1. 1.HESS’S LAW
2. BORN HABER CYCLE

2.  ΔHθ – The amount of heat absorbed/evolved
when the molar quantities of reactants as
stated in the equation react together under
std conditions.
 Std conditions : P = 1 atm
T = 298K
Substances in their normal
physical states

3.  ΔH f
θ - The heat change required to produce
1 mol of the substance from its elements
under standard conditions.
 Ag(s) + ½ Cl2 (g)  AgCl (s) ; ΔH f
θ = -127
kJmol-1
 ΔH c
θ - The heat energy evolved when 1 mol
of the substance is completely burnt in
excess oxygen under standard conditions.
 C2 H4(g) + 3O2 (g)  2CO2 (g) +2 H2O (l) ;
ΔH c
θ = -1411 kJmol-1

4.  Hess’s law states that overall heat change in a
chemical reaction is constant and not
dependent on the route taken.
X
A + B C + D
(initial) (final)
Y Z
ΔHθ Route 1 = ΔHθ Route 2
X = Y + Z

5.  C(s) + O2 (g)  CO2 (g) ; ΔH f
θ = -406 kJmol-1
 H2(g) + ½O2 (g)  H2O (l) ; ΔH f
θ = -285 kJmol-1
C2 H5OH(l) + 3O2 (g)  2CO2 (g) +3H2O (l) ; ΔH c
θ
= -1423 kJmol-1
 Find ΔH f
θ C2 H5OH (l).
 2C(s) + 3H2(g)+ ½O2 (g)  C2H5OH(l)
 Working, ΔH f
θ = 2 ΔH1 + 3ΔH2 – ΔH3
= 2(-406) + 3(-285) + 1423
= -244 kJmol-1


6. ΔH 4
2C + 3H2 C2 H5OH
ΔH1 ΔH2 ΔH3
2CO2 + 3H2O
By Hess’s law
ΔH1 + ΔH2 = ΔH4 + ΔH3
ΔH4 = ΔH1+ ΔH2- ΔH3
= 2(-406) + 3(-285) –(-1423)
= -244 kJmol-1

7.  Def : The enthalpy change when 1 mol of
an ionic compound is broken apart into
its constituent gaseous ions under
standard conditions.
 NaCl(s)  Na+(g) +Cl-(g) ;
ΔH latt
θ =+771 kJmol-1

8. E.g.1
Draw a Born Haber cycle for RbI. Hence use the
data below to calculate the ΔH
f
of RbI.
ΔH a Rb = +86 kJmol-1
ΔH a I = +107 kJmol-1
ΔH latt Rbl = +609 kJmol-1
IE Rb = +402 kJmol-1
I(g) + e I- (g) ; ΔH = -314 kJmol-1

9.  Rb(s) + ½ l2(s) Rbl(s)
◦ ΔH2 ΔH4 ΔH1
 Rb(g) l(g) ΔH6
◦ ΔH3 ΔH5
 Rb+(g) I-(g) Route 2

10. By Hess’s law,
Heat change in route 1=Heat change in route 2
ΔH1= ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
= +86+402+107+(-314)+(-609)
= -328 kJmol-1
Hence ΔHf RbI = -328 kJmol-1