Ammonia dissociates into $N_{2}$ and $H_{2}$ such that degree of dissociation $α$ is much less than 1 and equilibrium pressure is $P_{0}$ then the value of $α$ is, if $K_{p}$ for $2 {NH}_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$ is $27 \times 10^{- 8} P_{0}^{2} :$

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0.02

$10^{- 4}$

can't be calculated

$4 \times 10^{- 4}$

answer is C.

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Detailed Solution

In the following reaction, we have;

$\begin{array}{l} 2 {NH}_{3} ⇌ N_{2} + 3 H_{2} \\ 1 - α \frac{α}{2} \frac{3 α}{2} \\ ≃ 1 \\ K_{P} = \frac{(\frac{α}{2} P_{0}) \times {(\frac{3 α}{2} P_{0})}^{3}}{P_{0}^{2}} \\ P_{0}^{2} \times 27 \times 10^{- 8} \\ = α^{4} \times \frac{27}{2^{4}} \times P_{0}^{2} \\ \Rightarrow α = 2 \times 10^{- 2} \end{array}$