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Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Not helpful? See similar books
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Principles Of Chemical Reactivity: Energy And Chemical Reactions. 58PS

The enthalpy change for the formation the following reaction has to be determined. Concept Introduction: If a reaction proceeds in two or more other reaction, Δ r H 0 for the overall process is the sum of Δ r H 0 values of all those reactions-Hess’s law. The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as: Δ r H= enthalpy change number of moles Δ r H= Δ H number of moles

Question
Chapter 5, Problem 58PS

(a)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,ΔrH0 for the overall process is the sum of ΔrH0 values of all those reactions-Hess’s law.

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

  ΔrH=enthalpy changenumber of moles 

  ΔrH=ΔHnumber of moles 

(a)

Expert Solution

Answer to Problem 58PS

The enthalpy change for the formation is -4.6kJ

Explanation of Solution

Given mass is 1.0g

The enthalpy of formation is  -296.8 KJ/Mol 

S+O2SO2  ΔfH0= -296.8 KJ/Mol

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

  ΔrH=enthalpy changenumber of moles 

  ΔrH=ΔHnumber of moles 

ΔrH=1.0g(1mol÷64.066 g)(-296.8kJ/mol÷1mol) =-4.6kJ

So, the enthalpy change for the formation is -4.6kJ

(b)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,ΔrH0 for the overall process is the sum of ΔrH0 values of all those reactions-Hess’s law.

(b)

Expert Solution

Answer to Problem 58PS

The enthalpy change for the formation is 18.14kJ

Explanation of Solution

Given mass is 0.20mol

The enthalpy of formation is 90.7kJ 

  HgO2Hg+O2  ΔfH0= 90.7kJ

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

  ΔrH=enthalpy changenumber of moles 

  ΔrH=ΔHnumber of moles 

  ΔfH0= 90.7kJ/mol (reverse rxn)

  (90.7kJ1mol)(0.20mol) =18.14kJ

So, the change in enthalpy is 18.14kJ

(c)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,ΔrH0 for the overall process is the sum of ΔrH0 values of all those reactions-Hess’s law

(c)

Expert Solution

Answer to Problem 58PS

The change in enthalpy is -7.3kJ

Explanation of Solution

Given mass is 2.4g

The enthalpy of formation is ΔfH0=-411.12kJ/mol

  2Na +Cl2NaCl  ΔfH0=-411.12kJ/mol

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

  ΔrH=enthalpy changenumber of moles 

  ΔrH=ΔHnumber of moles 

  ΔfH0=-46.0 kJ/mol

  2.40g×1mol15.01464g=0.159mol

Energy change = -46.0 kJ/mol×0.159mol=-7.3kJ

So, the change in enthalpy is -7.3kJ

(d)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,ΔrH0 for the overall process is the sum of ΔrH0 values of all those reactions-Hess’s law.

(d)

Expert Solution

Answer to Problem 58PS

The change in enthalpy is -4.13 kJ

Explanation of Solution

Given mass is 1.05×10-2mol

The enthalpy of formation is ΔfH0=-393.5 kJ/mol 

  C +O2CO2  ΔfH0=-393.5 kJ/mol 

The change in enthalpy, ΔH in kJ per mole of a given reactant for the reaction can be calculated as:

  ΔrH=enthalpy changenumber of moles 

  ΔrH=ΔHnumber of moles 

  ΔfH0=-393.5 kJ/mol 

Energy change =(-393.5 kJ/mol 1mol)(1.05×10-2mol)=-4.13 kJ

So, the change in enthalpy is -4.13 kJ

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Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Not helpful? See similar books
Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
Principles Of Chemical Reactivity: Energy And Chemical Reactions. 58PS
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Chapter 5 Solutions

