   Chapter 13, Problem 27E

Chapter
Section
Textbook Problem

# At a particular temperature, a 3.0-L flask contains 2.4 moles of Cl2, l.0 mole of NOCI, and 4.5 × 10−3 mole of NO. Calculate K at this temperature for the following reaction: 2 NOCI ( g ) ⇌ 2 NO ( g ) + Cl 2 ( g )

Interpretation Introduction

Interpretation: The number of moles of Cl2 , NOCl and NO at a particular temperature are given. The value of the equilibrium constant (K) for the given reaction 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 K .

To determine: The concentrations of NO , Cl2 and NOCl for the given reaction.

Explanation

Given

The stated reaction is,

2NOCl(g)2NO(g)+Cl2(g)

The number of moles Cl2 is 2.4moles .

The number of moles NOCl is 1.0mole .

The number of moles NO is 4.5×103mole .

The given volume of the flask is 3.0L .

The concentration of a reactant is calculated by the formula,

Concentration=MolesVolume(L)

For NO ,

The concentration of NO is calculated by the formula,

Substitute the values of the number of moles of NO and the volume of the flask in the above expression in the above

ConcentrationofNO=4.5×103mole3.0L=1.5×10-3M_

For Cl2 ,

The concentration of Cl2 is calculated by the formula,

Substitute the values of the number of moles of Cl2 and the volume of the flask in the above expression in the above

ConcentrationofCl2=2.4mole3.0L=0

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