Answers to Taz review for prelim 1

Answers to Taz review for prelim 1 - PRELIM 1 REVIEW (SAT...

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Unformatted text preview: PRELIM 1 REVIEW (SAT FEB 23rd ANSWER KEY) Syud Ahmed Sma88@corneil.edu 1. 44g of propane is completely combusted in a homemade aluminum calorimeter. (3) Use the table of bond dissociation energies to find the enthalpy change of the reaction. Is the process exothermic or endothermic? (Table on last page) (b) Suppose there is no heat loss to the surroundings. Given that the mass of the aluminum calorimeter is 141.5 kg, and that the temperature increases from 285°C to 550°C, find the molar heat capacity of aluminum. IMO, = Irma! H H l l o u—c—a—-c,——H + 5‘0; ———> 33’1"" L5H?“ l H H H (0 so) (‘5 zcm) C lick a) was twice» gator; M k: :1 rial ’W’X 8XC—H I—f/BILO/ \wwtfiéxfi O Anus! W1 o—rH 366 U/W (Mal xzxc—c Bag [(37%] llel 18x 772:2.th “MD! XSX 0:0 9‘73 k—J/w-Ei -> (’R’m€)C9)GS-Z&.§>tc = lag-Lig- $5: AMOIMML‘PQ—TLLL‘A 3.1T W 2 033078" x [Ia/mil Wm“; 2. You are given a bottle of an unknown liuid. The label reads: 100mL Density = 0.7138 g/mL, molar mass = 74.0 g/mol Boiling point = 500°C A Hvap= 26.5kJ/mol Specific heat capacity of liquid = 2.20 Jg'lK'l Specific heat capacity of gas = 0.75 J g'lK"l It’s late at night and you decide to heat 0.415 kg of iron metal to 400°C in a container. Trying to cool the hot iron, you reach for the bottle of unknown liquid and pour it all into the container. (a) Calculate the heat absorbed to convert all the unknown liquid to vapor. (b) Given that the unknown liquid was initially at room temperature, 26°C, and also that the specific heat capacity of iron is 0.45 Jg'lK'l, calculate the final temperature of the iron metal. (0) Calculate the work done by the unknown substance. Is work being done on or by the system? (d) Calculate the AU for the unknown liquid, assuming an atmospheric pressure of latm (760 mmHg). (6) Would it have been smarter to use water instead to have cooled the iron in the container? Why? 1 o fiéqgml 7qa/‘Wtbl :— ” 7/“? ' AH“? *(“6 M‘U‘bwkw) ,ZS‘SLL3 — (é) al’unJcnbw‘n "P 1:54?" ’ O Aki— Zwknow = 2"“? + 25°” + :0 9o, 731m +75m-1 +3“? 4-?643 .. t: "(58.3 T isoil = 77.383 (Q-aoé‘fémxfd as): 3 2 ,ZS'S-EO :1- ‘2’? 18M 2 7’l338(o.75.J/a K)(1—F,go) K a”, : 9/53 (o-es'czwm-m) K am + 26"” = "QM +7/WW‘1”) % 1/ 156 (a «mfg LXTf-m)n+ 71. $86 (a 75547-90? '9) K 2 ~293z%r«83 1. 240mm 7‘}; = 443097.95” .7" 3M C (a) M: —Pa+AU =* —AV\P~T An=©~fi飣5 awn! 2 —(o .aeqsmt)Ce.31Lt 37ml (Jean; L) :2 - 37733" TKR' aJ? TM We”) (do AU; junkhovo‘fl +0”) J 7va++7/bd;/ + $3”) +oq =‘ 25110 + 376828 + 8’030.ZS ' 37C? 3 3. (a) Given that the A H°f of NH3(g), NO(g) and H200) are -46.1 kJ/mol, 90.2 kJ/mol, and -285.8 kJ/mol respectively. Balance the equation below and use the information given to find the enthalpy change that occurs in the reaction. Make sure that the stoichiometric coefficients of the balanced equation are integers. NH3 (9 + 02(g) 9 N0(g) + H200) (b) Using the information from (a) and the equations below, calculate the A Huf for aqueous nitric acid, HN03 (aq). 3N02(g) + H20 (1) -> 2 HN03(aq) + N0 (g) A H° = -138.4 k] 2N0(g) + 02(g) a 2N02(g) A H" = -114.0 kJ (at) Jinn-{3* 5731(3) —-> #Mocj) + 6HLO(4) (a) H = 1+ m:— —- a CY¢e9=WH¥ZH°3“KWHWU A fCP ) F flqgfimfl a 2 a(ao.t)+éC—zss’ta/) -’+CQL~l) : -nm.ebJ AHéb \ftafi—“k 2. Ca) Ca) 438'” No AH = Wkl— 3/;v6‘DL + below—a abroad:- y La) ) z (3) 501m b 3’ AH: k3- % N0(a) + 3/q01C37 '9 /2’sz(‘6) T” "M". Var’JZ—Ca) + gqoLca) -% 100(5)) * $4.