Q.
The equilibrium constant for the reaction A2(g)+B2(g)⇌2AB(g) at 100°C is 50. If a one liter flask containing one mole of A2 is connected to a 2 litre flask containing two moles of B2and allowed to react, the amount of AB formed at equilibrium at 100°C is
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a
0.168 moles
b
2.218 moles
c
5.122 moles
d
1.868 moles
answer is D.
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Detailed Solution
A2(g)+B2(g)⇌2AB(g) Initial moles 1 2 0 -x -x +2x At equilibrium 1-x 2-x 2x Equilibrium constants Kc=[AB]2A2B2 KC=2x321-x32-x3 50=(2x)2(1-x)(2-x) 50x2-150x+100=0 x=0.934 Number of moles AB = 2x = 1.868 moles
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