1.

2NH_(3)hArrN_(2)+3H_(2).the vessel is such that the volume remains effectively constant where as pressure increases to 50 atm. Calculate the percentage of NH_(3) actually decomposed

Answer»

`65%`
`61.3%`
`62.5%`
`64%`

Solution :`2NH_(3)hArrN_(2)+3H_(2)`
`{:("Initial mole",a,0,0),("Mole at equlibrium",(a-2x),x,3x):}`
Initial pressure of `NH_(3)` of mole = 15 atm at `27^(@)C`
The pressure of 'a' mole of `NH_(3)=P atm at 347^(@)C`
`therefore15/300=P/620`
`thereforep=31atm`
At constant volume and at `347^(@)C, `mole `prop` pressure
`a prop31` (before equlibrium)
`thereforea+2x prop50` (after equlibrium)
`THEREFORE(a+2x)/(a)=50/31`
`thereforex=(19)/(62)a`
`therefore% of NH_(3)"decomposed"=(2x)/(a)xx100`
`=(2xx19a)/(61xxa)xx100=61.29%`


Discussion

No Comment Found

Related InterviewSolutions