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Unformatted text preview: (IIIIBIVI 1:31
FINAL EXAM K. Lindenberg . Fall 2007
NAME g, L NDENE 6'6? 63
STUDENT ID
1 /18 4. /16
2 /17 5 /16
3 /16 6 /17 TOTAL FINAL EXAM/100 MID/35 __ + HW/10 __ + FNL/55 /100 COURSE LETTER GRADE NOTE: Your REASONING and ANSWERS MUST be clear to us. Make
sure to offset your answer clearly — if it is lost amidst a lot of other stuff we may not ﬁnd it. If you have more than one answer we won’t know which one you
want us to use. TOTAL NUMBER OF PAGES INCLUDING THIS COVER PAGE: 16. It is your responsibility to make sure that you have them all  if not, get. another
exam. You should recognize a lot of these problems from homeworks. 1. 18 points CHEM 131 This problem has ﬁve parts, one part for each oflChapters 6—10. In each part you have
to choose the noted number of correct statements or answers by placing a check mark
in the appropriate spaces. Be careful: pure guessing is a very bad idea because
you lose points for incorrect choices. Numbers in parentheses indicate, for each
part, points for (correct answer, blank, incorrect answer, excess answer). Chapter 6  Reaction Equillbirum in Ideal Gas Mixtures. Pick the TWO
correct ones (+1, 0, 0.25, 0.25) ' / The chemical potential of ideal gas 2' in an ideal gas mixture
at temperature T and partial pressure Pi equals the chemical
potential of pure gas i at temperature T and pressure Pi. Kg for the reverse reaction is the negative of K2 for (the forward
reaction. Doubling the coefﬁcients in a reaction doubles K3.
If in a a gas—phase closed system all the N2 and H2 come from the dissociation of NH3 according to 2NH3 : N2 + 3H2, it follows
that at any time 331m2 = 23112. ' When the idealgas reaction A + B (2‘ C + D has reached equilibrium, it must hold that no + up = nA + n13. In any closed system with P — V work only, G is always minimized
at equilibirum. CHEM 131 Chapter 7  One Component Phase Equilibrium. Pick the TWO correct
ones (+1, 0, O.25, 0.25) The number of degrees of freedom f is the number of variables needed
to specify the thermodynamic state of a system. The minimum possible value of f in the phase rule is 1. The enthalpy of vaporization of a liquid becomes zero at the critical
point. When three phases coexist in equilibrium in a one—component system;
one must be a gas, one must be a liquid, and one must be a solid. For a onecomponent system, the most stable phase at a given T
and P is the phase with the lowest Gm. Solid H20 cannot exist at 100°C as a stable phase. Chapter 8  Real Gases. Pick the ONE correct one (+1, 0, —0.75, 0.75) / The parameter a in the van der Waals equation has the same value
for all gases. The parameter a in the van der Waals equation is different
for different gases. CHEM 131 Chapter 9  Solutions. Pick the TWO correct ones (+1, 0, 0.25, 0.25) The volume of a solution at T and P equals the sum of the volumes of
its pure components at T and P. M is a partial molar quantity. Intermolecular interactions are negligible in an ideal solution. If B is a component of a solution, ,uB cannot be greater than [1%. For equilibrium between an ideal solution and an ideal vapor, the
mole fraction .7313 of component B in the solution must be equal to the mole fraction yB of B in the vapor. In an ideally dilute solution, [M = M: + RTln 1:1 for all components. Chapter 10  Nonideal Solutions. Pick the TWO correct ones (+1, 0, 0.25, 0.25) / When a solution component is in its standard state, its activity is 1.
ai and 71 are extensive properties. For a solute in a solution, 71113 = 71,13. The activity ai is never negative. Activity coefﬁcients may be positive or negative. The ionic charge 2+ in an electrolyte solution does not affect the value
of the mean ionic activity coefﬁcient 7i. CHEM 131 2. 17 points  Problems about ideal gas equilibria (a) An experimenter places 15.0 mmol of A and 18.0 mmol of B in a container.
The container is heated to 600 K, and the gas—phase equilibrium A + B : 2C +
3D is established. The equilibrium mixture is found to have pressure 1085 torr
and to contain 10.0 mmol of C. Find K3 and AGO at 600 K, assuming ideal gases.
R = 8.314 J/mol—K. _ :&_5l5
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We (3W =00?” P ﬁvxKo: _. [34° —_ «(lﬂg'Dcd/rmrl) 3—4.3? (mg } cal/WM) (ma/O CHEM 131 3. 16 points  Problems about onecomponent phase equilibria (a) Find the melting point of ice expressed in degrees Kelvin at 100 atm. You will
need the following information: AfusHm for ice is 1.437 kcal/mol, the density of
liquid water is 1.000 g/cm3, that of ice is 0.9166 g/cm3, and the molecular weight of water is 18.02 g/mol. Remember that 0°C = 273.15 K. The gas constant
R = 1.987 cal mol‘1 K’1 = 82.06 cm3 atm mol‘1 K‘l. (g'e’x AQmsH (12,17) Shirl: W m MW W" 5M alrrm 4‘4 W 3
(913.1501) (, (laqlw ) inbred/”MK 3
, :' .. 0,“? KM >‘ ,f
A Ar 3 am gQ'04m3m1'm/7ndk :zﬁ'S—K :.:> 12: a?3"5'l<‘0~7'5"< CHEM 131 (b) The normal boiling point of diethyl ether is 346°C and its AvapHm at the
normal boiling point is 6.38 kcal/mol. Find the vapor pressure of ether at 250°C
using appropriate approximations. R = 1.987 cal mol‘1 K‘l. Agw Jul» mm
VM)°( V4151 1‘! Vin)? MW ‘V 4W 4+ AWN“
l vi/
airwaieﬁ>
1": mic—3077K Jwﬂ ~Lesgoauww i: ,7 K
ﬂgil :50}
(1.42? ’ECTAB 1 \ (9:74”) 17‘“ W CHEM 131 (c) The vapor pressure of water at 25°C is 23.76 torr. Suppose that 0.360 g of
H20 is placed in an empty rigid containet at 25°C. What phase(s) are present
at equilibrium if the volume of the container is V = 10.0 L? What phase(s) are
present at equilibrium if the volume of the container is V = 20.0 L? Assume the vapor to be an ideal gas. Make sure I understand how you get ’your answers.
