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ChemistryChemical EquilibriumProblems involving KpMedium2 minQB

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For the equilibrium $NH_{4} HS (s) ⇌ NH_{3} (g) + H_{2} S (g)$ , the expression of $K_{p}$ in terms of equilibrium pressure ' $p$ ' is

Answer:

Solution:

It is given that equilibrium pressure is $p$ .

∴ $P_{N H_{3}} + P_{H_{2} S} = p$ .

$P_{N H_{3}} = \frac{p}{2}$ ,

$P_{H_{2} S} = \frac{p}{2}$

For the given reaction

$K_{p} = \frac{P}{2} \times \frac{P}{2}$

$K_{p} = \frac{p ^{2}}{4}$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp

⏱ 2mℹ️ Source: QB

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