The correct option is
B 0.5×(1.5)3atm2Solution:-
InitiallyAtequillibriumN21(1−x)+3H23(3−3x)⟶2NH302x
Initially-
Total no. of moles, ni=1+3=4
Pressue, Pi=4 atm
At equillibrium-
Total no. of moles, nf=(1−x)+(3−3x)+2x=4−2x
Pressure, Pf=3 atm
From ideal gas law,
PV=nRT
At constat temperature,
P∝n
PiPf=ninf
⇒43=44−2x
⇒4−2x=3
⇒x=0.5
Mole fraction =no. of molestotal no. of moles
∴ At equillibrium-
Partial pressure of N2,PN2=XN2×P=0.5
Partial pressure of H2,PH2=XH2×P=1.5
Partial pressure of NH3,PNH3=XNH3×P=1
KP for the above reaction is given by-
KP=(PNH3)2PN2(PH2)3=10.5×(1.5)3atm−2
Hence KP=10.5×(1.5)3atm−2.
Hence, option (D) is correct.