The reaction of 2N2O5(g)→4NO2(8) + O2(g) is experimentally determined to be second order. What experimental data would support this conclusion? I. Plotting 1/ [N2O5] vs. time would be linear II. The half-life is independent of the initial concentration Doubling the initial concentration of N205 would yield a quadrupled III. initial rate. I. and III. Only I. Only II. O II. and III.

The reaction of 2N2O5(g)→4NO2(8) + O2(g) is experimentally determined to be second order. What experimental data would support this conclusion? I. Plotting 1/ [N2O5] vs. time would be linear II. The half-life is independent of the initial concentration Doubling the initial concentration of N205 would yield a quadrupled III. initial rate. I. and III. Only I. Only II. O II. and III.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

The reaction AB+ 2C is known to be zero order with a rate constant of 5.0 x 10 mol/L/s at 25°C. An experiment was run at 25°C with (Ale=1.0 x 10 M. The concentration of C after 5 x 10 sis A. 2.5 x 10 M B. 7.5 x 10 M x C. 5.0 x 10-M D. 2.5x 10- M E. None of these since there are no such thing as zero order reaction Your answer is incorrect.
The half-life of a first order reaction is constant and equal to 0.693/k. CA first order reaction with k= 3.465 x 10-2 s-1 is started with [A]o=0.800 M. What is the concentration of A after 60 seconds? Oa. 0.100 M Ob. 0.400 M Oc. c. 0.200 M Od. 0.0800 M
7. DETAILS The first-order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 101 s ¹. C4H8 → 2 C₂H4 A. What is the half-life of the reaction at 1000."C? [ B. If you started with 3.350 g of cyclobutane, how long would it take to consume 0.4020 g of it? C. How much of a 2.50 g sample of cyclobutane would remain after 0.03430 s? Useful Information: 1.013 bar = 760 torr = 1 atm = 760 mm Hg h = 6.62607x10-34 3.s c = 2.99792 x 108 m/s R = 8.314472 3/mol.K NA = 6.02214 x 1023 1/mol R = 0.08205746 L.atm/mol.K F = 9.648534 x 104 c/mol Equations: K= K (RT)An AG=AHⓇ - TASⓇ MY NOTES pH = pk +log. E = E [A] [HA] 7=45 0.05916 n ASK YOUR TEACHER PRACTICE ANOTHER log Qrxn AG=-RTINK AG=-nFE 1 [A]t In(2) k [A] = [A]okt 1 =kt + [A] 9 [A] t= k[A] [A] 2k -kt -Ea k Ae RT
How many half-lives are required for theconcentration of reactant to decrease to 1.56% ofits original value?A. 4 D. 7B. 5 E. 8C. 6
The decomposition of N2O5 to NO2 and O2 is first order with respect to the reactant N₂05 and first order overall. The rate constant for the reaction at 70.°C is 6.82x10-3 s-1. The reaction is started with 0.300 mol of N205 in a 0.500L container. What will the concentration of N2O5 be after 1.50 minutes? O a. 0.594M O b. 0.438M O c. 0.297M O d. 0.325M O e. 0.162M O f. 0.253M
The decomposition of N205 in solution in carbon tetrachloride proceeds via the reaction: 2 N205 (soln) -----> 4 NO2 (soln) + O2 (soln) The reaction is first order. The rate law for the reaction is rate = a. rate = k [N2O5] b. rate = k [N205]? rate = k [N205]3 с. d. rate = k [NO2] [O2]
Hydrogen peroxide, H₂O₂ (aq), decomposes to H₂O₂(l) and Q₂ (g) in a reaction that is first-order in H₂O₂ and has a rate constant k = 1.06 × 10-³ min-¹ at a given temperature. a. How long will it take for 10.% of a sample of H₂O₂ to decompose? min b. How long will it take for 86% of the sample to decompose? min
Molecular iodine, I2 (g), dissociates into iodine atoms at 652 K, with a first order rate constant of 0.452 s-1. I2(g) -----> 2I a. What is the half life for this reaction? b. If you start with 0.35 M I2 at this temperature, how long will it take for the concentration of I2 to reach 0.12 M?
The reaction of 2 N2O5(g) → 4 NO2(g) + O2(g) is experimentally determined to be fırst oder. What experimental data would support this conclusion? I. Plotting 1/[N2O5] vs. time would be linear. II. The half-life is independent of the initial concentration. III. Doubling the initial concentration of N205 would yield a quadrupled initial rate. Only I. Only II. II. and III. I. and III.