Derive the relation between $K_{p}$ and $K_{c}$ for a general chemical equilibrium reaction.

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Solution

Let the gaseous reaction is a state of equilibrium is-
$a A_{(g)} + b B_{(g)} ⇌ c C_{(g)} + d D_{(g)}$
Let $p_{A}, p_{B}, p_{C}$ and $p_{D}$ be the partial pressure of $A, B, C$ and $D$ repectively.
Therefore,
$K_{c} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}} . . . . . (1)$
$K_{p} = \frac{{p_{C}}^{c} {p_{D}}^{d}}{{p_{A}}^{a} {p_{B}}^{b}} . . . . . (2)$
For an ideal gas-
$P V = n R T$
$\Rightarrow P = \frac{n}{V} R T = C R T$
Whereas $C$ is the concentration.
Therefore,
$p_{A} = [A] R T$
$p_{B} = [B] R T$
$p_{C} = [C] R T$
$p_{D} = [D] R T$
Substituting the values in equation $(2)$ , we have
$K_{p} = \frac{{[C]}^{c} {(R T)}^{c} {[D]}^{d} {(R T)}^{d}}{{[A]}^{a} {(R T)}^{a} {[B]}^{b} {(R T)}^{b}}$
$\Rightarrow K_{p} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}} {(R T)}^{[(c + d) - (a + b)]}$
$\Rightarrow K_{p} = K_{c} {(R T)}^{Δ n_{g}} (From (1)]$
Here,
$Δ n_{g} =$ Total no. of moles of gaseous product $-$ Total no. of moles of gaseous reactant
Hence the relation between $K_{p}$ and $K_{c}$ is-
$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

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