`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.

Answer

A, C, D

Explanation

Pressure of `NH_(3)` at `27^(@)C=15 "atm"`
Pressure of `NH_(3)` at `347^(@)C=P "atm"`
Using relation `P_(1)/T_(1)=P_(2)/T_(2)`
`P/620=15/300`
`P=31 "atm"`
Let a mol of ammonia be present. Total pressure at equilibrium`=50` atm.
`{:(,2NH_(3)(g),hArr,N_(2)(g),+,3H_(2)(g)),("At equilibrium",(a-2x),,x,,3x),("Total moles" =,a-2x+x+3x=a+2x,,,,):}`
`("Initial number of moles")/("moles at equilibrium")=("Initial pressure")/("Equilibrium pressure")`
`a/((a+2x))=31/50`
`x=16/62 a`
Amount of ammonia decomposed `=2x=2xx19/62a=19/31a`
`%` of ammonia decomposed `=(19xxa)/(31xxa)xx100=61.3`