Suppose the value of the equilibrium constant for the given reaction H 2 ( g ) + I 2 ( g ) ⇌ 2 HI ( g ) at 717 K is 64 , then calculate the same (eq. const.) for the reaction HI ⇌ 1 2 H 2 + 1 2 I 2 .

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ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

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Suppose the value of the equilibrium constant for the given reaction $H_{2} (g) + I_{2} (g) \overset{}{⇌} 2 HI (g)$ at $717 K$ is $64$ , then calculate the same (eq. const.) for the reaction $HI ⇌ \frac{1}{2} H_{2} + \frac{1}{2} I_{2}$ .

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Answer:

Solution:

As the reaction is reversed and divided by $2,$

the new equilibrium constant will be :

${K^{'}}_{C} = \frac{1}{\sqrt{K_{C}}}$

$= \sqrt{\frac{1}{64}} = \frac{1}{8}$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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