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Reaction of Benzoyl Peroxide - General Chemistry - Solved Quiz, Exercises of Chemistry

This is short quiz. Answers are given in empty space. This solved quiz of chemistry includes: Reaction of Benzoyl Peroxide, Rate Equation, Changes in Concentrations, Average Rate Constant, Decomposition Reaction, Acne Medication, Second Order Process, Rate Law of Reaction

Typology: Exercises

2011/2012

Uploaded on 12/23/2012

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Download Reaction of Benzoyl Peroxide - General Chemistry - Solved Quiz and more Exercises Chemistry in PDF only on Docsity!

  1. Write a rate equation for this reaction in terms of the changes in the concentrations of the species present.

3W + 2H → P + 2M + 4B

  1. Use the reaction below and the initial rate data in the table at right to determine the order and average rate constant of the decomposition reaction of benzoyl peroxide (an acne medication).
  2. The decomposition of 1,3-butadiene is described in the reaction below. The rate constant of the reaction is 0.013 L·mol -1·s -1^ , meaning it is a second order process. Determine the concentration that remains after 1.00 hour when you begin the reaction with an initial concentration of 0.0172 M.

C 4 H 6 (l) → C 4 H 4 (l) + H 2 (g)

Exp. [BP] Initial Rate (mol·L-1^ ·s -1) 1 1.00 2.22×10- 2 0.70 1.64×0- 3 0.50 1.12×10-

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  1. Use the initial concentration and rate data to determine the rate law of this reaction:

S 2 O 8 - (aq) + 3 I - (aq) → 2 SO 4 2- (aq) + I 3 (aq)

  1. The half-life of the decomposition of hydrogen peroxide is 10.7 hr at 20 °C. What is the rate constant (in units involving seconds) for the process if it is known to follow first-order kinetics?
  2. Azomethane (CH 3 N 2 CH 3 ) decomposes at 600 K to C 2 H 6 and N 2 following a first- order reaction. Initially you set-up an experiment using 0.00166 M azomethane. After 2000 s you observe the azomethane concentration to be 8.10×10-4^ M. Determine the half-life of the reaction based on this information.

CH 3 N 2 CH 3 (g) → C 2 H 6 (g) + N 2 (g)

Answers: 1) 4 t

Δ[B] 2 t

Δ[M] t

Δ[P] 2 t

Δ[H] 3 t

Rate Δ[W] Δ

= Δ

= Δ

= Δ

=− Δ

= − 2) order = 1,^ k^ = 2.27×10^ -4^ s-

  1. 0.00952 M 4) Rate = 365[S 2 O 8 -^ ][I 3 -^ ]^4 5) k = 1.80×10-5^ s-1^ 6) t (^) 1/2 = 1930 s

Trial [ S 2 O 8 -^ ] [I - ] Initial Rate ( M ·s -1^ ) 1 0.27 0.38 2. 2 0.40 0.38 3. 3 0.40 0.33 1.

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