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Ammonia under a pressure of 15 atm at 27...

Ammonia under a pressure of 15 atm at `27^@C` is heated to `347^@C` in a closed vessel in the presence of a catalyst. Under these conditions `NH_3` is partially decomposed according to the equation
`2NH_3 Leftrightarrow N_2 +3H,`
The vessel is such that the volume remains effectively constant, whereas pressure increases to 50 atm. Calculate the percentage of `NH_3` actually decomposed. Pressure of `NH_3" at "27^@C" or 300 K = 15 atm."`

A

`65%`

B

`61.3%`

C

`62.5%`

D

`64%`

Text Solution

Verified by Experts

The correct Answer is:
B
`2NH_3 hArr N_2 + 3H_2`
`{:("Initial mole",alpha,0,0),("Mole at equilibrium", (alpha -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 =31` atm
At constant volume and at `347^@ C` , mole `prop` pressure
`alpha prop 31` (before equilibrium)
`:. alpha + 2x prop 50` (after equilibrium)
`:. (a+2x)/a = 50/31`
`:. x = (19)/62 a`
`=(2xx19a)/(62xxa)xx100 = 61.29%`
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