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 $\left(\Delta H_{\text{balanced rxn}}^{\text{ο}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$

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

Answer to Problem 5.120AP

Solution

The standard enthalpy change for the balanced chemical reaction is $\underset{\_}{-5,118.4kJ}$.

Explanation of Solution

Explanation

The given reactions are stated as,

$\begin{array}{c}\text{Zn}\left(s\right)+\frac{1}{8}{\text{S}}_8\left(s\right)\to \text{ZnS}\left(s\right)\\ \text{ZnS}\left(s\right)+2{\text{O}}_2\left(g\right)\to {\text{ZnSO}}_{\text{4}}\left(s\right)\\ \frac{1}{8}{\text{S}}_8\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)\end{array}$

Refer Appendix $4$.

The standard enthalpy of formation of ${\text{SO}}_2\left(g\right)$, $\left(\Delta H_{{\text{f, SO}}_2\left(g\right)}^{\text{ο}}\right)$ is $-296.8\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{ZnSO}}_4\left(s\right)$, $\left(\Delta H_{{\text{f, ZnSO}}_4\left(s\right)}^{\text{ο}}\right)$ is $-982.8\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{O}}_2\left(g\right)$, $\left(\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}\right)$ is $0\text{kJ}/\text{mol}$.

The standard enthalpy of formation of $\text{Zn}\left(s\right)$, $\left(\Delta H_{\text{f, Zn}\left(s\right)}^{\text{ο}}\right)$ is $0\text{kJ}/\text{mol}$.

The standard enthalpy of formation of ${\text{S}}_8\left(s\right)$, $\left(\Delta H_{{\text{f, S}}_8\left(s\right)}^{\text{ο}}\right)$ is $0\text{kJ}/\text{mol}$.

Add the above reaction stated as given.

$\begin{array}{c}\text{Zn}\left(s\right)+\frac{1}{8}{\text{S}}_8\left(s\right)\to \text{ZnS}\left(s\right)\\ \text{ZnS}\left(s\right)+2{\text{O}}_2\left(g\right)\to {\text{ZnSO}}_{\text{4}}\left(s\right)\\ \frac{1}{8}{\text{S}}_8\left(s\right)+{\text{O}}_2\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)\end{array}$

$\overline{\underset{\_}{\text{Zn}\left(s\right)+\frac{1}{4}{\text{S}}_8\left(s\right)+3{\text{O}}_2\left(g\right)+\text{ZnS}\left(s\right)\to {\text{ZnSO}}_{\text{4}}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)+\text{ZnS}\left(s\right)}}$ (1)

Eliminate the common terms of equation (1).

$\text{Zn}\left(s\right)+\frac{1}{4}{\text{S}}_8\left(s\right)+3{\text{O}}_2\left(g\right)\to {\text{ZnSO}}_{\text{4}}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)$ (2)

The overall equation (2) is not in balanced form as fraction $1/4$ is the coefficient of ${\text{S}}_8\left(s\right)$.

To remove the fraction coefficient of ${\text{S}}_8\left(s\right)$, multiply the equation (2) with four.

$4\times \left(\text{Zn}\left(s\right)+\frac{1}{4}{\text{S}}_8\left(s\right)+3{\text{O}}_2\left(g\right)\to {\text{ZnSO}}_{\text{4}}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)\right)$

$\overline{\underset{\_}{\text{4Zn}\left(s\right)+{\text{S}}_8\left(s\right)+12{\text{O}}_2\left(g\right)\to 4{\text{ZnSO}}_{\text{4}}\left(s\right)+4{\text{SO}}_{\text{2}}\left(g\right)}}$ .(3)

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

The standard enthalpy change for a balanced reaction $\left(\Delta H_{\text{balanced rxn}}^{\text{ο}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (4)

Where,

• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)$ is standard enthalpy of formation of the products.
• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)$ 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 ${\text{ZnSO}}_4\left(s\right)$ is $4$.
• Number of moles of product ${\text{SO}}_2\left(g\right)$ is $4$.
• Number of moles of reactant $\text{Zn}\left(s\right)$ is $4$.
• Number of moles of reactant ${\text{S}}_8\left(s\right)$ is $1$.
• Number of moles of reactant ${\text{O}}_2\left(g\right)$ is $12$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the balanced chemical reaction is calculated by the formula,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=4\text{mol}\times \left(\Delta H_{{\text{f, ZnSO}}_4\left(s\right)}^{\text{ο}}\right)+4\text{mol}\times \left(\Delta H_{{\text{f, SO}}_2\left(g\right)}^{\text{ο}}\right)$ (5)

Substitute the value of $\Delta H_{{\text{f, ZnSO}}_4\left(s\right)}^{\text{ο}}$ and $\Delta H_{{\text{f, SO}}_2\left(g\right)}^{\text{ο}}$ in equation (5).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=4\text{mol}\times -982.8\text{kJ}/\text{mol}+4\text{mol}\times -296.8\text{kJ}/\text{mol}\\ =-3,931.2\text{kJ}-1,187.2\text{kJ}\\ =-5,118.4\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the balanced chemical reaction is calculated by the formula,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=4\text{mol}\times \left(\Delta H_{\text{f, Zn}\left(s\right)}^{\text{ο}}\right)+1\text{mol}\times \left(\Delta H_{{\text{f, S}}_8\left(s\right)}^{\text{ο}}\right)+12\text{mol}\times \left(\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}\right)$

Substitute the values of $\Delta H_{\text{f, Zn}\left(s\right)}^{\text{ο}}$, $\Delta H_{{\text{f, S}}_8\left(s\right)}^{\text{ο}}$ and $\Delta H_{{\text{f, O}}_2\left(g\right)}^{\text{ο}}$ in above expression.

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=4\text{mol}\times 0\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}+12\text{mol}\times 0\text{kJ}/\text{mol}\\ \text{=0}\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (2).

$\begin{array}{c}\Delta H_{\text{balanced rxn}}^{\text{ο}}=-5,118.4\text{kJ}-\text{0}\text{kJ}\\ =-5,118.4\text{kJ}\end{array}$

Hence, the standard enthalpy change for the balanced chemical reaction is $\underset{\_}{-5,118.4kJ}$.