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aA+bB hArr cC + dD here A,B, C , D are in gaseous phase. Derive the relation between K_p and K_c. |
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Answer» SOLUTION :General reaction, `aA_((g)) + bB_((g)) hArr cC_((g)) +dD_((g))` `K_c=([C]^C [D]^d)/([A]^a[B]^b)` Here , concentration of component in [ ] is in mol `L^(-1)` . If partial pressure of GAS A,B,C,D is `p_A, p_B,p_C` and `p_D` at equilibrium , so Equilibrium constant = `K_p` and `K_p =((p_C^c)(p_D^d))/((p_A^a )(p_B^b))` but p=cRT So, `p_C^c=([C]RT)^c =[C]^c (RT)^c` `p_D^d = ([D]RT)^d =[D]^c (RT)^d` `p_A^a =([A]RT)^a = [A]^c (RT)^a` `p_B^b=([B]RT)^b=[B]^b(RT)^b` `THEREFORE K_p=([C]^c (RT)^c [D]^d (RT)^d)/([A]^a (RT)^a [B]^b (RT)^b)` `=([C]^c [D]^d)/([A]^a[B])(RT)^((c+d)-(a+b))` `therefore K_p = K_c(RT)^(DELTAN)`....(Eq.-i) where , `Deltan`=(c+d)-(a-b) `Deltan`=(Addition of coefficient of gases mole of products )- (Addition of coefficient of gases mole of reactant ) For equilibrium equation , `(n_p-n_r)=(c+d)-(a+b)` |
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