Midterm 1 Key

Midterm 1 Key - CHEM 14B YOUR NAME Instructor: Dr. Laurence...

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Unformatted text preview: CHEM 14B YOUR NAME Instructor: Dr. Laurence Lavelle ID# ............................. .. WINTER 2002 IST MIDTERM (Total number of pages = 7) (Total points = 50) (Total time = 50 mins) YOUR DISCUSSION SECTION .................................... .. YOUR TA IS: Andrew Clark Carlos Hernandez Franklin Ow Write in pen. Show all your work. Check your units and significant figures. Think clearly. - Good Luck Constants and Formulas Planck constant, h = 6.63 x 10‘34 J l- s Avogadro constant, NA = 6.02 x 1023 mol‘1 Gas constant, R = 8.314 J.K'1.mol'1 = 8.206 x 10‘2 L.atm.K'1.mol‘1 = 62.364 L.Torr.K“.moI‘1 Speed of light, 0 = 3.0 x 108 m.s‘1 ‘ Water, specific heat capacity = 4.18 J.°C.g‘1 0°C=273.15K 1L=1dm3 1atm=101.325 kPa 7t=3.14 E=hv AE=q+w q=nCAT W=-PxAV S=kBInW PV=nRT w=-v1lV2PdV = -nFITln%1a AS = 9&3?! 2 .dflflEM IR 0 o AST1-.->T2 =T1JT T = nCln T1 AG =AH 'TASo AG°=-_FITInK AG=AG°+RTInQ 0.0592 AG°=-nFE° ECELL= Eo- Qt. When 25.0 mL of 0.700 M NaOH(aq) was mixed in a calorimeter with 25.0 mL of 0.700M HCl(aq), both initially at 20.00 °C, the temperature increased to 31 .34°C. Calculate the enthalpy of neutralization in kJ per mole of HCl. ‘ (10pt) dffl: _% t - 50. 07(¢./¢? Ty‘zaoo 0C : - 23%. (9/ :7' (/M/ : '2'” “N3: 2’1-37/KT film/(AH [wnflH/CZ). WK 0. 025,5 = 0. (om—{7,1,6 in the above calculation two assumptions were made to solve the problem. What are they ? 02A. Calculate the reaction enthalpy for the reduction of hydrazine to ammonia N2H4(i) + H2(g) —) NH3(3) from the following data: (5pt) N2(g) + 2 H2(g) 9 N2H4(l) AH° = 50.63 kJ N2(g) + 3H2(g) e 2NH3(9) AH°=-92.22 kJ (W) Wet/mm; AC; HM H/Z/fi 91% W (M) M W a 4/4/32 *yz/(a/ M”: mm (M) MHZM -v—> Z/V/z/JW d#‘=”7z.zz%T r xv: #445) + 54%) ——s z #436,] 46/50; —/¢z,554€7- (w B. For a certain reaction at constant pressure, AE = ~95 kJ, and 56 kJ of expansion work is done by the system. What is AH for this process ? ' (5pt) (M) <\E : 4/ +h/ (My 454;: 4/,4 «as/Mr Qfity/f: ‘“15’+5’0/ ‘5“37/43— ea 03. initially a sample of ideal gas at 412 K occupies 12.62 L at 0.6789 atm. The gas is allowed to expand to 19.44 L via two pathways. Calculate Aston Assys, ASSlJrr for both pathways. A. isothermal, reversible expansion / ,’pt) éTsfl 55:4 E f: 7 +A/ ' W = “14/ __ W . —— w -(.. 14K ,4 Kg) _ A V; d .- *" : ~"‘ : r *- 14/? “‘ 07/17 5 7' 7" r V/ 17 1 47° W 11.4th rm 7‘ M: AC. —_— 0.675%;5 flab/2%,) KT (9-0320! 4%.k'.;.¢f'/ (¢/L/f (M7 h: 045’} {4% (W As: hKflv = (azrfzxga/gz U./£/m2/)’Z' QM} Air? 0.705 31k" Q4A. Using AH° = 1.96 kJ.mol'1 and 138° = -109.58 J.K'1.mol'1 for the production of formaldehyde. Determine the spontaneity of the reaction under standard conditions at 25 °C. (4m) <54~°= «$7415 — - -I —.— MM 1.“,{1— gar/((404.57 I./T.M) G; > : 315/44“: TM" < : 3.54:3“.M’) B. The temperature dependence of the equilibrium constant of the reaction, N2(9) + 02(9) = 2N0(Q) which makes an important contribution to atmospheric nitrogen oxides, can be expressed as In K = '217190 K + 2.5 What is the standard enthalpy of the forward reaction ? (6pt) ’ g4-°=4//'~Tar'=’KT/M’ (1%] Akgfiémv K Mr) M,» = *2/Wok -.- 1% 7 ~/ (M) 1 AH = (2,900 k)[5’ T/FJ-frl-M ': /504‘/3-5 377ter (Ma 2 Axoéélxxo‘rwrl «Ingram/Jmfl 6 05. Calculate the standard free energy of each of the following reactions: A- |2(g) = 2 l(g) K=6.8 at1200K (5m) (W <4“ = 'KTAK 7 hi) = ‘(50/gzy./(T’M){/€M /r)/€v {)7 ((77427 :’/fl (-2 44.7'5311‘44 (27¢) : ’M k J Mch 22 [W ‘ ‘B. AggCrO4(s) = 2Ag+(aq) + CrO42'(aq) K: 1.1 x10“2 at 298K (5pt) (M) a4” : «7 AK fin (/Wt) 7— ‘(3.J/¢ 1/? .’ ME/Xz fl/r)/4 fl/X/a : ~+ (521/.7 (PM / (2M 2: éX/kf.7w’[/ 47L4~0¢ ...
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This test prep was uploaded on 04/03/2008 for the course CHEM 14B taught by Professor Lavelle during the Winter '01 term at UCLA.

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Midterm 1 Key - CHEM 14B YOUR NAME Instructor: Dr. Laurence...

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