   Chapter 13, Problem 58E

Chapter
Section
Textbook Problem

# At a particular temperature, K = 4.0 × 10–7 for the reaction N 2 O 4 ( g ) ⇌ 2 NO 2 ( g )   In an experiment, 1.0 mole of N2O4 is placed in a 10.0-L vessel. Calculate the concentrations of N2O4 and NO2 when this reaction reaches equilibrium.

Interpretation Introduction

Interpretation:

The equilibrium constant for the stated reaction and the initial number of moles of N2O4 is given. The equilibrium concentration of N2O4 and NO2 is to be calculated.

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 as K .

Explanation

To determine: The equilibrium concentration of N2O4 and NO2 .

Given

The stated reaction is,

N2O4(g)2NO2(g)

The initial number of moles of N2O4 are 1.0mole .

The volume of the vessel is 10.0L .

The value of equilibrium constant (K) is 4.0×107 .

The concentration of a reactant is calculated by the formula,

Concentration=MolesVolume(L)

The initial concentration of N2O4 is calculated by the formula,

Substitute the given values of the number of moles of N2O4 and the volume of the flask in the above expression.

ConcentrationofN2O4=1.0mole10.0L=0.1M

The concentration of N2O4 consumed is assumed to be x .

The equilibrium concentrations are represented as,

N2O4(g)2NO2(g)Initialconcentration0.100Changex+2xEquilibriumconcentration0.10x2x

At equilibrium, the equilibrium ratio is expressed by the formula,

K=ConcentrationofproductsConcentrationofreactants

Where,

• K is the equilibrium constant.

The equilibrium ratio for the given reaction is,

K=[NO2]2[N2O4] (1)

According to the formulated ICE table,

The equilibrium concentration of N2O4(g) is (0.10x)M

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