The Foundations of Chemistry

EXAMPLE 15-10 Using Hf^0 Values: Hess’s Law

Use the following information to determine
Hf^0 for PbO(s, yellow).

PbO(s, yellow)CO(g)88nPb(s)CO 2 (g) 
H^0 rxn65.69 kJ


H^0 ffor CO 2 (g)393.5 kJ/mol and 
H^0 ffor CO(g)110.5 kJ/mol

Plan

We again use Hess’s Law in the form
H^0 rxnn
Hf products^0 n
H^0 f reactants. The stan-
dard state of lead is Pb(s), so
H^0 f Pb(s)0 kJ/mol. Now we are given
H^0 rxnand the
Hf^0
values for all substances exceptPbO(s, yellow). We can solve for this unknown.

Solution
We list the known
H^0 fvalues:

PbO(s, yellow) CO(g) Pb(s) CO 2 (g),


H^0 f, kJ/mol: 
H^0 f PbO 2 (s, yellow) 110.5 0 393.5

Hrxn^0 n
H^0 f products n
H^0 f reactants


Hrxn^0 
H^0 f Pb(s)
H^0 fCO 2 (g)[
H^0 f PbO(s, yellow)
H^0 f CO(g)]

65.69 0 (393.5) [
H^0 f PbO(s, yellow)(110.5)]

Rearranging to solve for
H^0 f PbO(s, yellow), we have

Hf PbO(s, yellow)^0 65.69393.5110.5217.3 kJ/mol of PbO

You should now work Exercise 42.

For all practical purposes, the bond

energy is the same as bond enthalpy.

Tabulated values of average bond

energies are actually average bond

enthalpies. We use the term “bond

energy” rather than “bond enthalpy”

because it is common practice to

do so.

15-9 Bond Energies 609

BOND ENERGIES

Chemical reactions involve the breaking and making of chemical bonds. Energy is always
required to break a chemical bond. Often this energy is supplied in the form of heat.

The bond energy (B.E.)is the amount of energy necessary to break one moleof

bonds in a gaseous covalent substance to form products in the gaseous state at

constant temperature and pressure.

The greater the bond energy, the more stable (stronger) the bond is, and the harder it is
to break. Thus bond energy is a measure of bond strengths.

15-9

We will consult Appendix K, only after

working the problem, to check the

answer.

Problem-Solving Tip:Remember the Values of 
Hf^0 for Elements

In Example 15-10, we were not given the value of 
Hf^0 for Pb(s). We should know

without reference to tables that 
H^0 ffor an elementin its most stable form is exactly 0

kJ/mol. But the element mustbe in its most stable form. Thus, 
Hf^0 for O 2 (g) is zero,

because ordinary oxygen is gaseous and diatomic. We would notassume that 
Hf^0 would

be zero for oxygen atoms, O(g), or for ozone, O 3 (g). Similarly, 
Hf^0 is zero for Cl 2 (g)

and for Br 2 (
), but not for Br 2 (g). Recall that bromine is one of the few elements that is

liquid at room temperature and 1 atm pressure.