What is the rate law implied by the mechanism given below? CH3COCH3(aq) + H+(aq) ←→ CH3C(OH)CH3+(aq) (fast, reversible) CH3C(OH)CH3+(aq) → CH3C(OH)=CH2(aq) + H+(aq) (slow) CH3C(OH)=CH2(aq) + Br2(aq) → CH3C(OH)CH2Br+(aq) + Br-(aq) (fast) CH3C(OH)CH2Br+(aq) → CH3COCH2Br(aq) + H+(aq) (fast) A. Rate = k[CH3COCH3][H+] B. Rate = k[CH3COCH3] C. Rate = k[CH3COCH3][Br2] D. Rate = k[CH3COCH3]2 E. Rate = k[CH3COCH3][Br2]/[H+] (Answer is A, looking for explanation why!).

What is the rate law implied by the mechanism given below? CH3COCH3(aq) + H+(aq) ←→ CH3C(OH)CH3+(aq) (fast, reversible) CH3C(OH)CH3+(aq) → CH3C(OH)=CH2(aq) + H+(aq) (slow) CH3C(OH)=CH2(aq) + Br2(aq) → CH3C(OH)CH2Br+(aq) + Br-(aq) (fast) CH3C(OH)CH2Br+(aq) → CH3COCH2Br(aq) + H+(aq) (fast) A. Rate = k[CH3COCH3][H+] B. Rate = k[CH3COCH3] C. Rate = k[CH3COCH3][Br2] D. Rate = k[CH3COCH3]2 E. Rate = k[CH3COCH3][Br2]/[H+] (Answer is A, looking for explanation why!).

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter12: Chemical Kinetics

Section: Chapter Questions

Problem 12Q: The plot below shows the number of collisions with a particular energy for two different...

See similar textbooks

Similar questions

The plot below shows the number of collisions with a particular energy for two different temperatures. a. Which is greater, T2 or T1? How can you tell? b. What does this plot tell us about the temperature of the rate of a chemical reaction? Explain your answer.
Using the mechanism provided, determine a possible rate law for the following overall reaction: Q2 + 2 AJ2 → 2AJ2Q Step #1. AJ2 + Q2 → AJ2Q + Q slow Step #2. Q + AJ2 → AJ2Q quick Group of answer choices Rate = k [AJ2]2 [Q2] Rate = k [AJ2Q]2 Rate = k [AJ2] [Q] Rate = k [AJ2Q] [Q]2 Rate = k [AJ2] [Q2]
What is the molecularity of the rate limiting step for the mechanism below? Step 1: H2O2 (aq) + I–(aq) → H2O(l) + OI–(aq) (slow)Step 2: H2O2 (aq) + OI–(aq) → H2O(l) + O2 (g) + I–(aq) (fast)
What is the rate law for this elementary reaction?2NO2 ⇌ N2O4A. rate = k[NO2]2B. rate = k[NO2]xC. rate = [N2O4]/[NO2]2D. rate = k[NO2]x[N2O4]yE. rate = k[NO2]2[N2O4]
2A -> A2 (Fast) A2 + B= A2B (slow) predict the rate law from the following two step mechanism
. The proposed reaction mechanism is as follows:i. BrO3- (aq) + H+ (aq) à HBrO3 (aq) [Fast]ii. HBrO3 (aq) + H+ (aq) à H2BrO3+ (aq) [Medium]iii. H2BrO3+ (aq) + Br- (aq) à Br2O2 (aq) + H2O (ℓ) [Slow]iv. Br2O2 (aq) + 4H+(aq) + 4Br-(aq) à 3Br2 (ℓ) + H2O (ℓ) [Fast]Evaluate the validity of this proposed reaction. Justify your answer.
The proposed reaction mechanism is as follows:i. BrO3- (aq) + H+ (aq) = HBrO3 (aq) [Fast]ii. HBrO3 (aq) + H+ (aq) = H2BrO3+ (aq) [Medium]iii. H2BrO3+ (aq) + Br- (aq) = Br2O2 (aq) + H2O (ℓ) [Slow]iv. Br2O2 (aq) + 4H+(aq) + 4Br-(aq) = 3Br2 (ℓ) + H2O (ℓ) [Fast]Evaluate the validity of this proposed reaction. Justify your answer.
The second-order rate constant for the reaction CH3COOC2H5(aq) + OH-(aq)→ CH3CO2-(aq) + CH3CH2OH(aq) is 0.11 dm3 mol-1 s- 1. What is the concentration of ester after(a) 15 s, (b) 15 min when ethyl acetate is added to sodium hydroxide so that the in it ial concentrations are [NaOH] 0.055 mol dm-3 and [CH3COOC2H5] = 0.150 mol dm-3?
Overall order and unit of rate constant. Rate = k [CH3CHO]3/2
The following mechanism is proposed: C6H5COOH ⇌ C6H5CO2H2+ (fast) C6H5CO2H2+ ⇌ C6H5CO+ + H2O (fast) C6H5CO+ + HX + H2O ⟶ C6H6NH2 + N2 + CO2 (slow) The overall rate constant for this reaction is 0.200x 10-3L mol-1 s-1. Is this value consistent with rate law r = kobs*[C6H5COOH]*[HX] ?
An irreversible unimolecular reaction (A -> B) has a rate constant of 3.0 per second. What is the half-time (t1/2)? Type in your answer. t1/2 =
Given the following reaction and table of data, determine the rate law for this reaction: CH3X + 3X2 --> CX4 + 3 HX [CH3X] (M) [X2] (M) Initial Rate (M/s) 0.050 0.050 0.014 0.100 0.050 0.028 0.100 0.100 0.056 0.100 0.200 0.112 Group of answer choices rate = k[CH3X][X2]2 rate = k[CH3X]2[X2] rate = k[CH3X] rate = k[CH3X][X2] rate = k[X2]