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Derive a general expression for the equilibrium constant K_(P) and K_(C) for the reaction. 3H_(2)(g)+N_(2)(g)hArr2NH_(3)(g) |
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Answer» Solution :Synthesis of ammonia : `square`Let us consider the formation of ammonia in which, 'a' moles nitrogen and 'b' moles HYDROGEN gas are allowed to react in a container of volume V. Let 'x' moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`  `square` Applying law of mass ACTION, `K_(C)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))` `=(((2x)/(V))^(2))/(((a-x)/(V))((b-3x)/(V))^(3))` `(((4x^(2))/(V^(2))))/(((a-x)/(V))((b-3x)/(V))^(3))` `K_(C)=(4x^(2)V^(2))/((a-x)(b-3x)^(2))` The equilibrium constant `K_(p)` can also be calculated as FOLLOWS : `K_(P)=K_(C)(RT)^((Deltan_(g)))` `Deltan_(g)=n_(P)-n_(r)=2-4=-2` `K_(P)=(4x^(2)V^(2))/((a-x)(b-3x)^(3))(RT)^(-2)` Total number of moles at equilibrium, `n=a-x+b-3x+2x=a+b-2x` `K_(P)=(4x^(2)V^(2))/((a-x)(b-3x)^(3))xx[(PV)/(n)]^(-2)` `K_(P)=(4x^(2)V^(2))/((a-x)(b-3x)^(3))xx[(n)/(PV)]^(2)` `K_(P)=(4x^(2)V^(2))/((a-x)(b-3x)^(3))xx[(a+b-2x)/(PV)]^(2)` `K_(P)=(4x^(2)(a+b-2x)^(2))/(P^(2)(a-x)(b-3x)^(3))` |
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