The relation between equilibrium constant K_p and K_c is

MCQ

The relation between equilibrium constant ${K_p}$ and ${K_c}$ is

✓

Answer

Correct option: B.

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

b
Let the gaseous reaction is a state of equilibrium is-

$aA _{( g )}+ bB _{( g )} \rightleftharpoons cC _{( g )}+ dD _{( g )}$

Let $p _{ A }, p _{ B }, p _{ C }$ and $p _{ D }$ be the partial pressure of $A , B , C$ and $D$ repectively. Therefore,

$K _{ c }=\frac{[ C ]^{ c }[ D ]^{ d }}{[ A ]^{ a }[ B ]^{ b }} \ldots . .(1)$

$K _{ p }=\frac{ pC ^c p _{ D }^{ d }}{ p _{ A }^{ a } p _{ B }^{ b }} \ldots \ldots .(2)$

For an ideal gas-

$PV = nRT$

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

Whereas $C$ is the concentration.

Therefore,

$p _{ A }=[ A ] RT$

$p _{ B }=[ B ] RT$

$p _{ C }=[ C ] RT$

$p _{ D }=[ D ] RT$

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

$K _{ p }=\frac{[ C ]^{ c }( RT )^{ c }[ D ]^{ d }( RT )^{ d }}{[ A ]^{ a }( RT )^{ a }[ B ]^{ b }( RT )^{ b }}$

$\Rightarrow K _{ p }=\frac{[ C ]^{ c }[ D ]^{ d }}{[ A ]^{ a }[ B ]^{ b }}( RT )^{[( c c )-( a + b )]}$

$\left.\Rightarrow K _{ p }= K _{ c }( RT )^{\Delta n_{ g }} \quad \text { (From (1) }\right]$

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-

$K _{ p }= K _{ c }( RT )^{\Delta n _{ g }}$

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