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a2 - Martin Robert Homework 2 Due 5:00 pm Inst Deb Walker...

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Martin, Robert – Homework 2 – Due: Jan 25 2007, 5:00 pm – Inst: Deb Walker 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Thermo Review 001 (part 1 of 1) 2 points Calculate the heat of formation of butane (C 4 H 10 ) using the following balanced chemi- cal equation and information. Write out the solution according to Hess’s law. C(s) + O 2 (g) -→ CO 2 (g) Δ H 0 f = - 393 . 8 kJ / mol H 2 (g) + 1 2 O 2 (g) -→ H 2 O( ) Δ H 0 f = - 282 . 7 kJ / mol 4 CO 2 (g) + 5 H 2 O -→ C 4 H 10 (g) + 13 2 O 2 (g) Δ H 0 c = 2877 . 6 kJ / mol Correct answer: - 111 . 1 kJ. Explanation: Δ H f of butane (C 4 H 10 ) = ? The balanced equation is 4 C(s) + 5 H 2 (g) C 4 H 10 (g) Reaction Δ H 0 f (kJ / mol) 4 C(s) + 4 O 2 (g) 4 CO 2 (g) 4( - 393 . 8) = - 1575 . 2 5 H 2 (g) + 5 2 O 2 (g) 5 H 2 O( ) 5( - 282 . 7) = - 1413 . 5 4 CO 2 (g) + 5 H 2 O C 4 H 10 (g) + 13 2 O 2 (g) 2877 . 6 4 C(s) + 5 H 2 C 4 H 10 (g) - 111 . 1 kJ Δ H f C 4 H 10 = ( - 111 . 1 kJ / mol)(1 mol) = - 111 . 1 kJ 002 (part 1 of 1) 2 points Would more heat be released when 1.00 gram of CH 4 (g) is burned in excess O 2 (g) to form CO 2 (g) and H 2 O( ) in a constant-pressure calorimeter or in a constant-volume calorime- ter? 1. It would be the same either way. 2. It could be either one, depending on the pressure. 3. More heat would be released in a constant volume calorimeter. 4. More heat would be released in a constant pressure calorimeter. correct Explanation: m CH 4 = 1.00 g A constant pressure calorimeter measures Δ H while a constant volume calorimeter mea- sures Δ E . These quantities are related by the equation: Δ E = Δ H - Δ n R T The balanced equation here is CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O( ) As 3 moles of gaseous reactants (1 of CH 4 and 2 of O 2 ) become 1 mol of gaseous product, Δ n must be - 2. Note that this reaction is EXOTHERMIC; i.e. , Δ H is negative. Δ E must be more positive than the Δ H (or the magnitude of Δ H is larger), and con- sequently the heat given o ff by the constant pressure calorimeter would be greater. 003 (part 1 of 1) 2 points A 1.00 g sample of n -hexane (C 6 H 14 ) under- goes complete combustion with excess O 2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64 C to 29.30 C. The heat capacity of the calorimeter is 4042 J/ C. What is Δ U for the combustion of n -C 6 H 14 ? One mole of n -C 6 H 14 is 86.1 g. The specific heat of water is 4.184 J / g · C. 1. - 7 . 40 × 10 4 kJ/mol 2. - 5 . 92 × 10 3 kJ/mol correct 3. - 1 . 15 × 10 4 kJ/mol

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Martin, Robert – Homework 2 – Due: Jan 25 2007, 5:00 pm – Inst: Deb Walker 2 4. - 4 . 52 × 10 3 kJ/mol 5. - 9 . 96 × 10 3 kJ/mol Explanation: m C 6 H 8 = 1.00 g m water = 1502 g SH = 4.184 J/g · C HC = 4042 J/ C Δ T = 29 . 30 C - 22 . 64 C = 6 . 66 C The increase in the water temperature is 29.30 C - 22.64 C = 6.66 C. The amount of heat responsible for this increase in tempera- ture for 1502 g of water is q = (6 . 66 C) 4 . 184 J g · C (1502 g) = 41854 J = 41 . 85 kJ The amount of heat responsible for the warm- ing of the calorimeter is q = (6 . 66 C)(4042 J / C) = 26920 J = 26 . 92 kJ The amount of heat released on the reaction is thus 41.85 kJ + 26.92 kJ = 68.77 kJ per g of n -hexane.
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a2 - Martin Robert Homework 2 Due 5:00 pm Inst Deb Walker...

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