Question

Derive the relation between $K_p$ and $K_c$ for a general chemical equilibrium reaction.

Answer 1

Let the gaseous reaction is a state of equilibrium is-

$a{A}_{\left(g\right)}+b{B}_{\left(g\right)}\rightleftharpoons c{C}_{\left(g\right)}+d{D}_{\left(g\right)}$

Let $p_A,p_B,p_C$ and $p_D$ be the partial pressure of $A,B,C$ and $D$ repectively.

Therefore,

$K_c=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}.....\left(1\right)$

$K_p=\frac{{p_C}^c{p_D}^d}{{p_A}^a{p_B}^b}.....\left(2\right)$

For an ideal gas-

$PV=nRT$

$\Rightarrow P=\frac{n}{V}RT=CRT$

Whereas $C$ is the concentration.

Therefore,

$p_A=\left[A\right]RT$

$p_B=\left[B\right]RT$

$p_C=\left[C\right]RT$

$p_D=\left[D\right]RT$

Substituting the values in equation $\left(2\right)$, we have

$K_p=\frac{{\left[C\right]}^c{\left(RT\right)}^c{\left[D\right]}^d{\left(RT\right)}^d}{{\left[A\right]}^a{\left(RT\right)}^a{\left[B\right]}^b{\left(RT\right)}^b}$

$\Rightarrow K_p=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}{\left(RT\right)}^{[(c+d)-(a+b)]}$

$\Rightarrow K_p=K_c{\left(RT\right)}^{\Delta n_g}\left(\text{From}\left(1\right)\right]$

Here,

$\Delta n_g=$ Total no. of moles of gaseous product $-$ Total no. of moles of gaseous reactant

Hence the relation between $K_p$ and $K_c$ is-

$K_p=K_c{\left(RT\right)}^{\Delta n_g}$