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14b2m1k ex 3 hmk - CHEM 14B-2 YOUR NAME Instructor Dr...

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Unformatted text preview: CHEM 14B-2 YOUR NAME Instructor: Dr. Laurence Lavelle |D# ............................. .. WINTER 2006 CLASS TEST 1 (Total points = 40) (Total time = 40 mins) YOUR DISCUSSION SECTION .................................... .. YOUR TA ....................................... .. Write in pen. Show all your work. Check your units and significant figures. Check that all reactions are balanced. Think clearly. Good Luck Constants and Formulas Planck's constant, h = 6.63 x 10—34 J~s Boltzmann's constant, k = 1.38 x 10-23 J-K-1 Avogadro's constant, NA = 6.02 x 1023 mol-1 Faraday's constant, F = 96,485 C-mol'1 Gas constant, R = 8.314 J-K-l-mol'1 = 8.206 x 10-2 L-atm-K'l-mol-1 = 8.314 x 10'2 L-bar-K'l-mol‘1 = 62.364 L-Torr-K'l-mol'1 Speed of light, 0 = 3.0 x108 mis-1 Specific heat capacity of water = 4.184 J-°C'1-g'1 Iog(X)= 2.303 |n(x) 1 kcal = 4.18 kJ 1 A = 1 C-s-1 _ 5 For water: AHfus = 6.01 kJ-mol-1 AHvap = 40.7 kJ-mol‘1 Cp(vap) = 36.4 J~K‘1-mol‘1 Cp(|iq) = 75.3 J-K‘1-mol‘1 Cp(so|id) = 36.0 J~K‘1-mol'1 O°C=273.15K 1L=1dm3 1atm=101.325 kPa 11:3.14 E=hv C=AV AE=q+w q=nCAT En Wz-PXAV S=kBInW Pnock—— v ET 2 V PV = nRT w= IPdV=-nRTIn(V—2) AS: q?“ v1 1 T2 d T O 0 O O ASTMEI =nC.n(T_f) AG =AH -TAS AG =-RTInK T1 0.0592 AG:AG°+RTInQ AG°=-nFE° EceH=E°- n Iog(K) 9&1 —- _ m _d_[A1 _ [A] =-kdt In [A]_ kt+|n [A10 t1?_ k [A —kdt 1 1 1 _= _ ___.. dA=-kdt A=-kt A [Ab EA EA t‘3=-2k- k=Ae1R—T) In(k)=--R7+In(A) F93+(aq) + e' —> Fe2+(aq) E° = 0.77 V 12(3) + 2 e- —> 2 |'(aq) E° = 0.54 V V2+(aq) + 26' ——> V(s) E° = -1.18 V Sn2+(aq) + 26‘ -—> Sms) E° = -O.14 V 2 H+(aq) + 2 e' —> H2(g) E0 = V QUESTION SCORE 1 1. Welders have been hired by the United Nations to work on the International Space Station. Below is some pertinent data they took with them. Answer these questions to help the welders get home as soon as possible. M11W'ht Mlt' T. Substance 0 ecu ar ég Cm (J.K—l.m01-l) 60 mg emp 'm01 1 C, latm 63.55 24.44 1083.0 65.41 25.40 419.0 (a) The welders wish to form 50.0 g of high brass, which is 65.0% Cue) and 35.0% 2%). That is, a 50.0 9 sample of high brass contains 32.5 g (0.511 moles) of Cue) and 17.5 g (0.268 moles) of 2mg). The temperature of the welder’s workroom is a chilly 20.0 °C. What is the total heat required to raise the temperatures of enough Cu and Zn to their respective melting points? (6 pt) q : n-Cm-AT : n-Cm-(Tf— T1) 0.511 mol Cu - 24.44 J‘K'Lmol“ - [(1083 + 273) — (20 + 273)] = 13300 J 0.268 mol Zn . 