1.

Derive the ralation between K_P and K_C.

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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