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Unformatted text preview: 6.4 (a) Let us suppose that we have exactly 1.0000 L of the solution. We then must have 205.0 g
I of acetic acid and 820.0 g of water. The total mass of the solution is 10250 g. Density : 410250 —— 10250 g cm‘3 1000 cm3 _
(b) We use the molar masses of acetic acid and water to convert the two masses to chemical
amounts 205.0g _ 820.0g : 3.4136 mol acetic acid and = 45.517 mol water 60.05gmol'1 18.015gmol""1 The molarity of the acetic acid is the number of moles of acetic acid per liter of solution. Our 1.000 L of solution contains 3.4136 mol of acetic acid, so the answer is 3.414M. The molality of the acetic acid is the number of moles of acetic acid per kilogram of solvent. We have 34136
mol of acetic acid dissolved in 0.8200 kg of water. The molality of the acetic acid is therefore 3.4136 mar/0.8200 kg = 4.103 mol kg“1 The mole fraction of acetic acid is 3.4136 mol
3.4136 mol —I— 45517 mol and the mass percentage of the acetic acid is 205.0g
1025.0g : 0.06976 >< 100% = 20.00% (c) The molarity of the solute water in this solution is the number of moles of water per liter
of solution. Our 10000 L of solution contains 45.517 mol of water so the answer is 45.52 M. The molality of the water is 45.517 mol/0.2050 kg acetic acid : 222.03 mol kgml. The mole
fraction of water is I 45.51? mol 3.4136 + 45.517 _mol
Finally, the mass percent of the water—"is (820.0 g/10250 g) X 100% = 80.00% = 0.9302 6.18 At 200C. and a pressure of 0.970 atm, the chemical amount of NO is readily calculated by
assuming ideal gas behavior and writing 50 :. PV/RT. It is 0.2016 mol. This amount of NO(g) requires g X 02016 = 0.3024 mol of N aNOg for its generation. according to the balanced ' equation. 0.3024 mol W = 0.468 Lsolution
. mo r. —— ——  H 6.32 In acidic solution. H30+ and H20 may be added either as reactants or as products to achieve balance . (a) 8 MnO; (sq) + 5 HgS(uq) + 14 H30+(uq) —> 8 Mn2+(oq) —I— 5 SOi‘(og) + 26 1120(1)
(b) 4 Zn(s) + N0; ((19) + 10 H30+(oq) —> 4 Zn2+(aq) —— NHqu) + 13 1120(1) (c) 5H202(uq) — 2 MnO; (sq) + 6 H30+(aq) —> 5 02(9) —— 2 Mn2+(ng) + 14 H200) (d) 2811(3) —— 2 NO§(aq) +10 H30+(cq) ~+ 2 Sn4+(aq) + 1220(9) +15 H200) (e) 3 U02+(aq) + Te(s) + 4 H30+(aq) ——> 3 U4+(aq) + Tao.. (g) + 0 1120(2) \_._ . —— ___ .42 The chemical amount of maleic acid is 006203 mol, and that of diethyl ether is 1.349 mol.
The mole fraction of maleic acid (the solute) is then 0.06203 mol on e I M = . 1“ 1349+ 00620311101. 00 X5 \. The vapor pressure lowering is AP = “Pinon... 2 “(0.8517 atm)(0.04396) = ——0.0374 atm so the vapor pressure of ether above the solution is .. . ———— P 2 0.8517 ——— 0.0374 atm = 0.8143 atm 6.50 There is 1/8 kg = 125 g NaCl per kg of solvent. This corresponds to 2.14 mol NaCl. so m z
2 x 2.14 = 4.28 mol kg"1 because each NaCl gives two ions in solution. AT = as} m = ——(1.55 K kg mol"1)(4.28 mol kg'l) : "8.0 K This gives a freezing point of 8.0°C. or 18°F. 6.54 6.56 6.58 c —— i — 00319 mm —— 0001320 1L—1
"' RT “" (0.05200 L atm moi1 K‘1)(293.15 K) " ' . m0
__ 23.7 g L‘1 4 __1
M "" 0.001325 mol L1 “1'79 X10 g “ml a = pgh : (0759 X 103 kg m3)(9.50065 m a"2)(0.253 m)
2 2.035 ><1O3 Pa : 2.01><10_2 atm = 5.35 X 10"4 mol L‘i ZRT C One liter contains 125 g of protein, or 8.35 X 10“1 mol. Dividing gives the molar mass: M :125 g/8.35 X 104 mol : 1.50 X 104 g mai1 (3.) The partial pressures in air are PM 2 0.78 atm and P02 2 0.21 atm. Thus, from Henry’s
law Pa: = leaXa 1.: z 0.75 atm/8.57 >< 10"1 atm = 0.10 X 10—5 we have __.PN2
._ km X02 2 ><104 {1.13111 1: X XN 2 One liter of water contains 55.5 mol water. Because the mole fractions of gases dissolved are
so small, we can write nixlg 2 X112 (nNg + n03 + m XNﬂnI—IQO : (0.10 x10"6)(55.5 moi) .—.._ 5.124 10'"4
[N2] 2 5.1 X 1011 M In the same way, we calculate [02] : 2.7 >< 10*4 M (b) For a given partial pressure. less helium dissolves in the diver’s blood than nitrogen because
helium has an intrinsically lower solubility (larger Henry’s law constant) in water than nitrogen.
