100218 1032 recitation

100218 1032 recitation - nd order d plot data to...

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13.30 .Half life problem t 1/2 of C14 is 5730 years, not f([A] 0 ) a. how long for 25% of C14 to decay?? quick answer, ½ of t 1/2 = 2865 so incorrect longer answer, t 1/2 = 0.693/k, so k = 0.693/t = 0.693/5730 yr = 1.209 x 10-4 y -1 so ln (A/A0) = - kt - ln (0.75)/(1.209 x 10 -4 ) = t = 2379 years b. if C14 initially 1.5 mmole, how much after 2255 years? ln (A/A0) = - kt, or ln A = ln A0 – kt ln A = ln (1.5) – (0.0001209 y -1 ) x 2255 = 0.1328 , so A = 1.142 mmole 13.53 half life from time, conc data for CH3CN CH3NC t(h) {CH3CN, M 0 1.000 5.0 0.794 10 0.631 15 0.501 20 0.398 25 0.316 a. order, value of k? b. value of t1/2 at initial [ ] c. how long for 90% to convert? t1/2 = 15 hr, by inspection, a. [ ] not constant over constant time intervals, so not 0 order b. if half life not constant over constant time interval, not 1 st order c. if t1/2 increasing as [ ] drops, 2
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Unformatted text preview: nd order? d. plot data to check/verify independent of [ ]0, so first order 15 h = 0.693/k, so k = 0.0462 h-1 ln (A/A0) = - kt, so t = - (ln (A/A0))/k = ln(0.1)/(0.0462 h-1) = 49.8 hr 13.32 . everyone to do at seats reaction endothermic, Eact = 2x ∆ H rxn. draw rxn progress diagram, label position of reactants and products, Eact and ∆ H rxn 13.17 reworked 2 NO2 + F2 = 2 NO2F [NO2], M [F2], M init rate, M/s 0.100 0.100 0.026 1. 0.200 0.100 0.074 2. 0.200 0.200 0.148 3. 0.400 0.400 0.837 4. rate = k[NO2] m [F2] n rate 2/rate 1 = 2 m = 74/26 = 2.85 2 m = 2.85; mlog2 = log2.85, or m=log2.85/log2 = 1.51 = 3/2 rate 3/rate 2 = 2 = 2 n , so n = 1 a check, rate 4/rate 3 = 5.66 = 2 1.5 x 2 = 2 2.5 = (5.66), so OK get k...
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