Moles of NH4HS initially taken =51g/mol5.1g=0.1mol
volume of vessel =2L
NH4HS(s)⇌NH3(g)+H2S(g)
t=00.1mol
t=∞0.1(1−0.2)0.1×0.20.1×0.2
Partial pressure of each gaseous component
P=VnRT=20.1×0.2×0.082×300
=0.246atm
⇒kP=PNH3×PH2S=(0.246)2=0.060516
=6.05×10−2
⇒6