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Derive the ralation between K_P and K_C. |
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Answer» SOLUTION :Consider a general reaction in which all reactants and PRODUCTS are ideal gases. `xA + yB HARR lC +m D` The equilibrium constant `K_C` is `K_C = ([C]^L[D]^m)/([A]^x[B]^y)`…(1) `K_P = (p_C^lxxp_D^m)/(p_A^x xx p_B^y)`…(2) The ideal gas EQUATION is `PV= nRT or P = n/VRT` Since, Active mass =molar concentration `= n/V` P = Active mass `xx RT` Based on the above expression, the partial pressure of the reactant andproducts can beexpressed as, `p_A^x = [A]^x. [RT]^x` `p_B^y = [B]^y. [RT]^y` `p_C^l = [C]^l.[RT]^l` `p_D^m = [D]^m . [RT]^m` On substitution in equation(2), `K_P = ([C]^l[RT]^l[D]^m[RT]^m)/([A]^x[RT]^x[B]^y[RT]^y)` `K_P = ([C]^l[D]^m)/([A]^x[B]^y) xx RT^((1+m)-(x+y))` By comparing equation (1) and (4), we get `K_P = K_C (RT)^((Deltan_g))` where `Deltan_g` is the difference between the sum of number of moles of products and the sum of number of moles of reactans in the gas phase. (i) If `Deltan_g =0, K_P = K_C(RT)^0` `K_P = K_C` Example : `H_2(g)+ I_2(g) hArr 2HI(g)` (ii) When `Deltan_(g) = +ve` `K_P= K_C(RT)^(+ve)` `K_P = K_C` Example : `2NH_3(g) hArr N_2(g)+ 3H_2(g)` (iii) When`Deltan_g = -ve` `K_P = K_C (RT)^(-ve)` `K_P lt K_C` Example : `2SO_2(g)+ O_2(g) hArr 2SO_3(g)` |
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