   # Gaseous azomethane, CH 3 N=NCH 3 , decomposes in a first-order reaction when heated: CH 3 N=NCH 3 (g) → N 2 (g) + C 2 H 6 (g) The rate constant for this reaction at 600 K is 0.0216 min −1 . If the initial quantity of azomethane in the flask is 2.00 g, how much remains after 0.0500 hour? What mass of N 2 is formed in this time? ### 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 26PS
Textbook Problem
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## Gaseous azomethane, CH3N=NCH3, decomposes in a first-order reaction when heated:CH3N=NCH3(g) → N2(g) + C2H6(g)The rate constant for this reaction at 600 K is 0.0216 min−1. If the initial quantity of azomethane in the flask is 2.00 g, how much remains after 0.0500 hour? What mass of N2 is formed in this time?

Interpretation Introduction

Interpretation: The amount of Azomethane after 0.0500hrs and the mass of Nitrogen formed in this time 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 first order reaction:

Consider A as substance, that gives the product based on the equation,

aAproducts

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

Rate=-Δ[A]Δt=k[A]

The integrated rate law equation can be given as,

ln[A]t[A]o=-kt

The above expression is called integrated rate law for first order reaction.

### Explanation of Solution

The amount of Azomethane after 0.0500hrs and the mass of Nitrogen formed in this time is calculated as,

Reactionrate = k [Azomethane]1.Given:k=0.0216 min-1;[Azomethane]0=(2.0g58.08g/mol)1L=0.0344mol/LTherefore,First order rate law,ln[Azomethane]t=-kt+ln[Azomethane]0ln[Azomethane]t=-kt - ln(0.0344mol/L)ln[Azomethane]t=-(0

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