Answer :
[tex]K_{p} =\frac{K_{c} }{[RT]^{2} }[/tex]
The equilibrium constants for a perfect gaseous mixture are[tex]K_{p} \:and\:K_{c}[/tex]When equilibrium concentrations are stated in atmospheric pressure, the equilibrium constant is [tex]K_{p}[/tex], and when they are expressed in molarity, the equilibrium constant is [tex]K_{c}[/tex]
[tex]N_{2}(g)\rightarrow+3H_{2} (g)\Longleftrightarrow2NH_{3}(g)[/tex]
we need to find the relation between [tex]K_{p} \:and\:K_{c}[/tex] for this balanced equation:
[tex]K_{p} =\frac{[P_{NH_{3} }]^2}{[P_{N_{2} }][P_{H_{2} }]^3}\:\rightarrow(1)[/tex]
where[tex]P_{x}[/tex] means partial pressure of gas [tex]x[/tex].
By ideal gas equation
[tex]PV=NRT\\P=\frac{N}{V} RT[/tex]
And [tex]\frac{N}{V}=\frac{moles}{volume} \\[/tex] denotes concentration
[tex]P=[C]RT \rightarrow[/tex] where C means the (1)
[tex]K_{p} =\frac{[NH_{3}] ^{2}.[RT]^{2} }{[N_{2} ] ^{2}.[RT]^{}\times[H_{2} ]^{3} [RT ]^{3} }[/tex]
[tex]K_{p} =\frac{[NH_{3}] ^{2}. }{[N_{2} ] ^{}.[^{}\[H_{2} ]^{3} } \:[/tex][tex]\times\frac{1}{RT} \rightarrow(2)[/tex]
we know for the reaction equilibrium constant in terms of concentration-
[tex]K_{p} =\frac{[NH_{3}] ^{2}. }{[N_{2} ] ^{}.[^{}\[H_{2} ]^{3} } \:[/tex]
Replacing this in equation no (2),
[tex]K_{p} =\frac{K_{c} }{[RT]^{2} }[/tex]
[tex]K_{p} ={K_{c} }{[RT]^{\triangle n} }\\\triangle n=(moles\:of \:gaseous\: products)-(moles \:of \:gaseous\:reactant)[/tex]
Learn more about the equilibrium constants
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