Exams › JEE Advanced › Chemistry
Which reason below does not play a role in roasting sulphide ores to oxides instead of directly reducing them with carbon?
- Metal sulphides are more thermodynamically stable compared to CS$_2$.
- CO$_2$ has greater thermodynamic stability than CS$_2$.
- Metal sulphides are not as stable as their oxide counterparts.
- CO$_2$ is less volatile than CS$_2$.
- Metal sulphides are not as stable as their oxide counterparts.
Correct answer: Metal sulphides are not as stable as their oxide counterparts.
Solution
Roasting sulphide ores converts them into oxides, which are generally easier to reduce. The option stating that metal sulphides are not as stable as their oxide counterparts is not a reason for roasting; it is a broad statement and not the specific thermodynamic/practical basis used here.
Related JEE Advanced Chemistry questions
- During the electrolysis process used to extract aluminium from alumina, what is the primary purpose of adding cryolite?
- Find the number of the following reactions that are involved in roasting: i. $\mathrm{S + O_2 \rightarrow SO_2}$ ii. $\mathrm{P_4 + 5O_2 \rightarrow P_4O_{10}}$ iii. $\mathrm{4As + 3O_2 \rightarrow 2As_2O_3}$ iv. $\mathrm{2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2}$ v. $\mathrm{ZnCO_3 \rightarrow ZnO + CO_2}$
- Out of the listed ores, how many can be enriched using the magnetic separation technique?
- Sulfide ores are common for the metals:
- When Cu$_2$S is heated, which of the following substances produces copper metal?
- Galena, a lead ore, undergoes partial oxidation when air is passed through it at elevated temperatures. Afterward, the air flow is stopped, and heating continues in a sealed furnace, allowing self-reduction to occur. What is the mass of lead formed per kilogram of oxygen used? (Atomic masses in g mol\(^{-1}\): O = 16, S = 32, Pb = 207)
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →