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.

Explanation

`2NH_(3)(g)hArr3H_(2)(g)+N_(2)(g)`
`{:(NH_(3),,rarr,,NH_(3)),(at 27^(@)C,,,,at 327^(@)C),((15 "atm"),,,,(P=?) "V remains constant"):}`
First, let us find the initial pressure of `NH_(3)` at `327^(@)C`.
`rArr P prop T` (V is constant)
`rArr P_(1)/T_(1)=P_(2)/T_(1) rArr P_(2)=(P_(1)T_(2))/T_(1)=(15xx600)/300=30 "atm"`

Now final equilibrium pressure `=50 "atm"`
`rArr 50=30-x+3/2x+x/2rArr x=20 "atm"`
rArr `% NH_(3)` decomposed `=20/30xx100=66.7%`
Alternative method:
`2NH_(3)(g)hArr3H_(2)(g)+N_(2)(g)`
Let `alpha` be the degree of dissociation

`rArr` Total mole`=a+aalpha`
`("Initial moles")/("Final moles")=("Initial pressure")/("Final pressure")`
`a/(a+aalpha)=30/50 rArr alpha=20/30`
rArr `%` dissociation `=20/30xx100=66.7%`