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Derive the relatioshhip between K_p and K_c |
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Answer» Solution :Consider a gaseous reaction `aA+aBhArrcC+dD` `K_(p)=(P_(C)^(c)P_(D)^(d))/(P_(A)^(a)P_(B)^(b))""...(1)` where `P_(A),P_(B),P_(C)andP_(D)` represent the PARTIAL pressure of A, B, C and D in the equilibrium mixture. For an ideal gas, PV = nRT or `P=(n/V)RT,n/V=("NUMBER of moles")/("Volume in litre")` represents molar concentrations. `thereforeP_(A)=[A]RT,P_(B)=[B]RT,P_(C)=[C]RT,P_(D)=[D]RT` EQ(1) becomes `K_(p)=({[C]RT}^(c){[D]RT}^(d))/({[A]RT}^(a){[B]RT}^(b))=([C]^(c)[D]^(d))/([A]^(a)[B]^(b))xx(RT)^((c+d)-(a+b))` `K_(p)=K_(c)(RT)^(Deltan)` where `Deltan=(c+d)-(a+b)orDeltan` = Number of moles of products - Number of moles of reactants in that gaseous reaction. |
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