   # The decomposition of nitrogen dioxide at a high temperature NO 2 (g) → NO(g) + ½ O 2 (g) is second-order in this reactant. The rate constant for this reaction is 3.40 L/mol · min. Determine the time needed for the concentration of NO 2 to decrease from 2.00 mol/L to 1.50 mol/L. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 14, Problem 20PS
Textbook Problem
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## The decomposition of nitrogen dioxide at a high temperatureNO2(g) → NO(g) + ½ O2(g)is second-order in this reactant. The rate constant for this reaction is 3.40 L/mol · min. Determine the time needed for the concentration of NO2 to decrease from 2.00 mol/L to 1.50 mol/L.

Interpretation Introduction

Interpretation: The time required for the reaction has to be calculated.

Concept introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

aA + bBxXRate of reaction = k [A]m[B]n

Integrated rate law for second order reactions:

Taking in the example of following reaction,

aAproducts

And the reaction follows second order rate law,

Then the relationship between the concentration of A and time can be mathematically expressed as,

1[A]t=kt+1[A]0

The above expression is called as integrated rate for second order reactions.

### Explanation of Solution

The time required is calculated as,

-Δ[R]Δt = k[R]2,The relation can be transformed into,1[R]t - 1[R]0 = ktGiven:[NO2]t = 1.50 mol/L[NO2]0 = 2.00 mol/Lk = 3.40  L/mol.min t = ?Therefore,1[R]t - 1

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