#Chem 161-2007 Homework 7th week

#Chem 161-2007 Homework 7th week - Chem 161-2007 Homework...

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Chem 161-2007 Homework 7 th Week Chapter 6: 25,27,33,37,39,41,43,45,49,51,55,59,63,65,67,69,71,73,75,79,88,90,97,98,101 Enthalpy and Enthalpy Changes 25. At high temperatures, water is decomposed to hydrogen and oxygen. H 2 O(g) → H 2 (g) + ½ O 2 (g) Decomposition of 67.3 g H 2 O at constant pressure requires that 902 kJ of heat be absorbed by the system. Is the reaction endothermic or exothermic? What is the value of q for the reaction, per mole of water? Is the value of q equal to ∆U or ∆H? Explain. mol = g/MW 67.3g/(18.02gmol -1 ) = 3.735 mol If heat is absorbed by the system the reaction is endothermic. 902 kJ/3.735 mol = 242 kJ/mol of water. At constant volume, ∆U = q v . At constant pressure, ∆H = q p Since this reaction is being done at constant pressure, then q = ∆H. 27. When 0.1500 mol of solid calcium oxide is mixed with 0.1500 mol of liquid water, solid calcium hydroxide is formed and 9.78 kJ of heat is released. Write a chemical equation for the reaction producing 1 mol of calcium hydroxide, including the physical states of all the substances and the value of ∆H. 0.1500CaO(s) + 0.1500H 2 O(l) 0.1500Ca(OH) 2 (s) ∆H = -9.78J 1CaO(s) + 1H 2 O(l) 1Ca(OH) 2 (s) ∆H = -65.2J Heats of Reaction and Reaction Stoichiometry 33. Calcium oxide (lime) reacts with carbon dioxide to form calcium carbonate (chalk). CaO(s) + CO 2 (g) → CaCO 3 (s) ∆H = -178.4 kJ How many kilojoules of heat are evolved in the reaction of 0.500 kg CaO(s) with an excess of carbon dioxide? ∆H of -178.4 kJ is for the reaction of 1 mol of CaO. MWcao = 56.08g/mol 500g/56.08gmol -1 = 8.916 mol CaO CaO(s) + CO 2 (g) → CaCO 3 (s) ∆H = -178.4kJ x 8.916 mol = -1590 kJ 8.916mol 37. How many liters of ethane, measured at 17 o C and 714 Torr, must be burned to give off 2.75 x 10 4 kJ of heat? 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O(l) ∆H = -3.12 x 10 3 kJ 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6H 2 O(l) ∆H = -2.75 x 10 4 kJ ?L 273+17K 1
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714Torr It takes 2 moles of ethane to give of 3.12 x 10 3 kJ. Therefore, it will take (((2.75x10 4 )/(3.12x10 3 )) x 2 =) 17.6 mol ethane to give off 2.75 x 10 4 kJ of heat. PV = nRT V = nRT/P V = 17.6 mol x 0.08205Latmdeg -1 mol -1 x (273+17K)/(714 Torrx(1 mm/Torr)x(1 atm/760 mm) = 446L 39. How many grams of CaO(s) must react with an excess of water to liberate the same quantity of heat as does the combustion of 24.4 L of CH 4 (g), measured at 24.7 o C and 753 Torr? CaO(s) + H 2 O(l) → Ca(OH) 2 (s) ∆H = -65.2 kJ CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆H = -890.3 kJ The given delta H’s are for one mole of CaO and one mole of CH 4 . Convert the CH
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This homework help was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

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#Chem 161-2007 Homework 7th week - Chem 161-2007 Homework...

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