Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Question
Chapter 16, Problem 116IL

(a)

Interpretation Introduction

Interpretation:

Each prduct as a lewis acid or a lewis base is to be stated.

Concept Introduction:

A Lewis acid is a substance that contains an empty orbital which is capable of accepting an electron pair. A Lewis base is a substance that has a filled orbital containing an electron pair which is not involved in bonding but may form a dative bond with a lewis acid.

(a)

Expert Solution
Check Mark

Answer to Problem 116IL

On the product side, BF3 can act as a lewis acid and (CH3)2O can act as a lewis base.

Explanation of Solution

A lewis acid can accept a pair of electrons from a lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it as a lewis acid.

Methyl ether ((CH3)2O) is a lewis base, it can donate it’s lone pair of electrons. Chemical reaction between ((CH3)2O) and BF3, the lone pair from methyl ether will form a dative bond with the empty orbital of BF3 to form an adduct.

    (CH3)2O+BF3[(CH3)2O:BF3]

On the product side, BF3 can act as a lewis acid and (CH3)2O can act as a lewis base.

(b)

Interpretation Introduction

Interpretation:

Total pressure at equilibrium, partial pressure of lewis acid and lewis base has to be calculated.

Concept Introduction:

A Lewis acid is a substance that contains an empty orbital which is capable of accepting an electron pair. A Lewis base is a substance that has a filled orbital containing an electron pair which is not involved in bonding but may form a dative bond with a lewis acid.

Ideal gas equation,

PTV=nRT (1)

Here,

PT is total pressure of the gas.

V is volume of the gas container.

n is total number of moles of gas particles.

R is gas constant.

T is temperature.

(b)

Expert Solution
Check Mark

Answer to Problem 116IL

The total pressure in the flask at Equilibrium is 0.404atm and the partial pressure of BF3,(CH3)2O and (CH3)2O:BF3 are 0.14atm,0.14atm and 0.12atm respectively.

Explanation of Solution

From ideal gas equation,

PTV=nRT (1)

Here,

PT is total pressure of the gas.

V is volume of the gas container.

n is total number of moles of gas particles.

R is gas constant.

T is temperature.

An equilibrium constant (KP) can be written in terms of partial pressure (P) of gases.

    [(CH3)2O:BF3](g)(CH3)2O(g)+BF3(g) (2)

An equilibrium constant (KP) for the above reaction is,

KP=PBF3×P(CH3)2OP(CH3)2OBF3 (3)

Where, partial pressure of gases is,

PBF3=xBF3×PTP(CH3)2O=x(CH3)2O×PTP(CH3)2OBF3=x(CH3)2OBF3×PT (4)

Where x is the mole fraction of the species which equals to the ratio of moles of the substance to the total number of moles,

xBF3=nBF3nBF3+n(CH3)2O+n(CH3)2OBF3

Given:

The mass of boron ether complex is 1g.

The volume of the flask is 0.565L.

The temperature is 398K.

The gas constant R is 0.0821atmLK1mol1.

Molecular weight of complex is 141.98gmol1.

Total number of moles is equal to the mass of the complex divided by the molecular weight of the complex,

n=1g141.98gmol1=0.007mol

Total pressure at equilibrium is calculated from equation (1) as follows,

PT=0.007(mol)×0.0821(atmLK1mol1)×398(K)0.565(L)=0.404atm

Set up an ICE table for the chemical reaction given in equation (2)

Equilibrium[(CH3)2O:BF3](g)BF3(g)+(CH3)2O(g)Initialmoles0.00700Changey+y+yAtEquilibrium0.007yyy

Total number moles at Equilibrium is equal to 0.007+y.

Substitute these values in equation (3) to calculate change in moles at Equilibrium,

KP=(y0.007+y)2×PT(0.007y0.007+y)0.17=y2×0.404(4.9×105)y2y=3.81×103

Now substitute the value of y in equation (4) to calculate partial pressure of given gases,

PBF3=(3.81×103)(7×103)+(3.81×103)×(0.404)=0.14

Similarly,

P(CH3)2O=0.14P(CH3)2OBF3=0.12

The total pressure in the flask at Equilibrium is 0.404atm and the partial pressure of BF3,(CH3)2O and (CH3)2O:BF3 are 0.14atm,0.14atm and 0.12atm respectively.

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Chapter 16 Solutions

Chemistry & Chemical Reactivity

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