Drive the following relation:
$K p = K c (R T)^{n}$

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Solution

Consider the following reversible reaction
$a A + b B ⇌ c C + d D$
$K_{C} = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$
If the reaction involves gaseous compounds, then
$K_{P} = \frac{P_{C}^{c} P_{D}^{d}}{P_{A}^{a} P_{B}^{b}}$
we know that, $P V = n R T \Rightarrow P = \frac{n R T}{V}$
$[\frac{n}{V}] = c o n c e n t r a t i o n = C \Rightarrow P = C R T$
$P_{A} = [A] R T P_{B} = [B] R T P_{C} = [C] R T P_{D} = [D] R T$
$K_{P} = \frac{{[[C] R T]}^{c} {[[D] R T]}^{d}}{{[[A] R T]}^{a} {[[B] R T]}^{b}} = \frac{{[C]}^{c} {[D]}^{d} {(R T)}^{c + d}}{{[A]}^{a} {[B]}^{b} {(R T)}^{a + b}} = K_{C}$
$K_{P} = K_{C} {(R T)}^{(c + d) - (a + b)}$
where $Δ n =$ no. of moles $= (c + d) - (a + b)$
$K_{P} = K_{C} {(R T)}^{Δ n}$

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