} H30 (,1) AH: “Tit: H10 (4) —--> VzoL—r (rib—g) AH :39:ng Vzhca) 4' VLNLC6)I+577/OLC.3) flame-o; 4. Dissolution of the anti-cancer drug cis—platin, Pt(NH3)2C12, in water is described by the equation Pt(NH3)2C12 (39) 9 Pt(NH3)2C1+ (3C1) + (3113(1) The rate of this reaction increases by a factor of 15 on raising the temperature from 25°C to 50°C. (a) Find the value of the rate constant at 50°C, if the rate constant at 25°C is 4.75x10'4. (b) Find the Activation Energy of the reaction in kJ/moi. (c) If the reaction time at 25°C is 20.85, find the reaction time at 50°C. (a) Rambo/grail 5E vuci-t'm,¥ 04 K Y7, ' KL % -3;ng H 75%“) G we, Tm ,kv "‘00 = lg i532. : 7 G b C, (Y1— ) K i5” " __|E_3—'-—-—— L’ Y; 2— '” ,... '- zr. K! > 5 _ KL 7. 7.:3xto' T :25"+?—73)i< —E°°/°’T' l 27; K- (a) :4. M T256“ 752: r Ae”EO‘/PTL Ea i __ ,1— —> in EL): -—-—;l€' T1 T2, 1/ 5. You walk into a room and you find a balloon labeled “Property of Taz.” Taz walks into the room and hands you the balloon along with a sheet which says: This balloon is filled with a strange form of C02. The balloon was initially at 1.00 atm, but the pressure is changing as the C02 decomposes. C02 (g) 9 CO (e) + 1/2 02 (g) The following data was obtained for the gas in the (a) With the information given on the sheet, write a rate law for the decomposition of C02 (9- (b) Determine the order of the reaction with respect to C02 and write the value of the rate constant. (c) Determine the overall pressure inside the balloon after 3000s in terms of mmHg. (d) Determine the time at which 70% of C02 (g) has decomposed. (EL) gal-e (m3 :— If you have a graph then plot it: [C9,] vs 1'ima [co:1(10“ M) P P P P H >- r‘ u a: as m H is is a- O D 0.5 1 1.5 2 Tim- (II-II) In [C01] vs Time (ms) D 0.5 1 1.5 1 11m. (ms) w Q" 7 my 2 a . ,. ' .- 25.73390113 24.05 547035 25.17761313 MM gm; “Mua 193%.? Votmugxc&,ng:Aj—w @4w—m£ vhfmsaw‘ mt—nobm‘lngmt .,.> PL=%LT) 9° I'VE-:th ?[F1t2[mz_1t L46 fifiu;m°~m 3T V '57 Mshflh‘ “Ah” co] “8 CPHL [Pjfi [hf t'L :flk} FLT :_LL£_>IV\,__.;-(d‘ [Cola Epo] [375 A .pr': : ZQ\G‘=7-WLMH-6. 7’02. = v7, (2 gums) = Saarmbg Tel-M raw/54m a (24.67%— 733 + 34c.f)m.u.t4§. TOW»! P‘YW a HZGWL—j Cal) If) 70% {a CwsmaQ,—H.QM 130/. M 1479/” G? SGM£,EAjt:o.stA-aj a qutzo-3szjb Ew‘VJO I 42 B—ozla (“b—Io a; In [6023" ,_ (I // x/o"3)(f) ’46 1" funk—+342— [CO’JE Gwfiwfiw 6. The following are a set of elementary reactions for a reaction mechanism: 03 (g) 4—.——”'02 (g) + o k co (g) + o —3——+ €02 (g) (a) Write the overall reaction taking place and identify any intermediates. (b) Use steady state approximation to find the rate law of the reaction. (c) The overall reaction rate is found to be first order with respect to O3 and zero order with respect to CO. The rate law also has a rate constant kobs. Under what conditions does the rate law you found in (b) agree with the actual rate law? What is the value of kobs? (d) Which step is the slowest step in the reaction mechanism? CC) Adrenal? YaQSC {mo U K" O o. (n F? O u) LA 85 @ MAM 2 lcz. [10L] + [<3 [1‘93 Cakefih‘m 3 i? KzEOl] << m n it t [-053 kg? 912601 = was ...
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This note was uploaded on 03/02/2008 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell University (Engineering School).

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Answers to Taz review for prelim 1 - PRELIM 1 REVIEW (SAT...

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