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The reaction 2NOB → 2NO + Br2 exhibits the rate law Rate = k[NOB1]²= A[NOB1] At where k= 1.0 × 10-°M'·s¯l at 25°C. This reaction is run where the initial concentration of NOBr ([NOB1]o) is 1.00 x 10-1M. The [NO] after 1.00 hour has passed is

The reaction 2NOB → 2NO + Br2 exhibits the rate law Rate = k[NOB1]²= A[NOB1] At where k= 1.0 × 10-°M'·s¯l at 25°C. This reaction is run where the initial concentration of NOBr ([NOB1]o) is 1.00 x 10-1M. The [NO] after 1.00 hour has passed is

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The reaction 2NOBR → 2NO + Br2 exhibits the rate law Rate = k[NOB1]² A[NOB1] At where k= 1.0 × 10³M. s at 25°C. This reaction is run where the initial concentration of NOBR ([NOB1]0) is 1.00 x 10-1M. The [NO] after 1.00 hour has passed is
Transcribed Image Text:The reaction 2NOBR → 2NO + Br2 exhibits the rate law Rate = k[NOB1]² A[NOB1] At where k= 1.0 × 10³M. s at 25°C. This reaction is run where the initial concentration of NOBR ([NOB1]0) is 1.00 x 10-1M. The [NO] after 1.00 hour has passed is
Expert Solution
Step 1

Introduction:

The given reaction is 

                            2 NOBr →2 NO + Br2

Given data ;

  • k is rate constant 1.0 ×10-5 M-1.s-1
  • NOBr0 is intial concentration of NOBr  1×10-1 M
  • t is reaction time 1 hr

from stochiometry of reaction Concentration of NOBr is equal to concentration of NO

  • NOBr =NO

The unit of rate constant is M-1.s-1 so given reaction is second order reaction .

Integration of the second-order rate law is 

                                   1A=1A0+ kt

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