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# chapter15 - Homework Chapter 15 15.16(a Comparing...

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Homework Chapter 15 15.16 (a) Comparing experiment 1 to experiment 2, [H 2 PO 4 ] remains constant, [OH ] doubles, and the rate increases by a factor of four. The reaction is second order in OH . Comparing experiment 1 to experiment 3, [OH ] is constant, [H 2 PO 4 ] triples, and the rate triples. The reaction is first order in H 2 PO 4 . Rate = k [H 2 PO 4 ][OH ] 2 (b) [ ][ ] [ ][ ] min 10 2 . 4 00040 . 0 0030 . 0 min / 0020 . 0 2 6 2 2 4 2 × = = = M M OH PO H Rate k (c) [ ] [ ] ( ) [ ] M M M M OH k Rate PO H 0044 . 0 00033 . 0 min 10 2 . 4 min / 0020 . 0 2 1 6 2 4 2 = × = = 15.24 When three-fourths of the sample has decomposed, the fraction remaining is 1 / 4 ( ) s t t s 580 10 4 . 2 4 1 ln 1 3 = × =

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15.34 The plot of ln[phenyl acetate] versus time is linear, so the reaction is first order in phenyl acetate. k = –slope = –(–0.0207 min –1 ) = 0.0207 min –1 y = -0.0207x - 0.5618 -2.5 -2 -1.5 -1 -0.5 0 20 40 60 80 100 Time (min) 0 2 4 6 8 10 12 0 20 40 60 80 100 Time (min) 15.40 (a) The plot of 1/[C 4 H 6 ] versus time is linear, indicating that the reaction is second order in C 4 H 6 .
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chapter15 - Homework Chapter 15 15.16(a Comparing...

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