`N_(2)+3H_(2)hArr 2NH_(3)`. This is gaseous phase reaction taking place in 1L flask at `127^(@)C`. Starting with 1 mole `N_(2)` and 3 moles `H_(2)`, equilibrium required 500mL of 1M HCl. Hence `K_(c)` is approximately

Options

A. 0.06
B. 0.08
C. 0.03
D. 2.05

Answer

C

Explanation

`NH_(3)` produced `=` 500 mL of 1 M HCl
`:. NH_(3)` produced `=0.5 mol`
`[NH_(3)]=0.5 "mol"L^(-1)("As" V=1L)`
`{:(N_(2),+,3H_(2),hArr,2NH_(3)),(1"mol",,3"mol",,-),(1-x,,3-3x,,2x):}`
`2x=0.5`
`:. X=0.25`
`[N_(2)]=(1-x)/(1)=1=1-0.25=0.75"mol"L^(-1)[H_(2)]`
`=(3-3x)/(1)=3-0.75=2.25"mol"L^(-1)`
`K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((0.5)^(2))/(0.75xx(2.25)^(3))`
`=0.02926~~0.03`