25.40 J-K‘1-mol'1- [(419 + 273) ~ (20 + 273)] z 2720 J Total Hcat = 1.60 x 104] or 16.0 k] (b) The high brass is then cooled from 1000.0 °C to 70.0 °C using water. The water absorbs 1250.0 kJ of heat in order to cool the high brass. What is the specific heat capacity of high brass? (4 pt) (1151-1155 : "qwalel‘ or q = C5- mass - AT 50.0 g ~ C1 ~ (70.0 —1000.0)= -1250000 J or —1250000 J = 50.0 g - CS - (70.0 —1000.0) -46500 g-K'l 4C5 2 —1250000 J -46500 g-K" -C5 = —1250000 J CS = 26.9 J-K‘1-g'l 2. When 229 J of energy is supplied as heat at constant pressure to 3.0 mol Ang), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas. (10 pt) There are a few methods that can be used. I will share two of them here. q I n C AT Use this equation to solve for Cp directly, since q is given at constant pressure. _ _q_ - __22_9J__ _ K1. 1-1 Cr)‘ n'AT ‘ (3.0 molX2.55 K) ‘30-] “‘0 Now solve for CV. q n'AT Begin with the same formula as before: C = q is equal to AE at constant volume: _ AB C"_ n'AT Find AE, then plug into above equation. Because Change in energy is a state variable, it is the same whether the system undergoes a constant pressure or constant volume process. AB = q + w AB = 229 J — PAV AE : 229 J — nRAT AE : 229 J — (3.0 mol) - (8.314 J-K"-mol'1)-(2.55 K) AE : 229 J — 63.6021 J AE =165.3979 J Now plug this into the original formula. 165.3979 J _—_—— . -1, -1 CV — (3-0 mol)(2.55 K) — 22 J K mol Alternatively, you can use the relation of Cp = CV + R, first solving for either Cp or CV, then plugging in to get the other. 3. Use two different methods to determine the enthalpy for the following reaction using thermochemical equations and data supplied below: (10 pt) 3 F8203(s) + 00(9) —> 2 Feso4(s) + 002(9) (1) F9203(s) + 3 CO(g) ——> 2 Fe(s) + 3 C02(g) AH = -7.7 kJ (2) 3 FeO(s) + CO2(g) —+ Fe304(s) + CO(g) AH = 4.2 kJ (3) Fe(s) + C02(g) —> FeO<s> + CO(g) AH = -O.7 kJ AHOf (Fe203) = -824.2 kJ mol'1 AH°f (F9304) = -1104.2 kJ mol'1 AHof (CO) = -110.5 kJ mol'1 AHof (C02) = -393.6 kJ mol'1 1. Hess’s Law: ° Multiply rxn (1) and AH by 3 gives AH: - 23.1 k] ‘ Multiply rxn (2) and AH by 2 gives AH: 8.4 K] ‘ Multiply rxn (3) and AH by 6 gives AH: -4.2 kJ ' Sum the equations and simplify ' Sum the corrected enthalpies of rxns (1—3) gives Aern= -18.9 kJ 2. Standard enthalpy of formation: . Aern = Z AHproducts ' Z AHreactants ' Aern = [(2 x -1104.2) + (—393.6)] — [(3 x -824.2) + (-110.5)] . Aern = '18.9 kJ 4. (a) Which substance of each of the following pairs has the higher residual entropy? Circle your answer (or answers if they are equal). (3 pt) i. CH4 or iii. or 02 (b) A 30. L compressed gas cylinder develops a leak, slowly filling the lab with 165 mol of flammable but undetectable methane gas. If the entropy of the methane increases by 11.7 kJ/K, what is the volume of the lab? Assume the leak is slow enough that the temperature remains constant and the methane acts as an ideal gas. (7 pt) _1 g. L_ i AS— 1.7 K 1000 k] —11,700 K V2=(30L)'e K'mol 11,700 'K- J (165 mol)'(8.3l451 ) V2 = 150,000 L ...
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