$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$ . This is a gaseous phase reaction taking place in 1 L flask at 127° C. Starting with 1 mole of $N_{2}$ and 3 mole of $H_{2}$ , the equilibrium mixture obtained is such that if it is titrated requires 500 mL of 1.0 M $H_{2} {SO}_{4}$ for neutralisation. Which of the following is the most appropriate $K_{c}$ ?

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0.03

0.59

0.27

0.11

answer is B.

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Detailed Solution

The reaction is:

$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3} 1 3 — 1 - x 3 - 3 x 2 x = 0.5 X 1 X 2$

$\begin{array}{l} \Rightarrow x = \frac{1}{2} \end{array}$

$K_{C} = \frac{1^{2}}{\frac{1}{2} \times {(\frac{3}{2})}^{3}}$

$K_{C} = 0.59$

Thus, the correct option is B.

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