Wright CHM 1220 - CHM 1220_Kinetics_Fall2017_answers (2 pages)

CHM 1220 ANSWERS 1 In Class Activity Kinetics Fall 2017 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 OH 2 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 6 20 1 86 10 1 00 10 3 6 30 1 71 10 1 00 10 3 6 80 1 00 10 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 Note You could approximate the pseudo firstActual Rate law order rate constant using HO2 and time from the table however you CANNOT use them to figure Rate k HO2 O3 overall second order out the second order rate constant However experimental conditions make it pseudo first order Rate k HO2 Where k k O3 0 Integrated rate law for a first order reaction Given in problem ln HO2 k t ln HO2 0 ln HO2 0 0103 t 12 991 Therefore pseudo first order rate constant k 0 0103 ms 1 Second order rate constant can be determined directly based on previous statements k k O3 0 2 The reaction k k 0 0103 ms 1 10 3 M 1 ms 1 3 O3 0 1 00 10 M CO g NO2 g CO2 g NO g was studied at 25 C and the following data were obtained Experiment 1 2 3 4 CO 0 M 5 0 10 4 5 0 10 4 1 0 10 3 1 5 10 3 NO2 0 M 0 36 10 4 0 18 10 4 0 36 10 4 0 72 10 4 Initial Rate M hr 3 4 10 8 1 7 10 8 6 8 10 3 a What is the rate law for this reaction at 25 C including the value of the rate constant 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 Rate law Rate k CO NO2 Using data from any experiment k b What is the initial rate of the reaction for experiment 4 Rate 3 4 10 8 M hr 1 1 89 M 1hr 1 4 4 CO NO2 5 0 10 M 0 36 10 M Rate k CO NO2 1 89M 1hr 1 1 5 10 3 M 0 72 10 4 2 04 10 7 M hr 2 Given following rate constants determine the activation energy of the decomposition reaction shown below T C CH3CHO CH4 CO k M 1 2 s 1 0 00105 486 Arrhenius equation 0 0214 563 k1 Ea 1 1 relating temperature ln Knowns Unknowns k rate constants and R T2 T1 2 activation energy T 486 k 0 00105 M 1 2 s 1 k2 0 0214 M 1 2 s 1 Plug in values solve for Ea 0 00105 M 1 2s 1 Ea ln 1 2 1 0 0214 M s 8 314 J mol K Ea 2 06 105 J mol 1 1 836 K 759 K 206 kJ mol C 759 K T2 563 C 836 K 1 1 Ea 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 transcycloheptene is 0 035 M a What is the concentration of trans cycloheptene after a reaction time of 1600 s 2 Second order Rate k tCHP rate law 1 1 kt tCHP tCHP 0 Second order integrated rate law 1 1 0 030 M 1s 1 1600 s tCHP 0 035 M Plug in values solve trans cycloheptene tCHP 0 0135 M b At what time will the concentration drop to one twentieth of its initial value Calculate time to reach 1 20 remaining tCHP 1 1 tCHP 0 20 1 kt tCHP 0 cis cycloheptene 1 tCHP 0 20 1 1 0 030 M 1s 1 t 0 035 M 1 0 035 M 20 t 18 100 s 5 03 h c What is the half life of trans cycloheptene at an initial concentration of 0 75 M Calculate time to reach 50 remaining tCHP 0 5 tCHP 0 1 1 kt tCHP 0 0 5 tCHP 0 1 1 0 030 M 1s 1 t 0 75 M 0 5 0 75 M t 44 4 s 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 Reaction is first order Finally calculate time for reaction to proceed to 19 5 left C rate law Rate k integrated rate law ln C kt ln C 14 14 14 0 Need to calculate rate constant using half life definition C 0 50 C 14 14 kt ln C ln 0 50 14C 14 0 k 0 ln 0 5 1 21 10 4 years 1 5730 years 14 0 Substitute kt ln C ln 0 195 14C t 0 Substitute C 0 195 C 14 14 0 0 ln 0 195 13 510 years 1 21 10 4 years 1