Question

According to Le Châtelier’s Principle, the amount of solid reactant or product present does not have an impact on the equilibrium. Why? The solid will not impact the temperature, and thus not the equilibrium The solid does not appear in the equilibrium constant, so adding or removing solid has no effect The solid will not impact the pH and thus not the equilibrium Solids are unreactive and generally ignored when looking at equilibrium due to this

Answer 1

Answer: Option (b) is the correct answer.

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

For example,   $CaCO_{3}(s) + CO_{2}(aq) + H_{2}O(l) \rightleftharpoons Ca^{2+}(aq) + 2HCO^{-}_{3}(aq)$

Hence, expression for equilibrium constant will be as follows.

        $K_{eq} = \frac{[Ca^{2+}][HCO^{-}_{3}]^{2}}{[CO_{2}][H_{2}O]}$

Since, the concentration for a solid substance is considered as 1 or unity. Therefore, adding or removing a solid will not affect the equilibrium.

Thus, we can conclude that according to Le Châtelier’s Principle, the amount of solid reactant or product present does not have an impact on the equilibrium because the solid does not appear in the equilibrium constant, so adding or removing solid has no effect.