Chapter 14, Problem 20PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# 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

The time required is calculated as,

â€‚Â -Î”[R]Î”tÂ =Â k[R]2,TheÂ relationÂ canÂ beÂ transformedÂ into,1[R]tÂ -Â 1[R]0Â =Â ktGiven:[NO2]tÂ =Â 1.50Â mol/L[NO2]0Â =Â 2.00Â mol/LkÂ =Â 3.40Â Â L/mol.minÂ tÂ =Â ?Therefore,1[R]tÂ -Â 1

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