2 mole of AB(s) is taken in close container of volume of 2 litre. If 1 mole of AB(s) has taken part in reaction to attain the equilibrium state and equilibrium moles of $D_{2} (g)$ is 0.75.
$AB (s) ⇌ A (g) + B (g) K_{C} = a {(m o l / l)}^{2} A (g) ⇌ \frac{1}{2} C_{2} (g) + \frac{3}{2} D_{2} (g) K_{C} = b (m o l / l^{2})$

at equilibrium $[A] = x (mol/litre) [C_{2}] =y (mol/litre)$
Calculate the value of $\frac{b \sqrt{3}}{a (x +y)}$ is ……

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answer is 12.

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Detailed Solution

$\underset{2 - 1}{A B (s)} ⇌ \underset{1 - x_{1}}{A (g)} + \underset{1}{B} (g) \underset{1 - x_{1}}{A (g)} ⇌ \underset{\frac{x_{1}}{2}}{\frac{1}{2} C_{2} (g)} + \frac{3}{2} \underset{\frac{3 x_{1}}{2}}{D_{2} (g)} \frac{3 x_{1}}{2} = 0.75 x_{1} = \frac{1}{2} a = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} {(m o l e / l i t r e)}^{2} b = \frac{{(\frac{3}{8})}^{3 / 2} {(\frac{1}{8})}^{1 / 2}}{(\frac{1}{4})} = \frac{3 \sqrt{3}}{16} x = [A] = \frac{1}{4} M y = [C_{2}] = \frac{1}{8} M \frac{b \sqrt{3}}{a (x + y)} = \frac{\frac{3 \sqrt{3} \times \sqrt{3}}{16}}{\frac{1}{8} (\frac{1}{4} + \frac{1}{8})} = \frac{\frac{9}{16}}{\frac{1}{8} \times \frac{3}{8}} = \frac{9}{16} \times \frac{64}{3} = 12$