Chapter 13, Problem 28E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# At a particular temperature a 2.00-L flask at equilibrium contains 2.80 × 10−4 mole of N2, 2.50 × 10−5 mole of O2, and 2.00 × 10−2 mole of N2O. Calculate K at this temperature for the reaction 2 N 2 ( g ) + O 2 ( g ) ⇌ 2 N 2 O ( g ) If [N2] = 2.00 × 10−4 M, [N2O] = 0.200 M, and [O2] =0.00245 M, does this represent a system at equilibrium?

Interpretation Introduction

Interpretation: The number of moles of N2 , O2 and N2O at a particular temperature are given. The value of the equilibrium constant (K) for the given reaction is to be calculated. A comment on the equilibrium state for the reaction with the reactants having the given concentration values is to be determined.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of concentration, it is represented K .

To determine: The concentrations of N2 , O2 and N2O for the given reaction.

Explanation

To determine: The equilibrium constant (K) for the given reaction.

Given

The stated reaction is,

2N2(g)+O2(g)2N2O(g)

The number of moles N2 is 2.80×104mole .

The number of moles O2 is 2.50×105mole .

The number of moles N2O is 2.00×102mole .

The given volume of the flask is 2.00L .

The concentration of a reactant is calculated by the formula,

Concentration=MolesVolume(L)

For N2 ,

The concentration of N2 is calculated by the formula,

Substitute the values of the number of moles of N2 and the volume of the flask in the above expression in the above

ConcentrationofN2=2.80×104mole2.00L=1

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