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Unformatted text preview: l95 '  riiihrr'trm
Herrrogtvreutrs trad horritigtnerao gm tip d narnic calculations in:
mm if dfi”. iiii”, and d3'. Look for occasional ' ' her of simple the
.1, cu wtll ﬁnd a num _ I I I 1 we“
' aliri'iig donversion of K,, to no". tht. relation b;t;:m
and the variation of ac" and of K, with temp I‘ll 1 1 .tortcuts.
HETEROGENEOUS
AND HOMOGENEOUS GAS EQUILIBRIUM EQU ATIUNFi AND Ct) NC EPTS l: uilihrium—Constant Expressions ' " ' ‘ urc.
olving only ideal gases at a gtvcn tcntpcrat ltiriven the general rea etiort In v ill1]:
ad + hB + = mili t— Iiiv' + III in _
’“i'd’i' "'_ "ii""*:"'llili = iii 2:
K” : tit—tit”? ' spit,“ N a e total pressure, a the moles of ts rnin us the moles of reactants.
I terms that total moles present. F tli ‘ otes the L
Here. n den Art the moles of produc  i i. I and '   ' t the
a “will” WELme ' = v ~' wIll contact on y
I pﬁ‘mllds are lnwhw m thin :dLllillLbJ: IJthe diil'erettce between moles of
i ' ' ' 'ies. and I: wt 1 I I I mu“. Dr
rarer m Edger1m wed moles of gaseous reactants. The solid species CU MMENTS  s roduets an . . id,
There are three general types ofproblems in this chapter. First. you will find If: present in the actual system 11' K,, Le t0 W
a number that involves equilibria between gaseous substances and requires F
that you calculate K,1 l'rom the data, or. given it”. determine the extent of Thermodynamic Relationshth H 14]
reaction at equilibrium. In most cases sloichiometry is very important— that ' u a
.   . . . . not=aHTﬁ5 4
ts. you must utilize the equation for the overall process to determine the ill 1
mole ratios in which reactants disappear and products form. Often it will be Mp” = _ ET in KP
convenient to set up stoichiometric relationships between the various molc
numbers. and insert the results in the expression for Ha. Ha then follows, if palft i—lofl equation:
the total pressure is known {and N, the total moles. can be obtained from the H L“
sttm of the various mole numbers]. If the amount of initial substance is not Kr Mi '1
speciﬁed. it is ol'ten less confusing to assume some convenient amount as a ' ii l“ 31? '1 iii"F’
basis.‘ your choice should cancel out in the course of the calculation. Mic. I 1 {1 Hit
The second group of problems consists of those dealing with equilibria In K_z : _ ._ _[— — between solids and gases. Here. it is important to remember that. although the . K1 R T1 ’ solids do not appear in the Kn expression, they must be present in the physical
systetn for KP to hold. Remember. too, that in a reaction of the type .
nits] + Big} = products, the amount of B needed to convert all of ﬁt to , PROBLEMB
products is the sum of the equilibrium amount required by K”. plus the 
amount needed by stoiehiornetry to convert a to products. Finally, ifa solid dissociates into gaseous products, these will be formed in stoichiometric
ratio: hence they, and KP, can be expressed in terms of the total pressure it
no other gases were present initially. ' l " much that
" ' G =Slh+zﬂzlss I
at mg minimum 5at btitJT; the equilibrium pTcs '  tiasit '
“ducal! mm a respectively. [at Calculate it ,,.
to be introduced into the Hit mint Suppose th  t1'
when ti g of ﬁt}, an. In Til
sure and density are Lit atm and L6 glitter. {bi Calculate the moles o'I‘ helium that would have I94 r_____ _ __. 1915 [trialsntmno‘iug physical chemistry 2. [30 min] K, is I03i atm‘l at 3"00°K For the reaction ‘l" D} = eqlLZiits’ET; ppgsxgl ﬁéntrodueed into a 30liter ﬂask at TWK until the
N01. is . atm. Calculate a. the degree of dissociation of the and :13” is —40.? calldeg and dC: is zero. to} if er were 0.3 under the above conditions, what would be molecular weight of the '
1 gas mntture'.l Sh ' '
[c] Lalcuiate em” for the reaction at lg‘llClrum calmldlmn‘ [d1 Calculate the temperature coeﬂicieut of K at ‘i‘tFK that is the percent change in K per de
—_ d _ _ gree temperature chan e.
N03 N10,, cqmlihrium, that is, assume no N20,? is gistll]? neglml the 3. {24 min} K, is it >< 10'°atm at 100°C for the equilibrium:
CDCiyt‘g] = COig] + Clglgl or???” is 30 calr'deg.
