CHEMISTRY TEXTBOOK

4.10.9 Hess’s law of constant heat summation

The law states that, “Overall the
enthalpy change for a reaction is equal to
sum of enthalpy changes of individual steps
in the reaction”.

The enthalpy change for a chemical
reaction is the same regardless of the path by
which the reaction occurs. Hess’s law is a direct
consequence of the fact that enthalpy is state
function. The enthalpy change of a reaction
depends only on the initial and final states and
not on the path by which the reaction occurs.

To determine the overall equation
of reaction, reactants and products in the
individual steps are added or subtracted like
algebraic entities.

Consider the synthesis of NH 3

i. 2H 2 (g) + N 2 (g) N 2 H 4 (g),

∆rH 10 = + 95.4 kJ

ii. N 2 H 4 (g) + H 2 (g) 2 NH 3 (g),

∆rH 20 = -187.6 kJ

3 H 2 (g) + N 2 (g) 2 NH 3 (g),

∆rH^0 = -92.2 kJ

The sum of the enthalpy changes for
steps (i) and (ii) is equal to enthalpy change
for the overall reaction.

Application of Hess’s law

The Hess's law has been useful to calculate the
enthalpy changes for the reactions with their
enthalpies being not known experimentally.

Example 4.14 : Calculate the standard

enthalpy of the reaction,

2Fe(s) + 3/2 O 2 (g) Fe 2 O 3 (s)

Given :

i. 2Al(s) + Fe 2 O 3 (s) 2Fe(s) + Al 2 O 3 (s),

∆rH^0 = -847.6 kJ

ii. 2 Al(s) + 3/2 O 2 (g) Al 2 O 3 (s),

∆rH^0 = -1670 kJ

Example 4.15 : Calculate the standard

enthalpy of the reaction,

SiO 2 (s) + 3C(graphite) SiC(s) + 2 CO(g)

from the following reactions,

i. Si(s) + O 2 (g) SiO 2 (s),

∆rH^0 = -911 kJ

ii. 2 C(graphite) + O 2 (g) 2CO(g),

∆rH^0 = -221 kJ

iii. Si(s) + C(graphite) SiC(s),

∆rH^0 = -65.3kJ

Solution : Reverse the Eq. (i)

iv. SiO 2 (s) Si(s) + O 2 (g),

∆rH^0 = -911 kJ

Add equations (ii), (iii) and (iv)

ii. 2 C(graphite) + O 2 (g) 2 CO(g),

∆rH^0 = -221 kJ

iii. Si(s) + C(graphite) SiC(s),

∆rH^0 = -65.3 kJ

iv. SiO 2 (s) Si(s) + O 2 (g),

∆rH^0 = +911 kJ

SiO 2 (s)+3 C(graphite) SiC(s) +2 CO(g),

∆rH^0 = +624 kJ

Solution :

Reverse Eq.(i) and then add to Eq. (ii)

2Fe(s) + Al 2 O 3 (s) 2 Al(s) + Fe 2 O 3 (s),

∆rH^0 = +847.6 kJ

2 Al(s) + 3/2 O 2 (g) Al 2 O 3 (s),

∆rH^0 = -1670 kJ

2Fe(s) + 3/2 O 2 (g) Fe 2 O 3 (s),

∆rH^0 = -822.4 kJ