If 'a' is the fraction of ammonia present by volume in an equilibrium mixture obtained from reaction between $N_{2}$ and $H_{2}$ gases, initially taken in the stoichiometric ratio and $K_{eq}$ for the reaction.
$(N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}) is \frac{K^{'} a^{2}}{(1 - a)^{4} P^{2}}$ where 'P' is the total pressure of gaseous mixture of equilibrium, then calculate the value of 27 K '.

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answer is 256.

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Detailed Solution

The fraction of ammonia present by volume in an equilibrium mixture obtained from reaction between $N_{2}$ and $H_{2}$ gases,

$\begin{array}{l} N_{2} + 3 H_{2} ⇌ 2 {NH}_{3} \\ \frac{1}{4} (1 - a) \frac{3}{4} (1 - a) \\ K_{P} = \frac{(a \times P)^{2}}{\{\frac{1}{4} (1 - a) \times P\} \times {\{\frac{3}{4} (1 - a) \times P\}}^{3}} \\ = \frac{4^{4}}{27} \times \frac{a^{2}}{(1 - a)^{4} P^{2}} \\ \Rightarrow K^{'} = \frac{4^{4}}{27} \Rightarrow 27 K^{'} = 256 \end{array}$