ta} Calculate the dog
total pressure. lb] Calculate eat,“ for the reaction. {c} Assuming dCﬂ to be zero. at what dissociation of phosgene be 0.03:5;l
calculations. l rec of dissociation of phosgene at l00°C and 2 atm temperature would the degree ol‘ again at 2 atm total pressure‘.J Show your 4. [24 min} [ﬁ nal examination ue '
catalyst at l q “10an aseous COF is a
,000°C and comes to equilibriu : p 553d mm a m according to the equation :ziiﬁlspﬁligdmaiptlined at I0 atm. a sample of the equilibrium '
. w ic sto s a ' ' ' that, out ol‘ 500 cc [STP] ol‘lil'tenll'Slllft In mnmnlrﬂlmn
CIDF2 and CCIIJ. [This is don hydroxide solution which
‘ r absorbs C01: d
[at Calculate rep for [he Equilibrium 2 an Ct)J but not CF” to} tilt, increases by 1% per degree around Lil00° and titG" at this temperature. C, calmlale IllH01 $301 Helerugermtms and homogeneous gas equilibrium Ill"! 5. {25 min} The following questions pertain to the equilibrium: 2 NDCI =
2 NO + C11. [a] A certain amount of NClCl is introduced into a ﬂask at
200°C. At equilibrium the total pressure is I aim, and the partial pressure of
NUCI is 0.64 atm. Calculate KP. {bl KP increases by 1.5% per degree around
200°C. Calculate dtH° for the reaction. Assumin g that K, at 200°C is 0.1 atm,
calculate est. in} Ass timing that K p is 0.1 atm at 330°C, calculate the pressure
at which the degree of dissociation of NDCi will be 0.2. b. [20 min} {ﬁnal examination question] One and onetenth grams of
NOBr is placed in an evacuated IIiter ﬂask at —55°C. The flash: is then
warmed to 0°C, at which temperature the contents are gaseous and exert a
pressure oi" 0.] atm. On further warming to 25°C, the equilibrium total pres
sure rises to 0.35 atm. At both 0°C and 25°C the equilibrium snosr = 2ND + Er: is present. Calculate Ki, for the reaction at 0°C, and MI". The atomic weight
of Br is 80. J [20 min} [ﬁnal examination question] In a study of the equilibrium,
H: + I: = 2H], a certain number of moles x of Hl are formed when 1 mole
of H1 and 3 moles ol [1 are introduced into a ﬂask oi" volume V at temperature T. On introducing 2 additional moles of H1, the amount of H] formed is
found to be lit. Calculate KP. E. [20 min} {ﬁnal examination question} For the reaction below KF is
0.05atm and so” is 5.35 lrcal at 900°K: C2 = C3H4lgl ‘l' Hzlgl' If an initial mixture comprising 20 moles of Clllﬁ and 30 moles of bl 2 [inert]
is passed over a dehydrogenation catalyst at l:3'00"l~»’.. what is the equilibrium
percent composition of the efﬂuent gas misture'l The total pressure is kept
at 0.5 atm.C1iyen that :33“ is 31.3 cal"deg at 900°K, calculate so“ at 300°K
{assume as": = 0}. 9. [IE min} {ﬁnal examination question} A E—liter ﬂask maintained at
T00°K contains 0.1 mole of CD and a catalyst for the reaction cots} + 2H1tsl = CHJOngl Hydrogen is introduced until the equilibrium total pressure is 'latm. at
which point 0.06 mole of methanol is formed. {a} Calculate KP. 193 Understanding physimi' chemistry {b} 1tlv'hat would the ﬁnal H2 “sud hm With no (mam re the same amounts of CO and I no reaction occurs?
for the equilibrium and d3“ do not change much with es of H1. and 3 moles of C
liter flask at 25°C. Calculate ta} no“ at 25'": Err: . c '
_ i i the moles oi‘ each species present at equilibrium 3t] ' a '
i[ mm} at? rs 3th] cal at 4WC for the reaction: stvo2 = N304 [a] Calculate K, for this
lb} The densin ol‘ an eq
Found to be 5.35 gﬂiter at 4
[a] For Ki, calculate
molecular weight of 1 reaction at one.
uilibrium mixture of NC}
UT and a certain pressu
the degree of dissociation of t
he mixture, and the total : and. N204 gases is
re. Using the value in he N204. the average
pressure of the mixture. 2 moles of PC13, and 2 moles of C1 the temperature is “PK.
A . . .
penalrilertatn amount olIClz is now introduced+ kee '
rc constant1 until the equilibrium volume is 2V moles of C1: that were adde HIE”. Cﬂlcuratc me '3 I d, and the value of K, For the equilibrium.
. [13 min} atn equilibrium mixture for the reaction
CD + 2H2 = CHJJH
at Still] 1 arm of CO, and [Ham of H1.
riginal volume, still 14. [12min ﬁ. ' ' ' introduced iritti :1;:::n;bnsa:ilmlgiii1ﬂmnl Ir a 5m
50335103 is 10“ at LWC.
to a volume Four times
new value oi" the SO 15. Press h all amount oi‘ SID1 is
_ are, t e equilibrium t' h if this gaseous mixture is now allowed tat:1: 1:1.
