`NH_(3)` is heated at `15` atm, 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 moles",a,,0,,0),("moles at equilibrium",(a-2x),,x,,3x):}`
Initial pressure of `NH_(3)` of a moles `=15 "atm"` at `27^(@)C`
The pressure of a moles of `NH_(3)=P "atm"` at `347^(@)C`
`:. 15//300=P//620`
`:. P=31` atm
At constant volume and at `347^(@)C` moles `prop` Pressure
`aprop31` 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%`