The substitution of helium for nitrogen in breathing mixtures for divers means that a smaller amount of gas dissolves in the divers’ blood and moderates the risk of “beads” 6.170 (a) The oxidation is Mn(ll) —+ Mn(VII) + 5 e‘. Hence 8.0 x 10"2 mol Mn(ll)) X ( 5 mol e ne 2 0.0183 L >< ( L m ) = 7.32 X 10"”3 mol e— If m mol of electrons is needed to oxidize 1 mol of acid} then 0.293 g A represents (7.32 X
10—3/m) mol or 0.293 g MA = 7.32 x 10‘3 mol rn=£10.0mg:mol"1 m21,2.... (1)) mol OH“ = 0.0157 L X 0.490 mol L"1 = 7.69 X 10"3 mol. If there are n acidic H atoms
per molecule of A, then 0.385 g A represents 7.69 X 10‘3/n mol, or 0.385 g MA I 7.69 X 10‘3 mol  350.0ngmol“l 37121.2)... (c) Since A can have only one molar mass, 4 500?: : 40.0w or n : —m
Since it and m are both integers nz4p, m=5p $921.12.... l and
r '  ...____ _ ._._______. _ ..
6 78 The molality of the required anti—freeze solution of ethylene glycol is
m: vé—E :w(—5 0) = 2 69 mollig"1
In 1.86 Such a solution has 269 mol of CHQOHCHQOH (M :2 6féll67gg 111:1‘31 dissolgle; 13%;: ' ' ls . g o e y ene .
t . Tlns chemical amount of ethylene glycol equa _
3331669 g of ethylene glycol for every 1166.9 g of solution. The percentage by mass oftflh:
ethylene glycol in the solution is 166.9/1166.9 X 100% = 14.3%. Note the assumptioin t. a
ethylene glycol neither dissociates into two or more particles nor assoc1ates 1n aqneons so 11 ion. 7.10 (al We identify the battery (not the toy truck) as the system of interest and apply the ﬁrst law
to 1t. The battery performs work on the surroundings so to = “117.0 J . It also absorbs negative heat (evolves heat to the surroundings) so 9* = ——3.0 J. By the ﬁrst law, AE : —120.0 J. (b) Recharging the battery puts it back the way it was before it was discharged. The ﬁnal
state 1s indistinguishable from the original; state. so AEDVEHH :— 0. But AEoverall I 0 : AEdischarge ‘l‘ AElucharge I (q + w)discharge + (q + wlcharge Substituting the known quantities in joules gives 0 = —3.0 + (—1170) + qchmge + 210.0 which means qcharge : —90.0 J The sign of 9 during the charge cycle is negative because the battery evolves heat to the
surroundings as it is charged. ' I products reactants Enthalpies of formation are always tabulated on a “per amount” basis. The AHf’s from
Appendix D are given per mole, so each must be multiplied by the number of moles appearing in the balanced equation.
(3»)
AHO : 2 mol<33.18—1::i) — 1 mol — 2 Incl <9025£> = —114.14 kJ mol mol Inol 00)
AH” : 2 mol (“1105233) 1mol<——393.51—1—{—J—) : +1724? kJ mol mol (c) We omit the somewhat repetitious listing of the units: use : 3(——241.82)+ 2(33.18) — 7/2(0) — 2(—46.11) = —566.88 kJ (d) AH” =1(0)+1(——~110.52)—1(—241.82)~—1(0)= +131.30 kJ I 7.38 (a) The reaction is . i l F6203(S) Ti 2 F€(5) + Al203(3) The standard enthalpy of reaction is computed by combining the AH? is from Appendix D: AH“: (0) +1(—1675.7)——2 (0) _1(_s24.2)=—851.5kJ
V \—v——" V \—v—’
F€(S) Al203(8) .Al(5) F6203(3) (b) According to the result in part (a), 851.5 kJ of heat is given off at constant pressure when
one mole of Fe303(s) reacts. One mole of FegOg is 159.69 g, so the amount of heat given off by the reaction of 3.21 g of Fe203(s) is 3.21 g
:... ,_kJ 1—"1 #__—————:—17.1 kJ.
