   # At 573 K, gaseous NO 2 (g) decomposes, forming NO(g) and O 2 (g). If a vessel containing NO 2 (g) has an initial concentration of 1.9 × 10 −2 mol/L, how long will it take for 75% of the NO 2 (g) to decompose? The decomposition of NO 2 (g) is second-order in the reactant and the rate constant for this reaction, at 573 K, is 1.1 L/mol · s. ### 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 21PS
Textbook Problem
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## At 573 K, gaseous NO2(g) decomposes, forming NO(g) and O2(g). If a vessel containing NO2(g) has an initial concentration of 1.9 × 10−2 mol/L, how long will it take for 75% of the NO2(g) to decompose? The decomposition of NO2(g) is second-order in the reactant and the rate constant for this reaction, at 573 K, is 1.1 L/mol · s.

Interpretation Introduction

Interpretation: The time required for the reaction has to be given

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of mol/(L.s).

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]0 = 1.9×10-2 mol/LInitially,100%of sampleispresent,duringreaction75%ofsamplehasdecomposed.So,100%-75%=25%of[NO2]tFinalconcentration,[NO2]t=25100×1.9×10-2 mol/L=0

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