For the equilibrium, AB(g) ⇌ A(g) + B(g) , K_p is equal to four times…

The correct answer is option 2. \(\frac{2}{\sqrt{5}}\).

Let the total equilibrium pressure be \(= P\)

Given, \(K_p = 4P\)

Let the start be made with \(1\) mole of \(AB(g)\) and the degree of dissociation be \(x\)

	\(AB(g)\)	\(⇌\)	\(A(g)\)	\(+\)	\( B(g)\)\)
At equilibrium	\(1 - x\)		\(x\)		\(x\)

Total moles at equilibrium \(= 1 + x + x + x = 1 + x\)

Thus,

\(p_A\) = Partial pressure of \(A\) = \(\frac{x}{1 + x}.P\)

\(p_B\) = Partial pressure of \(B\) = \(\frac{x}{1 + x}.P\)

\(p_{AB}\) = Partial pressure of \(AB\) = \(\frac{1 - x}{1 + x}.P\)

Applying the law of mass action,

\(K_p = \frac{p_A \times p_B}{p_{AB}}\)

or, \(K_p = \frac{\left(\frac{x}{1 + x}.P\right)\left(\frac{x}{1 + x}.P\right)}{\left(\frac{1 - x}{1 + x}.P\right)}\)

or, \(4P = \frac{x^2}{1 - x^2}.P\)

or, \(4 - 4x^2 = x^2\)

or, \(5x^2 = 4\)

or, \(x = \frac{2}{\sqrt{5}}\)

Hence, number of moles of \(A\) formed \(= \frac{2}{\sqrt{2}{5}}\) times initial moles of \(AB\) taken