   # At 35°C, K = 1.6 × 10 −5 for the reaction 2 NOCl ( g ) ⇌ 2 NO ( g ) + Cl 2 ( g ) If 2.0 moles of NO and 1.0 mole of Cl 2 arc placed into a 1.0-L flask, calculate the equilibrium concentrations of all species. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 12, Problem 101CP
Textbook Problem
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## At 35°C, K = 1.6 × 10−5 for the reaction 2 NOCl ( g ) ⇌ 2 NO ( g ) + Cl 2 ( g ) If 2.0 moles of NO and 1.0 mole of Cl2 arc placed into a 1.0-L flask, calculate the equilibrium concentrations of all species.

Interpretation Introduction

Interpretation: The decomposition reaction of NOCl to form NO and Cl2 gas is given. A tiny hole is made in the side of the flask when the system reaches equilibrium. The equilibrium concentration of all the species involved in 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 as K .

To determine: The equilibrium concentration of all the species involved in the given reaction.

### Explanation of Solution

Explanation

Given

The stated reaction is,

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

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

K=ConcentrationofproductsConcentrationofreactants

Where,

• K is the equilibrium constant.

The equilibrium constant expression for the given reaction is,

K=[NOCl]2[NO]2[Cl2] (1)

The amount of Cl2 that reacted till the equilibrium stage is 0.975M_ .

The initial number of moles of NO is 2.0moles .

The initial number of moles of Cl2 is 1.0mole .

The volume of the flask is 1.00L .

The given equilibrium constant value is 1.6×105 .

The initial concentration of NO is 2.0moles/L that is 2.0M .

The initial concentration of Cl2 is 1.0mole/L that is 1.0M .

The amount of Cl2 that reacted till the equilibrium stage is assumed to be x .

The equilibrium concentrations are represented as,

2NO(g)+Cl2(g)2NOCl(g)Initialconcentration2.01.00Change2xx+2xEquilibriumconcentration2.02x1.0x2x

The equilibrium concentration of NO is (2.02x)M .

The equilibrium concentration of Cl2 is (1.0x)M .

The equilibrium concentration of NOCl is 2xM .

Substitute these values in equation (1).

K=[2x]2[2.0x]2[1.0x]

The equilibrium constant for this reaction will be the reciprocal of the given equilibrium constant value (as this reaction is reverse of the given reaction).

Substitute the value of the equilibrium constant in the above expression.

11.6×105=[2x]2[2.0x]2[1.0x]0

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