The formation of ammonia from nitrogen and hydrogen is represented by the equation,
`N_(2(g))+3H_(2(g)) Leftrightarrow 2NH_(3(g))`
If we start with 'a' moles of `N_2`, and b'moles of `H_2 and` at equilibrium (a - x) moles of `N_2,` are left then the value of `K_p,` will be given by the expression (Total pressure at equilibrium =P)

Options

A. `K_(p) =(4x^2 (a+b-2x)^2)/((a-x) (b-3x)^3 p^2)`
B. `K_(p)=x^2/((1-x)^2)`
C. `K_(p)=(4x^2 p^2)/(27(1-x)^4)`
D. `K_(p)=0`

Answer

A

Explanation

`"Initial moles "underset(a)(N_(2(g)) +underset(b)(3H_(2(g)) Leftrightarrow 2NH_(3(g))`
At equilibrium `"a-x b-3x 2x"`
Total moles at equilibrium =a-x+b-3x+2x=a+b-2x
`p_(N_2)="Moles fraction of N"_2 xx P =((a-x)/(a+b-2x))P`
`p_(H_(2))=((b-3x)/(a+b-2x))P , p_(NH_(3)) =((2x)/(a+b-2x)) P`
`K_(p)=(p_(NH_(3))^2)/(p_(N_(2)) xx p_(H_(2))^3)=((2x)/(a+b-2x))^2 (P^2)/((a-x)/(a+b-2x) P)" "((b-3x)/(a+b-2x)^3 P^3)`
`=(4x^2 (a+b-2x)^2)/((a-x) (b-3x)^3 P^2)`