Explanation

Let the initial number of moles of NH₃ be 1.

Let the degree of dissociation of NH₃ be α.

Then, at equilibrium:

NH₃ = 1 − α
N₂ = α/2
H₂ = 3α/2

Total number of moles at equilibrium:

= (1 − α) + (α/2) + (3α/2)

= 1 − α + 2α

= 1 + α

Total pressure = √3 atm

Hence, partial pressures are:

P_NH3 = (1 − α)/(1 + α) × √3

P_N2 = (α/2)/(1 + α) × √3

P_H2 = (3α/2)/(1 + α) × √3

Expression for K_p:

$$ K_p = \frac{(P_{N_2})^{1/2} (P_{H_2})^{3/2}}{P_{NH_3}} $$

Substitute the values:

$$ K_p = \frac{\left(\frac{\alpha \sqrt{3}}{2(1+\alpha)}\right)^{1/2} \left(\frac{3\alpha \sqrt{3}}{2(1+\alpha)}\right)^{3/2}} {\frac{(1-\alpha)\sqrt{3}}{(1+\alpha)}} $$

Simplifying step by step:

$$ K_p = \frac{3^{1/4} \alpha^{1/2}}{(2^{1/2})(1+\alpha)^{1/2}} \times \frac{3^{9/4} \alpha^{3/2}}{(2^{3/2})(1+\alpha)^{3/2}} \times \frac{(1+\alpha)}{(1-\alpha)\sqrt{3}} $$

Combining terms:

$$ K_p = \frac{3^{(1/4+9/4-1/2)} \alpha^2}{4(1-\alpha)(1+\alpha)} $$

$$ K_p = \frac{3^2 \alpha^2}{4(1-\alpha^2)} $$

Given K_p = 9:

$$ 9 = \frac{9\alpha^2}{4(1-\alpha^2)} $$

$$ 4(1-\alpha^2) = \alpha^2 $$

$$ 4 = 5\alpha^2 $$

$$ \alpha^2 = \frac{4}{5} $$

$$ \alpha = \left(\frac{4}{5}\right)^{1/2} $$

Given:

$$ \alpha = (x \times 10^{-2})^{-1/2} $$

Comparing,

$$ (x \times 10^{-2}) = \frac{5}{4} $$

$$ x = \frac{5}{4} \times 10^2 = 125 $$

Therefore,

x = 125

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