t e original one, still at Lilﬂﬂ‘C, calculate whali ii
3,502 ratio should be at equilibrium. I E [15 min] The value 1' ' '
e K, is l X It! 3 at 25“C For the reaction: Heterogeneous and homogeneous gen: equiir'hnmrr I99 JNt t.2 + C12 2 2 NUCLA ﬂask containstlﬂ2 atn'ioil*~olIC.“II at 2S”C.Calculate Iltv moles of Cl; that must be added if. at equilibrium, 1% ol the N02 is to be
t't'IIivorlod to HDCI. The volume oi" the ﬂash is such that [1.2 mole of gas lumltlcc 1 atrn pressure at 25°C. {Ignore the possible association of NO; to
NJ 34.} to. ill; mini The degree of dissociation, a, of PCIS into PEI; and C12 is
ll. itl 250°C and 2 atm pressure Also. or increases by 2% per degree tag,
In l].i12 at 251°C} with pressure constant at 2atnt. Calculate did” and d3"
fur the reaction at 25WC. 12. [1121.5 tnin} K, is 9 atm2 for the reaction:
l..iCl3HHJts1= LiC1HH3is} + 2NHaig} tll WT. How many moles ol' ammonia must be added at this temperature In il 51jtcr ﬂask containing [1.1 mole of LiClHH3 in order to completer
convert the solid to LiCl  Hillll? Show your calculations. I it. {12 min] Ferrous sulfate undergoes a thermal decomposition as follows:
2Fe504ts1 = Fe203is] + Sﬂllg] + Sﬂglg} i‘tt ll2ﬁt°K the total gas pressure is [1.9 atrn with both solids present. in} Calculate K, [or this temperature. ibl Calculate the equilibrium total pressure that will be obtained il'exeess
I'crrous sulfate is placed in a ﬂask at 92TH, which contains an initial SD:
pressure of 0.6 atm. ll [24 min] [ﬁnal examination question] lCine—tenth mole of H2 and
[1.2 mole of C01 are introduced into an evacuated ﬂask at 4512]“(3, and the
reaction [a] occurs to give an equilibrium pressure of [1.5 atm. Analysis of the mixture
shows that it contains ll} rnole ‘29 water. a mixture of Coﬂis] and [lots]: is then introduced so that the additional
equilibtia lb} and [cl are established : {a} CoCtis} + H; = Cots} + H10 ic] Cools} + CD = Cots] + Ci.)I ")3 Linda rs is riding physical exam or r y {bi 1i'i'hat would the ﬁnal pressure be were the same amounts of CG and
H2 used but with no catalyst present so that no reaction occurs'l Ill. [13 min} KP has the value lit—5 for the equilibrium
co2 + H2 2 co + Hzﬂlg} at 25°C, and as“ is  lIZII cali'deg. [dH‘l and as“ do not change much with
temperature] Cine mole of CO, 2 moles of H1. and 3 moles of CD2 are
introduced into a 5—1iter ﬂask at 25°C. Calculate [a] so” at 25°C, {b} the equilibrium pressure, {c} the moles of each species present at equilibrium,
and id} K” at lllﬂ°C_ ll. {Jill min] dG” is Bill! cal at 40°C For the reaction: = N104 [a] Calculate K, for this reaction at 4ti°C.
lb] The density of an equilibrium mixture of NC!I and Nail, gases is
found to be 5.35 gi'iiter at 40°C and a certain pressure. Using the value in la} for K”, calculate the degree of dissociation of the N204, the average
molecular weight of the mixture, and the total pressure of the mixture. 12. [ll min]: {ﬁnal examination question} A container whose volume is
V liters contains an equilibrium mixture that consists of 2 moles of PCI,.
2 moles of Pcla , and 2 moles ol'ClJ [all as gases}. The pressure is 3 atm. and
the temperature is T°K. A certain amount of Ci2 is now introduced, keeping the pressure and tem« perature constant. until the equilibrium volume is 2V liters. Calculate the
moles of Cl1 that were added, and thc value of K” for the equilibrium. 13. [IS min]: An equilibrium mixture for the reaction CD + 2H2 2 {31—13014 at ’lﬂITK consists of 2atm ol‘ CH3DH, latm of C0, and Ill atm of H3. The above mixture is allowed to expand to twice its original volume, still
at TWK. Calculate the new equilibrium pressures. l4. {l2 min] {ﬁnal examination question] ll‘a small amount of SDI is
introduced into oxygen gas at latm pressure, the equilibrium ratio of
SDy’SDZ is lil° at l,ﬂﬂt}°C. [lthis gaseous mixture is now allowed to expand to a volume four times the original one, still at 1,EH21}°C, calculate what the
new value of the $033502 ratio should be at equilibrium. 15. [IS min} The value of K, is l X I'll"J at 25°C for the reaction: Hercmgannnur and homogeneous gals equilibrium 199 r 5:0 + CI; = 2 NOCI. A ﬂask contains 13132 atm oil‘iCu1 at 25°C. Calculage
iiii: moles oi oi, that must be added if, at equilibrium, iss ol‘ the no, is to e converted to NUCI. The volume of the ﬂask is such that I212 mollqm‘lcgI git:
produce I atm pressure at 25°C. [Ignore the possible association o 2 N_.t‘i,,_] 1 I
In {iii mini The degree of dissociation, a, of FC,, into PCI3 and Cl2 is ill at 25ﬂ°C and 2atm pressure. Also, a increases by 2% per degreediiga
to ii. 102 at 251°C] with pressure constant at 2 atm. Calculate an an li II1' the reaction at 250°C.
__—_—___—_——— IT. [10.5 mini K, is 9 atm2 lot the reaction:
LiCl  ENHﬂS} = LiCl  NHﬁs] + 2NH3{g] iiI 4ti°C. How many moles of ammonia must be added at this temperatulre
in ii 5liter ﬂask containing {11 mole of LiCl +Nll3 in order to complete y
convert the solid to LiCl 3NH3‘? Show your calculations. IH. l I 2 min} Ferrous sulfate undergoes a thermal decomposition as follows: 2FeSCi4ts] = Fe3o3isi + soggi + soggi oi ‘J'ELPK the total gas pressure is 0.9 atm with both solids present. iii] Calculate K For this temperature. I I I
[lit Calculate this equilibrium total pressure that Will be obtained if excess leii'oiis sulfate is placed in a flask at 929°K, which contains an initial 501 pressure of 0.6 atm.
