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
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

(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

(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

(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

(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
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
Principles Of Chemical Reactivity: Energy And Chemical Reactions. 58PS
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