HW-Solns-Chap-7-3530 (1)

HW-Solns-Chap-7-3530 (1) - Chapter 7: Homework Solutions...

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Chapter 7: Homework Solutions 7.6 (1/2) N 2 (g) + (3/2) H 2 (g) NH 3 (g) r G o = -16.5 kJ P NH3 = 4.0 bar P N2 = 3.0 bar P H2 = 1.0 bar T = 298 K kJ kJ kJ J kJ bar bar bar K K mol J mol kJ P P P nRT G Q nRT G G H N NH o r o r r 4 . 14 1 . 2 5 . 16 2070 5 . 16 ) 1 ( ) 3 ( 4 ln ) 298 )( / 314 . 8 )( 1 ( 5 . 16 ln ln 2 / 3 2 / 1 2 / 3 2 / 1 2 2 3 Because r G < 0, under these conditionsm, the reaction is spontaneous towards products. 7.9 A + B = 2 C 4 2 1 10 4 . 3 ] ][ [ ] [ x B A C K (a) 2 A + 2 B = 4 C 9 2 4 2 1 2 2 4 2 10 2 . 1 ) 10 4 . 3 ( ] [ ] [ ] [ x x K B A C K (b) (1/2) A + (1/2) B = C 2 4 1 2 / 1 2 / 1 3 10 8 . 1 10 4 . 3 ] [ ] [ ] [ x x K B A C K 7.11 G o = -RTlnK 10 . 1 ) 400 )( / 314 . 8 ( / 10 67 . 3 ln 3 K K mol J mol J x RT G K o K = e +1.10 = 3.01 7.19 H o = +224 kJ/mol (indep. of T) G o = +33 kJ/mol at 1280 K We’ll use data at 1280 K to determine S o . o o o S T H G K mol kJ K mol kJ mol kJ T G H S o o o / 149 . 0 1280 / 33 / 224 K = 1 corresponds to G o = 0. Larger K’s correspond to negative G o ’s. We’ll find temperature where G o = 0 0 = H o - T S o K K mol kJ mol kJ S H T o o 1503 / 149 . 0 / 224
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7.23 In each case, calculate r S o and r H o from data in Appendix 1 and then calculate r G o from r G o  r H o T r S o (a) r S o (94.6 186.91 192.45) J K 1 mol 1  284.8 J K 1 mol 1 r H o ( 314.43 92.31 46.11) kJ mol 1  176.01 kJ mol 1 r G o  176.01 298 ( 0.2848) kJ mol 1
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This note was uploaded on 04/12/2011 for the course CHEM 3520 taught by Professor Staff during the Fall '08 term at North Texas.

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HW-Solns-Chap-7-3530 (1) - Chapter 7: Homework Solutions...

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