9' 8515 m X 159.69 g mol—l 7.44 (a) The combustion of benzoic acid is represented C6H5COOH(S) + E 02 (g) —> 7C02(g) + 3 (b)
as : qv : CAT = (9382 J K”) X (2.15 K): 2.02 x104 J The heat absorbed by the reaction Within the calorimeter is the negative of this, We therefore
have for the reaction of benzoic acid: (—2.02 x104 J) X (12.12 g _ 6 —1
0.800% 1m01)—3.08><10 Jmol _ The answer is therefore AB” : —3.08 X 106 J .mol'"1 X 1.000 mol :: —3.08 X 103k]. (c) Write AH” : AE” + AngRT where Ana is the change in the number of moles of gas on
the two sides of the equation. Substituting gives E
i
l
E
E
l
l l 1 .
am = —3_08 X 106 J + (—5 mol)(8.3145 J K"lmol"1)(298.15 K) = —3.08 >< 103 M (d) The necessary relationship is am: 2 AH?—— 2 AH;
products ' reactants '—'—n.
—.__n_ __ where the quantity on the left is ——3080 hi and all but one of the terms on the right are known: ... ——3080 M = 7 (—393.51)+3'(—s'85.83)——1moMAHﬂosmcoonn
w w
002(9) H20“) AHHCSHgCOOH) : —532 kJ marl q = AH : (33 J K"1mol"1)(0 K —— —30 K)(1.00 moi) + (1.00 m51)(0007 J K—lmorl) + (75 J K1m511)(100 K _ 0 K)(1.00 mol) + (1.00 m51)(40050 J mol“1)
+ (30 J K"1m511)(140 K — 100 K)(1.00 5551) = 5.075 >< 104 J 2 50.7 M AB .7..." — : "' (Pﬁnalvﬁnal — Pinitiall/initial) m "‘ Pﬁnalvﬁnal because the volume of ice is much smaller than that of steam. PV = nRT :_ (1.00 m51)(3.3145 J K1m511)(413 K) ': 3.43 X 103 J as = 5075 x1011 —— 3.43 ><103 = 5.33 ><:10‘1 J = 53.3 M
a 2 as “ q = 53.3 —— 55.7 = —3.4 M 7.60 (a) Substituting in PV : nRT gives P : 1.000atm.
(b) q I +3000 k]. (c) The strong rigid container neither expands nor contracts during the heating. There is therefore no pressureﬂvolume work. Because there is no other mechanical linkage to the sur— roundings, w = 0. (d) The temperature change of the “gas is given by
q 3000 J _
—l I 24 . K
in cV (1.000 mol)(12.47 J mol‘lK‘l) 0 6 AT: so the ﬁnal T is 513.7 K .
(e) We use the ideal gas equation: P = 1.881 atm. (r) AE=q+w=3000+0=+3000J. (g) AH : AE + A(PV). We know the pressure and volume before the heating and we also
know them both after the heating. Therefore we can compute A(PV). It is 19.73 L atm, which is 2000 J. Hence, AH : 3000+ 2000 : +5000 J. (h) The law of conservation of energy applies to energy, not enthalpy. The change in enthalpy
is equal to the heat absorbed only if the pressure is held constant. —I—. 7.72 (a) AH” : 2(393.5) —— 502418) —— 2(—112) = —1288 Id: This is for the reaction as written,
which is for the burning of 2 mol. For1 mol, AH“ would be ——644 kJ . The reaction is highly exothermic. (b) If CH3N02(g) were burned the reaction would be more exothermic by the molar heat of
vaporization of nitrornethane is. the gas has a higher enthalpy than CHBNOQU). Formally, AH“ for the reaction CHgNOg(g) a» CHgNOgU) (a negative number) is added to AH” for
burning 1 mol of CH3N02(2), since addition of the corresponding equations gives the equation for the desired reaction. in the compounds in parts (b) through C20 double bonds; H——O—H has two
CgHs OH has ﬁve C——H single" OHH single bOﬂd; Cngg hELS (b) Cp 5/2R 5 7 2
:———:———:— d ————:—
7 CV 3/23 3 an 7 5
P2 2/5 10313111 WE
T 2T — :40 K :
2 1(131) 0 (5051111) 84K (C) The change in random thermal energy per molecule is (3/3531? — (3/2)kBiE : (3/2)(1.381>< 10"23 J 31*) x (84 — 400 K) : “555 X 10.51 J
This energy must go into net directed motion 1/ 2 171172 1/2 m5? = 6.55 X 10*"21 J '52 * 2(555 >< 1021 J)(6.022 X 1023 mol'l)
"" (4.0026 X 103 kg moi1) 21.97 >~<106 1’112 8—2 8.16 T i
routon s rule states that the molar entropy of vaporization of most substances is near 88 J 1‘7:_*]‘J‘11."1(23'1"—1 Gas and 11 ' u *
' quid forms of a s bstan ‘ ' ' . . .
