358 ❯ STEP 5. Build Your Test-Taking Confidence- B—In general, gases have much higher entropy
than either liquids or solids. For this reason, the
predictions depend primarily upon which reac-
tion results in the greatest decrease in the number
of moles of gas. A and D both result in an
increase in the number of moles of gas, so there
is an increase in entropy. C shows no change in
the number of moles of gas; therefore, the change
in entropy will be small. - D— The relationship, given on the equations page
of the exam is ΔG = ΔH - TΔS. Nonspontaneous
under standard conditions means: ΔG > 0. To
become spontaneous, ΔG must be less than zero.
Increasing the temperature will change the TΔS
term (entropy). If ΔH is greater than zero and ΔS
is also greater than zero, the combination will be
positive as long as the ΔH is greater than TΔS.
As the temperature increases, TΔS will eventu-
ally become larger than ΔH, making the process
spontaneous. - C—In general, the more oxygen atoms present
not attached to hydrogen atoms, the stronger the
oxyacid. If two oxyacids have the same number
of oxygen atoms not attached to hydrogen atoms,
the acid with the more electronegative central
atom is the stronger acid. The number of oxygen
atoms without hydrogen atoms attached are
HClO = 0, HBrO = 0, H 2 SeO 3 = 1, and H 2 SO 3
= 1. The electronegativities increase in the order
Se < S and Br < Cl. - D—The melting points of ionic compounds
increase with increasing lattice energy. Lattice
energy increases with increasing ionic charge and
with decreasing sum of ionic radii. It is apparent
from comparing NaF to CaO that charge is more
important than small changes in radii. The charges
are Na+, Ca^2 +, Sr^2 +, Ba^2 +, F-, Cl-, and O^2 -. - C—The melting points of ionic compounds
increase with increasing lattice energy. Lattice
energy increases with increasing ionic charge and
with decreasing sum of ionic radii. The oxide ion
radius is a constant, while the metal radii decrease
in the order Ba^2 + > Sr^2 + > Ca^2 +. The decrease in
metal radii is due to the smaller ions having fewer
electron shells.
45. B—The balanced chemical equation is:
2 HgO(s) → 2 Hg(l) + O 2 (g)
The calculation is:
(4.32gHgO) =1mol HgO
216 gHgO1moleO
2moleHgO2
=
(4.32)
1
216
1 mole O
2(2.16)
1
216
(^2) ×
1 mole O
1(^2) = 0.0100 mole O 2
- A—The substance with the highest melting
point has the strongest intermolecular forces. All
four molecules are nonpolar; therefore, the inter-
molecular forces are London dispersion forces. In
general, London dispersion forces, for molecules
with similar structures, London dispersion forces
increase with increasing molar mass. - D—Since phenol has a Ka values given, it is a
weak acid; as such, the equilibrium expression is:
C 6 H 5 OH(aq) ^ H+(aq) + C 6 H 5 O-(aq)
Use the Ka expression:
Ka =[H+][CB]
[CV]
= 1 × 10 -^10 =
[][]
[1.0]
xxThis leads to:
[H+] = (1.0 × 1 × 10 -^10 )1/2 = (1 × 10 -^10 )1/2 =
1 × 10 -^5 M- C—The higher the average number of bonds
between the nitrogen atoms, the shorter the
bond is. For diazene there are two bonds, for
triazene the average is 1.5 bonds, and for tetra-
zene the average is 1.33 bonds. The length of
the average bond length increases in the order
2 < 1.5 < 1.33. - D—The carbon is a solid and the water is a
liquid; therefore, neither of these will be in the
calculation. Since the volume of the container is
1.00 L, the molarities of the other two substances
are 4.00 M NH 3 and 2.00 M CO 2.
Kc = [NH 3 ]^2 [CO 2 ] = (4.00)^2 (2.00) = 32.0
21-Moore_PE02_p341-370.indd 358 31/05/18 1:54 pm