# 1441answertest3 - Answer Key 1441 Test 3 chapters 6,7 and 8...

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Answer Key 1441 Test 3 chapters 6,7 and 8 Silberberg 1. D 2. C ( by definition heat of formation is producing one mole of a compound from its elements in their standard state. Since you are producing one mole of the compound, you may have to use fractions for coefficients on the reactant side.) 3. ∆H rxn = H produces - H reactants = [ H 2 S(g) + 2H 2 O(l)] – [SO 2 (g) + 3H 2 (g)] = [ -4.8kcal + (2) -68.3 kcal] – [ -71kcal + (3) 0] = -70.4 kcal 4. 41.8 kcal ( see website help topics Hess’ Law) 5. 36kJ ( see website help topics Hess’ Law) 6. (32gCH 4 /1)(1 mole CH 4 / 16gCH 4 )( 4.34X 10 5 J/1 mole CH 4 ) = 8.6 X 10 5 J Heat is released since the H value was negative which means exothermic. Generally, answers are expressed without sign if the question is, what is the energy change or how much heat is liberated; however, if -8.6 X 10 5 J had been a choice, select that for your answer. 7. ∆E = q + w q is heat, if the volume changes, then the work is p∆V work and w = -P∆V. So the equation becomes: ∆E = q - P∆V = -92.2kJ - 40atm(-1.12L) = -92.2kJ + 44.8 L atm (conversion factor is 101J = 1 Latm) = -92.2kJ + (44.8Latm/1)(101J / 1 Latm) = -92.2kJ + 4524 J (1 kJ/1000J) = -87.7 kJ 8. ∆E = q – P∆V q = + 1.3 X 10 8 J because it is endothermic = 1.3 X 10 8 J – (1 atm) ( 4.5 X 10 6 L – 4.00 X 10 6 L) ∆ means final – initial = 1.3 X 10 8 J - .5 X 10 6 Latm(101.3J/1 Latm) = 1.3 X 10 8 J – 5.1 X 10 7 J = 8 X 10 7 J 9. Energy = ∆T ( heat capacity) = (2.25 o C)(11.3 kJ/ o C) = 25.4 kJ (25.4 kJ / 0.5269 g)(114.2 g C 8 H 18 / 1 mole C 8 H 18 ) = 5.5 X 10 3 kJ/mole Since heat is released during combustion, the sign would be negative - 5.5 X 10 3 kJ/mole.

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10. Mass of the solution : D = m/v m = D V = (1 g/ml)(100ml) = 100g q = (specific heat)(mass of solution)(∆T)
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## This note was uploaded on 10/13/2009 for the course CHEM 1441 taught by Professor Tanizaki during the Spring '08 term at UT Arlington.

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1441answertest3 - Answer Key 1441 Test 3 chapters 6,7 and 8...

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