None of the above.
The answer given to you is correct, you cannot increase the concentration of a solid, even if dealing with a chemical equilibrium.
In your case, the decomposition of calcium carbonate,
#"CaCO"_text(3(s]) rightleftharpoons "CaO"_text((s]) + "CO"_text(2(g][) uarr#
As you know, solids and pure liquids are not included in the expression of the equilibrium constants
In this example, you would have
#K_c = ["CO"_2]" "#and #" "K_p = ("CO"_2)#
So just by looking at the expressions for the equilibrium constants that the concentration of the two solids has no impact on the position of the equilibrium.
Moreover, what's interesting here is that you an only change the concentration of carbon dioxide by changing the temperature at which the reaction takes place.
This is the case because equilibrium constants are temperature dependent.
So, why are the concentrations of solids and pure liquids said to have no impact on equilibrium reactions?
This is because solids and pure liquids have chemical activities equal to
The chemical activity of a substance is simply the ratio between the concentration of that substance under a specific set of conditions and the concentration of that substance under standard conditions.
The concentration (sometimes given as molar density) of solids is independent of the conditions used for pressure and temperature.
This means that a solid will have the same concentration under certain conditions for pressure and temperature and under standard conditions
Moreover, the volume of a solid is proportional to its mass. Adding more moles of a solid will increase its volume proportionally
What this means is that adding or removing solid from this equilibrium will not affect its position, since the concentrations of the two solids are considered constant.
As long as you have some calcium carbonate and some calcium oxide present, the position of the equilibrium will not change.