Question

Why do solids not affect equilibrium?

Answer 1

Solid:

1. Solid is one of the four fundamental states of matter.
2. The molecules in it are tightly packed together.

Solids do not affect equilibrium as:

1. Solids do not affect equilibrium as the effective concentration of it remains the same throughout the complete reaction.
2. Only the concentration which undergoes change during the process of reaction is included in the expression of the equilibrium.
3. The concentrations of pure solids and pure liquids do not change during the course of the reaction.
4. As the concentration of solids does not undergo any change, these concentrations are not included in the expression for the equilibrium constant.
5. So, solids do not affect equilibrium.

For pure liquids and solids, their concentration can be given by the formula,
$M={\displaystyle \frac{n}{V}}$
Where ‘M’ is the molarity or molar concentration of the substance
‘n’ is the number of moles of substance.
‘V’ is the volume of the solution.
Also,
A number of moles are the ratio of the given mass of the substance and its molar mass.
$\Rightarrow$$n={\displaystyle \frac{m}{M}}$
Where ‘m’ is the given mass and ‘M’ is the molar mass of the substance.
Substituting the expression for ‘n’ in the molarity expression,
$\Rightarrow$$M={\displaystyle \frac{m}{V\times M}}$
Also,

$\Rightarrow$$d={\displaystyle \frac{m}{V}}$
$\Rightarrow$$M={\displaystyle \frac{d}{M}}$

Since both density and molar mass of solid remains the same throughout the reaction at the same temperature and pressure, therefore, the molarity or concentration of pure solids and liquids remains constant throughout the reaction.

Now for writing the equilibrium constant expression of a reaction,

Let us consider the following reaction:
$C(s)+C{O}_2(g)\rightleftharpoons 2CO(g)$
Equilibrium constant expression,
$\Rightarrow$ $K_c={\displaystyle \frac{{[CO]}^2}{[C][C{O}_2]}}$
The concentration of C(s) is constant throughout the reaction.

Therefore, its concentration or activity can be considered as a unity, and it can be ignored in the equilibrium constant expression.
$\Rightarrow$$K_{eq}={\displaystyle \frac{{[CO]}^2}{[1][C{O}_2]}}$
$\Rightarrow K_{eq}={\displaystyle \frac{{[CO]}^2}{[C{O}_2]}}$

Hence, solids can be ignored while writing the equilibrium constant expression because it does not affect it.