Q7 / 200 · Manipal (MET) 2013 · Chemistry
Following method of extracting Zn is based on thermodynamics
$A: 2 \mathrm{ZnS}+3 \mathrm{O}_3 \longrightarrow 2 \mathrm{ZnO}+2 \mathrm{SO}_2$
$B: \mathrm{ZnO}+\mathrm{C} \longrightarrow \mathrm{Zn}+\mathrm{CO}$
If $\Delta G_f^{\circ}$ (standard free energies of formation, in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of $\mathrm{ZnS}=-205.4$,
$\begin{aligned} & \mathrm{ZnO}=-318.0, \mathrm{SO}_2=-300.4 \\ & \mathrm{CO}=-137.3\end{aligned}$
Free energy changes of the above reaction $A$ and $B$ (respectively) will be
- Correct answerA$-826.4 \mathrm{~kJ},+180.9 \mathrm{~kJ}$
- B$+826.4 \mathrm{~kJ},-180.9 \mathrm{~kJ}$
- C$-826.4 \mathrm{~kJ} \mathrm{~J}_1-180.9 \mathrm{~kJ}$
- D$+826.4 \mathrm{~kJ},+180.9 \mathrm{~kJ}$
Solution
For reaction $A$
$\begin{aligned} \Delta G^{\circ} & =2 \Delta G_f^{\circ}(\mathrm{ZnO})+2 \Delta G_f^{\circ}\left(\mathrm{SO}_2\right)-2 G_f^{\circ}(\mathrm{ZnS}) \\ & =2[-3182-300.4+205.4] \\ & =-826.4 \mathrm{~kJ} \quad\left(\Delta G_f^{\circ}(\text { element })=0\right)\end{aligned}$
For reaction $B$
$\begin{aligned} \Delta G^{\circ} & =\Delta G_f^{\circ}(\mathrm{CO})-\Delta G_f^{\circ}(\mathrm{ZnO}) \\ & =-137.3+328.2 \\ & =+180.9 \mathrm{~kJ}\end{aligned}$
Hence, the correct answer is option: 'A'.