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Kerfi: afturkræf (e. reversible) vinna er mesta framkvæmanleg vinna.

1a. Kerfi: afturkræf (e. reversible) vinna er mesta framkvæmanleg vinna. óafturkræf vinna (e. irreversible) er minni en afturkræf vinna. þ.e. W rev ,(kerfi) > W irrev. (kerfi) Umhverfi: Afturkræf vinna er minnsta framkvæmanleg vinna.

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Kerfi: afturkræf (e. reversible) vinna er mesta framkvæmanleg vinna.

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  1. 1a Kerfi: afturkræf (e. reversible) vinna er mesta framkvæmanleg vinna. óafturkræf vinna (e. irreversible) er minni en afturkræf vinna. þ.e. Wrev,(kerfi) > Wirrev.(kerfi) Umhverfi: Afturkræf vinna er minnsta framkvæmanleg vinna. óafturkræf vinna er meiri en afturkræf vinna þ.e. Wrev.(umhv.) < Wirrev. (umhv.)

  2. 1 b-c) • Ag+ + Cl- AgCl(s); DH < 0 • ii – iii) • mole AgNO3 / Ag+ = (2g/(0.5L 169.9 g mole-1)) x 10-1L= 0.00235 mole • mole HCl / Cl- = 0.20 mole L-1 x 10-2 L = 0.0020 mole • 0.002 mole AgCl are formed = n • DH = DfH(AgCl(s)) –DfH(Ag+(aq)) – DfH(Cl-(aq)) • DH = (-127.0) – (+105.90) – (-167.2) = -65.7 kJ mole-1 • q = DH x n = (-65.7) x 0.002 = -0.131 kJ • Cp = heat capacity = -q/DT = 0.131 kJ / 0.254K = 0.516 kJ K-1

  3. 2a) Efnahvarf (e. reaction): Bruni á glókosa: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) - Óreiðuaukning Fasabreyting (e. phase change): uppgufun vatns: H2O(l) H2O(g) -Óreiðuaukning

  4. 2 b-c) (sbr. fyrirlestur 20.09.05) • System / kerfi = Hg; surrounding / umhverfi = water + vessel • Final temperature (Tx) determined: • 4.18 (80+20) (tx - 20) = 0.140 200 (100 – tx) => Tx = 25.02oC => Tx = 298.17 K • i) DSHg = 0.140 x 200 x ln(298.17 / 373.15) = - 6.28 J K-1 • ii) DSu = 4.18 x (80 + 20) ln(298.17 / 293.15) = +7.10 JK-1 • iii) DStotal = -6.28 +7.10 = +0.82 JK-1

  5. CH4(g) + 2O2(g) CO2(g) + 2H2O(g); P = 1 bar • DH = (-393.5) + 2(-241.8) –(-74.85) = -802 kJ mole-1 • DS = 213.6 +2x188.7 -186.2 – 2x205 = -5.2 JK-1mole-1 • DG = DH – TDS = b + a T; b<0, a>0 • DG negative for low T / DG positive for high T • The “switch over” temperature (i.e. T for DG = 0) is • T = -DH/DS = 802 / 5.2x10-3 = 3666 K

  6. (see lecture 06.10.05) Water = w; solute = B; Mwi = molecular weight of i; • gi = mass of i; Pw = vapor pressure of solution; P0w = vapor pressure of solvent • Pw = P0w xw = P0w (1-xB) = P0w (gw/Mww)/((gw/Mww)+(gB/MwB)) => • MwB = 10 / ((100x0.025/(0.0247x18) – (100/18)) = 147 g mole-1 • NB!: xB = 1 – xw = 1 – (Pw/P0w) = 1 - (0.0247/0.0250) = 0.012 • If 50 % (unknown) dissociation (AB = A + B) => • 50 % increase in number of particles => • xB = 1.5 * xB(a) = 1.5x0.012 = 0.018 • Pw = P0w(1 - xB) = 0.0250x (1 - 0.018) = 0.02455 • Mw = 10 / ((100x0.025/(0.02455x18) – (100/18)) = 98 g mole-1 • Hence molecular mass would be underestimated.

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