______—_—_—_— I”. {24 min} {ﬁnal examination question] One—tenth HIGligﬂaﬂtl: Eggs:
Ii .'. mole of C01 are introduced into an evacuated ﬂask at 4 . Ii'iit'llon
iii] H3 + CD: = H20 + iiiiiirs to give an equilibrium pressure of 05 atm. Analysis of the mixture .I that itcontains Ill mole lilo water. I d I I
I I‘t'ltllllliﬂlLll'C of Cools] and Cole] is then introduced so that the additional I I1llll2illill lb] and {ciare established: [hi CoCHs} i H; = Cots] + H30 it'] CoClis]: + CO = Coils} + CD; Mil Understanding physical chemistry Analysis of the new equilibrium mixtu to shows it to contain 30 mole 13:3 water.
Calculate the three equilibrium constants. if K, increases by it}; per
degree for temperatures around 450°C, calculate dHﬁ'. 2t}. [15 min} K, is 0.05 atm2 at 20°C for the reaction:
NHaﬂslSi = NHsigl + Hsﬂlgl 0.36 mole of solid NH4HS is introdueed into a 2.4liter ﬂask at 20°C. [a] Calculate the percent of the solid that will have decomposed into NH;
and H25 at equilibrium. {b} Calculate the number of moles of ammonia that would have to be
added to the ﬂash to reduce the decomposition of the solid to I%.
[c] Having reached the state of equilibrium described in {b}, explain whether the addition of more NH4HStfs] would increase, decrease, or leave
unchanged the ammonia pressure. 2]. [30 min] Four and fourten tbs grams of CC:2 are introduced into a liiter
ﬂask containing excess solid carbon, at LDDIITC, so that the equilibrium C0; + Cls‘l = 2CD is reached. The gas density at equilibrium corresponds to an average mole
cular weight oi‘EIIS. Ila] Calculate the equilibrium pressure and the valtte el' K . {is} El, now, an additional amount of He tinertt is lntroduceii until the total
pressure is doubled, the equilibrium amount oi‘ CD will be lincreased, de
creased, unchanged, insufﬁcient information to tell}. II‘, instead. the volume
ol the ﬂask were doubled, with He introduced to maintain the same total
pressure, the equilibrium amount of CD would {increase decrease, be
unchanged, insufﬁcient data to tell). [c] If in is] there were actually 1.2 g of Cis] present, how many moles of CCI2 would have to be introduced so that at equilibrium Bnly a trace of
carbon remained? id} Il'the K, for the equilibrium doubles with a Itl°C increase in temperau
ture, what is aill” for the reaction? 22. [12 min} Calculate it:p for the reaction
Sis] + ZCCrt‘g} = Sthg] + 2Ctsj At the temperature in questiott, 2 atm of CD are introduced into a vessel containing excess solid sulfur, and a ﬁnal equilibrium pressure of It}?! atm II
observed. Heterogenenut and [rearrangement gas sq'ttt'ilt'bn'turr 2D] 23. [24 mini [ﬁnal examination question} K, has the value lﬂ'° atm} and lit"1 atm3 at 25°C and 50°C. respectively, for the reaction
CuSGpEHZt'J = CuSO4 + 3Hlﬂig} {all What is the minimum number of moles of water vapor that must be
ttitrod uoed into a 2—1iter ﬂask at 25°C in order to completely convert llﬂl mole of CuSCI4 to the trihydrate. Show your calculations.
[b] Calculate as“ for the reaction. 21. “5 mini Ammonium chloride vaporizes according to the process: The vapor pressure of NHsCltgl is negligible. It is found that at 52D°K the
equilibrium dissociation pressure [3",] is iltlStl atm. In a second experiment,
lltt2 mole of NHyClt'si and £11.02 mole of NH; is introduced into a 42.2—liter
Iliislt maintained at 52ﬂ°K. Calculate the amount of each substance present at
equilibrium and the partial pressures oi" the two that are gaseous. IRS. til min] The standard free energy of formation of AgCI is —2ti Itt'iiI.'mole. and dH” is St] ltcal for the reaction AgCl —— Ag + {H31 1, both
values for 25°C. [a] Calculate ﬂﬁgggax for the above reaction. Lb} Calculate KI, and the dissociation pressure of chlorine over AgCl. [cl Obtain the value
ttl'rii[T]n KFIMT.
1b. 112 minillittalexaminalion question] iii,1 lorthe equilibrium H: + I1 = J ill is 2t] at 410°C. 25 g [[12 mole} of I2 solid are placed in a 25.‘l~liter ﬂask at
tilt”: ': the vapor pressure of solid iodine being [LI atm at this temperature.
