5.6. (a)1.5 mol (b)cannot equal 3 in this case, be-
cause H 2 will act as a limiting reagent at 1.66 mol.
5.9.False. p° is the standard pressure, defined as 1 atm or
1 bar.
5.10. (^) rxnG°514.38 kJ; (^) rxnG489.49 kJ
5.11. (b) G°68 kJ (c)K8.2 1011
5.12.The system would not necessarily be at equilibrium, be-
cause the pior pjvalues in equation 5.9 now have different
values. Only if there were the same number of moles on either
side of the chemical reaction would these partial pressures
cancel mathematically and the equilibrium constants have the
same value.
5.14. (a) G°32.8 kJ (b) (^) rxnG29.4 kJ
5.15. Gwill be zero when all partial pressures are approxi-
mately 1.29 10 ^3 atm.
5.16.p(H 2 ) 0.4167 atm, p(D 2 ) 0.0167 atm, p(HD)
0.1667 atm, 0.0833 mol
5.18. (a)K6.96 (b)0.393
5.19. G°10.2 kJ, K1.63 10 ^2
5.21. (a)K
(b) (c)
5.22.K0.310
5.23.p1.49 104 atm
5.24.K6.3 10 ^5
5.25. (a) G°10.96 kJ (b)mHmSO 42 6.49 10 ^3
molal, mHSO 4 3.51 10 ^3 molal
5.26. H°77 kJ
5.27.A 5-K temperature drop, to 293 K, increases Kby a fac-
tor of 2. Lowering the temperature to 282 K, a drop of 16 K,
increases Kby a factor of 10. For H20 kJ, the tempera-
tures necessary are 274 K and 232 K, respectively.
Chapter 6
6.1. (a) 1 (b) 2 (c) 4 (d) 2 (e) 2
6.3.FeCl 2 and FeCl 3 are the only chemically stable, single-
component materials that can be made from iron and chlo-
rine. Note that we are identifying single components as the
compound, not the elements that make up the compound.
6.6. (a)The equilibrium shifts toward the liquid phase. (b)
The equilibrium shifts to the gas phase. (c)The equilibrium
shifts to the solid phase. (d)No change in phase is expected
(unless there is a stabler solid allotrope or crystal form; metal-
lic tin is one example).
ppCO°^2
H^2 C^2 O^4 m°mH^2 C^2 O^4
Hm°mHNO^2 m°mNO^2
HNO^2 m°mHNO^2Pb^2
m
°mPb^2
Cl
m
°mCl
2
HNO^2 m°mHNO^26.7.By definition, every pure substance has only one normal
boiling point.
6.8.dnliquiddnsolid
6.12./T214 J/K
6.13. S87.0 J/mol
6.14.MP (Ni) 1452°C
6.15.MP (Pt) 3820°C
6.17.Assumptions are that Hand Vare invariant over the
temperature range involved.
6.19.A pressure of ~7.3 atm will make the rhombic form of
sulfur the stable phase at 100°C.
6.21. (a)yes (b)yes (c)no (d)no (e)no (f)no (g)no
(h)yes
6.23.A pressure of ~7.3 atm will make the rhombic form of
sulfur the stable phase at 100°C. This is very similar to the
pressure predicted in equation 6.10.
6.26.The Clausius-Clapeyron equation predicts decreasing va-
por pressures in the following order: tert-butanol (44.7 mmHg),
2-butanol (20.5 mmHg), isobutanol (13.6 mmHg), and
1-butanol (9.6 mmHg). This order is also the lowest-to-
highest in normal boiling points of the isomers.
6.27.dp/dT7.8 10 ^6 bar/K, or about 10600 mmHg per de-
gree K
6.28.p0.035 bar
6.30.p97 atm, corresponding to approximately 800 m
beneath the ocean surface.
6.38.Higher pressures are needed to have a stable liquid
phase for a compound that sublimes under normal pressure.
CO 2 is an example.Chapter 7
7.1.Degrees of freedom 3
7.3.There would have to be five distinct phases (for example,
you could have three different solid phases).
7.7.Minimum amount of H 2 O is 6.39 10 ^3 mol (0.115 g);
minimum amount of CH 3 OH is 3.36 10 ^2 mol (1.08 g).
7.8.yH 2 O0.0928; yCH 3 OH0.907
7.9.a0.984
7.12.Total vapor pressure equals 124.5 torr.
7.13.pEtOH0.0693 torr
7.14.xMeOH0.669, xEtOH0.331
7.16.yC 6 H 14 0.608, yC 6 H 12 0.392
7.18.Equation 7.24 is written in terms of the vapor-phase
composition, not the liquid-phase composition.7.19. (^) mixG3380 J (for 2 mol of material), (^) mixS
11.5 J/K (for 2 mol of material)
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