Thermochemistry
Section 3: Hess’ Law
Hess’ Law
since H is a state function, the change in H
is independent of pathway
Hess’ Law- when going from a set of
reactants to a set of products, the ∆H is
the same whether it happens in one step
or a series of steps
Example 1
Example 1
N2(g) + 2O2(g)  2NO2(g) ∆H = 68 kJ
OR
N2(g) + O2(g)  2NO(g)
∆H = 180 kJ
2NO(g) + O2(g)  2NO2(g)
∆H = -112 kJ
N2(g) + 2O2(g)  2NO2(g)
∆H = 68 kJ
Rules
1. If a reaction is reversed, the sign of ∆H
must be reversed as well.
– because the sign tells us the direction of
heat flow as constant P
2. The magnitude of ∆H is directly
proportional to quantities of reactants
and products in reaction.
If coefficients are multiplied by an integer,
the ∆H must be multiplied in the same
way.
– because ∆H is an extensive property
Example 2
Using the enthalpies of combustion for
graphite (-394 kJ/mol) and diamond (-396
kJ/mol), find the ∆H for the conversion of
graphite to diamond.
Cgraphite (s)  Cdiamond (s) ∆H=?
Example 2
(1) Cgraphite(s) + O2(g)  CO2(g)
(2) Cdiamond(s) + O2(g)  CO2(g)
∆H= -394kJ/mol
∆H= -396kJ/mol
to get the desired equation, we must reverse 2nd
equation:
(1) Cgraphite(s) + O2(g)  CO2(g) ∆H= -394kJ/mol
(2) CO2(g)  Cdiamond(s) + O2(g) ∆H= (-1)-396kJ/mol
Cgraphite (s)  Cdiamond (s)
∆H= -394 + 396
∆H= 2 kJ/mol
Example 3
Find ∆H for the synthesis of B2H6, diborane:
2B(s) + 3H2(g)  B2H6(g)
Given:
(1) 2B(s) + 3/2O2(g)  B2O3(s)
∆H1=-1273kJ
(2) B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g)
∆H2=-2035kJ
(3) H2(g) + ½O2(g)  H2O(l)
(4) H2O(l)  3H2O(g)
∆H3=-286kJ
∆H4=44 kJ
Example 3
Start by paying attention to what needs to be
on reactants and products side
(1) 2B(s) + 3/2O2(g)  B2O3(s)
∆H1= -1273kJ
(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g)
∆H2= (-1)-2035kJ
(3) H2(g) + ½O2(g)  H2O(l)
∆H3= -286kJ
(4) H2O(l)  3H2O(g)
∆H4= 44 kJ
Example 3
Underline what you want to keep- that will help you
figure out how to cancel everything else:
(1) 2B(s) + 3/2O2(g)  B2O3(s)
∆H1= -1273kJ
(2)
B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g)
∆H2=(-1)-2035kJ
(3)
H2(g) + ½O2(g)  H2O(l)
∆H3= -286kJ
(4)
H2O(l)  3H2O(g)
∆H4= 44 kJ
Example 3
Need 3 H2 (g) so 3 x (3)
Need 3 H2O to cancel so 3 x (4)
(1) 2B(s) + 3/2O2(g)  B2O3(s)
(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g)
(3) 3H2(g) + 3/2O2(g)  3H2O(l)
(4) 3H2O(l)  3H2O(g)
∆H1= -1273kJ
∆H2=(-1)(-2035kJ)
∆H3=(3)(-286kJ)
∆H4=(3)(44 kJ)
2B(s) + 3H2(g)  B2H6(g)
∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