M5 Exam

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Chemistry

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Jan 9, 2024

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QuesƟon 1 (5.): (1): In the reacƟon of 0.200 M gaseous N2O5 to yield NO2 gas and O2 gas as shown below: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) the following data table is obtained: Time (sec) [N2O5] [O2] 0 0.200 M 0 300 0.180 M 0.010 M 600 0.162 M 0.019 M 900 0.146 M 0.027 M 1200 0.132 M 0.034 M 1800 0.110 M 0.045 M 2400 0.096 M 0.052 M 3000 0.092 M 0.054 M (a) Use the [O2] data from the table to calculate the average rate over the measured Ɵme interval from 0 to 3000 secs. (b) Use the [O2] data from the table to calculate the instantaneous rate early in the reacƟon (0 secs to 300 sec). (c) Use the [O2] data from the table to calculate the instantaneous rate late in the reacƟon (2400 secs to 3000 secs). (d) Explain the relaƟve values of the average rate, early instantaneous rate and late instantaneous rate. Answer: (a) The average rate over the measured Ɵme interval from 0 to 3000 secs is: rate = Δ[O2] / Δt = (0.054 - 0) / 3000 - 0 = 1.80 x 10^-5 mol/L ∙ s (b) The instantaneous rate early in the reacƟon from 0 to 300 secs is: rate = Δ[O2] / Δt = (0.010 - 0) / 300 - 0 = 3.33 x 10^-5 mol/L ∙ s (c) The instantaneous rate late in the reacƟon from 2400 to 3000 secs is: rate = Δ[O2] / Δt = (0.054 - 0.052) / 3000 - 2400 = 3.33 x 10^-6 mol/L ∙ s (d) It can be seen that the early instantaneous rate is the largest since the concentraƟons of reactants is highest during the earliest stages of the reacƟon and the late instantaneous rate is smallest since the concentraƟons of reactants is lowest during the late stages of the reacƟon.

(2): In the reacƟon of gaseous CH3CHO to yield CH4 gas and CO gas as shown below: CH3CHO (g) → CH4 (g) + CO (g) the following data table is obtained: Time (sec) [CH3CHO] 0 0.0500 M 1200 0.0300 M 2000 0.0240 M 6000 0.0120 M 10000 0.0080 M 15000 0.0056 M 20000 0.0043 M (a) Use the [CH3CHO] data from the table to calculate the average rate over the measured Ɵme interval from 0 to 20000 secs. (b) Use the [CH3CHO] data from the table to calculate the instantaneous rate early in the reacƟon (0 secs to 1200 sec). (c) Use the [CH3CHO] data from the table to calculate the instantaneous rate late in the reacƟon (15000 secs to 20000 secs). (d) Explain the relaƟve values of the average rate, early instantaneous rate and late instantaneous rate. QuesƟon 2: (1): The following rate data was obtained for the reacƟon: 2 ClO2 + 2 OH- → ClO3- + ClO2- + H2O the following data table is obtained: Experiment # [ClO2] [OH-] rate 1 0.060 0.030 2.48 x 10^-2 2 0.020 0.030 2.76 x 10^-3 3 0.020 0.090 8.28 x 10^-3

Determine the reacƟon order with respect to (1) ClO2 and (2) OH-, (3) write the rate law and then (4) determine the value of the rate constant, k. Answer: Rate = k [ClO2]x [OH-]y If data from experiments 1 and 2 was subsƟtuted into the equaƟon we would obtain (2.48 X 10^-2)/(2.76 X 10^-3) = (k[.060]^x X [.030]^y)/(k[.020]^x X [.030]^y) and if we cancel all common terms we would obtain (2.48 X 10^-2)/(2.76 X 10^-3) = ([.060]^x)/([.020]^x) 9=(3)^x (1) which yields x = 2 as the order of reacƟon with respect to ClO2 If data from experiments 2 and 3 was subsƟtuted into the equaƟon we would obtain (2.76 X 10^-3)/(8.28 X 10^-3) = (k[.020]^x X [.030]^y)/(k[.020]^x X [.090]^y) and if we cancel all common terms we would obtain (2.76 X 10^-3)/(8.28 X 10^-3) = ([.030]^y X [.090]^y) .333=(.333)^y (2) which yields y = 1 as the order of reacƟon with respect to OH-. (3) the overall rate law can now be wriƩen as follows: Rate = k [ClO2]^2 [OH-]^1 (4) and using the data from experiment 1 we can determine the rate constant as follows: 2.48 x 10^-2 = k [0.060]^2 [0.030] k = 2.48 x 10^-2 / [0.060]^2 [0.030] = 229.6 (2): The following rate data was obtained for the reacƟon which takes place in a soluƟon of OH-: ClO- + I- → IO- + Cl- the following data table is obtained: Experiment # [I-] [ClO-] [OH-] rate 1 0.0030 0.0010 1.00 1.8 x 10^-4 2 0.0030 0.0020 1.00 3.6 x 10^-4 3 0.0060 0.0020 1.00 7.2 x 10^-4

4 0.0030 0.0010 0.50 9.0 x 10^-5 Determine the reacƟon order with respect to (1) ClO-, (2) I- and (3) OH-, (4) write the rate law and then (5) determine the value of the rate constant, k. Answer: 1) Rate = k [ClO-]^x [I-]^y [OH-]^z If data from experiments 1 and 2 was subsƟtuted into the equaƟon we would obtain: (1.8 X 10^-4)/(3.6 X 10^-4)= (k[.0010]^x [.0030]^y [1.00]^z)/ (k[.0020]^x [.0030]^y [1.00]^z) (1.8 X 10^-4)/(3.6 X 10^-4)=[.0010]^x / [.0020]^x .50=(.5)^x which yields x = 1 as the order of reacƟon with respect to ClO- If data from experiments 2 and 3 was subsƟtuted into the equaƟon we would obtain: (3.6 X 10^-4)/(7.2 X 10^-4)= (k[.0020]^x [.0030]^y [1.00]^z)/ (k[.0020]^x [.0060]^y [1.00]^z) (3.6 X 10^-4)/(7.2 X 10^-4)= [.0030]^y/[.0060]^y .50=(.5)^y 2) which yields y = 1 as the order of reacƟon with respect to I-. If data from experiments 1 and 4 was subsƟtuted into the equaƟon we would obtain: (1.8 X 10^-4)/(9.0 X 10^-5)= (k[.0010]^x [.0030]^y [1.00]^z) / (k[.0010]^x [.0030]^y [.50]^z) (1.8 X 10^-4)/(9.0 X 10^-5)= [1.00]^z / [.50]^z 2.0=(2.0)^z 3) which yields z = 1 as the order of reacƟon with respect to OH-. The overall rate law can now be wriƩen as follows: 4) Rate = k [ClO-]1 [I-]1 [OH-]1 5) and using the data from experiment 1 we can determine the rate constant as follows: 1.8 x 10-4 = k [0.0010] [0.0030] [1.00] k = 1.8 x 10-4 / [0.0010] [0.0030] [1.00] = 60

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