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2 moles each of A and B were taken in a flask at some temperature so as to establish the following equilibrium $A + B ⇌ 2 D$ It was observed that $\frac{1}{3}$ of A has reacted. After attainment of equilibria 2 mole of C was added which caused establishment of another equilibria $B + C ⇌ E + D$ If at new equilibria mole fraction of E is $\frac{1}{6}$ , then calculate % of A dissociated after both equilibria has established. Given: $\sqrt{61} = 7.81$ . Give answer after multiplying by 100.

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Detailed Solution

In the reaction of A and B;

$\begin{array}{l} A + B ⇌ 2 D \\ (2 - \frac{2}{3}) (2 - \frac{2}{3}) \frac{4}{3} \end{array}$

$K_{C_{1}} = 1$

Final moles of E = 1., as $Δ n_{g} = 0$ for both reactions

At new equilibrium

$\begin{array}{l} A + B ⇌ 2 D \\ (2 - x) (2 - x - y) (2 x + y) \\ B + C ⇌ E + D \\ (2 - x - y) (2 - y) y (2 x + y) \end{array}$

Clearly, y = 1

=> In 1 st equilibrium,

$\begin{array}{l} \frac{(2 x + 1)^{2}}{(1 - x) (2 - x)} = 1 \\ \Rightarrow 3 x^{2} + 7 x - 1 = 0 \\ x = \frac{- 7 \pm \sqrt{61}}{6} \\ % diss. of A = \frac{\frac{0.81}{6}}{2} \times 100 = 6.75 % \end{array}$

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