Sol3 - Name: Problem Set 3 Undergrad. TA: Lab Day: 1. . 1e...

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Unformatted text preview: Name: Problem Set 3 Undergrad. TA: Lab Day: 1. . 1e steam reforming of methane is an important industrial process: CH4(g) + H20(g) <2 C0(g) + 31-12(g) K = 0.260 at 1200K You mix 1.00 mole of CH4 with 1.00 mole of H20 in a 1.00L flask at 1200K, and allow the reaction to come to equilibrium. What are the equilibrium concentrations of each component, in molesz’L? What is the percent conversion of CH4 to CO? cum“ HLO fl Cocw 34cm \niHM) [.00 1.00 o 0 Ch (m) - ~2< a: 22min) 2.50% zoo-x 2c 3K (7.06%) (Mo-2:) ((04%): 24997.0; 27 2.. 4‘1. 3 0.260 _ awry-z. Lao-A 27 4 24": “int/0"L’- 9.91:» A’ /0-sz‘+X-/=0 = — I 1V I +9/m.z)m : —/ 34.94: 5% r a.“ 30.9 20:51 Z014 : _-;. 5' ,2, = , Qiéj [GHQ [Hep] 0.73M [co] 0 7M EHJ ragga”? 31;? Starting with the equilibrium mixture in part a), if the following changes are made, will the equilibfiu shift to the left, to the right, or remain unchanged? In each case, justify your answer in one sentence. i. The total volume, V is increased. SETS-km will 4'0 increase flumbcr“ 0": 3a.“; I'm \ecule-IT . . . o The temperature 13 increased. For this reaction, AH = 206.1 kJ Q'W’ Since regatta" t'S WAD-’mevc, Rr‘ma‘l’w‘On U¥Pfdclvoé' W!” ;“C‘0/\£‘Ume” (game beagl- iii. Some platinum catalyst is ad 5-}; Adm: “0 ended Equi’tLr'me. NO iv. Some argon is added, causing the total pressure to double. “0 eméi' ' AJA'I'ROA a")? Af‘ A“; m, eggéflb,‘ Concédfiulwkj and [nrjldfa M4 appear” in equdfbnium expmfiim. ii. Problem Set 3 .-'1 You evacuate a 1.00 L flask, and then add 1.00 mole of C0(g). How many moles of H2 would you need to add to the flask so that at equilibrium, exactly ?0.0% of the C0(g) has been converted to CH4(g)? C0£33+3H2M3 :3 044(9) +HLO<9) K=3,9$ Ier (m) Loo X *3 0 030 0.70 I one 0.10 3 ._.._ 3‘85 (0.30)(x-2. Io) :- (X—‘Z.IO)3= 1. o‘qzq (0.30) (3.815) x—ZJo = Warm/2.9 = 0.750 X: 2,85 Nashua 2.215 male“? “1.- 2. '. 0 an empty 1.00 L container at 600 K, you add 1.00 gram of C(s). The container is then pressurized " ith 1.00 atmosphere of C02 (g) and 2.00 atmospheres of C12. Calculate the final equilibrium pressures [fall gaseous species in the container at 600 K. Justify any assumptions that you make. PU : (1 I97 C(s) + C02(g) + 2012 (g) 4: 200012 (g) KP = 4.32 x 104 at 600 K. ni‘l' Pfa-lm) Lao 2.00 O a J “5 (film) -x -2,“ 2x 400?: 932an was .____.___———-—-—-—---‘—"‘_' ate—0,972. 4124:”. (“SW |_oa-:- 200-2: 2* checkflxmfim-mhxl Kf= _£Z2<_J_——Z -.= 443mm" 2—0.97axmw2fié‘m (zoo 49/20”) Z I .-.= 2.8% ‘2. x 0K. .. __i__._...= ——-—-— . .BZK/OV' K fljgo new/7‘0 3nme ' _ 4 gm» A Cr?) W05 1.00 J : 14.3mm? n“ -— t. u. (3.00%)]3: I ~ 2' + amt) (1.00 -x Hamw i , =mam cm , mad/Ms 0"» Problem Set 3 3. 