The standard enthalpy change for the balanced reaction is to be calculated. Concept introduction: The standard enthalpy change for a balanced reaction ( Δ H balanced rxn ο ) is calculated by the formula, Δ H balanced rxn ο = ∑ n × Δ H f ο ( Products ) − ∑ m × Δ H f ο ( Reactants ) To determine: The standard enthalpy change for the balanced chemical reaction .

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Answer 1

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Chapter 5, Problem 5.120AP

Interpretation Introduction

Interpretation: The standard enthalpy change for the balanced reaction is to be calculated.

Concept introduction: The standard enthalpy change for a balanced reaction $(ΔHbalanced rxnο)$ is calculated by the formula,

$ΔHbalanced rxnο=∑n×ΔHfο(Products)−∑m×ΔHfο(Reactants)$

To determine: The standard enthalpy change for the balanced chemical reaction.

Expert Solution & Answer

Answer to Problem 5.120AP

Solution

The standard enthalpy change for the balanced chemical reaction is $-5,118.4 kJ_$ .

Explanation of Solution

Explanation

The given reactions are stated as,

$Zn(s)+18S8(s)→ZnS(s)ZnS(s)+2O2(g)→ZnSO4(s)18S8(s)+O2(g)→SO2(g)$

Refer Appendix $4$ .

The standard enthalpy of formation of $SO2(g)$ , $(ΔHf, SO2(g)ο)$ is $−296.8 kJ/mol$ .

The standard enthalpy of formation of $ZnSO4(s)$ , $(ΔHf, ZnSO4(s)ο)$ is $−982.8 kJ/mol$ .

The standard enthalpy of formation of $O2(g)$ , $(ΔHf, O2(g)ο)$ is $0 kJ/mol$ .

The standard enthalpy of formation of $Zn(s)$ , $(ΔHf, Zn(s)ο)$ is $0 kJ/mol$ .

The standard enthalpy of formation of $S8(s)$ , $(ΔHf, S8(s)ο)$ is $0 kJ/mol$ .

Add the above reaction stated as given.

$Zn(s)+18S8(s)→ZnS(s) ZnS(s)+2O2(g)→ZnSO4(s) 18S8(s)+O2(g)→SO2(g)$

$Zn(s)+14S8(s)+3O2(g)+ZnS(s)→ZnSO4(s)+SO2(g)+ZnS(s)_¯$ (1)

Eliminate the common terms of equation (1).

$Zn(s)+14S8(s)+3O2(g)→ZnSO4(s)+SO2(g)$ (2)

The overall equation (2) is not in balanced form as fraction $1/4$ is the coefficient of $S8(s)$ .

To remove the fraction coefficient of $S8(s)$ , multiply the equation (2) with four.

$4×(Zn(s)+14S8(s)+3O2(g)→ZnSO4(s)+SO2(g))$

$4Zn(s)+S8(s)+12O2(g)→4ZnSO4(s)+4SO2(g)_¯$ .(3)

The equation (3) is now in overall balanced form.

The standard enthalpy change for a balanced reaction $(ΔHbalanced rxnο)$ is calculated by the formula,

$ΔHbalanced rxnο=∑n×ΔHfο(Products)−∑m×ΔHfο(Reactants)$ (4)

Where,

$ΔHfο(Products)$ is standard enthalpy of formation of the products.
$ΔHfο(Reactants)$ is of standard enthalpy of formation of the reactants.
$n$ is number of moles of products.
$m$ is number of moles of reactants.

In the balanced chemical equation (3) the,

Number of moles of product $ZnSO4(s)$ is $4$ .
Number of moles of product $SO2(g)$ is $4$ .
Number of moles of reactant $Zn(s)$ is $4$ .
Number of moles of reactant $S8(s)$ is $1$ .
Number of moles of reactant $O2(g)$ is $12$ .

The $∑n×ΔHfο(Products)$ for the balanced chemical reaction is calculated by the formula,

$∑n×ΔHfο(Products)=4 mol×(ΔHf, ZnSO4(s)ο)+4 mol×(ΔHf, SO2(g)ο)$ (5)

Substitute the value of $ΔHf, ZnSO4(s)ο$ and $ΔHf, SO2(g)ο$ in equation (5).

