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Ammonia under a pressure of 1.5 atm at 2...

Ammonia under a pressure of 1.5 atm at `27^(@)`C is heated to `374^(@)`C in a closed vessel in the presence of a catalyst. Under the conditions, `NH_(3)` is partially decomposed according to the equation. `2NH_(3) hArr N_(2) + 3H_(2)` the vessel is such that the volume remains etfectively constant where as pressure increases to 50 atm. Calculate the percentage of `NH_(3)` actually decomposed

A

0.065

B

0.613

C

0.625

D

0.64

Text Solution

Verified by Experts

The correct Answer is:
B

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`
`(P_(1))/(T_(1))=(P_(2))/(T_(2)), (15)/(300)=(P)/(629) implies p =31` atm
At constant volume and at `347^(@)C` mole `alpha` pressure a `alpha` 31 (before equilibrium)
`(a+2alpha)alpha50` (at equilibrium) `(a+2alpha)/(a)=(50)/(31) implies x=(19a)/(62) implies %NH_(3)` decomposes
`=(2x)/(a) xx 1000, (2 xx 19a)/(62 xx a) xx 100=61.3%`
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