At `450^(@)C` the equilibrium constant `K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` was found to be `1.6xx10^(-5)` at a pressure of `200` atm. If `N_(2)` and `H_(2)` are taken in `1:3` ratio. What is `%` of `NH_(3)` formed at this temperature?

Explanation

`N_(2) + 3H_(2) rarr 2NH_(3)`
`K_(p) = (2x)^(2)/((1-x)[3(1-x)]^(3)) xx [4-2x/P]^(2)`
`K_(p) = 16x^(2) xx (2-x)^(2)/(27(1-x)^(4) xx p^(2))` OR `1.6 xx 10^(-5) = 16x^(2) xx (2-x)^(2)/(27(1-x)^(4) xx (200)^(2))`
`x^(2)(2-x)^(2)/(1-x)^(4) = 1.6xx10^(-5)xx27xx(200)^(2)/16 = 1.08`
`x(2-x)/(1-x)^(2) = 1.039 implies x = 0.301`
`"Moles of " NH_(3) " formed " = 2 xx 0.301 = 0.602`
Total mole at equilibrium `= 4-2xx0.301 = 3.398`
`%` of `NH_(3)` at equilibrium `= 0.602/3.398 xx 100 = 17.72%`