   Chapter 13, Problem 43E

Chapter
Section
Textbook Problem

# For the reaction 2 H 2 O ( g ) ⇌ 2 H 2 ( g ) + O 2 ( g ) K = 2.4 × 10−3 at a given temperature. At equilibrium in a 2.0-L container it is found that [H2O (g)] = 1.1 × 10−1 M and [H2(g)] = 1.9 × 10−2 M. Calculate the moles of O2(g) present under these conditions.

Interpretation Introduction

Interpretation: The equilibrium constant value for the stated reaction is given 2.4×103 . The equilibrium concentrations of H2O and H2 are given. The moles of O2 present are 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 K .

To determine: The moles of O2 present under the given conditions.

The concentration of O2 present under the given conditions

Explanation

Given

The stated reaction is,

2H2O(g)2H2(g)+O2(g)

The value of equilibrium constant K is 2.4×103 .

The concentration of H2O is 1.1×101M .

The concentration of H2 is 1.9×102M .

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=[H2]2[O2][H2O]2

Substitute the given values of the equilibrium constant and the concentration values of H2O and H2 in the above expression.

K=[H2]2[O2][H2O]22.4×103=[1.9×102]2[O2][1.1×101]2

Simplify the above expression.

2.4×103=[1.9×102]2[O2][1.1×101]2[O2]=[2.4×103]×[1

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