Look at the first reactant:
PbCl
2
(s).
The only given equation that has PbCl
2
(s) is the first
one, but PbCl
2
(s) is a product in the equation, so the equation must be reversed.
Therefore,
we write the reverse of equation with the sign of its
H° changed.
Look at the second
reactant:
Cl
2
.
This chemical is in both of the given reactions, so let’s skip it.
Look at the
product:
PbCl
4
(s).
The only given equation that has PbCl
4
(s) is the second one, so you must
include that equation, as written, with its given
H°.
We have now planned how we will use
all the given reactions, so let’s do it:
Add all the reactions as planned, eliminate reactants that also show up as products:
one of
the Cl
2
(g) molecules.
Add all the remaining reactants and products to form the net equation
for the desired reaction, and add all the individual
H° values to get the
H° for the reaction:
PbCl
2
(s)
Pb(s) + Cl
2
(g)
H° = – (–359.4 kJ)
+
Pb(s) + 2 Cl
2
(g)
PbCl
4
(s)
+
H° = –329.3 kJ
PbCl
2
(s) + Cl
2
(g)
PbCl
4
(s)
H° =
30.1 kJ
Reasonable Answer Check:
The net equation is the desired reaction.
The enthalpy of
formation of these two lead chloride compounds are similar to each other, so forming one
from the other should have a small
H°.
58.
Answer:
H° =
–905.47 kJ; exothermic
Strategy and Explanation:
Given a balanced chemical equation for a reaction and a table of
molar enthalpies of formation, determine the enthalpy change of the reaction and whether it
is exothermic or endothermic.
Use the stoichiometric coefficient of the balanced equation to describe the moles of each of

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