${NH}_{3}$ is heated at 15 atm from $27^{\circ} C$ to $347^{\circ} C$ assuming the volume constant. The new pressure becomes 50 atm at equilibrium for the reaction $2 {NH}_{3} ⇌ N_{2} + 3 H_{2}$ . Calculate % of moles actually decomposed.

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61.3%

80.5%

59%

71%

answer is C.

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Detailed Solution

Pressure:

$\begin{array}{l} \frac{15}{300} = \frac{P_{0}}{620} \\ \Rightarrow P_{0} \\ = 31 atm \\ 2 {NH}_{3} ⇌ N_{2} + 3 H_{2} \\ 31 (1 - α) 31 \frac{α}{2} 31 \times \frac{3 α}{2} \\ 31 (1 + α) = 50 \\ \Rightarrow α = 0.613 \end{array}$

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