Ammonia under a pressure of `15 atm`, at `27^(@)C` is heated to `327^(@)C` in a vessel in the pressure of catalyst. Under these conditions, `NH_(3)` partially decomposes to `H_(2)` and `N_(2)`. The vessel is such that the volume remains effectively constant, whereas the pressure increases to `50 atm`. Calculate the precentage of `NH_(3)` actually decomposed.

Options

A. `65%`
B. `61.3%`
C. `62.5%`
D. `64%`

Answer

B

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 = 3` 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)/(62xx1)xx100=61.33%`