At a certain temperature, only 50% HI is dissociated into H 2 and I 2 at equilibrium. The equilibrium constant is :

2 H I g ⇌ H 2 g + I 2 g Moles 1 - α α 2 α 2 Concentration 1 - α V α 2V α 2V K C = H 2 I 2 H I 2 = α 2V α 2V 1 - α V 2 ∴ α = 0.50 α 2 = 0.25 = α 2 4 1 - α 2 = 0 .50 × 0 .50 4 ×0.50 × 0 .50 = 1 4 K C = 0.25

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ChemistryChemical EquilibriumProblems involving KcMedium2 minPYQ_2014

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At a certain temperature, only 50% HI is dissociated into H₂and I₂at equilibrium. The equilibrium constant is :

Options:

Answer:

Solution:

$\begin{matrix} 2 H I (g) & ⇌ & H_{2} (g) & + & I_{2} (g) \\ Moles & 1 - α & \frac{α}{2} & \frac{α}{2} \\ Concentration & \frac{1 - α}{V} & \frac{α}{2V} & \frac{α}{2V} \end{matrix}$

$K_{C} = \frac{[H_{2}] [I_{2}]}{{[H I]}^{2}} = \frac{[\frac{α}{2V}] [\frac{α}{2V}]}{{[\frac{1 - α}{V}]}^{2}}$

$(\begin{matrix} ∴ α & = & 0.50 \\ \frac{α}{2} & = & 0.25 \end{matrix})$

$= \frac{α^{2}}{4 {(1 - α)}^{2}} = \frac{0 .50 \times 0 .50}{4 \times0.50 \times 0 .50} = \frac{1}{4}$

$K_{C} = 0.25$

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

⏱ 2mℹ️ Source: PYQ_2014

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