NH_(3) " is heated at 15 atm from "27^(@)C " to " 347^(@)C keeping the volume constnt . The new pressure becomes 50 atm at equilibrium of the reaction 2 NH_(3) hArr N_(2) + 3H_(2) Calculate % of mole of NH_(3) actually decomposed.

NH_(3) " is heated at 15 atm from "27^(@)C " to " 347^(@)C keeping the

1.	NH_(3) " is heated at 15 atm from "27^(@)C " to " 347^(@)C keeping the volume constnt . The new pressure becomes 50 atm at equilibrium of the reaction 2 NH_(3) hArr N_(2) + 3H_(2) Calculate % of mole of NH_(3) actually decomposed.
Answer» Solution :` {: ( ,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 % `