How many moles oi" hydrogen gas should be introduced into the flask in
urder to convert all oFthe solid iodine present to Hl'il Were: The volume
21.? liter is chosen to facilitate calculation '. at dﬂ°C this is the molar volume
iIi'II pits at 1 atm] i't NH “12115 1. Lil The E g corresponds to (1.1 mole. [act :c = moles 503, then
it  t  moles $03,. :t,“2 = moles Oz, and the total moles is [L] + .v."2.
llu equilibrium constant is then r.~ot.~:..;.+it'1 P )“i
" tut — .11}— e.I'+ xiii 101 Hittitestanding physical chemistry Also, M3... = FRTi'P = 1.6 x 114332 x STEELE = 63.8 but from which x = lltlﬁl. K1, is then K =io.5iiiti.o25_si1'I Ls “1
F one uﬁ = 0'“ [In] Adding inert gas at constant volume does not shift the equilibrium so the molesofheliurn needed isjustill + 2 [i 
double the moles Present. If or 'us’ that 151 "13 Emeunt to 2. {a} [t is convenient to write K, in the form: 1
K, = um, N
P 2 ' _
FIND X "a! P {I
bet am, he the number of moles of N02 before dissociation: then "ND; = H — lingo:
__ o
"No — 5mm.
or
_ e
"a: — illNﬂl N = it + sizing“, it.  ,_n ‘ilii’ﬂaii._ usean“
at {1:30,}! x isﬂjnﬂm {12
Sinoe K, is so large, it should be safe to neglect I a in com arison with u '
and we can then write p mm or 1‘3 = Itl'4andot = H.546. As a second approximation we write it — a} as H.954 "‘35' This leads to at = ties x ttl" and o: = ones. m H + m} as Heterogeneous and homogeneous: go's equii't'brium 2'93 [h] On substituting ct = (13 into the above equations. em, = ﬂ'iuﬂm.
and so on, and the mole fractions become MmI = 0.?! 1.15 = I161,
NM, = il.3i‘l.15 = [L215 and N0, = 13.13. The average molecular weight of
the mixture is then MI, = [let x 46 + {1.26 x 3H + [113 x 32 = 40. to] Mi" at i‘llt}°K is —.93 x ".I'ﬂﬂ x 2.3 log lﬂ'i' = —r5.9I keel; hence
Mi” = — 15.9 + Tﬂﬂ x i—4tl.'i].il,l]fltl = —44.4 koal. Then at 25°C. dG” = 44.4 — 293 x [4ﬂ.'i]i‘l,ﬂi}ﬂ = —32.4 litetal+ til]: It is eonuenient here to use the vanit Hoff equation in differential
form: ,Hn KPMT z nﬁﬂjnl“ = —44,4t]ﬂ,t'l.93 x tee2 = 4.43445 Hinee iii in KP is the same as dei'IKP, the above is the desired temperature
onetiieicnt. or. in it} per degree: —4.ti'iii.
.t. {a} K, = ME.iﬂtssuming I mole ot‘phosgene initially, new. =
"ml, N '
I — or, I'l'co = rim2 = a, and N =1 + tr. Then 3 x104 = [trifil — all] x
Ilt'il + at] and, sineeet will be small, 3 x ill—9 2 2a: orn = 6.3 x I'll—5.
ihi £621.31 = — 1.98 x 313 x 2.3 log 8 x lit—'3 = 13.? lteal. dH“ =
1.1.? + 3?} x EililJltltl = 14.3ltcal.
tei If a is to be [LED], then X, = 2 it it)!"i and, from the van‘t Hoff
equation. logtti x til—“3'2 x lit—“i .—— —[24,iitllili2_3 x LQtitiili'Bil — UT] or
[LISTS — tiTi = 4.43 ii iii" or T = 448°}; 4. iaJ There must have been 500 — 300 or Hillier: [STPi of CF“ and hence
also 200 cc {STP} of C01 sinoe this is Formed in equimolar amounts. There
were then 100 ee {STP} of CGFE. The mole fractions are then 0.4, [L4, and
{1.2 for these substances, respectively. and the partial pressures are then 4,
4. and 2 atm. KP is thus i4li4iifli" = 4 atm. {hi We use the equation: iiln KnidT= dﬂni’RTz, 1ii'here din Khan?" is
simply the fractional change in KF per degree and is equal to [Lil]. Then
fiti“ = ilﬂl x 1.93 at Ll??? = 32.1 Iteal. Also, d5” = ART In K, = HIE x I,2‘i'3 x 2.3 log 4 = —3.4'i' ltcal. Therefore d3” = iaH‘J — not“);
'i' = 27.9 ealideg. S. ta] The pressure of the combined products must be I — 0.54 or [136
Mm; sinoe PM, = 21%,. it follows that Pm = (124 and Pm = ELIE. K, is
theniiil.2¢i}2i::ll.l2,1,"I[{}.31'§i}I = truss sen. tbi The I.5% increase per degree means that ii in Kid?" 2 Gilli, and
therefore dH” [tilts =
RTa IllI oroH“ = 1.93 >< H32 493 where itﬂ is the initial moles as NOEL The total given ll‘eis 6. The 1.1 gol‘ NDEr corresponds to 9.191 mole an
of dissociation. then at equilibrium I am... it  elﬂlﬂl “N0 = n1,”2 —— [1.00511
N = 1.1.131“ + e52}
But
N = Fiﬁ“RT = 0.3 x tiEIﬂEE x 2713 = 01113::
hence
1.34 = I + no? or at = 9153
it1 n P
K : NO Er2_ = '3
