A reaction carried out by `1` mol of `N_(2)` and `3` mol of `H_(2)` shows at equilibrium the mole fraction of `NH_(3)` as `0.012` at `500^(@)C` and `10` atm pressure. Calculate `K_(p)` Also report the pressure at which "mole" `%` of `NH_(3)` in equilibrium mixture is increased to `10.4`.

Answer

A, C, D

Explanation

`{:(,N_(2),hArr,3H_(2),+,2NH_(3)),("moles before",1,,0,,0),("reaction",,,,,),("moles at",1-x,,(3-3x),,2x),("equilibrium",,,,,):}`
Given "mole" fraction of `NH_(3)=0.012` at `P=10` "atm"
`:. (2x)/((4-2x))=0.012 rArr x=0.0237`
`:. K_(p)=((P'_(NH_(3))))/(P'_(N_(2))xx(P'_(H_(2)))^(3))=([(2x.P)/((4-2x))]^(2))/(((1-x).P)/((4-2x))[((3-3x)P)/((4-2x))]^(3))`
`K_(p)=(4x^(2)(4-2x)^(2))/((1-x)(3-3x)^(2)P^(2))`
`=(4(0.0237)^(2)[4-2(0.0237)]^(2))/((1-0.0237)[3-3(0.0237)^(3)]xx100)`
`K_(p)=1.431xx10^(-5) "atm"^(-2)`
Let "mole" `%` of `NH_(3)` in equilibrium mixture be increased to `10.4` at pressure P.
`:. (2x)/(4-2x)=10.4/100` or `x=0.1994`
Now again using equation (i)
`K_(p)=(4x^(2)(4-2x)^(2))/((1-x)(3-3x)^(3).P^(2))`
`1.431xx10^(-5)=([4xx(0.1884)^(2)][4-2(0.1884)]^(2))/([1-0.1884][3-3(0.1884)]^(3)xxP^(2))`
`P=105.41 "atm"`