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Find the percentage dissociation of ammo...

Find the percentage dissociation of ammonia into `N_(2) "and' H_(2)` if the dissociation is carried out at constant pressure and the volume at equilibrium is `20%` greater than initial volume. (Initially. Equal moles of `NH_(3) "and" N_(2)` are present with no hydrogen)

Text Solution

Verified by Experts

The correct Answer is:
40
`2NH_(3)hArrN_(2)+3H_(2)`
`{:(t=0,a,a,0),(t_(eq),a(1-alpha),(aalpha)/(2)+a,(3aalpha)/(2)):}`
`n_(r)=2a+aalpha=a(2+alpha)=2a+(20)/(100)xx2a`
`2+alpha=2.4`
`alpha=0.4`
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