F “Ema: N {9.6312(9341{13.31.19.331}.34}=11.345 atm. Repeating the calculation for 25°C. N Then l9§i3.9?iﬁ.345] = _. 9H” and an“ K [enolaiomm“EW etli’ZP Understanding physimi chemistry x {HHS = ﬁ.t§5ﬂcal.ﬁG° = —RTln K x23 xi..1].=21]5[}cal .uh J_= —t.93>c
to Let "m that M" — “5:95” — 11501913 = 9.5 e.u. rnoles at equilibrium is h N =
it it... for K... becomes + entail. and the expression is ——— ﬂ] ‘ tu — sinners + ennﬂl = fl — at] n + c2: = m be 0.2. then P ={ﬂ.l]lttl.311tt + animate; = 116 atm. d it‘ct denotes the degree 2 news. e = ass. and K. = set. L93 x 25i1i298 — 1.3293} = 1.961 x 1.98 x 2.3 x 298 x 293.95 =' 15.? ltcal. Heterogeneous and homogeneous gas equiiibritrm 295 '1. lly material balance. "In = 5‘ "m = ll:
t'l‘ll; = l — XE and "H: = 3 a x
"12 = 3 _ xii: Hiuce there are an equal number of moles of gaseous species on both sides of
the equation. K, = K.1 '. hence "I:=3—X x1 _ 12x}:
Hijﬁjfffgﬁ ' {3 — it]3 On solving. it = 3;? and K, = 33it3i213 = 4. 11. Let 3: denote the moles of 1:2114 present at equilibrium, then HE'2lI4 = I "EzHo = m _ x N = 1m + I
l“!1I2 = X “N1 = 80 P = EllS ﬁlm
l'lien
K = lIll‘Cl 35 = {1.05 " 2D—xlllﬂ+JZ and on solving the quadratic. x = 113.3 moles. The various mole fractions
are then N". = We.“d = lﬂ.3i11tl.3 = [1.993 {9.3%}. When. = 9.'t_."tlﬂ.3 =
11.933 {3.8%}. and N N. = [1.92 {9293]. i119” = out?” + Th3“ = 5.35 + 9013 x 32.3il.[t[lt1 = 34.4 ltcal.‘ hence at
31311”K. not = 31.4 — 3111] x 313,1 1.991] = 24.? Itcal. 9. [a] The ﬁnal total moles is T x Efﬂﬂﬂl' x 9119 = 13.244. of which
[tile is of methanol. so [1.134 is the moles of CO and H1. The moles of CD.
however. must be [1.1 — [1.t'1ﬁ = t'md. so the moles of H2 present is 11.144.
Then K, = tﬂﬂoltﬂltltllli‘tﬂiHl x {(1.144FETF = 11.033 atm' 1. [121} Had no reaction occurred. the moles oiCD would remain {1.1. and the
moles of H3 would be 13.144 plus twice the moles of methanol in {a}, or
[1.144 + {112 = 13.264. The total moles are then 13.364 and P —— 111.4 atm. II]. is} so“ = —RTln K. = 1.93 x 293 x 2.3log.1tl'5 =6.'l'ﬂlteal. {b} Since no change in the number of moles occurs on reaction, there will
still he s moles of gas. and F = 19.932 x 298 x oij = 29.3 atm. {e} Let x = moles of H20. then sinoe K... = K. in this ease. lt’l'5 =
[l + xe."I[3 — xltl  x]. As a ﬁrst approximation. then, x = s 5: 119—5 [no
Further approximations are needed}. Then the moles of C02. H 2. C0. and
H20 are 3. 2. 1. and 6 x 113—5. respectively. 20ft Umfersrrrmﬂ'ng pfu'xr'eef cheering id} as“ = so” + res”
El [00“(2', ago 2 380
3.3 x 10"”. I]! {3] den = or KP a [1.5; any I_ [h] From the ideal
x gas law. PM = R
Let n” denote the number of moles iftbe NSOT and s. the degree ofdisso ' '
. . eiat .
N E I“ + amu Th ion Then n",I = eis + ass it {—~ new one
_ I _. . = 3.30
3?}. x {mlﬂlrlﬂﬂﬂ = 153 heal. From nK, or log KP = —3l}f}rl.93 x 3l3 x 2.3 = —I[l.2l = 5.35 x 0.032 x 3l3 = 15!). 4 were completely undissociated
e av :04 L‘ H  2hr", mm = Earn“ a d
erage molecular weight M is I l i n H = at .
J ﬁNha: + QENMm {N denotes mole fraction] CI!" 21 ].
M=46__ '1
Maggi—i +—m'92ft‘l Ht} Since P =
Iﬁﬂy‘M, we ﬁnd P = fl + stHlSilr'Ql}. Finally, CﬂﬂStant is the equilibrium
in N
K = “10°_ 
B “an: F _ ‘ + elffinal” or 952 = 92H  eirise x 432 Thﬁn — all}? 3 __ ._ I 4 E . 15W“ = 2.13 am. 0r or 0.39. M is then 92nd? = 5.51 and P l
12. Since K ___ _Prh (“L Heteregerreons and ileumgenroes gas equfiihrr'imr 1]? lln introduction of more C1], the qualitative cﬁ'ect will be to shift the equi
Illunnn to the left. so let x denote the moles of FCl: formed as a result of
lint shiFt. Then lim~=2+x nc11=1+ﬂhh—X armllél—x N=ﬁ+n;:h—.t: where elm denotes the rnoles ofiIII2 added. Since the new equilibrium volume
is 2“ liters, with P and Teonstant, it follows that the addition of Clz doubled
IllL‘ number of moles present. so N = 12 and therefore n'ctl — x = 45. We
run nonr write IllIllﬂlﬁ
K’_l_ l2+irl E from which 3: = i. The moles of Cl2 added were then net: = ti + if}: = Ill’3.
