Question

Derive the relation $K_p=K_c(RT{)}^{\Delta n_g}$ for a general chemical equilibrium reaction.

Answer 1

Consider a general chemical equilibrium reaction in which the reactants and products are in gaseous phases.
$aA+bB+cC+.......\rightleftharpoons L+mM+nN+.....$
then $K_p=\frac{P_L^L\cdot P_M^M\cdot P_N^N....}{p_n^A\cdot P_B^b\cdot P_C^c....}$
Where p is a partial pressure of the respective gases. In terms of normal concentrations of reactants and products.
$K_c=\frac{\left[L{]}^l\right[M{]}^m[N{]}^n....}{\left[A{]}^a\right[B{]}^b[C{]}^c....}$
For any gaseous component 'i' in a maxture, its partial pressure, ${}^{\prime }{p}_i^{\prime }$ is related to its molar concentration 'ci' as
$C_1=\frac{Pi}{RT}since{p}_i=\left(\frac{n_i}{V}\right)RT$
When $\left(\frac{n_i}{V}\right)=C_i=$ number of moles of 'i' per litre and
V= volume in litres.
Substituting concentration terms by partial pressures,
$K_c=\frac{\left(P_L/RT{)}^l\right(P_M/RT{)}^m(P_n/RT{)}^n....}{\left(P_A/RT{)}^a\right(P_b/RT{)}^b(P_C/RT{)}^c....}$
$=\frac{P_L^l\cdot P_M^m\cdot P_N^n}{P_A^a\cdot P_B^b\cdot P_c^c}{\left(\frac{1}{RT}\right)}^{(l+m+n....)-(a+b+c+...)}$
$=\frac{K_p}{(RT{)}^{\Delta n,}}$ and
$\therefore K_p=K_c(RT{)}^{\Delta n_g}$
Where $\Delta n_g=$ Total number of stoichiometric moles of gaseous products - Total number of stoichimetric moles of gaseous reactants.