adamson equilibria

adamson equilibria - l95 ' - riiihrr'trm Herrrogtvreutrs...

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Unformatted text preview: l95 ' - riiihrr'trm Herrrogtvreutrs trad horritigtnerao gm tip d narnic calculations in: mm if dfi”. iii-i”, 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 ATIUNF-i AND Ct) NC EPTS l: uilihrium—Constant Expressions ' " ' ‘ urc. olving only ideal gases at a gtvcn tcntpcrat ltiriven the general rea etiort In v ill-1]: ad + hB + = mil-i 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 Edger-1m 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=aH--Tfi5 -4 ts. you must utilize the equation for the over-all 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 pal-ft 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 [trials-ntmno‘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 30-liter 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 gist-ll]? 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‘l-il'tenll'Sl-llft 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 Lil-00° and tit-G" at this temperature. C, calmlale Ill-H01 $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 one-tenth grams of NOBr is placed in an evacuated I-Iiter 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 Cll-lfi 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..iCl-3HHJts1= LiC1-HH3is} + 2NHaig} tll WT. How many moles ol' ammonia must be added at this temperature In il 5-1jtcr flask containing [1.1 mole of LiCl-HH3 in order to completer convert the solid to LiCl - Hill-ll? 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 iii-i: 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 {ii-i 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 5-liter flask containing {11 mole of LiCl +Nl-l3 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 + iii-i-iirs 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 I||1llll2illill 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 2.4-liter 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 four-ten tbs grams of CC:2 are introduced into a l-iiter 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‘E-IIS. 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 tempera-u ture, what is ail-l” 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. 2-1. “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 “1-2115 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 Hittite-standing 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._ use-an“ 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' = —r|5.9I 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 fit-i“ = 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 Ill-I 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 1:21-14 present at equilibrium, then HE'2l-I4 = I "EzHo = m _ x N = 1m + I l“!1-I2 = X “N1 = 80 P = Ell-S film l'lien K = lIll-‘Cl 3-5 = {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 = 3-1.4 — 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 3-80 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 _ ‘ + elf-final” or 9-52 = 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 arm-llél—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 Ill-L‘- number of moles present. so N = 12 and therefore n'ctl — x = 45. We run nonr write Ill-Illfllfi 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 szPé-f 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 iii-3 = 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—2dlna-dT=[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 LiCi-SNHS, 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 dill-i"? = 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 1-115 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 1|iai 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 1-93 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... = FNHJ-PHCI 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" hit-13l 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 15-5” = — 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 1-11. 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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adamson equilibria - l95 ' - riiihrr'trm Herrrogtvreutrs...

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