Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 18, Problem 108AE

(a)

Interpretation Introduction

Interpretation : The enthalpy of the formation of NI3(s) is to be calculated.

Concept Introduction :

The enthalpy is defined as the product of the pressure and volume added up with the system internal energy. It is the property of thermodynamic system. The heat can be absorbed or released is the change in enthalpy.

(a)

Expert Solution
Check Mark

Answer to Problem 108AE

The enthalpy of formation of NI3 is 287kjmol-1

Explanation of Solution

Nitrogen trifloride is colorless, inorganic thermally stable compound. It is non- flammable in nature. The stability is due to comparable size of nitrogen and fluoride. While the nitrogen iodide is an explosive compound due to the difference in the size of the atoms.

Consider the following reaction

  BN(s)+3IF(g)BF3(g)+NI3(g)

Consider the enthalpy of formation of NI3 as x

  enthalpyofreaction=enthalpy of product-enthalpy of reactantenthalpyofreaction=[ΔHofBF3+ΔHofNI3]-[ΔHofBN+3(ΔHofIF3)]

Subtitling ΔHofBF3=-1136kjmol-1,ΔHofNI3=x,ΔHofBN=-254kjMol-1

  ΔHofIF3=-96kjmol-1 and the enthalpy of reaction as -307kjmol-1 in the above reaction

  307kjmol-1=[1136kjmol-1+x][(254kjmol-1)+3(96kjmol-1)]x=1136kjmol-1307kjmol-1354kjmol-1288kjmol-1307kjmol-1=287kjmol-1

Thus, the enthalpy of formation of NI3 is 287kJmol-1

(b)

Interpretation Introduction

Interpretation : The molecular geometries of the species are determined.

Concept Introduction :

The molecular geometry is the three dimensional shape of the atom that contains the molecule. It consists of the general shape which contains bond lengths, bond angles, transition angles and other parameters determining the position of atoms.

(b)

Expert Solution
Check Mark

Answer to Problem 108AE

It will have 4 bond pairs and the geometry will become tetrahedral with sp3 hybridization

  Chemical Principles, Chapter 18, Problem 108AE , additional homework tip  1

Explanation of Solution

According to the vesper theory the molecular geometry are predicted as follows

Total valance shell electrons of the [IF2]+ can be calculated as

  [IF2]+=7eofI+2(7eofF)(netchargeofmolecule)=7+141[IF2]+=20e

Thus it will have 2 lone pairs and 2 bond pairs and the geometry will become V-shaped geometry with sp3 hybridization.

It is represented as follows:

  Chemical Principles, Chapter 18, Problem 108AE , additional homework tip  2

Total valance shell electrons of the [BF4] can be calculated as follows:

  [BF4]=3eofB+4(7eofF)(net charge of molecule)=3+28+1[BF4]=32e

Thus it will have 4 bond pairs and the geometry will become tetrahedral with sp3 hybridization.

  Chemical Principles, Chapter 18, Problem 108AE , additional homework tip  3

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Chapter 18 Solutions

Chemical Principles

Ch. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26ECh. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Prob. 35ECh. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Prob. 40ECh. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - The synthesis of ammonia gas from nitrogen gas...Ch. 18 - Prob. 51ECh. 18 - Prob. 52ECh. 18 - Prob. 53ECh. 18 - Prob. 54ECh. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Prob. 58ECh. 18 - Prob. 59ECh. 18 - Prob. 60ECh. 18 - Prob. 61ECh. 18 - Prob. 62ECh. 18 - Prob. 63ECh. 18 - Prob. 64ECh. 18 - Prob. 65ECh. 18 - Prob. 66ECh. 18 - Prob. 67ECh. 18 - Prob. 68ECh. 18 - Prob. 69ECh. 18 - Prob. 70ECh. 18 - Prob. 71ECh. 18 - Prob. 72ECh. 18 - Prob. 73ECh. 18 - Prob. 74ECh. 18 - Prob. 75ECh. 18 - Prob. 76ECh. 18 - Prob. 77ECh. 18 - Prob. 78ECh. 18 - Prob. 79ECh. 18 - Prob. 80ECh. 18 - Prob. 81ECh. 18 - Prob. 82ECh. 18 - Prob. 83ECh. 18 - Prob. 84ECh. 18 - Prob. 85ECh. 18 - Prob. 86ECh. 18 - Prob. 87ECh. 18 - Prob. 88ECh. 18 - Prob. 89ECh. 18 - Prob. 90AECh. 18 - Prob. 91AECh. 18 - Prob. 92AECh. 18 - Prob. 93AECh. 18 - Prob. 94AECh. 18 - Prob. 95AECh. 18 - Prob. 96AECh. 18 - Prob. 97AECh. 18 - Prob. 98AECh. 18 - Prob. 99AECh. 18 - Prob. 100AECh. 18 - Prob. 101AECh. 18 - Prob. 102AECh. 18 - Prob. 103AECh. 18 - Prob. 104AECh. 18 - Prob. 105AECh. 18 - Prob. 106AECh. 18 - Prob. 107AECh. 18 - Prob. 108AECh. 18 - Prob. 109AECh. 18 - Prob. 110AECh. 18 - Prob. 111AECh. 18 - Prob. 112AECh. 18 - Hydrogen gas is being considered as a fuel for...Ch. 18 - Prob. 114AECh. 18 - Prob. 115AECh. 18 - Prob. 116AECh. 18 - Prob. 117AECh. 18 - Prob. 118AECh. 18 - Prob. 119AECh. 18 - What is the molecular structure for each of the...Ch. 18 - Prob. 121AECh. 18 - Prob. 122AECh. 18 - Prob. 123CPCh. 18 - Prob. 124CPCh. 18 - Prob. 125CPCh. 18 - Prob. 126CPCh. 18 - Prob. 127CPCh. 18 - Prob. 128CPCh. 18 - Prob. 129CPCh. 18 - Prob. 130CPCh. 18 - Prob. 131CPCh. 18 - Prob. 132CPCh. 18 - Prob. 133CP
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