Explanation The chemical equation for the formation of P I 3 ( s ) can be represented as: 2 P ( s ) + 3 I 2 ( s ) → 2 P I 3 ( s ) , Δ H ( f , P I 3 ( g ) ) = 2 ( – 24.7 kJ/mol ) = − 49.4 k J / m o l Reverse the above equation, to get, the equation for the dissociation of P I 3 molecule: 2 P I 3 ( s ) → 2 P ( s ) + 3 I 2 ( s ) , Δ H = + 49.4 k J / m o l ..... ( 1 ) The formation of gaseous iodine can be represented as: 3 I 2 ( s ) → 3 I 2 ( g ) , Δ H ( f , I 2 ( g ) ) = 3 ( 62 k J / m o l ) = 186 k J / m o l ..... ( 2 ) The formation of gaseous phosphorus can be represented as: 2 P ( s ) → 2 P ( g ) , Δ H ( f , P ( g ) ) = 3 ( 334 k J / m o l ) = 668 k J / m o l ..... ( 3 ) The formation of iodine atom from a iodine molecule can be represented as: 3 I 2 ( g ) → 6 I ( g ) Δ H ( b o n d e n e r g y o f I 2 ) = 3 ( 151 ) k J / m o l = 453 k J / m o l ....... ( 4 ) P I 3 has 3 P − I bonds, thus, for 2moles of P I 3 : 2 P I 3 ( g ) → 2 P ( g ) + 6 I ( g ) Δ H = 1104 k J / m o l Reverse the above equation, to get, the equation for the gaseous P I 3 molecule: 2 P ( g ) + 6 I ( g ) → 2 P I 3 ( g ) Δ H = − 1104 k J / m o l ...... ( 5 ) Add equations from 1 to 6, to get, 2 P I 3 ( s ) → 2 P ( s ) + 3 I 2 ( s ) , Δ H = + 49

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

4th Edition

ISBN: 9780134162430

Author: Nivaldo J. Tro

Publisher: PEARSON

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Question

Chapter 9, Problem 121E

Interpretation Introduction

To calculate: The heat of sublimation of PI3(s) using the following data:

The standard heat of formation of PI3(s) is –24.7 kJ/mol.

The bond energy of PI is 184 kJ/mol.

The standard heat of formation of P(g) is 334 kJ/mol.

The standard heat of formation of I2(g) is 62 kJ/mol.

The I₂ bond energy is 151 kJ/mol.

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Chapter 9, Problem 120E

Chapter 9, Problem 122E

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Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

Ch. 9 - Prob. 1SAQ Ch. 9 - Q2. Which set of elements is arranged in order of...Ch. 9 - Q3. Which is the correct Lewis structure for...Ch. 9 - Q4. Which compound is likely to have an incomplete...Ch. 9 - Prob. 5SAQ Ch. 9 - Prob. 6SAQ Ch. 9 - Prob. 7SAQ Ch. 9 - Prob. 8SAQ Ch. 9 - Prob. 9SAQ Ch. 9 - Prob. 10SAQ

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