Ch 14 #82 RATE = [NO] = [O2] = 9.3 * M/s = 4.7 * M/s 2 t t PART B,C)

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Presentation transcript:

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 rate increases 3.24 times

RATE = k [I-] [OCl-]/[OH-] 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!

Ch 14 #88-a ln[c]t = -kt + ln[c]i 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 * 10-4 M (at 30 min)

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

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