I]. PEH DH 2 —
K = ~ = — = see i 3
P ramps: 1 x em a m a little reﬂection at this point shortens the problem considerably. The qualitative eﬁeet of expanding the mixture will be to shift the equilibrium
to the left, but the shift cannot affect the CD and CHJOH pressure very much since there is not much H2 present. As a ﬁrst approximation, then,
consider that the expansion halves the CD and CHJOH pressures, and ﬁnd the HI pressure required by KP: I
_ __.._ __ p = _ of H2 0 I. as before. Had no shift in equilibrium occurred, PH:t would have been 0.05.
so evidently as a second approximation we should take the CG pressure as
0.525 and the CHJDH pressure as DIETS. On doing this, and recalculating
PHI. we get 0.096 aim. 14. K], is given by K = PHD]
F szPéf Under the conditions given, oxygen is in great exoess, and therefore at some
constant pressure P“ so that Kp = ltl","P°”1. On expanding the mixture. 203 Understanding physical" chemisrrj' the oxygen pressure drops to F014. and since KP does not change, the nevir value of 3011502 must be 2 x 11]". 15. The equilibrium pressure of NDCI is to be 131.1311 >< 111.112. so the pressure of N02 remaining is 121.99 x 121.112. We thus have KP =1 x iii3 = Lin; _ 1‘10” “3112 Final)“: _ was s. ecziwm or Pm = 11.1112 atm This pressure corres
I 1_ _ . ponds to I]. 1131211312 or 111151 mole:
111 addition 1111] x 11.11222 or 1131‘“L moles have reacted, so the total moles needed would be 11.0511. 16. Let nu be the initial moles of PC15; then Hm: = _ mjnﬂ
nPC'la = ’1112']: = “no
11.... = {1 + 0!]11‘0 Then =_ tenure at"?
" 1I—sins11+xin..‘1_s== orK =iﬂ.1}2x2fofl—ﬂlz}=ﬂ[l7[12 1"
P _. . . _ atm. at} t —
x 2.31ogfﬂ.1}21}2]= 4.1201231]. IS hen 1'98 x 523
Since it is small K "—“azP so d] " — '
. P_ . nK.dT—2dlnadT=[l.[l4. The
[1.04 = diHE’fRi"2 or 2315'” = 11134 >< 1.93 £55233 = 21,?011ca1. Also .113“ =11
[amen — 4.1212111523 = 33.1 em. " IT. Since KP = £34m. the equilibrium ammonia pressure must be 3 atm
and the amount of ammonia gas in the ﬂask would then be 3 x Sfﬂﬂiié
x 313 f 11.53. In addition. (1.2 mole of ammonia is needed to effect the
conversion to LiCiSNHS, so the total ammonia required is 11.73 mole. 18. lia} KP = P50213501. Since each gas is formed in equal amounts, their
partial pressures are each 11.45 atm. K I, is then 11.451 or 11.2113 atmz. [b] Lct P501 be the equilibrium $03 presﬁure.‘ then P902 = Pm + 116 so
cote 2 P3511152“. + as} or em. = 11.24, Pm: = 1134, and P... = 1.118 anti. 1%. Sinroe thelnumber of moles of gaseous species are the same on both
5: es o equation 1a]. KP = K... = nHlonmfnmnml, where n denotes number Heterogwreoars and Jinmogcnms gas equilibrium 1119 of moles. Then if x = um“. "use ‘—' 3‘ “as = I it“! = 131.1— x nml = 11.2 — I I he mole fraction of water is thus 1:11:13 = 11.1. where x = [1.113. K‘, is then
ntiniZ.'ro.nnic. i 2] = 11.111252. Neither reaction 1b] nor 1c} changes the total moles of gas present. which therefore remains at 11.3. According to the new equilibrium composition.
11.1111I mole of water are present. or an increase of [1.126. This must have come
about through reaction [b], so that 11.1215 mole of H2 were used up. and
H.112 — one or 0.131 are left. K5 is then 11119111111 = 9. Since reaction 1b} minus reaction {a} gives reaction 1:]. Kr —— libiKl51 = 1111+ l‘rom the van‘t Hoff equation, :1 1n 519,"ch = dH‘Jlr'RTl. where d In KﬁidT
In the fractional change in K, per degree and is given as 11.1211. Then dilli"? = 11.111 x 1.93 x {12311 = 1113 [res].
2.11. is] K, = 13‘NH_,‘JF'H2S = i132. since PNHJ = P12. Therefore P1 = 12.2 and l" = 121.442 atm. The moles of gas present are then it = 121.44? x 2.4.:‘111132
 293 = 12.121442 mole. The moles of NH; or of 1115 are then 1113223. and the
moles of solid remaining are 13.116 — 11.11223 = 1111322 mole1 or 32.2 151: de rmnposed.
