CHM 1220 In-Class Activity: Kinetics Fall 2017 ANSWERS 1. The reaction of hydroxyl radical with ozone, HO2(g) + O3(g) OH(g) + 2 O2(g), is a rapid, second order reaction. The rate of reaction was measured in milliseconds for different OH2 concentrations in the presence of a large excess of O3. Time (ms) [HO2] (M) [O3] (M) 0 2.28  10–6 1.00  10–3 10 2.07  10–6 1.00  10–3 20 1.86  10–6 1.00  10–3 30 1.71  10–6 1.00  10–3 80 1.00  10–6 1.00  10–3 From this data, a plot of ln[HO2] versus time has a slope of 0.0103 and an intercept of 12.991. What is the pseudo-first-order rate constant, and what is the second-order rate constant? 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 × 10-3 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? 2. Given following rate constants, determine the activation energy of the decomposition reaction shown below, CH3CHO CH4 + CO k (M1/2 s1) T (°C) 0.00105 486 0.0214 563 Actual Rate law: overall second order However, experimental conditions make it pseudo-first order: Where: Integrated rate law for a first-order reaction: Given in problem: Therefore pseudo-first order rate constant: Second-order rate constant can be determined directly based on previous statements: k1 = 0.00105 M-1/2 s-1 T1 = 486 °C = 759 K Note: You could approximate the pseudo-first-order rate constant using [HO2] and time from the table, however you CANNOT use them to figure out the second-order rate constant. T2 = 563 °C = 836 K Ea = ? Plug-in values, solve for Ea, k2 = 0.0214 M-1/2 s-1 Arrhenius equation relating temperature, rate constants, and activation energy: Knowns & Unknowns:  K 7591K 8361KmolJ314.8M 0214.0M 00105.0ln11/2-1-1/2aEssFrom 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: CO (g) + NO2(g) → CO2(g) + NO(g) 122111lnTTREkkamolkJ 206molJ1006.25aE  32O HORate k 2HORate k 03Okk    022HOlnHOln  tk 991.120103.0HOln2 t-1ms 0103.0k 03Okk  113-103ms 3.101000.1ms 0103.0O-MMkk ]CO][NO[Rate2k  1144182 89.11036.0100.5 104.3]CO][NO[Rate hrMMMhrMk   hrMMhrMk7431121004.21072.0105.189.1]CO][NO[RateCHM 1220 In-Class Activity: Kinetics Fall 2017 3. Trans-cycloheptene, a strained cyclic hydrocarbon, converts to cis-cycloheptene at low temperatures. This molecular rearrangement is a second-order process with a rate constant of 0.030 M-1s-1 at 60 °C. If the initial concentration of trans-cycloheptene is 0.035 M: (a) What is the concentration of trans-cycloheptene after a reaction time of 1600 s? (b) At what time will the concentration drop to one-twentieth of its initial value? (c) What is the half-life of trans-cycloheptene at an initial concentration of 0.75 M? 5. Anthropologists can estimate the age of a bone or other sample of organic matter by its carbon-14 content. The carbon-14 in a living organism is constant until the organism dies, after which carbon-14 decays with first-order kinetics and a half-life of 5730 years. Suppose a bone from an ancient human contains 19.5% of the C-14 found in living organisms. How old is the bone? trans-cycloheptene cis-cycloheptene Second-order integrated rate law Second-order rate law Plug-in values, solve: Calculate time to reach 1/20 remaining: Calculate time to reach 50% remaining: Reaction is first-order: rate law Need to calculate rate constant using half-life definition: integrated rate law Finally, calculate time for reaction to proceed to 19.5% left: Substitute: Substitute:    0CHP1CHP1tktt 2CHPRate tk   MssMt 035.01 1600 030.0CHP111 Mt 0135.0CHP    0CHP201CHP tt   00CHP1CHP2011tktth 5.03 s ,10018 t   MtsMM 035.01 030.0 035.0201111   0CHP5.0CHP tt    00CHP1CHP5.01tktt s 44.4t   MtsMM 75.01 030.0 75.05.0111 Ck14Rate     014014ln0.50ln CktC 1-4years 1021.1years 5730)5.0ln(k   01414lnln CktC    0141450.0 CC    01414195.0 CC     014014ln0.195ln CktC years 510,13years