Question

A sulphide ore is rarely reduced directly with carbon.
A. If roasted in air, the products are $CO,C{O}_2$ and $S{O}_2$. These being gases escape from the reaction and thus, entropy change is positive and reaction is thermodynamically favourable.
B. If directly reduced with carbon, $C{S}_2$ (liquid) is formed which remain in reaction mixture. Entropy change is less positive and reaction is thermodynamically less favourable.
1. A is correct
2. B is correct
3. A and B are correct
4. A and B are wrong

Answer 1

In practice, sulphide ore is first roasted in air to the corresponding oxide which in turn is reduced to metal with carbon.

$2ZnS+3O_2\to 2ZnO+2S{O}_2\uparrow$
$ZnO+C\to Zn+CO\uparrow$
$2ZnO+C\to 2Zn+C{O}_2\uparrow$

$CO,C{O}_2$ and $S{O}_2$ being gases escape from the reaction and thus, entropy change is positive and reaction is thermodynamically favourable.

If $ZnS$ is directly reduced with carbon, $C{S}_2$ (liquid) is formed which remain in reaction mixture. Entropy change is less positive and reaction is thermodynamically less favourable.

$2ZnS+C\to 2Zn+C{S}_2$ (liquid)