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

Ammonia under a pressure of 1.5 atm at 27^(@)C is heated to 374^(@)C i

1.	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
Answer» 0.065 0.613 0.625 0.64 Solution : 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%`