1e03-tut3-a

1e03-tut3-a - CHEM 1E03 TUTORIAL #3 Solutions _ 1. Use...

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CHEM 1E03 TUTORIAL #3 Solutions _____________________________________________________________________ Page 1 of 4 1. Use enthalpies of formation (found in Appendix D of your textbook) to calculate the standard enthalpy change for the following reaction at 298K: 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(l) H rxn ° = Σ (# mol product)( H f ° (product)) – Σ (# mol reactant)( H f ° (reactant)) H rxn ° = 4 H f ° (CO 2 , g) + 6 H f ° (H 2 O, l) – 2 H f ° (C 2 H 6 , g) – 7 H f ° (O 2 , g) H rxn ° = [4(–393.5) + 6(–285.8) – 2(–82.6) – 7(0)] kJ = –3123.6 kJ 2. Coal and natural gas are both fuels used for home heating. Assuming that coal is pure graphite and that natural gas is pure methane, and that graphite and methane have the same cost per gram, decide which of these two fuels is the more economical for heating a home. (Assume that gaseous water is produced when methane burns, and that both combustion reactions are complete.) Use standard formation enthalpies from Appendix D of your textbook. The balanced equations are: graphite (coal) combustion: C(graphite) + O 2 (g) CO 2 (g) for which Hc omb ° = H f ° (CO 2 , g) = –393.5 kJ mol –1 C methane combustion: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) for which H comb ° = H o (CO 2 , g) + 2 H o (H 2 O, g)) H o (CH 4 , g) H comb ° = [–393.5 + 2(–241.8) – (–74.81)] kJ = – 802.3 kJ mol –1 CH 4 On a gram basis, H ° graphite reaction = (–393.5 kJ/mol) / (12.01 g/mol) = –32.76 kJ g –1 On a gram basis, H ° methane reaction = (–802.3 kJ/mol) / (16.04 g/mol) = –50.02 kJ g
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This document was uploaded on 03/29/2012.

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1e03-tut3-a - CHEM 1E03 TUTORIAL #3 Solutions _ 1. Use...

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