Question

Following method of extracting $Zn$ is based on thermodynamics
A : $2ZnS+3O_2⟶2ZnO+2S{O}_2$
B : $ZnO+C⟶Zn+CO$

If $△G_f^o$ (standard free energies of formation, in $kJmo{l}^{-1}$ ) of $ZnS=-205.4,ZnO=-318.0,S{O}_2=-300.4$ and of $CO=-137.3$

Free energy changes of the above reaction A and B (respectively) will be :

• A. $-826.4kJ,+180.7kJ$
• B. $+826.4kJ,-180.7kJ$
• C. $-826.4kJ,-180.7kJ$
• D. $+826.4kJ,+180.7kJ$

Answer 1

The correct option is A $-826.4kJ,+180.7kJ$

For the reaction A :

$2ZnS+3O_2⟶2ZnO+2S{O}_2$

$\Delta G_{rxn}^0=2\Delta G_{ZnO}^0+2\Delta G_{S{O}_2}^0-2\Delta G_{ZnS}^0-3\Delta G_{O2}^0$

$\Delta G_{rxn}^0=2\times (-318.0)+2\times (-300.4)-2\times (-205.4)-3\times 0$

$\Delta G_{rxn}^0=-826kJ/mol$
For the reaction B :

$ZnO+C⟶Zn+CO$

$\Delta G_{rxn}^0=\Delta G_{Zn}^0+\Delta G_{CO}^0-\Delta G_{ZnO}^0-\Delta G_C^0$

$\Delta G_{rxn}^0=0+(-137.3)-(-318.0)-0$

$\Delta G_{rxn}^0=+180.7kJ/mol$