Question

Derive the relation between $\bf{K_p}$ and $\bf{K_c}$.
⇒ Don't just post the answer as $\sf{K_p = K_c(RT)^{\Delta n}}$
⇒ Please derive with proper explanation.
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⇒ Thanks for helping me and Good Luck!

Answer 1

Explanation:

Solution :-

Let us Consider a reaction :-

$\sf mA_{(g)}+nB_{(g)}\rightleftharpoons pC_{(g)}+qD_{(g)}$

The equilibrium Constant [K_c] for the above reaction is :-

$\sf K_c = \dfrac{[C]^{p}[D]^{q}}{[A]^{m}[B]^{n}}$

[Let it be equation -1 ]

Partial Pressure[K_p] for the above reaction is :-

$\sf K_p = \dfrac{P_C^{p}p_D^{q}}{p_A^{m}p_B^{n}}$

We know that :-

Pv = nRT [Ideal gas equation]

P = nRT/v

$\sf K_{p} = \dfrac{\left(\dfrac{n_{C}RT}{v}\right)^{p}\left(\dfrac{n_{D}RT}{v}\right)^{q}}{\left(\dfrac{n_{A}RT}{v}\right)^{m}\left(\dfrac{n_{B}RT}{v}\right)^{n}}$

$\sf K_{p} = \dfrac{\left(\dfrac{n_{C}}{v}\right)^{p}(RT)^{p}\left(\dfrac{n_{D}}{v}\right)^{q}(RT)^{q}}{\left(\dfrac{n_{A}}{v}\right)^{m}(RT)^{m}\left(\dfrac{n_{B}}{v}\right)^{n}(RT)^{n}}$

$\sf K_p = \dfrac{\left(\dfrac{n_C}{v}\right)^{p}\left(\dfrac{n_D}{v}\right)^{q}(RT)^{p+q}}{\left(\dfrac{n_A}{v}\right)^{m}\left(\dfrac{n_B}{v}\right)^{n}(RT)^{m+n}}$

We know that :-

Number of moles per volume = concerntration

c = n/v

$\sf K_p= \left(\dfrac{[C]^{p}[D]^{q}}{[A]^{m}[B]^{n}}\right)(RT)^{p+q-m-n}$

Substituting value of equation 1 :-

$\sf K_p = K_c\times (RT)^{p+q-m-n}$

We know that  :-

$\sf\Delta n$ = Total number of moles of gaseous products - Total number of moles of gaseous reactants.

Since , In the above chemical Reaction :-

$\sf\Delta n$ = p+q-m-n

$\sf K_p = K_c(RT)^{\Delta n}$.