   # Consider the isomerization of butane with an equilibrium constant of K = 2.5. (See Study Question 13.) The system is originally at equilibrium with [butane] = 1.0 M and [isobutane] = 2.5 M. (a) If 0.50 mol/L of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? (b) If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 15, Problem 29PS
Textbook Problem
391 views

## Consider the isomerization of butane with an equilibrium constant of K = 2.5. (See Study Question 13.) The system is originally at equilibrium with [butane] = 1.0 M and [isobutane] = 2.5 M. (a) If 0.50 mol/L of isobutane is suddenly added and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas? (b) If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

(a)

Interpretation Introduction

Interpretation: The concentration of butane and isobutane when the system comes to equilibrium under the given conditions has to be determined.

Concept introduction:

• Equilibrium constant: At equilibrium the ratio of products to reactants has a constant value. And it is represented by the letter K.

For a general reaction, aA+bBcC+dD

The equilibrium constant Kc = [C]c[D]d[A]a[B]b, where a, b, c and d are the stoichiometric coefficients of reactant and product in the reaction. Concentration value for solid substance is 1.

If the value of Kc and the concentration of any of the reactant of a reaction is known then the concentration of product can be determined by multiplying Kc with the concentration of reactant.

• ICE (reaction initial concentration equilibrium) table is mainly used to calculate the value of K for a reaction. This table contains the concentration of reactant and product in various stage of reaction.

### Explanation of Solution

The given equilibrium set up contains butane and isobutene. And the value of Kc for the interconversion of butane and isobutene is 2.5at250C

The system is originally at equilibrium with [butane]=1.0M and [isobutane]=2.5M

If  0.50mol/L of isobutane added to the equilibrium system the equilibrium concentration of butane and isobutane can be calculated as follows,

By using these concentration ICE table for this reaction can be constructed.

 Reaction                               Butane              ⇄          Isobutane Initial concentration(mol/L) 1.0 2.5  - 0.5 Change in concentration (mol/L) +x -x Equilibrium  (mol/L) 1.0  + x 3.0 −x

The equilibrium constant for the reaction can be written as,

Kc  = [Isobutane][Butane]

2

(b)

Interpretation Introduction

Interpretation: The concentration of butane and isobutane when the system comes to equilibrium under the given conditions has to be determined.

Concept introduction:

• Equilibrium constant: At equilibrium the ratio of products to reactants has a constant value. And it is represented by the letter K.

For a general reaction, aA+bBcC+dD

The equilibrium constant Kc = [C]c[D]d[A]a[B]b, where a, b, c and d are the stoichiometric coefficients of reactant and product in the reaction. Concentration value for solid substance is 1.

If the value of Kc and the concentration of any of the reactant of a reaction is known then the concentration of product can be determined by multiplying Kc with the concentration of reactant.

• ICE (reaction initial concentration equilibrium) table is mainly used to calculate the value of K for a reaction. This table contains the concentration of reactant and product in various stage of reaction.

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