test1s - Chemistry 2000 Spring 2006 Test 1 Solutions...

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Chemistry 2000 Spring 2006 Test 1 Solutions Marginal notes indicate problems assigned during the term which draw on similar themes. 1. (a) [ S 2 O 2 - 8 ] doubles from experiment 1 to experiment 2 while [ I - ] is held Chapter 15 #11 constant, causing the rate to double. The reaction is therefore first- order with respect to [ S 2 O 2 - 8 ] . In experiments 2 and 3, [ S 2 O 2 - 8 ] is held constant while [ I - ] is doubled, causing a doubling of the rate. The reaction is therefore also first-order with respect to [ I - ] . The rate law is v = k [ S 2 O 2 - 8 ][ I - ] . (b) We can use the data from any of the experiments. From experiment 1, we have k = v [ S 2 O 2 - 8 ][ I - ] = 1 . 4 × 10 - 5 molL - 1 s - 1 ( 0 . 038mol / L )( 0 . 060mol / L ) = 6 . 1 × 10 - 3 Lmol - 1 s - 1 . (c) i. For an elementary reaction, the rate law would agree with the Chapter 15 #47 and additional problems on elementary vs complex reactions molecularity. In this case, we would have a third-order reaction with respect to [ I - ] . Since the order with respect to
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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test1s - Chemistry 2000 Spring 2006 Test 1 Solutions...

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