`N_2 and H_2` in 1:3 molar ratio are heated in a closed container having a catalyst. When the following equilibrium `N_(2(g))+ 3H_(2(g)) Leftrightarrow 2NH_(3(g))` is attained, the total pressure is 10 atm and mole fraction of `NH_3` is 0.60. The equilibrium constant `K_p` for dissociation of `NH_3` is

Options

A. `1.333" atm"^(-2)`
B. `0.75" atm"^(2)`
C. `0.75" atm"^(-2)`
D. `1.333" atm"^(2)`

Answer

D

Explanation

`"Initially "underset(1)(N_2) _underset(3)(3H_(2)) Leftrightarrow underset(0)(2NH_3)`
`"At equilibrium 1-"alpha" "3-3alpha" "2lpha`
Total moles `=1-alpha+3-3alpha+2alpha=4-2alpha`
`x_(NH_(3))=(2alpha)/(4-2alpha)=0.6`
`alpha/(2-alpha)=0.6 rArr =1.2 -0.6 alpha`
`rArr 1.6 alpha=1.2 rArr alpha=3/4=0.75`
`x_(N_2)=(1-alpha)/(4-2alpha)=(0.25)/(2.5)=(1)/(10)=0.1`
`x_(H_(2))=(3-3alpha)/(4-2alpha)=(3-3 xx 0.75)/(4-2 xx 0.75)=0.3`
`K_(p)` for the reverse reaction.
`2NH_(3) Leftrightarrow N_2+3H_2`
`K_(p). =(p_(N_2))(p_(H_(2))^3)/((p_(NH_(3)))^2)=((0.1 xx 10) (0.3 xx 10^(-3)))/((0.6 xx 10)^2)=3^3/6^2=0.75"atm"^2`
`2NH_(3) Leftrightarrow N_2+3H_2`
`K_(p). =((p_(N_2))(p_(H_(2)))^3)/((p_(NH_(3)))^2)=((0.1 xx 10) (0.3 xx 10^(-3)))/((0.6 xx 10)^2)=3^3/6^2=0.75"atm"^2`