so we ca write: ce are 1n equilibrium at the normal bolhng pomt. __ AHWP 16.15 X 103 J mol"jL Tb ._ : _.__.___.__.._..___________ _.
Astra}: J K‘lmol—l — assume that the gas of sod   ' 5 ' " ' _'
atom is in the left half of the optical trap is not inﬂuenced by the presence of the ﬁrst. It is also 1/2. The chance that both by these two atoms are the left half is 1/2 X 1/2 = 1/4. By
extension, p, the chance that all 500 sodium atoms are in the left half. is 1 5110 (a) To compute p we take the logarithm of both sides of this equation 1
logp : 500 log2 :— 500 X (#030103) :: —150.51 "Fa—NﬁurWu—ur—q—"m m1
' ' "I: "l'u: :l'qLI' '""'m and then the antilog of both sides : 10—15051. : 100.49 X 10—151 : 3 X 10—15]. —.a——. 8.52 (a) The process takes 2.00 mol of an ideal monatomic gas from 1/ = 19.15 L to 38.15 L and
from T : 350 K to 272 K. To calculate A533,. We connect the same two states by a reversible path: ﬁrst expanding reversibly at constant temperature as. 2 ann Vg/Vl = (2.00 m51)(5.3145 J mol"1 K"11n2):11.5 J K“1 and then cooling at constant volume:
A52 2 nCV ln Tg/Tl = (2.00 mol)[3/2)(8.3145 J K"1mol‘l) 11:1[272/350) : “6.3 J K"1
Thus as... = as. + as. = 5.2 J Krl. Because 9 = 0, 83...... = 0. and 413...... = 5.2 J K“ (b) AG : AH ... MTS) :: napAT — (T1551. — TiSi) z (2.00 m51)(5/2)(s.3145 J K_1mol—l)(272 ~— 350 K) _ (272 K)(158.2 + 5.2 J K‘1)+ (350 100582 J K“)
4— —3243 J — 44,445 J + 55. 370 J = +7680 J 1— Note that AG > O in this process even though it is spontaneous. AG < 0 is a criterion for
spontaneity only at constant temperature and pressure. ASH...V > 0 is a more general criterion. is a temperature above which any given molecule will not be stable at the given pressure, ale, a temperature above which the free energy of dissociation at that pressure is negative. That
temperature T is such that TAS > AH, or AH _ 1.05 x103 kJ __. —————————————= .5 103K
as 0.110 kJ K1 9 x The value of AS quoted is pressure—dependent. It increases by Rln 10 z: 19 J K“1 (in accord with Eq. 8.9) for each 10fold decrease in pressure. Thus at pressures much lower than 1 atm}
AS is even larger and the highest temperature at which a molecule will be stable is even lower. 8.62 (a) The value for AH" computed by the student is for a process at 2500. The value in the ge, but at a different temperature. the boiling point of the carbon tetrachloride. Interestingly. the AH" of vaporization of a liquid substance decreases with
increasing T and becomes zero at TC, the critical temperature. (b) At the boiling point of the CCla We have AH 30.0 kJ mol“1
Asva 2 vap : , k _l 1—1 r: 1 ‘1 "'1
p T 34965 K 0 0858 J K mo 85 8 J K mol ...
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This note was uploaded on 04/30/2008 for the course CSE 3 taught by Professor Ord during the Fall '08 term at UCSD.
 Fall '08
 Ord

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