134 ❯ STEP 4. Review the Knowledge You Need to Score High
❯ Answers and Explanations
- D—You may wish to review the Kinetics chap-
ter if you have forgotten what the activation
energy is. - A—The free energy is the minimum energy
required for a nonspontaneous reaction and the
maximum energy available for a spontaneous
reaction. - B—The Law of Conservation of Energy (First
Law of Thermodynamics) says the amount of
heat lost must be equal to the heat gained. - B—This is the reverse of the lattice energy
definition. - D
2[C(s) + O 2 (g) → CO 2 (g)] 2 (–393.5 kJ)
H 2 (g) + (1/2) O 2 (g) → H 2 O(l) –285.8 kJ
2 CO 2 (g) + H 2 O(l) → C 2 H 2 (g) + (5/2) O 2 (g)
- (–1,299.8 kJ)
2 C(s) + H 2 (g) → C 2 H 2 (g) 227.0 kJ
Simple rounding to the nearest 100 kJ gives
200 kJ.
- D—The system is insulated, and no work can be
done on or by the system (rigid container); thus,
the energy is constant. At the melting point,
some of the mercury will spontaneously melt;
changing from a solid to a liquid increases the
entropy. - A—The process is endothermic (the ammonium
chloride is absorbing heat to cool the water).
Endothermic processes are “helped” by higher
temperatures. C and possibly D would give an
increase in temperature. There is insufficient
information about B. - B—The reaction showing the greatest increase
in the number of moles of gas will show the
greatest entropy increase. If no gases are pres-
ent, then the greatest increase in the number of
moles of liquid would yield the greatest increase. - D—The reaction occurs readily; therefore, it
must be a spontaneous reaction. If the reaction is
spontaneous, then ΔG° < 0. If the free energy is
negative free energy, then K must be large (> 1).
10. A—Heat is required to melt something (ΔH > 0).
A transformation from a solid to a liquid gives
an increase in entropy (ΔS > 0).
11. D—This equation has an overall decrease in
the amount of gas present (decreases entropy).
The other answers produce more gas (increasing
entropy).
12. C—Nonspontaneous means that ΔG > 0. Since
the reaction becomes spontaneous, the sign
must change. Recall: ΔG = ΔH – TΔS (given
in the exam booklet). The sign change at higher
temperature means that the entropy term (with
ΔS > 0) must become more negative than the
enthalpy term (ΔH > 0).
13. C—The calculation is [2(436 kJ) + 499 kJ] –
{2[2(464 kJ)]} = –485 kJ.
14. B—This is the reaction that has the greatest
increase in the number of moles of gas.
15. C—The decomposition of water is the reverse
of the reaction shown; therefore, the enthalpy
change is positive instead of negative. The
amount of water decomposing is 2.00 moles,
which is the same amount of water in the reac-
tion.
16. B—Dissolving almost always has ΔS > 0. A
decrease in temperature means the process has
ΔH > 0 (the system is absorbing energy from the
surroundings).
17. A—The increase in temperature of the solu-
tion indicates that the process is exothermic. An
exothermic process will shift toward the starting
materials (solid lithium sulfate and water) when
heated.
18. D—This is a thermodynamic problem, so a
kinetics answer (A) is not applicable. For the
reaction to occur, ΔG must be negative. It is pos-
sible to determine ΔG from ΔG = ΔH – TΔS.
Since heating causes the equilibrium to shift
to the right, ΔH must be positive. A positive
ΔH will not lead to a negative ΔG; therefore,
enthalpy cannot be the driving force (elimi-
nating B and C). Thus, the reaction must be
entropy driven because ΔS is positive owing to
the release of the carbon dioxide gas.