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Chem Hw #4 Answers - Chemistry 30.101 Homework#4 Chapter...

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Unformatted text preview: Chemistry 30.101 Homework #4 Chapter 12 {2.12 Assume that the water and zinc are thermally isolated from the rest of the world. Then, QWater + qzinc : 0 (Cs,watermwaterAtwater) + (Cs,zincmzincAtzinc) : O ‘ The At of the zinc is (if —- 20.0)°C and the At of the water is (tf — 100.0)°C. Substitute the given specific heat capacities,1 the masses, and the temperature changes into this equation (4.22 J (°C)‘l g’l) (200.0 g) (t; — 100.0°C) + (0.389 J (°C)‘1 g-l) (60.0 g) (1:f — 200°C): 0 Solving gives tf = 978°C. @ For the reversible cooling of a system consisting of 6.00 mol of H2 when the pressure is held constant at 2.00 atm and the system contracts from 100 L to 50.0 L a) _ fl _ . (2.00 atm)(100 L) _ RT _ (6.00 mol)(0.08206 L atm mol‘lK‘l) 50.0 L T2 _ (1007:) (406 K) _ 203 K T1 = 406 K b) w = «PAV = —(2.00 atm)(50.0 L — 100 L): +100 L atm = +1.01 ><104 J c) - AU : chAT : n(cP — R)AT = (6.00 mol)(29.3 — 8.3145 .1 K-1m01—1)(203 — 406 K) = -2.56 x 104 J d) 4': AU—w :~2.56 x 104 121.01 x 104 J = —3.57 x104 J. 2.20 For an ideal gas, AU 2 chAT = 0 if AT : 0 (isothermal). Because Pext : 0, w = — MAX/=0. Finallyq:AU—~w=O—0=0. a) . w = _ extAV = —2.00 atm(10.00 — 6.00 L): —8.00 L atm : ~811J AU=q+w=+500-811=—311 J b) AU 2 ——311 J because E is a function of state. However q : 0 and w:AU—q=—311J—0=—311J @ The reaction of interest can be constructed as three times the first reaction added to the second reaction. The enthalpy changes combine in the same way as the equations. Thus AH = AHz + 3AH1 : 461.05 + 3(520.9) = +2023.8 Id 69 In every case the applicable equation is AH° 2 AH; — 2 AH; products reactants ll Enthalpies of formation are always tabulated on a “per amount” basis. The AHf’s from Appendix D are given per mole of substance. Each must be multiplied by the number of moles appearing in the balanced equation. 3) AH§98 : 2 mol 33.18—kg— —1mol Ofi ~ 2 mol 90.25fl : —114.l4 kJ mol mol mol b) AH§98 : 2 mol —110.52—k—J — 1 mol —393.51-£J- -—1mol 0fl : +1724? kJ mol mol mol c) Omitting the units: A113,”, = 3(—241.82) + 2(33.18) — 7/2(0) — 2(—46.11): ~566.88 M. d) AH§98 = 1(0) + 1(—110.52) — 1(—241.82) — 1(0) = +131.30 1d The subscript “298” appears because the AHfo ’s of Appendix D are for formation reactions occurring at‘298.15 K. This subscript is often omitted when the context warrants it. w The combustion reaction is CGH10(1)+ 137 02(9) —> 6 002(9) + 5H20(l) AHggg = —3731.7 kJ Use Hess ’5 law —3731.7 M = 6(—393‘.‘51) + 5(—285.83) — 1(AH€(05H10)(1)) — Em) AH; (caning) : —5s.5 kJ mol"1 ...
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