Chap 6 Sect 6

# Chap 6 Sect 6 - Chap 6 Sect 6 - notes.doc 6.6 Standard...

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Chap 6 Sect 6 - notes.doc Chap 6 Sect 6 - notes.doc 11/09 Page 1 6.6 Standard Heats of Reaction Heats of Reaction calculated using Hf ( )values . The H of a reaction can be predicted by using Hf values. Eqn: " H rxn = n p " H f (products) # n r " H f (reactants) # H f values - small table on p.253 -Silberberg & Appendix B in the back . Example: 4 NH 3 (g) + 5 O 2 (g) -----> 4 NO(g) + 6 H 2 O(g) 4 (-45.9) 5 (0) 4 (+90.3) 6 (–241.8) 183 .6 +361 .2 1450 .8 –183 .6 –1089 .6 H rxn = –1089 .6 (–183 .6) = –1089 .6 + 183 .6= Given these values: ( Another source) NO(g) = 90 kJ NO 2 (g) = 34 kJ HNO 3 (aq) = –207 kJ H 2 O(l) = –286 kJ H 2 O(g) = –242 kJ Do this: 3 NO 2 (g) + H 2 O(l) -----> 2 HNO 3 (aq) + NO(g)

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Chap 6 Sect 6 - notes.doc Chap 6 Sect 6 - notes.doc 11/09 Page 2 Book Table Why does this work? There is an assumption that the pathway for the reaction is that the compounds break down to their elements (the – H f part) and then they re-form into the products (the + H f part). This is not the way it happens but because it is a state function, i.e. doesn’t depend on pathway, it doesn’t matter.
Chap 6 Sect 6 - notes.doc

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## This note was uploaded on 04/06/2010 for the course CHEM 1A taught by Professor Okamura during the Fall '08 term at UC Riverside.

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Chap 6 Sect 6 - Chap 6 Sect 6 - notes.doc 6.6 Standard...

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