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Question

1 mole N2 and 3 moles H2 are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the equilibrium is attained N2(g)+3H2(g)2NH3(g). The equilibrium constant Kp for dissociation of NH3 is:

A
10.5×(1.5)3atm2
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B
0.5×(1.5)3atm2
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C
0.5×(1.5)33×3atm2
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D
10.5×(1.5)3atm2
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Solution

The correct option is B 0.5×(1.5)3atm2
Solution:-
InitiallyAtequillibriumN21(1x)+3H23(33x)2NH302x
Initially-
Total no. of moles, ni=1+3=4
Pressue, Pi=4 atm
At equillibrium-
Total no. of moles, nf=(1x)+(33x)+2x=42x
Pressure, Pf=3 atm
From ideal gas law,
PV=nRT
At constat temperature,
Pn
PiPf=ninf
43=442x
42x=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)3atm2
Hence KP=10.5×(1.5)3atm2.
Hence, option (D) is correct.

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