R = 82.06 cm3 atm mol‘1 K‘1 and 1 atm = 760 torr. mJanu (Saba/(36w W
Vulrtli’ WWW ﬁm$ ‘ \ LSWWM
I‘la ’ WW (L5 f—Hl’m
thW)tmmW :13 J
so WW :5 a» W900
4) = 31.: (amt/cm/ﬂolw“ >( 0’")
‘1“ gr (21,“ cm’mw/MK) (16134.0() CHEM 131 4. 16 points  Problem about real gases (a) The gas equation of state P (Vm — b) = RT is an improvement over the ideal
gas law because it takes the ﬁnite volume of the molecules into account. Derive an
expression for the fugacity coefﬁcient for a gas that obeys this equation of state
at temperature T and pressure P.. P
W 80 erg—W 10 CHEM 131 (b) The Virial expansion leads to an approximation to the van der Waals equation
of state valid at low pressures, Vm = (RT/ P) + B Where the virial coefﬁcient
B = b — a/ RT. Derive an expression for the fugacity coefﬁcient for a gas that
obeys this equation of state at temperature T and pressure P. 11 CHEM 131 (c) If you have done parts (a) and (b) correctly, you should have found that the
fugacity coefﬁcient in part (a) is always greater than 1, but that of part (b) may be
greater than 1 or less than one. In a sentence or two, give a physical explanation
for these outcomes. 12 CHEM 131 5. 16 points  Problems about ideal and ideally dilute solutions (a) At 100°C the vapor pressures of hexane and octane are 1836 and 354 torr,
respectively. A certain liquid mixture of these two compounds has a vapor pressure
of 666 torr at 100°C. Find the mole fractions in the liquid mixture and also in
the vapor phase. Assume an ideal solution. (7 Ph‘)‘ 4" Paar
xv
Y“, phex ‘\' (\‘XRHB Poor ’— 1n
‘;—>> thi = P;?0cf* __ ({(K‘ZSLU tort
ﬂew ad” (luézmotm ‘0 ﬂ gartm 13 CHEM 131 (b) The average human with a body weight of 70 kg has a blood volume of 5.00 L.
The Henry’s law constant for the solubility of N2 in H20 is 9.04 x 104 bar at 298
K. Assume that this is also the value of Henry’s law constant for blood and that
the density of blood is 1.00 kg/ L. The molecular weight of water/blood is 18.02
g/mol. Assume that the solutions in this problem are ideally dilute. (b.1) Calculate the number of moles of nitrogen absorbed in this amount of blood in air of composition 80% N2 at sea level, Where the pressure is 1 bar, and at a
pressure of 40 bar. R = 8. 314 x 10—2 Lbar/mol—K ’“N ; “he (ON; (50am (agolm) KS‘: C '3 03‘?) )‘m'lb (ﬁoqxiaﬂm) (b2) Assume that a diver accustomed to breathing compressed air at a pressure of 40 bar is suddenly brought to sea level at 298 K. What volume of N2 gas is
released as bubbles in the diver’s bloodstream? V7ﬂ£1~ _ (anon/mal— 9,§uo’i,,,{) (sawmoﬁ Lion.) (2% K)
P mull" l. Oohm ">— 14 CHEM 131 6. 17 points  Problems about real solutions (a) At 35°C the vapor pressure of chloroform is 295.1 torr and that of ethanol is
102.8 torr. A chloroformethanol solution at 35°C with math = 0.200 has a vapor pressure of 304.2 torr and a vapor composition of yeth = 0.138. R = 8.314 J / mol
K. (a.1) Calculate 71 and a] for chloroform and for ethanol in this solution. "1‘ Kit 35w: = W
>09; (“0.200) (1457va ’5— \‘r7Uf‘k; C0.‘3&) (309.;1717»)
(0,900) CH9? 9 TM} ‘3
Obj2"“ 7‘ 253“.“ Xcu. Z C\.\\§(O.g00) ﬁne/M : 0,23? Ora—ﬂay g. (9.0HBC01M") ‘ 011 all,» : Oﬂog
I (a2) Calculate ,ui — pf for each component in this solution in J / mol. 352:3081K
Mi “Mi;9 3 grim and Marlin ._ (93 m: {Mug} (208.\K)J2M OM? "‘3
f “ * —’
—M:e+y : 2300 A ’m 15 CHEM 131 (a3) Calculate AG for the mixing of 0.200 mol of liquid ethanol and 0.800 mol
of liquid chloroform at 35°C. [1me ; Z «wk/111*)
7 (0.800anr0é301T/~nrﬂ+ (0,:200w{)(—230037wr4> % W (b) Consider the strong electrolyte MgClg. Give the following numerical values
(rather than a formula) for this electrolyte: 14, V_, 2+, 2_, Vi. Express 7i in
terms of 7+ and 7, and a,» in terms of 7i and mi, exhibiting explicit exponents
and coefﬁcients. 16 ...
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 Fall '08
 Lindenberg

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