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Derive the relation between K_(P) and K_(C'). |
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Answer» Solution :Let us consider the reaction in which all reactants and products are ideal gases. `xA+yBhArrIC+mD` The equilibrium constant, `K_(C)` is `K_(C)=([C]^(1)[D]^(m))/([A]^(X)[B]^(y))`. . . (1) and `K_(P)` is, `K_(P)=(P_(C)^(1)xxP_(D)^(m))/(P_(A)^(x)xxP_(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 reactants and products can be expressed as, `P_(A)^(x)=[A]^(x)[RT]^(x)` `P_(B)^(y)=[B]^(y)[RT]^(y)` `P_(C)^(1)=[C]^(1)[RT]^(1)` `P_(D)^(m)=[D]^(m)[RT]^(m)` On substitution in eqn. 2, `K_(P)=([C]^(1)[RT]^(1)[D]^(m)[RT]^(m))/([A]^(x)[RT]^(x)[B]^(y)[RT]^(y))`. . . (3) `K_(P)=([C]^(1)[D]^(m)[RT]^(l+m))/([A]^(x)[B]^(y)[RT]^(x+y))` `K_(P)=([C]^(1)[D]^(m))/([A]^(x)[B]^(y))[RT]^((l+m)-(x+y)`. . .(4) By comparing equation (1) and (4), we get `K_(P)=K_(C)(RT)^(Deltan_(s))`. . . (5) where, `Deltan_(g)` is the difference between the SUM of number of moles of products and the sum of number of moles of reactants in the gas phase. |
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