| 5-76 |
note: "DMH" = dimethylhydrazine moles of N2 = (150 g DMH) (1 mol DMH / 60.12 g DMH) (3 mol N2 / 1 mol DMH) = 7.5 mol pressure of N2 = nRT / V = (7.5 mol)(0.08206 atm L mol-1 K-1)(300. K) / (250 L) = 0.74 atm moles of H2O = (150 g DMH) (1 mol DMH / 60.12 g DMH) (4 mol H2O / 1 mol DMH) = 10. mol pressure of H2O = nRT / V = (1.0 mol)(0.08206 atm L mol-1 K-1)(300. K) / (250 L) = 0.98 atm moles CO2 = (150 g DMH) (1 mol DMH / 60.12 g DMH) (2 mol CO2 / 1 mol DMH) = 5.0 mol pressure of CO2 = nRT / V = (5.0 mol)(0.08206 atm L mol-1 K-1)(300. K) / (250 L) = 0.49 atm total pressure = 0.74+0.98+0.49 atm = 2.2 atm |
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| 5-86 |
(rate1) (MM11/2)
= (rate2) (MM21/2) MMunknown = [(ratemethane) (MMmethane1/2) / (rateunknown)]2 = [(47.8 mL/min)(16.05 g/mol)1/2 / (24.0 mL/min)]2 = 63.7 g/mol |
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| 5-88 |
(rate1) (MM11/2)
= (rate2) (MM21/2) (ratechlorine) = (rateHe) (MMHe1/2) / (MMchlorine1/2) = (1.0 L/4.5 min) (4.00 g/mol)1/2 / (70.90 g/mol)1/2 = 0.528 L/min Therefore, it would take 1.0 L / (0.528 L/min) = 19 min for chlorine to effuse under the same conditions |
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| 6-38 |
(a) if 1 mol of methane releases -891 kJ, and
1.00 g of methane is 1.00 g / (16.05 g/mol) = 0.0623 mol then heat released is -891 kJ/mol * 0.0623 mol = -55.5 kJ (b) n = PV / RT = (740. torr) (1.00 x 103 L) / (62.364 L torr / K mol) (298 K) = 39.8 mol heat released is -891 kJ/mol * 39.8 mol = -3.55 x 104 kJ |
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| 6-52 |
assuming no heat loss from the calorimiter apparatus,
then the heat capacity of the calorimeter is the same as that for the water
added (also assuming the s for water and the solution is identical,
and the mass of the solution is the mass of the solid and the solvent), Cp = (75.0 g + 1.60 g) (4.18 J / oC.g) = 320. J / oC [delta]H = CP [delta]T = (320. J / oC) / (25.00 - 23.34 oC) = 532 J n = 1.60 g / (80.05 g/mol) = 0.0200 mol therefore, [delta]H = 520. J / 0.0200 mol = 2.66 x 104 J/mol = 26.6 kJ/mol |
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| 6-62 |
I have labelled the 4 equations 1 - 4 to avoid re-writing
them over and over. Eq. 1: reactant on left, can probably leave Eq. 2: reverse to get reactant on left, [delta]H = +191.2 kJ Eq. 3: leave (product in gaseous form on right) Eq. 4: leave (product now switched to liquid form) both 3 and 4 must be multiplied by 2. This gives: [delta]Hrxn = 177.4 + 191.2 + 2(-241.8) + 2(-43.8) kJ = -202.6 kJ |
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| 6-64 |
I have labelled the 4 equations 1 - 4 to avoid re-writing
them over and over. Eq. 1: ??? Eq. 2: reverse to get reactant on left, [delta]H = +2967.3 kJ Eq. 3: reverse to get reactant on left, multiply by 6, [delta]H = 6(+84.2 kJ) Eq. 4: leave (product on right), multiply by 10, [delta]H = 10(-285.7 kJ) What do we do with Eq. 1? It needs to be arranged to cancel out the unwanted part of equations 2-4, so let's add those equations together to see what we get: Eq. 2 + 3 + 4: P4O10 (s) + 6PCl5 (g) + 4 PCl3 (g) ---> P4 (s) + 6 Cl2 (g) + 10 Cl3PO (g) Therefore, to get rid of the substances we don't want, add Eq. 1 as is to get the correct reaction. [delta]Hrxn = (-1225.6) + 2967.3 + 6(84.2) + 10(-285.7) = -610.1 kJ |
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| 6-72 |
[delta]Hrxn = [SUM][delta]Hproducts
- [SUM][delta]Hreactants = 12(-242) + 9(0) + 4(-393.5) - [5 (-20) + 4(54)] = -4594 kJ |
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| 8-56 |
I find it easiest to break every bond in the reactants
and form every bond in the products total bond energy on reactant side (Dreactants) = 3(C-H) + (C-O) + (O-H) + (C[triple]O) = 3(413) + 358 + 467 + 1072 = 3136 kJ total energy on product side (Dproduct) = 3(C-H) + (C-C) + (C-O) + (C=O) + (O-H) = 3(413) + 347 + 358 + 745 + 467 = 3156 kJ Therefore, according to equation (p. 352), [delta]H = [SUM]Dreactants - [SUM]Dproduct = 3136-3156 = -20. kJ |
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Last updated, October 7, 2006
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