{bi 1f the decomposition is to be kept to 1%. the moles of H15 present must be 1113131116. and PHJS = 11.012115. From KP. the pressure of HHS is then
11.115.111.1Clue or 3.33 atm. The moles of ammonia will then be 11.333. {Notice
1iai 2.4litersisjusttl.] molar volumeifPis I atm. so molesofgas = [1.] >< P.) 1c]: No change. Addition of more solid does not affect its thermodynamic activity.
21. [a] From the deﬁnition of average molecular weight. as = =14le + ast or arm: = arm =i the initial 4.4 g correspond to [1.1 mole of C01. and if .t denotes the moles
111C121 formed. 11.1 — 39.12 = moles of C02 remaining and 11.1 + x."2thc total
moles. Hence 3! = #111] + 1,121 or x = 13.121th and the total moles are [1.133.
'1'hc total pressure is thcn P = 11.133 x 11.033 x 1,223.11 or F = 13.9 aim,
:ititl Pm: = Pm = 15.95 atm and KP — 16.95121635 = 15.95 atm. 1b} Introducing inert gas at constant volume will not change the equi—
librium partial pressures and hcncc will not change the position of equilib
liltrl't. Doing so at constant total pressure. however. dilutes the miitture1 and
me equilibrium will shift in the direction of forming more CO. [c1 The moles of CD formed must be 131.2 since the 131.1 mole of carbon is
In be essentially used up. Pm is then (1.2 x 0.1132 x 1.33.11 = 213.9. and.
1mm the equilibrium constant. Fm: must be 120.913.16.95 = 62.9 atm. so there 2H} UH denture ding pt' I ysieat' chem fairy must be 62.9 x 110.032 )6 1.213 or 0.1302 mole of C01 present. The total. moles of C02 required will then be 0.1 + 11.602 or 0.202.
{d} From the van‘t Hoﬁ‘ equation, 3111'"
 = _._ _ 2
og 2 193 x 2_3[1,I‘1,233 I. I, 131 {II 13H” = 0.3 x 1.93 x 2.3 x L233 x 1.213310 = 22.31022]. 22. Let n he the traction of C0 reacted: then P301 = 21:12] and Pm =
211 — at}. The total pressure is 2“ — 1121': 1.03. whence e: = 0.971. Then
K, = 0.9111006}: = 2'10 arm". 23. {a} Sinoe KP = Film. the equilibrium H213] pressure at 25°C is 10'2
atm. The ﬂask must then contain 0.01 x 2.10.082 x 203 = 8.2 x 10" mole. The total amount of strata needed is then 3 x 0.01 + 3.2 x It)" = 3.013
>< 111‘2 males. to} From the van’t Hoﬂ‘ equation, 11H” I 111—4111” = — —
“g i 1.93 x 2.3 “3'3 23 — 112981
or dH” = 2 x 1.03 x 2.3 x 323 x 293.125 = 35.1keal. 24. Since K... = FNHJPHCI and, in the ﬁrst case. PM.3 2 P110, then KP =
{.1512}1 = 121.0253 = 11.25 x 10“atm1. The 0.02 mole o1" hit13l corresponds to
a pressure of 0.02 x 0.082 >< 5211342.? or 0.02 am. Then KP = PHCIEPHC] +
11.02} = 15.25 x 10" or PHC. —— 11.110159. and PM“ = 0.1131119. The moles of
HCI are then I 0.0169 x 42.71.10.032 >t 520 = 0.0"59 and similarly the moles of NH; are 11.113169. [Notice that the volume was
chosen so that the pressure and mole number are equal.} The moles of
NH4C1 remaining are then 0.1121]  0.0I69 = 11.11103]. 15. [a] d635,? K is just the negative of the standard free energy of formation.
or 215 kcal. [1:1] From Eq. 111—4} — 26,000 1 K =_ .._._
“g F' 2.3 x 1.93 x ass = —1'§'.t or KP = a x 1131'” arm“ 111 Heterogeneous and homogeneous gas equr'iihrr'um ' ' ' ' ' ' fK [sineeK = 11 ~ dissociation pressure of tila is JUST. the square e P J,
“rigid x 1111‘” aim. [c1 The desired derivative is {1.1121 diﬁGﬂHd‘Tor —.o.3”_.1R. i'lien = 13.5 caL‘deg AH“ _ .1113” _ 313,000 — 26,000
155” = — T — ' — ass The answer is then —13.5,I'R or —ﬁ.3. 26. As the last trace ol‘ solid 11 disappears. the [3 pressure will still be 11.1 atm and 11.4 mole of 111. and hence 0.4 atrn of H1 will be present. The
pressure of H2 required for equilibrium is then mar
PH2 x 0.1 2n = or Fl“1 = [1.03 atrn s gas to maintain the equilibrium. I‘ H ust he resent it
Then 1.1.1218 mole o i 111 F' 0'4 mole Dr H]. or 11.23 mole total. plus 0.2 mote for the reaction producing ...
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This note was uploaded on 01/27/2010 for the course CHEM 3321 taught by Professor Goux during the Fall '09 term at University of Texas at Dallas, Richardson.
 Fall '09
 Goux
 Physical chemistry, pH

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