In the following reaction, we start with 2 moles of N2 and 5 moles of H2 exerting a total pressure of 7 atm at a given temperature in a closed vessel. When 50% of N2 is converted into NH3N2+3H2⟶2NH3, partial pressure of NH3 is

N2+3H2⟶2NH3 Taken                                  2        5           0   Reacted (50% of N2)        1        3       Left                                       1         2     Total moles of N2, H2 and NH3 = 5 mol initial mol = 7pressure at start = 7 atmHence, pressure at equilibrium = 5 atm∴  Partial pressure of NH3=25×5=2atm