Hess’s Law
• A consequence of the 1 st of Thermodynamics
(conservation of energy).
• If a reaction is the sum of 2 or more reactions, then the

H of the overall reaction is the sum of the

H of the constituent reactions.
• The enthalpy change of a reaction is independent of the route provided the starting and final conditions and the reactants and products are the same.

Example 1
From the information below, calculate

H

, for the reaction
CO (g) + ½ O
2
(g)

CO
2
(g)
C (s) + O
2
(g)

CO
2
(g)

H

= -393 kJ
C (s) + ½O
2
(g)

CO (g)

H

= -283 kJ

Answer to example 1
C (s) + O
2
(g)

CO
2
(g)

H

= -393 kJ
CO (g)

C (s) + ½ O
2
(g)

H

= +283 kJ
CO (g) + ½ O
2
(g)

CO
2
(g)

H

= -110 kJ
Draw an enthalpy level diagram and an enthalpy cycle for this example.

Drawing and energy cycle
• Use the frameworks below to build your cycle.
• Usually the equation of the unknown enthalpy change is the horizontal equation.
• Either o all three arrows represent equations and therefore each equation should be balanced.
o Or all particles at each of the three points are equal; show them in boxes.
• Use arrows to write a formula to calculate unknown enthalpy change.

Example 2
From the information below, calculate

H

, for the reaction
S (s) + O
2
(g)

SO
2
(g)
SO
2
(g) + ½ O
2
(g)

SO
3
(g)

H

= - 98 kJ
S (s) + 1 ½ O
2
(g)

SO
3
(g)

H

= -395 kJ

Answer to example 2
S (s) + 1 ½ O
2
(g)

SO
3
(g)

H

= -395 kJ
SO
3
(g)

SO
2
(g) + ½ O
2
(g)

H

= +98 kJ
S (s) + O
2
(g)

SO
2
(g)

H

= -297 kJ
Draw an enthalpy level diagram and an enthalpy cycle for this example.

Example 3
Calculate the

H for
CaCO
3
(s)

CaO (s) + CO
2
(g)
From the following equations and enthalpy changes:
CaCO
3
(s) + 2HCl (aq)

CaCl
2
(aq) + CO
2
(g)+ H
2
O (l)

H

= -17 kJ mol -
1
CaO(s) + 2HCl (aq)

CaCl
2
(aq) + H
2
O (l)

H

= -195 kJ mol -1

Answer example 3
CaCO
3
(s) + 2HCl (aq)

CaCl
2
(aq) + CO
2
(g)+ H
2
O (l)

H

= -17 kJ mol -1
CaCl
2
(aq) + H
2
O (l)

CaO(s) + 2HCl (aq)

H

= + 195 kJ mol -1
CaCO
3

CaO (s) + CO
2
(g)

H

= +178 kJ mol -1

Example 4
Calculate
C (s) + 2H
2
(g)

CH
4
(g)

H = ? kJ when given the following
CH
4
(g) + 2 O
2
(g)

CO
2
(g) + 2H
2
O (l)

H = - 890.4 kJ
C (s) + O
2
(g)

CO
2
(g)

H = - 393.4 kJ
2 H
2
(g) + O
2
(g)

2 H
2
O (l)

H = - 571.6 kJ

Answer example 4
Reverse the first equation but leave the next two as they are
CO
2
(g) + 2H
2
O (l)

CH
4
(g) + 2O
2
(g)

H = + 890.4 kJ
C (s) + O
2
(g)

CO
2
(g)

H = - 393.4 kJ
2 H
2
(g) + O
2
(g)

2H
2
O (l)

H = - 571.6 kJ
C (s) + 2H
2
(g)

CH
4
(g)

H = - 74.7 kJ

Example 5
Calculate

H f for ICl (g) the following
½ I
2
(s) + ½ Cl
2
(g)

ICl (g) when given the following
Cl
I
2
I
2
2
(g)

2 Cl (g)

H = + 242 kJ
(g)
(s)


I
2 I (g)
2

H = + 151 kJ
(g)

H = + 63 kJ
ICl (g)

Cl (g) + I (g)

H = + 211 kJ

Answer example 5
Solution: take half a mole for each of the first equations and reverse the fourth.
½ Cl
2
(g)

Cl (g)

H = + 121 kJ
½ I
2
(g)

I (g)

H = + 75.5 kJ
½ I
2
(s)

½ I
2
(g)

H = + 31.5 kJ
Cl (g) + I (g)

ICl (g)

H = - 211 kJ
½ I
2
(s) + ½ Cl
2
(g)

ICl (g)

H = + 17 kJ