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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