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Exercise:
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The reaction betweeen
mercury and oxygen to form mercuric oxide is
written below.
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2 Hg(l) +
O2(g)
----> 2 HgO(s)
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The
DH
for the reaction is -43,000 calories/mole
and clearly exothermic. What is the sign
for the entropy change? Is the reaction
spontaneous? Justify your
answer.
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Answer:
- The
DS
for the reaction at 298 Kelvin is a small
negative number because the reactants gas and
liquid are forming an organized solid. Remember
DS is negative for changes that go from
-
- gas -->
liquid;
- liquid --->
solid
All of the calculations
assume T = 298 Kelvin unless the number is stated
differently.
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Substitution into the
definition for free energy is shown here. The
TDS
term is very small. It will not cancel the overall
large negative value from the enthalpy.
The
DG
will be large and negative.
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DG=-43,000
calories/mole
- (298 K)(small negative
DS
)
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Because
the entropy change is small and negative the
-47,000 calories dominates the free energy change.
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The free energy change is
a large negative number. The reaction is
spontaneous. The entropy change is negative.
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