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Standard XII

Chemistry

Effective Equilibrium Constant

The equilibri...

Question

The equilibrium constant of the reaction $A 2 (g) + B 2 (g) ⇌ 2 A B (g)$ at $100^{\circ} C$ is 50.

If one litre flask containing one mole of $A_{2}$ is connected to a two litre flask containing two moles of $B_{2}$ the number of moles of AB formed at 373 K will be -

1.886

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2.317

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0.943

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18.86

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Solution

The correct option is A
1.886

The equilibrium is represented as:

$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$

Initial connection

1 2 0

moles at equilibrium

1-x 2-x 2x

Total volume = 1 + 2 = 3 litres

$[A_{2}] = \frac{1 - x}{3}, [B_{2}] = \frac{2 - x}{3} [A B] = \frac{2 x}{3}$

$K = \frac{[A B]^{2}}{[A_{2}] [B_{2}]} = \frac{(\frac{2 X}{3})}{(\frac{1 - x}{3}) (\frac{2 - x}{3})} = 50$

On solving we get $23 x^{2} - 75 x + 50 = 0$

x = 2.31 or 0.943, since x can't be more than 1

so, x = 0.943

moles of AB formed $= 2 \times 0.943 = 1.886$

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The equilibrium constant of the reaction A2(g)+B2(g)⇌2AB(g) at 100∘C is 50. If one litre flask containing one mole of A2 is connected to a two litre flask containing two moles of B2 the number of moles of AB formed at 373 K will be -

The equilibrium constant of the reaction $A 2 (g) + B 2 (g) ⇌ 2 A B (g)$ at $100^{\circ} C$ is 50.

If one litre flask containing one mole of $A_{2}$ is connected to a two litre flask containing two moles of $B_{2}$ the number of moles of AB formed at 373 K will be -