Lesson 6.6 Hess's Law
Suggested Reading

Zundahl Chapter 6 Section 6.3
Essential Question

How is Hess's law used to obtain the ∆H of one reaction from the ∆H's
of other reactions?
Learning Objectives


State Hess's law.
Apply Hess's Law to determine the enthalpy change of a reaction that is
the sum of two or more reactions with known enthalpy changes.
Introduction
In the previous lessons we looked at basic properties of heats of reaction as
well as how to measure heats of reaction. Now we are going to look at how
we can use heats of reactions. In this lesson we will see that the ∆H for one
reaction can be obtained from the ∆H's of other reactions. This means that we
can tabulate values for ∆H and use them to calculate others.
Hess's Law
Recall that enthalpy is a state function. This means that the enthalpy change
for a chemical reaction is independent of the path by which the products are
obtained. In 1840, the Russian chemist Germain Henri Hess, a professor at
the University of St. Petersburg, discovered this result by experiment. Hess's
law of heat summation (or simply Hess's Law) states that for a chemical
equation that can be written as the sum of two or more steps, the enthalpy
change for the overall equation equals the sum of the enthalpy changes for
the individual steps. In other words, no matter how you go from given
reactants to product (whether in one step or several), the enthalpy change for
the overall chemical change is the same.
Hess's law is best learned by example, so please watch the following video
(Do not skip this video!!!).
Watch the following YouTube Video:
https://www.youtube.com/watch?v=iETCSFit-zA
Enthalpy Diagram
Enthalpy diagrams are used to illustrate Hess's law. This diagram shows two
different ways to from from methane and oxygen to carbon dioxide and water
in the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
You can see that the combustion of one mole of methane to give one mole of
carbon dioxide and 2 mol of water has an enthalpy change of -890 kJ. This
value can be obtained in one step or by adding ∆H2 and ∆H3.
Hess's law is very useful for obtaining the enthalpy change for a reaction
that is difficult to determine by direct experiment. Hess's law is more
generally useful, however, in that it allows you to calculate the enthalpy
change for one reaction from the values for others. This is a big time saver!
Lets look at one final textbook example:
What is the enthalpy of reaction, ∆H, for the formation of tungsten carbide,
WC, from the elements? Tungsten carbide is very hard and is used to make
cutting tools and rock drills.
W(s) + C(graphite) → WC(s)
The enthalpy change for this reaction is difficult to measure directly,
because the reaction occurs at 1400°C. However the heats of combustion
for the elements and of tungsten carbide can be easily measured. The
reactions are:
2W(s) + 3O2 → 2WO3(s); ∆H = -1680.6 kJ
C(graphite) + O2(g) → CO2(g); ∆H = -393.5 kJ
2WC(s) + 5O2(g) → 2WO3(s) + 2CO2(g); ∆H = -2391.6 kJ
(1)
(2)
(3)
Compare equations 1, 2, and 3 with the desired equation. Note that
equation 1 has 2W(s) on the left side when the desired equation only has
one. This means we are going to have to multiply equation 1 by 1/2.
Equation 2 has one C(graphite) on the left, so this is good for now.
Equations 3 has 2 WC(s) on the left when the desired equation has
1 WC(s) on the rights, so we need to not only multiply this equation by 1/2,
but also reverse it. Remember that whatever we do to each equation, we
must also do to the value for ∆H. Lets start.
Equation (1) 1/2 x (2W(s) + 3O2 → 2WO3(s); ∆H = -1680.6 kJ) gives W(s) +
3/2O2(g) → WO3(s); ∆H = -840.3 kJ
Equation (3) 1/2 x (2WC(s) + 5O2(g) → 2WO3(s) + 2CO2(g); ∆H = -2391.6
kJ) gives WC(s) + 5/2O(g) → WO3(s) + CO2(g); ∆H = -1195.5 kJ
and reversing (3) gives WO3(s) + CO2(g) →WC(s) + 2½O(g); ∆H = 1195.5
kJ
Adding gives (We'll use red to denote the factors that cancel:
W(s) + 3/2O2(g) → WO3(s); ∆H1 = -840.3 kJ
C(graphite) + O2(g) → CO2(g); ∆H2 = -393.5 kJ
WO3(s) + CO2(g) →WC(s) + 5/2O(g); ∆H3 = 1195.5 kJ
W(s) + C(graphite) → WC(s); ∆H = -38.0 kJ
where ∆H = ∆H1 + ∆H2 + ∆H3