESS LAW EXAMPLE 2
FROM THE FOLLOWING REACTIONS AND ENTHALPY CHANGES:
2 SO2 (G) + O2 (G)  2 SO3 (G) DH = -196 KJ
2 S (S) +3 O2 (G)  2 SO3 (G) DH = -790 KJ
FIND THE ENTHALPY CHANGE FOR THE FOLLOWING REACTION:
S (S) + O2 (G)  SO2 (G)
2 SO3 (G)  2 SO2 (G) + O2 (G) DH = +196 KJ
2 S (S) +3 O2 (G)  2 SO3 (G) DH = -790 KJ
--------------------------------------------------------------------------------------------------------------
2 SO3(G) +2 S(S) + 2 3 O2 (G)  2 SO3 (G)+2 SO2 (G) + O2 (G) DH = -594 KJ
2 S(S) + 2 O2 (G)  2 SO2 (G) DH = -594 KJ
S(S) + O2 (G)  SO2 (G) DH = -297 KJ

22. AHL
Reactivity
1.2.3
Standard enthalpy changes of combustion,ΔHc⦵, and
formation,ΔHf⦵, data are used in thermodynamic calculations.
AHL
Reactivity
1.2.4
An application of Hess’s law uses enthalpy of formation data or
enthalpy of combustion data to calculate the enthalpy change of a
reaction.
AHL
Reactivity
1.2.5
A Born–Haber cycle is an application of Hess’s law, used to show
energy changes in the formation of an ionic compound.

26. Sample calculation
Calculate the standard enthalpy change for the following reaction, given that the
standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,
+33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element
2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
PRODUCTS REACTANTS
[ 4 x DfH of HNO3 ] minus [ (2 x DfH of H2O) + (4 x DfH of NO2) + (1 x DfH of O2) ]
DH°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0
ANSWER = - 252 kJ
Enthalpy of reaction from enthalpies of formation
DH =  DfH of products –  DfH of reactants

27. Enthalpy of reaction from enthalpies of combustion
Sample calculation
Calculate the standard enthalpy of formation of methane; the standard enthalpies of
combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .
C(graphite) + 2H2(g) ———> CH4(g)
By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...
REACTANTS PRODUCTS
[ (1 x DcH of C) + (2 x DcH of H2) ] minus [ 1 x DcH of CH4]
DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890)
ANSWER = - 76 kJ mol-1
DH =  DcH of reactants –  DcH of products

30. SAMPLE PROBLEM 1
DHoCaCO3 -1207
DHo HCl (aq) -167
DHoCaCl2 -796
DHo H2O (l) -286
DHo CO2 (g) -394
Solution
DHo
products =(-796)+(-286)+(-394)
= -1476 kJ
DHo
reactants =(-1207)+(2)(-167)
= -1541 kJ
DHo
reaction = -1476-(-1541) = +65 kJ

31. SAMPLE PROBLEM 2
DHo C3H8 -104
DHo O2 (g) 0
DHo H2O
(g)
-242
DHo CO2 (g) -394
Solution
DHo
products =(3)(-394)+(4)(-242)
= -2150 kJ
DHo
reactants =(-104)+(5)(0)
= -104 kJ
DHo
reaction = -2150-(-104) = -2046 kJmol-1
Now 11 grams = 0.25 mole of propane (11 g/44 g mol-1)
(0.25 mol )(-2046 kJ mol-1) = - 511.5 kJ

39. Enthalpy of atomisation: Enthalpy change when one mole of gaseous atoms is
formed from its elements under standard conditions.
Note: if direct atomisation values are not given especially in case of non metals
than we may need to use half the value of bond energies.
Ex: Na(s) Na(g)
½ Cl2(g) Cl(g)
ΔH˚atm = 108 KJ/mol
ΔH˚atm = 121 KJ/mol
Electron affinity: Enthalpy change when one mole of electrons is added to one
mole of gaseous atoms to form one mole of negative ions under
standard conditions.
Note: First electron affinity is exothermic.
Second and third electron affinity are endothermic because when
electrons are added to negative ion repulsion increases therefore requires
energy to add electron.
Ex: Cl(g) + e- Cl-
(g) ΔH˚ea1 = -342 KJ/mol

40. ENTHALPY CHANGES
Lattice energy: Enthalpy change when one mole of an ionic compound is formed
from its gaseous ions under standard conditions.
Ex: Na+
(g) + Cl-
(g) NaCl(s)
Energy is always released in this process because of the strong electrostatic
force between oppositely charged ions. It is an exothermic process.
Product that is the ionic compound has less energy than the reactants, that is its
ions in their Gaseous state.
Lesser the energy of the compound, more stable is the compound
More the energy released i.e. more exothermic the process, stable is
the compound.
ΔH˚latt = - 787 KJ/mol
Lattice Dissociation energy: Enthalpy change when one mole of an ionic compound
is dissociated into its gaseous ions under standard
conditions.
Ex: NaCl(s) Na+
(g) + Cl-
(g)
ΔH˚ = + 787 KJ/mol
Enthalpy of formation: Enthalpy change when one mole of a compound is formed
from its elements under standard state and under standard conditions.
Ex: Na(s) + ½ Cl2(g) NaCl(s) ΔH˚f = - 412 KJ/mol
Ionisation energy: It is the energy needed to remove one electron from each
atom in one mole of atoms of the element in the gaseous state to
form one mole of gaseous ions under standard conditions.
Ex: Na(g) Na+
(g) + e- ΔH˚i1 = + 494 KJ/mol

41. Born - haber cycle
Born – Haber cycle: It is the enthalpy cycle used to calculate the lattice energy
of any given ionic compound.
Ionic
compound
Ions in gaseous
state
Elements in their
standard state
Lattice energy
Enthalpy of
formation
ΔH˚1
ΔH˚latt = ΔH˚f - ΔH˚1
ΔH˚1 = ΔH˚atm + ΔH˚i + ΔH˚ea

44. Example2: Calculate the lattice energy of CaO.
ΔH˚f[CaO] = - 635 KJ/mol
ΔH˚at [Ca] = +178
KJ/mol
ΔH˚atm [O] = +248
KJ/mol
ΔHi1 [Ca] = +590
KJ/mol
ΔHi2 [Ca] = +1150
KJ/mol
ΔHea1 [O] = -141
KJ/mol
ΔHea2 [O] = +844
KJ/mol
Ca(s) + ½ O2(g)
CaO(s)
ΔH˚f
Ca(g) + ½ O2(g)
ΔH˚at [Ca]
Ca(s) +
O(g)
ΔH˚at [O]
Ca1+
(s) + O(g)
ΔH˚i1 [Ca]
Ca2+
(g) + O(g)
ΔH˚i2 [Ca] Ca2+
(g) + O-
(g)
ΔH˚ea1 [O]
Ca2+
(g) + O2-
(g)
ΔH˚ea2 [O]
ΔH˚latt
ΔH˚latt = ΔH˚f - ΔH˚1
= [-635]
-
[ ΔH˚at [Ca] + ΔH˚atm
[O]
+ ΔHi1 [Ca] + ΔHi2 [Ca]
+
ΔHea1 [O] + ΔHea2 [O]
]
= [ - 635] - [ 178 + 248 + 590 + 1150 + ( - 141 )
+ 844 ]
= [ - 635] - [
2869 ]
= - 3504 KJ/ mol

54. •