Q47 P - Chemistry: Molecular Nature Of Matter And Change

Question

Metal sulfides are often first converted to oxides by roasting in air and then reduced with carbon to produce the metal. Why aren’t the metal sulfides reduced directly by carbon to yield${\mathbf{\text{CS}}}_{\mathbf{\text{2}}}$? Give a thermodynamic analysis of both processes for$\mathbf{\text{ZnS}}$ .

Step-by-Step Solution

Answer

The reactionsarenot spontaneous; however stepwise reduction is more favourable than the sulfide reaction

Step 1: Concept Introduction

Zinc sulphide has the chemical formula$\mathbf{\text{ZnS}}$  and is an inorganic compound. This is the most common form of zinc found in nature, mostly in the mineral sphalerite.

 

Step 2: Finding outif the Direct Reduction of Zinc Sulfide is Spontaneous or not

In order to compare the processes thermodynamically, Gibb's energy can be compared.

For the direct reduction of $\text{ZnS}$:

$\text{2ZnS(s)+C( graphite )}\to {\text{2Zn(s)+CS}}_{\text{2}}\text{(g)}$

Since ${\text{ΔG}}_{\text{Txn}}^{\text{°}}$is a state function, it depends only on the initial and final values, not the path in between. So,

$\Delta G_{rxn}^{°}=\Delta G_{products}^{°}-\Delta G_{reac\tan ts}^{°}$

From the reaction equation above,

$\Delta G_{rxn}^{°}=[2\times \Delta G_{Zn\left(s\right)}^{°}+\Delta G_{C{S}_2\left(g\right)}^{°}]-[2\times \Delta G_{ZnS\left(s\right)}^{°}+\Delta G_{C\left(graphite\right)}^{°}]$

Using Appendix B,

$\begin{array}{c}\Delta G_{rxn}^{°}=[2 mol\times 0.00\frac{kJ}{mol}+1 mol\times 66.90\frac{kJ}{mol}]-[2 mol\times -198.00\frac{kJ}{mol}+1 mol\times 0.00\frac{kJ}{mol}]\\ \Delta G_{rxn}^{°}=462.90 kJ\end{array}$

Since the ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$is positive, the reaction is not spontaneous at standard conditions $\text{(298 K and 1 atm).}$

Step 3: Finding out the Stepwise reduction of Zinc Sulfide is Spontaneous or not

For the stepwise reduction of $\text{ZnS}$:

Theat first is converted into$\text{ZnO}$, which is further reduced to metallic .$\text{Zn}$

Considering the final oxide reduction step,

$\text{2ZnO(s)+C( graphite )}\to {\text{2Zn(s)+CO}}_{\text{2}}\text{(g)}$

similarly, Gibb's energy can be calculated as

$\Delta G_{rxn}^{°}=[2\times \Delta G_{Zn\left(s\right)}^{°}+\Delta G_{C{O}_2\left(g\right)}^{°}]-[2\times \Delta G_{ZnO\left(s\right)}^{°}+\Delta G_{C\left(graphite\right)}^{°}]$

Using Appendix B,

$\begin{array}{c}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{2 mol×0.00}\frac{\text{kJ}}{\text{mol}}\text{+1 mol×-394.40}\frac{\text{kJ}}{\text{mol}}\right]\text{-}\left[\text{2 mol×-318.20}\frac{\text{kJ}}{\text{mol}}\text{+1 mol×0.00}\frac{\text{kJ}}{\text{mol}}\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=242.00kJ }\end{array}$

This stepwise reduction reaction is also not spontaneous, however, it is more favourable than the sulfide reaction.