Hess’s Law
Chapter 5.4
State Function
A state function is a
property of a system
that is independent of
the path taken
The net vertical
distance is the same
whether the car goes
up the series of ramps
or directly up the
elevator
Hess’s Law
The enthalpy change for the
conversion of reactants to
products is the same whether
the reaction occurs in one step
or in several steps.
Enthalpy is a State Function
Rules for Enthalpy Changes
• To use Hess’s Law to calculate enthalpy changes for chemical
reactions, you must apply the following rules:
1) If you reverse a chemical reaction, you must also reverse the
sign of ∆H
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
∆H = -802 kJ
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
∆H = + 802 kJ
2) If the coefficients in a balanced equation are multiplied by a
factor, the value of ∆H is multiplied by the same factor
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
∆H = -802 kJ
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g)
∆H = -1604 kJ
Enthalpy is a State Function
Some reactions cannot be analyzed using calorimetry.
Ex: rusting is an exothermic reaction but it happens too slowly to measure in a calorimeter
How can we use Hess’s Law to calculate the enthlapy change for the following reaction?
C(graphite) + ½ O2(g)
CO(g) + ½ O2(g)
Practice
• Consider the combustion of methane to form CO2 and liquid H2O
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
• This reaction can be thought of as occurring in two steps
• In the first step methane is combusted to produce water vapor:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = -802 kJ
• In the second step water vapor condenses from the gas phase to
the liquid phase:
2H2O(g)  2H2O(l)
∆H = -88 kJ
Practice
Find the enthalpy change for the target reaction
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
H = ?
Given the following reactions:
3 C(s) + 4 H2(g)  C3H8(g)
H = -103.85kJ/mol
C(s) + O2(g)  CO2(g)
H = -393.51kJ/mol
H2(g) + ½ O2(g)  H2O(g)
H = -241.83kJ/mol
Practice Makes Perfect!
Find the enthalpy change for the target reaction
C(graphite) + 2S(rhombic)
CS2 (l)
Given the following reactions:
C(graphite) + O2 (g)
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
CO2 (g)
SO2 (g)
CO2 (g) + 2SO2 (g)
H0 = -393.5 kJ
H0 = -296.1 kJ
H0 = -1072 kJ
HOMEWORK
Required Reading:
p. 314-318
(remember to supplement your notes!)
Questions:
p. 317 #1-3
p. 318 #1-8