If for $2 A_{2} B (g) ⇌ 2 A_{2} (g) + B_{2} (g), K_{p} =$ TOTAL PRESSURE (at equilibrium) and starting the dissociation from $4 mol$ of $A_{2} B$ then:

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at equilibrium the no. of moles of $A_{2} B$ are equal to the no, of moles of $A_{2}$

total no. of moles at equilibrium will be $(14 / 3)$

degree of dissociation of $A_{2} B$ will be ( 2/3)

at equilibrium the no. of moles of $A_{2} B$ are not equal to the no. of moles of $B_{2}$

answer is A.

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Detailed Solution

We have the following reaction;

$\begin{array}{l} 2 A_{2} B ⇌ 2 A_{2} + B_{2} \\ 1 - α α \frac{α}{2} \\ \frac{α^{2} \cdot \frac{α}{2} \times P}{(1 - α)^{2} (1 + \frac{α}{2})} \\ = P \Rightarrow \frac{α^{3}}{2} \\ = (1 - α)^{2} (1 + \frac{α}{2}) \\ \Rightarrow α = \frac{2}{3} \end{array}$