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Ch 14 #82

Ch 14 #82. RATE = - [NO] = - [O 2 ] = 9.3 * 10 -5 M /s = 4.7 * 10 -5 M /s 2 t t 2. PART B,C) k = RATE / [NO] 2 [O 2 ] k = 4.7 * 10 -5 M/s = 0.8393 = 0.84 1/ M 2 s (0.040M) 2 (0.035).

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Ch 14 #82

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  1. Ch 14 #82 RATE = - [NO] = - [O2] = 9.3 * 10-5 M/s = 4.7 * 10-5 M/s 2 t t 2 PART B,C) k = RATE / [NO]2[O2] k = 4.7 * 10-5 M/s = 0.8393 = 0.84 1/ M2s (0.040M)2(0.035) The reaction is second order for NO, if you increase NO by 1.8 : 1.82 = 3.24, the rateincreases 3.24 times

  2. Ch 14 #83 Part a) RATE = k [I-] [OCl-]/[OH-] Part b) Tripling [I-] triples rate, I- is first order! Part c) OH- is inverse to rate so doubling it halves the rate!

  3. Ch 14 #88-a BEER’S LAW: A=abc Where a= asorbtivity, b = path length, A= absorbance, c = concentration. 1st ORDER INTEGRATED RATE LAW TO SOLVE FOR THE TIME ln[c]t = -kt + ln[c]i BEER’S LAW: A=abc solve for concentration Then use BEER’S LAW: A=abc Where a= asorbtivity (5.6 * 103/cm M), b = path length( 1.00 cm), A= absorbance (0.605), c = concentration. A=abc 0.605 = (5.6 * 103/cm M)(1.00 cm)(c) c = 1.08 * 10-4M (at 30 min)

  4. Ch 14 #88-a,b A=abc 0.605 = (5.6 * 103/cm M)(1.00 cm)(c) c = 1.08 * 10-4M (at 30 min = 1800s) a ln[c]t = -kt + ln[c]i ln[1.08 * 10-4M]30= -k(1800s) + ln[4.464 * 10-3]0 k = 4.91*10-4 s-1 b FIRST ORDER HALF LIFE = t1/2 = 0.693/k t1/2 = 0.693/k t1/2 = 0.693/4.91*10-4 s-1=1.41* 103 s (23.5 min) c

  5. Ch 14 #88-d A=abc 0.100 = (5.6 * 103/cm M)(1.00 cm)(c) c = 1.786 * 10-5M d.1 ln[c]t = -kt + ln[c]i ln[1.786 * 10-5M ]= -(4.91*10-4 s-1)t + ln[4.464 * 10-3]0 T = 3.666* 103 s = 61.1 min d.2

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