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Le Chatelier's Principle for Del N Equals to 0

Ammonia gas a...

Question

Ammonia gas at 15 atm is introduced in a rigid vessel at $300 K$ . At equilibrium the total pressure of the vessel is found to be 40.11 atm at $300^{o} C$ . The degree of dissociation of ${N H}_{3}$ will be:

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Solution

$2 N H_{3} ⇌ N_{2} + 3 H_{2} O$
$T_{1} = 300 K 15 a t m = P_{1} T_{2} = 573 K \frac{15}{300} \times (573) a t m = P_{2}$ $∵$ $P_{2} = (\frac{P_{1}}{T_{1}}) T_{2}$
$∵$ Degree of dissociation $α$ is given by,
$(\frac{15}{300}) \times 573 \times (1 + α) = 40.11 1 + α = 1.4 α = 0.4$

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Q. Ammonia gas at

15

atm is introduced in a rigid vessel at

300

K. At equilibrium, the total pressure of the vessel is found to be

40.11

300^{0}

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N H_{3}

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P_{1}

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