Q.

1 mol. N2  and 3 mol 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 following equilibrium is attained.  N2(g)+3H2(g)⇌2NH3(g) .  The equilibrium constant Kp for dissociation of NH3  is :

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a

10.5×(1.5)3atm−2

b

0.5×(1.5)3atm2

c

0.5×(1.5)33×3atm2

d

3×30.5×(1.5)3atm−2

answer is A.

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Detailed Solution

2NH3(g)⇌N2(g)+3H2(g) At equilibrium          2x                    1-x            3- 3x                                    1-x+3-3x+2x= 3 (given)                                      4 – 2x        =   3                                            X          =   0.5                                   kp=[N2][H2]3[NH3]2=[0.5][1.5]3[0.5]2=[1.5]3[0.5]atm−2
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