Question

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer 1

Hint: For pure liquids and solids, their concentration can be given by the formula,
$\Rightarrow$ $M = \dfrac{n}{V}$
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.

Complete step by step answer:
As said above the concentration of a substance is given by the formula,
$\Rightarrow$ $M = \dfrac{n}{V}$
Number of moles is the ratio of given mass of the substance and its molar mass.
$\Rightarrow$So, $n = \dfrac{m}{M}$
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$$M = \dfrac{m}{{V \times M}}$
As, mass divided by volume is equal to density, that is,
$\Rightarrow$$d = \dfrac{m}{V}$
$\Rightarrow$So, $M = \dfrac{d}{M}$
Since, both density and molar mass of a pure liquid or 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,
$\Rightarrow$ ${K_c} = \dfrac{{{{[CO]}^2}}}{{[C][C{O_2}]}}$
As the concentration of C(s) is constant throughout the reaction, therefore, its concentration or activity can be considered as unity, and it can be ignored in the equilibrium constant expression.
$\Rightarrow$${K_{eq}} = \dfrac{{{{[CO]}^2}}}{{[1][C{O_2}]}}$
$ \Rightarrow {K_{eq}} = \dfrac{{{{[CO]}^2}}}{{[C{O_2}]}}$

Note:
The rate law expression should not be confused with the equilibrium constant expression. Rate law expression for a reaction,
$aA + bB \to cC + dD$
Is given as, $rate = k{[A]^m}{[B]^n}$
Here, the concentration of the reactants is not necessarily raised to the power equal to their stoichiometric coefficients. In this expression, ‘m’ and ‘n’ represent the order of reaction with respect to A and B respectively.
$m + n$ give the overall order of the reaction.