Chapter 9
(a) The kinetic energy of O2 molecule = (1/2)mv2 = (1/2)×[(32.0 g mol-1)/(6.02x1023
mol-1)]×(780m s-1)2 = 1.62 × 10-20 J < 8.22 × 10-19 J.
Therefore, the kinetic energy cannot break the O=O bond.
(b) 8.22 × 10-19 J = (1/2)×[(32.0 g mol-1)/(6.02x1023 mol-1)]×v2
v= 5560 m s-1.
1 Btu = 1054.35 J and 1 cal = 4.186 J, then 102.20 x 1015 Btu = 1.0775 x 1020 J =
2.5741 x 1019 J.
However, this can only be applied in the universe of constant mass.
ΔE = Q + W
(a) -43 J = Q + 40 J, Q = -73 J.
(b) 31 J = Q + 34 J, Q = -3 J.
9.31 ANS 3.50 x 105 J
In page 354, q = mc∆T , where m is mass of material and c is specific heat capacity of
material.
Thus, q is (26.0 x 103 g)x(0.449 J g-1oC-1)x(55.0 – 25.0 oC) = 3.50 x 105 J.
3.50 x 105 = m x (4.184 J g-1 oC-1)x(79.5-55.0), m = 3.41 kg.
(8.75 oC) x (1.3 kJ oC-1) = (0.367 g) x χ , χ = 31 kJ g-1.
(5.24 g) x (43.4 kJ mol-1)/(28.09 g mol-1) = 8.10 kJ.
ANS
(a) [226.7 + (5/2)x0]-[2x(-393.5) + (-285.8)] = -1280. kJ/mol.
(b) [(-304.6) + 0]-[(-398.9)] = 94.30 kJ/mol.
(c) [52.26 – 241.8] – [(-235.1)] = 45.56 kJ/mol.
(d) [(-824.2 + 2x0)] – [(-1676) + 2x0] = 851.8 kJ/mol.
9.69 ANS 31.9 g/mol
The mol of unknown compound is (0.0157g)/Mw, where Mw is the molecular weight
of unknown compound and the unit is mole.
The heat of combustion of known compound is -37.6 kJ mol.
The heat released from the combustion of unknown compound is 18.5 kJ = (mol of
unknown compound)x(heat of combustion of known compound) =
(0.0157g/Mw)x(-37.6 kJ mol-1).
Thus, the Mw is 31.9 g/mol.
9.75 ANS
These terms refer to the ability to recharge a battery. A primary battery cannot be
recharged, and a secondary battery can be, ordinary alkaline cells are primary
batteries, whereas car batteries are secondary batteries are secondary batteries. (Other
examples are also possible)
The molar # of tea is (500.0 mL)x(1.0g mL-1)/(18 g mol-1) = 28 mol.
(28 mol)x(75.3 JK-1 mol-1)x(293.0 – 273.0 K) = 42000 J
The amount of ice melted is (42000 J) x (18 g mol-1)/(6.0 x 103 J mol-1) = 126 g.
The amount of ice remaining is 134 – 126 = 8 g.
ANS
The relationships of some units:
1 mile = 1.609 km
1 J = 1 (kg m2 s-2)
24 miles = 38.6 km = 39 km
Since Work is (mass)x(acceleration)x(distance), W = (980 kg)x(2.3 m s-2)(39 x 103 m)
= 8.8 x 107 kJ
(2.43 mol)x(28.0 g mol-1) x (1.04 J g-1 K-1)x (Tfinal – 285 K) = [(5.44 x 103 g)/(55.9 g
mol-1)] x (0.444 J g-1 K-1) x (755 - Tfinal)
Tfinal = 743 K
P = nRT/V = (2.43 mol)x(0.08206 atm L mol-1 K-1) x (743 K)/(31.7 L) = 4.68 atm.