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Unformatted text preview: CHEM 14B2 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 1023 JK1
Avogadro's constant, NA = 6.02 x 1023 mol1 Faraday's constant, F = 96,485 Cmol'1 Gas constant, R = 8.314 JKlmol'1 = 8.206 x 102 LatmK'lmol1 =
8.314 x 10'2 LbarK'lmol‘1 = 62.364 LTorrK'lmol'1 Speed of light, 0 = 3.0 x108 mis1 Specific heat capacity of water = 4.184 J°C'1g'1 Iog(X)= 2.303 n(x) 1 kcal = 4.18 kJ 1 A = 1 Cs1
_ 5
For water: AHfus = 6.01 kJmol1 AHvap = 40.7 kJmol‘1 Cp(vap) = 36.4 J~K‘1mol‘1 Cp(iq) = 75.3 JK‘1mol‘1 Cp(soid) = 36.0 J~K‘1mol'1 O°C=273.15K 1L=1dm3 1atm=101.325 kPa 11:3.14
E=hv C=AV AE=q+w q=nCAT
En
WzPXAV 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.m01l) 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 : nCmAT : nCm(Tf— T1)
0.511 mol Cu  24.44 J‘K'Lmol“  [(1083 + 273) — (20 + 273)] = 13300 J
0.268 mol Zn . 25.40 JK‘1mol'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) (11511155 : "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 gK'l 4C5 2 —1250000 J 46500 gK" C5 = —1250000 J
CS = 26.9 JK‘1g'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. 11
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 JK"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 — (30 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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