`NH_(3)` is heated at `15` at, from `25^(@)C` to `347^(@)C` assuming volume constant. The new pressure becomes `50` atm at equilibrium of the reaction `2NH_(3) hArr N_(2)+3H_(2)`. Calculate `%` moles of `NH_(3)` actually decomposed.

Explanation

` {: ( ,2 NH_(3),hArr,N_(2),+,3H_(2),), ("Intial moles",a,,0,,0,), ("Moles at eqm.",a-2x,,x,,3x,"Total " = a+2 x ):} `
Pressure of a moles of `NH_(3) at 27^(@)C = 15 "atm".`
Pressure of a moles of `NH_(3) "at" 347^(@)C ` = P atm (say)
As volume remains constant , ` P_(1)/T_(1) = P_(2) /T_(2)`
` 15/300 = P/620 or P= 31 atm .`
Now, at `327^(@)C` and constant volume ,
Pressure ` prop ` No. of moles
` :. 31 prop a`
` 50 prop a + 2 x `
` :. (a + 2 x )/ a = 50/31 or x = 19/62 a `
` % "of " NH_(3) " decomposed " = (2x)/a xx 100 = 2 xx (19 a)/62 xx 1/a xx 100 = 61* 3 % `