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Wright CHM 1220 - CHM 1220_SI_Kinetics_Fall2017_answers

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CHM 1220 SI Activity: Kinetics Fall 2017 ANSWERS 1. Given the following data, determine the rate law and estimate the rate constant. BrO3(aq) + 5 Br(aq) + 6 H+(aq) → 3 Br2(aq) + 3 H2O( ) Experiment [BrO3] (M) [Br] (M) [H+] (M) Rate (M/s) 1 0.100 0.100 0.100 1.20 × 10–3 2 0.200 0.100 0.100 2.40 × 10–3 3 0.100 0.300 0.100 3.60 × 10–3 4 0.200 0.100 0.150 5.40 × 10–3 2. The reaction: was studied at 25 °C, and the following data were obtained: Experiment [CO]0 (M) [NO2]0 (M) Initial Rate (M/hr) 1 5.0 × 10-4 0.36 × 10-4 3.4 × 10-8 2 5.0 × 10-4 0.18 × 10-4 1.7 × 10-8 3 1.0 × 10-3 0.36 × 10-4 6.8 × 10-8 4 1.5 × 10-3 0.72 × 10-4 ? (a) What is the rate law for this reaction at 25 °C, including the value of the rate constant? ( (b) What is the initial rate of the reaction for experiment 4? CO (g) + NO2(g) → CO2(g) + NO(g) From Experiments 1 & 2: [CO] = constant rate halved, [NO2] halved first order with respect to [NO2] From Experiments 1 & 3: [NO2] = constant rate doubled, [CO] doubled first order with respect to [CO] Using data from any experiment: Rate law:   1144182 89.11036.0100.5 104.3]CO][NO[Rate hrMMMhrMk]CO][NO[Rate2k   hrMMhrMk7431121004.21072.0105.189.1]CO][NO[RateRate law: From Experiments 1 & 2: [Br-], [H+] = constant rate doubles, [BrO3-] doubles first order with respect to [BrO3-] From Experiments 1 & 3: [BrO3-], [H+] = constant rate triples, [Br-] triples first order with respect to [Br-] From Experiments 2 & 4: [BrO3-], [Br-] = constant rate times 2.25, [H+] times 1.5 second order with respect to [H+] Using data from any experiment:    23H Br BrORate k2176.0352.0150.0100.0 log1040510402log][H order,reaction 33--..      1321-323 0.12 .1000 .1000 .1000101.20H Br BrORate sMMMMsMkCHM 1220 SI Activity: Kinetics Fall 2017 3. A reaction of the form: gives a plot of ln[A] versus time (in seconds), which is a straight line with a slope of -7.35×10-3. Assuming [A]0 = 0.0100 M, calculate the time required for the reaction to reach 22.9% completion. 4. You and a coworker have developed a molecule that has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law: Rate = k[AV][V] You have been given the following data from your coworker where [V]0 is much greater than [AV]0: [V]0 = 0.20 M and [AV]0 = 1.0 × 10-4 M A plot of ln[AV] versus t (s) gives a straight line with a slope of -0.32 s-1. What is the value of the rate constant (k) for this reaction? aA → Products Plot information indicates that the reaction is first-order. rate law first order integrated rate law  0A]ln[Aln  ktCalculate time to reaction to proceed to 22.9% reacted: 77.1% remains: indicates A][Rate k0A][771.0A][  00A]ln[A][0.771ln  kt kt.7710lnsst 4.351035.7)771.0ln(13-3107.35 kslope1-3107.35 skPlot information indicates that the data was collected under pseudo first-order conditions. pseudo first order rate law where, first order integrated rate law indicates Plug in and solve, AV][Rate k0V][kk 0AV]ln[AVln  tk 1 32.0 skslope1 32.0sk0V][kk 1110 6.1 20.0


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Wright CHM 1220 - CHM 1220_SI_Kinetics_Fall2017_answers

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