`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.

Explanation

`{:(,2NH_(3),hArr,N_(2),+,3H_(2)),("Initial mole",a,,0,,0),("Mole at equilibrium ",(a-2x),,x,,3x):}`
Initial pressure of `NH_(3)` of `a` mole `=15 atm` at `27^(@)C`
The pressure of `a` mole of `NH_(3)=P atm` at `347^(@)C`
`:. (15)/(300)=(P)/(620)`
`:. P=31 atm`
At constant volume and at `347^(@)C`, Mole prop pressure
`a prop 31` (Before equilibrium)
` a+2x prop 50` (After equilibrium)
`:. (a+2x)/(a)=(50)/(31)`
`:. x=(19)/(62)a`
`:. % of NH_(3)` decomposed `=(2x)/(a)xx100=(2xx19a)/(62xxa)xx100`
`=61.3%`