Animated Solution for Chemistry - General Principles and Processes of Isolation of Metals: Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

Imagine you are a metallurgist standing in front of a massive pile of zinc sulphide ore. Your goal is simple: extract the pure zinc metal. You know that carbon is the cheapest and most abundant reducing agent available.

So, a very natural, intuitive question arises. Why can't we just mix the sulphide ore with carbon, throw it in a furnace, and get our metal? Why do we go through the extra, energy-intensive, and polluting step of roasting the ore into an oxide first?

To answer this, we have to look past the obvious and dive into the beautiful world of chemical thermodynamics.

Let's entertain the idea of direct reduction. If we react a metal sulphide ($MS$) with carbon ($C$), the carbon has to strip the sulphur away to form carbon disulphide ($C{S}_2$).

The chemical equation would look like this:

Here is where nature throws a wrench in our plans. Carbon disulphide ($C{S}_2$) is a highly unstable, endothermic compound. It simply does not want to form.

Because the byproduct ($C{S}_2$) is so unstable, the overall Gibbs free energy change ($\Delta G$) for this reaction is positive. It is a steep uphill battle. The metal sulphide is much more stable than carbon disulphide, making direct reduction a thermodynamic dead end.

Since we cannot climb the mountain directly, we take a detour. We roast the sulphide ore in the presence of oxygen.

This converts the metal sulphide into a metal oxide ($MO$), releasing sulphur dioxide gas. This reaction is highly exothermic and spontaneous. We are rolling right down the thermodynamic hill.

Now that we have a metal oxide, let's bring carbon back into the picture.

Carbon reacts with the oxide to form pure metal and carbon dioxide ($C{O}_2$). And this is the magic key! Carbon dioxide is incredibly stable.

The immense stability of $C{O}_2$ provides the thermodynamic driving force. It pulls the reaction forward, making the Gibbs free energy change negative. The real reason we roast is because $C{O}_2$ is vastly more stable than $C{S}_2$.

Now, let's look at the options provided in the question. Options (a) and (b) perfectly describe our thermodynamic journey. They highlight the instability of $C{S}_2$ and the stability of $C{O}_2$.

But what about option (c)? It states that metal sulphides are less stable than the corresponding oxides.

Even if this statement is factually true, it does not explain why carbon fails to reduce the sulphide. Think about it: if the sulphide were less stable, it should be easier to break apart!

The failure of direct reduction is entirely due to the byproduct, $C{S}_2$. Therefore, the relative stability of the ore itself is of no significance to this specific metallurgical choice.