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Q>8 mole of a gas AB3 are introduced into a 1l vessel.It dissociates as:
2AB3(g)=A2(g)+3B2(g)
At equilibrium 2 mol of A2 are found to be present.The equilibrium constant of the reaction is:
(= is reverse reaction above)
Moles of B2=9
Moles of AB3=2
K={81(18)}/2
729{mol/lit}2
at eq.A2=2 molesB2=6 molesAB3=8-4=4 molesfind partial pr. and then find Eq cons
2AB3(g) -----------> A2(g) + 3B2(g)
t=0 8 0 0
t=t 8 - x x/2 3x/2
Now
At equilibrium 2 mol of A2 are found
x/2 = 2
x = 4
t=t 4 2 6
thus,
equilibrium constant K = (A2) (B2)3/(AB3)2
Neglecting the volume 1 litre.
K = 22 x 63 / 42
K = 4 x 216 / 16
K = 54
plz approve1
mole of A2 are 2
mole of b2 are 6
mole of ab3 are 8-4=4
equlilibruim constant is equal to 2*6*6*6/4*4=27
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