   # Using the following data, calculate the standard heat of formation of ICl( g ) in kJ/mol: Cl 2 ( g ) → 2 Cl ( g ) Δ H ° = 242.3 KJ I 2 ( g ) → 2 I ( g ) Δ H ° = 151.0 KJ ICI ( g ) → I ( g ) + CI ( g ) Δ H ° = 211.3 KJ I 2 ( s ) → I 2 ( g ) Δ H ° = 62.8 KJ ### 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 7, Problem 108AE
Textbook Problem
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## Using the following data, calculate the standard heat of formation of ICl(g) in kJ/mol: Cl 2 ( g ) → 2 Cl ( g ) Δ H ° = 242.3   KJ I 2 ( g ) → 2 I ( g ) Δ H ° = 151.0   KJ ICI ( g ) → I ( g ) + CI ( g ) Δ H ° = 211.3   KJ I 2 ( s ) → I 2 ( g ) Δ H ° = 62.8   KJ

Interpretation Introduction

Interpretation: The standard heat of formation of  ICI(g) should be calculated.

Concept Introduction:

Standard Enthalpy of Formation:

• The one mole of the product is formed by its standard state via the reaction of its elements in standard state at standard conditions.
• The at constant pressure the standard enthalpy of formation is equal to standard heat of formation.

Hess's Law:

•  Hess's Law is saying that if you convert reactants to products the overall enthalpy change will be exactly the same whether you do it in one or multiple steps.

Formula:

ΔH(total)ΔH(number of steps)......(1)

Rules:

• The sign of ΔH reversed when the reaction is reversed.
• The quantities of reactants and products is directly proportional to the magnitude of ΔH in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer.

### Explanation of Solution

Explanation

Cl2(g)2Cl(g)ΔH°1=242.3KJ......(1)I2(g)2I(g)ΔH°2=151.0KJ......(2)ICl(g)I(g)+Cl(g)ΔH°3=211.3KJ......(3)I2(s)I2(g)ΔH°4=62.8KJ......(4)

To calculate: The standard heat of formation of  ICI(g) .

Explanation:

ΔH=ΔH1+ΔH2+ΔH3+ΔH4.....(2)

12Cl(g)Cl(g)ΔH1=12(242.3KJ)......(1)12I(g)I(g)ΔH2=12(151.0KJ)......(2)I(g)+Cl(g)ICl(g)ΔH3=211.3KJ......(3)12I(s)12I(g)ΔH2=12(62

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