Derive the relation of value of K_p and K_c of the following balance reaction. H_(2(g)) + I_(2(g)) hArr 2HI_((g))

Derive the relation of value of K_p and K_c of the following balance r

1.	Derive the relation of value of K_p and K_c of the following balance reaction. H_(2(g)) + I_(2(g)) hArr 2HI_((g))
Answer» Solution :In this `H_(2(g))+ I_(2(g))hArr 2HI_((g))`EQUILIBRIUM reaction all is in gases phase so to take partial pressure in `K_p`. At given temperature, `p_(H_2)` = Partial pressure of `H_2` `p_(I_2)`= Partial pressure of `I_2` `P_(HI)` = Partial pressure of HI `K_p =(p_(HI))^2/((p_(H_2))(p_(I_2)))` but `p_(HI)=[HI]RT` because, p=cRT `p_(H_2)=[H_2]RT` where c=mol `L^(-1)` concen. `p_(I_2)` =[I]RT This value put in `K_p`, `K_p = ([HI]RT)^2/([H_2](RT)[I_2]RT)` `therefore K_p=[HI]^2/([H_2][I_2])=K_c` THUS, `K_p=K_c`, so the value of both CONSTANT equal, because `Deltan`=zero.