$∑n×ΔHfο(Products)=4 mol×−982.8 kJ/mol+4 mol×−296.8 kJ/mol=−3,931.2 kJ−1,187.2 kJ=−5,118.4 kJ$

The $∑m×ΔHfο(Reactants)$ for the balanced chemical reaction is calculated by the formula,

$∑m×ΔHfο(Reactants)=4 mol×(ΔHf, Zn(s)ο)+1 mol×(ΔHf, S8(s)ο)+12 mol×(ΔHf, O2(g)ο)$

Substitute the values of $ΔHf, Zn(s)ο$ , $ΔHf, S8(s)ο$ and $ΔHf, O2(g)ο$ in above expression.

$∑m×ΔHfο(Reactants)=4 mol×0 kJ/mol+1 mol×0kJ/mol+12 mol×0kJ/mol=0 kJ$

Substitute the values of $∑n×ΔHfο(Products)$ and $∑m×ΔHfο(Reactants)$ in equation (2).

$ΔHbalanced rxnο=−5,118.4 kJ−0 kJ=−5,118.4 kJ$

Hence, the standard enthalpy change for the balanced chemical reaction is $-5,118.4 kJ_$ .

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Answer 2

Question

Chapter 5, Problem 5.120AP

Interpretation Introduction

Interpretation: The standard enthalpy change for the balanced reaction is to be calculated.

Concept introduction: The standard enthalpy change for a balanced reaction $(ΔHbalanced rxnο)$ is calculated by the formula,

$ΔHbalanced rxnο=∑n×ΔHfο(Products)−∑m×ΔHfο(Reactants)$

To determine: The standard enthalpy change for the balanced chemical reaction.

Expert Solution & Answer

Answer to Problem 5.120AP

Solution

The standard enthalpy change for the balanced chemical reaction is $-5,118.4 kJ_$ .

Explanation of Solution

Explanation

The given reactions are stated as,

$Zn(s)+18S8(s)→ZnS(s)ZnS(s)+2O2(g)→ZnSO4(s)18S8(s)+O2(g)→SO2(g)$

Refer Appendix $4$ .

The standard enthalpy of formation of $SO2(g)$ , $(ΔHf, SO2(g)ο)$ is $−296.8 kJ/mol$ .

The standard enthalpy of formation of $ZnSO4(s)$ , $(ΔHf, ZnSO4(s)ο)$ is $−982.8 kJ/mol$ .

The standard enthalpy of formation of $O2(g)$ , $(ΔHf, O2(g)ο)$ is $0 kJ/mol$ .

The standard enthalpy of formation of $Zn(s)$ , $(ΔHf, Zn(s)ο)$ is $0 kJ/mol$ .

The standard enthalpy of formation of $S8(s)$ , $(ΔHf, S8(s)ο)$ is $0 kJ/mol$ .

Add the above reaction stated as given.

$Zn(s)+18S8(s)→ZnS(s) ZnS(s)+2O2(g)→ZnSO4(s) 18S8(s)+O2(g)→SO2(g)$

$Zn(s)+14S8(s)+3O2(g)+ZnS(s)→ZnSO4(s)+SO2(g)+ZnS(s)_¯$ (1)

Eliminate the common terms of equation (1).

$Zn(s)+14S8(s)+3O2(g)→ZnSO4(s)+SO2(g)$ (2)

The overall equation (2) is not in balanced form as fraction $1/4$ is the coefficient of $S8(s)$ .

To remove the fraction coefficient of $S8(s)$ , multiply the equation (2) with four.

$4×(Zn(s)+14S8(s)+3O2(g)→ZnSO4(s)+SO2(g))$

$4Zn(s)+S8(s)+12O2(g)→4ZnSO4(s)+4SO2(g)_¯$ .(3)

The equation (3) is now in overall balanced form.

The standard enthalpy change for a balanced reaction $(ΔHbalanced rxnο)$ is calculated by the formula,

$ΔHbalanced rxnο=∑n×ΔHfο(Products)−∑m×ΔHfο(Reactants)$ (4)

Where,

$ΔHfο(Products)$ is standard enthalpy of formation of the products.
$ΔHfο(Reactants)$ is of standard enthalpy of formation of the reactants.
$n$ is number of moles of products.
$m$ is number of moles of reactants.

In the balanced chemical equation (3) the,

Number of moles of product $ZnSO4(s)$ is $4$ .
Number of moles of product $SO2(g)$ is $4$ .
Number of moles of reactant $Zn(s)$ is $4$ .
Number of moles of reactant $S8(s)$ is $1$ .
Number of moles of reactant $O2(g)$ is $12$ .