The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal:
Fe$$_2$$O$$_3$$(s) + 3CO(g) $$\rightleftharpoons$$ 2Fe(l) + 3CO$$_2$$(g)
Using the Le Chatelier's principle, predict which one of the following will not disturb the equilibrium?

We first write the balanced chemical equation involved inside the blast furnace:

$$Fe_2O_3(s)+3CO(g)\;\rightleftharpoons\;2Fe(l)+3CO_2(g)$$

According to Le Chatelier’s principle, whenever a stress (change in concentration, pressure, temperature, etc.) is applied to a system at equilibrium, the system shifts in the direction that tends to nullify that stress.

In the given equilibrium, we notice two important facts:

1. The solid $$Fe_2O_3(s)$$ and the liquid $$Fe(l)$$ do not appear in the equilibrium-constant expression because their activities are taken as unity. Hence, changing the amount of either solid or liquid ordinarily has no effect on the position of equilibrium.

2. Only the gaseous species $$CO(g)$$ and $$CO_2(g)$$ enter the equilibrium-constant expression. Thus alterations in the concentrations (or partial pressures) of $$CO$$ or $$CO_2$$ will shift the equilibrium.

Let us now examine each option one by one:

Option A - Addition of $$CO_2$$: Adding $$CO_2$$ increases the concentration of a product gas. Le Chatelier’s principle tells us the equilibrium will shift to the left to oppose this increase, producing more $$CO$$ and $$Fe_2O_3$$. Hence, the equilibrium is disturbed.

Option B - Removal of $$CO_2$$: Removing a product gas lowers its concentration. The system will therefore shift to the right to restore $$CO_2$$ by consuming $$CO$$ and $$Fe_2O_3$$. Thus, again the equilibrium is disturbed.

Option C - Addition of $$Fe_2O_3$$: Because $$Fe_2O_3$$ is a solid, its activity is constant and equal to unity in the equilibrium-constant expression. Increasing the amount of a pure solid does not change its activity; consequently, the equilibrium position as determined by the gases remains unaltered. Therefore, this action does not disturb the equilibrium.

Option D - Removal of $$CO$$: Removing a reactant gas decreases its concentration. The equilibrium will shift to the left to make more $$CO$$ by converting $$CO_2$$ back into $$CO$$, hence disturbing the equilibrium.

From the detailed analysis above, only the addition of the solid $$Fe_2O_3$$ leaves the gas-phase equilibrium untouched.

Hence, the correct answer is Option C.