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Calculate the expression for `K_(C) "and" K_(P)` if initially a moles of `N_(2) "and" B "moles of" H_(2)` is taken for the following reaction. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) (Deltanlt0)` (P,T,V given) |
Answer» Correct Answer - `K_(c)=(4X^(2)V^(2))/((a-x)(b-3x)^(3)):K_(P)=((a+b-2x)^(2).4x^(2))/P^(2(a-x)(b-3x)^(3)` `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) (Deltanlt0)` (P.T.V given) `{:(At t=0,a,b,0),(t=t_(eq),(a-x)(b-3x),2x,):}` `[N_(2)]=(a-x)/(V), [H_(2)] =(b-3x)/(V), [NH_(3)]=(2x)/(V)` `K_(C)=(((2x)/(V))^(2))/(((1-x)/(V))((b-3x)/(V))^(3))=(4x^(2)V^(2))/((a-x)(b-3x)^(3))` Total no. of moles at equilibrium`=a+b-2x` `[P_(N_(2)]]=(a-x)/(a+b-2x). P, [P_(H_(2)]]=((b-3x))/(a+b-2x).P, [P_(NH_(3)]]=((2x).P)/(a+b-2x)` `K_(P)=[P_(NH_(3)]]^(2)/([P_(N_(2)]][P_(H_(2)]]^(3))=(((2x)/(a+b-2x).P)^(2))/([((a-x)/(a+b-2x)).P][((b-3x)P)/(a+b-2x)]^(3))` `((4x^(2).P^(2))/((a+b-2x)^(2)))/(P^(4)((a-x)(b-3x)^(3))/((a+b-2x)^(4)))=((a+b-2x)^(2).4x^(2))/P^(2(a-x)(b-3x)^(3))` |
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