3': Jl'iC acid and glycerin form a complex: B(OH)3(aq) + glycerin (aq) c> B(OH)3 - glycerin (aq) 1e equilibrium constant K is 0.90 at room temperature. If the concentration of boric acid is initially 0.10 M, .; 1w much glycerin should be added so that in a total volume of 1.00 L, 60% of the boric acid is in the form of _eomplcx? + fin; Bforila‘jlj inii’fl“) 0.10 X 0 Ch 01) -—o.oc 90.06 <3.on 24mm) 0.0% x—wm 0.06. act a I K: 6?, 90 = 3L => x~0,0£= (0.?0)(aov) (act-IXx-ooé) A = A 73 ARM/via 0.0/0" [‘73 070/49 aFjécc/{n ' f 4. I 39?}; of PC15 was introduced into an evacuated 1.00 L flask, and then heated to 600.0 K. The equilibrium I: instant Kp for the following reaction at 600.0 K is 11.5: PCIS (g) 43’ P013 (9 + C12 (9 alculate the partial pressure of PC15 and the total pressure of gas in the fia at equilibrium. 4 inf-lino PrQSSUre oi: E’CIS = P 2 fléj—t Egg-g7;— kaogzocfizmé /.00C : 0. L/ ’7’?4¥m. I. * pals fi Pg'3+%7' XL: //5{ 0,9519%} ln'rl I' aim) (Wm 2. __ .. x A ){+//.5'x S/W—O Chi] X = 4/5 i V //.SL:‘/(S'.leYl 24.. 0.49% " X _ 0 9332 Z _/’ " ’ I )2 z “ha—J PPmS: 0:9?‘1-05‘33: 0.0145 calm 9"" "" 9mg: pal: 0.43M». _PTDT: You are a member of a research team of chemists discussing plans to operate an ammonia processing plant: Nata} + 3H2th <2“: ZNfiatg) a} The plant operates at close to 700K, at which K? = 1.00 x 104 for the reaction. The synthesis employs the stoiehiometrie 1:3 ratio ofN31H3. At equilibrium, the partial pressure of NH; is 50.0 :tttn. Calculate the partial pressures ot‘cach reactant, and the initial and equilibrium PM. Hawaii-aw” w PM...) x 3x 0 at; (ah-1) '23 ~73 50 - arr-7 S X 2—3. S 0 t L [(10 z 3%. : 3 : $00310"? ’PH = ‘4,“ Pi): Pk: (tr-2.578105) 3' — - 3 : 50 r. 2.50/07 PM = 50 film (x 33(3): 7?) - 3 lint/o ,1 - * M (>«-25){3)3{.~c—2.t)3 = 2'50"" 3 Pan - “t x - 21% at ()<_25)Li : 2%{31§Ol101}’9.2‘m P #23: crux/a: 231.0 “lbs? -— Lit-80 —t‘)‘t aJm >13:- S'fih . , , . . bi One mem er 0 0 team suggests that Since the partial pressure of Hz is cubed m the reaction quotient, the ptant could produce the same amount of NH; if the reactants were in a 1:6 ratio of Nle:. and could do so at a lewer pressure. thereby cutting operating costs. Calculate the partial pressure of each reactant and the initial and final PM under these conditions. assuming an unchanged equilibrium partial pressure of 50.0 atm for NH}. Is the team member‘s argument valid? “3+ lht‘i’P(a'l'm) 3! £8 0 Cl‘jialml --25' ~75 50 squint") ms ant—7S So L KP: SO atom/0"" (meow—75? Here we we lo’ haven» fluorite Ega- Use Mod a: successive. alarms/1s. ...
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Sol3 - Name: Problem Set 3 Undergrad. TA: Lab Day: 1. . 1e...

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