Chemistry & Chemical Reactivity

Ch. 5.8 - Prob. 1.1ACPCh. 5.8 - Prob. 1.2ACPCh. 5.8 - Prob. 1.3ACPCh. 5.8 - Prob. 2.1ACPCh. 5.8 - Prob. 2.2ACPCh. 5.8 - Prob. 2.3ACPCh. 5.8 - Prob. 2.4ACPCh. 5.8 - Prob. 2.5ACPCh. 5 - Prob. 1PSCh. 5 - Prob. 2PSCh. 5 - Prob. 3PSCh. 5 - Prob. 4PSCh. 5 - Prob. 5PSCh. 5 - Prob. 6PSCh. 5 - Prob. 7PSCh. 5 - Prob. 8PSCh. 5 - Prob. 9PSCh. 5 - Prob. 10PSCh. 5 - Prob. 11PSCh. 5 - Prob. 12PSCh. 5 - Prob. 13PSCh. 5 - Prob. 14PSCh. 5 - Prob. 15PSCh. 5 - Prob. 16PSCh. 5 - Prob. 17PSCh. 5 - Prob. 18PSCh. 5 - Prob. 19PSCh. 5 - Prob. 20PSCh. 5 - Prob. 21PSCh. 5 - Prob. 22PSCh. 5 - Prob. 23PSCh. 5 - Prob. 24PSCh. 5 - Prob. 25PSCh. 5 - Prob. 26PSCh. 5 - Prob. 27PSCh. 5 - Prob. 28PSCh. 5 - Prob. 29PSCh. 5 - Prob. 30PSCh. 5 - Prob. 31PSCh. 5 - Prob. 32PSCh. 5 - Prob. 33PSCh. 5 - Prob. 34PSCh. 5 - Prob. 35PSCh. 5 - Prob. 36PSCh. 5 - Prob. 37PSCh. 5 - Prob. 38PSCh. 5 - Prob. 39PSCh. 5 - Prob. 40PSCh. 5 - Prob. 41PSCh. 5 - Prob. 42PSCh. 5 - Prob. 43PSCh. 5 - Prob. 44PSCh. 5 - Prob. 45PSCh. 5 - Prob. 46PSCh. 5 - Prob. 47PSCh. 5 - Prob. 48PSCh. 5 - Prob. 49PSCh. 5 - Prob. 50PSCh. 5 - Prob. 51PSCh. 5 - Prob. 52PSCh. 5 - Prob. 53PSCh. 5 - Prob. 54PSCh. 5 - Prob. 55PSCh. 5 - Prob. 56PSCh. 5 - Prob. 57PSCh. 5 - Prob. 58PSCh. 5 - Prob. 59PSCh. 5 - Prob. 60PSCh. 5 - Prob. 61PSCh. 5 - Prob. 62PSCh. 5 - Prob. 63PSCh. 5 - Prob. 64PSCh. 5 - Prob. 65GQCh. 5 - Prob. 66GQCh. 5 - Prob. 67GQCh. 5 - Prob. 68GQCh. 5 - Prob. 69GQCh. 5 - Prob. 70GQCh. 5 - Prob. 71GQCh. 5 - Prob. 72GQCh. 5 - Prob. 73GQCh. 5 - Prob. 74GQCh. 5 - Prob. 75GQCh. 5 - Prob. 76GQCh. 5 - Prob. 77GQCh. 5 - Prob. 78GQCh. 5 - Prob. 79GQCh. 5 - Prob. 80GQCh. 5 - Prob. 81GQCh. 5 - Prob. 82GQCh. 5 - Prob. 83GQCh. 5 - Prob. 84GQCh. 5 - Prob. 85GQCh. 5 - Prob. 86GQCh. 5 - Prob. 87GQCh. 5 - Prob. 88GQCh. 5 - Prob. 89GQCh. 5 - Prob. 90GQCh. 5 - Prob. 91GQCh. 5 - Prob. 93ILCh. 5 - Prob. 94ILCh. 5 - Prob. 95ILCh. 5 - Prob. 96ILCh. 5 - Prob. 97ILCh. 5 - Prob. 98ILCh. 5 - Prob. 99ILCh. 5 - Prob. 100ILCh. 5 - Prob. 101SCQCh. 5 - Prob. 102SCQCh. 5 - Prob. 103SCQCh. 5 - Prob. 104SCQCh. 5 - Prob. 105SCQCh. 5 - Prob. 106SCQCh. 5 - Prob. 107SCQCh. 5 - Prob. 108SCQCh. 5 - Prob. 109SCQCh. 5 - Prob. 110SCQCh. 5 - Prob. 111SCQCh. 5 - Prob. 112SCQCh. 5 - Prob. 113SCQCh. 5 - Prob. 114SCQCh. 5 - Prob. 115SCQCh. 5 - Prob. 116SCQCh. 5 - Prob. 117SCQCh. 5 - Prob. 119SCQCh. 5 - Prob. 120SCQ
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