Ammonia gas at a pressure of 20 atm and 27°C is heated in a constant volume in a container to a temperature of 327°C at which new pressure becomes 50 atm and the following equilibrium 2NH3g⇌ N2g+3H2gis established. Percentage dissociation of ammonia at this temperature is X2. The value of  is

Given, Pressure, p1=20atmPressure, p2=50atmTemperature,T1=270C+ 2730C=300 kTemperature,T2=3270C+ 2730C = 600 k Reaction,  2NH3(g)  ⇌   N2(g)  + 3H2(g)After dissociation  1-α                                      α2                          3α2 Totalmoles = 1+αIdeal gas equation,⇒PV=nRT⇒20 V50 V  =  1  R  300n  R  600⇒n = 1.25⇒α = 0.25⇒% = 25 i.e. x=5