Ammonia gas at a pressure of 20 atm and $27^{0} C$ is heated in a constant volume in a container to a temperature of $327^{0} C$ at which new pressure becomes 50 atm and the following equilibrium $2 N H_{3 (g)} \underset{}{⇌} N_{2 (g)} + 3 H_{2 (g)}$ is established. The percentage dissociation of ammonia at this temperature is

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Detailed Solution

$2 N H_{3 (g)} \underset{}{⇌} N_{2 (g)} + 3 H_{2 (g)}$
$1 - α \frac{α}{2} \frac{3 α}{2}$ Total moles $= 1 + α$

$P V = n R T \frac{20 V}{50 V} = \frac{1 R 300}{n R 600} n = 1.25 α = 0.25 % = 25$

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