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If 2/9 of 1 mol of HI is dissociates, the equilibrium constant of disintegration of acid at same temperature will be
A.64
B.\[\dfrac{1}{{64}}\]
C.49
D.\[\dfrac{1}{{49}}\]

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Last updated date: 21st Jul 2024
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Answer
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Hint: Hydroiodic acid (or hydriodic acid) is a hydrogen iodide aqueous solution (HI). It's a strong acid that's fully ionised in aqueous solution. It has no hue. HIGH concentrations of concentrated solutions range from 48 to 57 percent. Iodine is produced when hydroiodic acid reacts with oxygen in the environment. Hydroiodic acid, and all hydrogen halides, reacts with alkenes to form alkyl iodides. It may also be used as a reducing agent, such as when aromatic nitro compounds are reduced to anilines.

Complete answer:
The dissociation constant is a direct result of the law of mass motion, which is a more general description of equilibria. As extended to salts, the dissociation constant is also known as the ionisation constant. The association constant is the reciprocal of the dissociation constant.
A dissociation constant is a type of equilibrium constant in chemistry, biochemistry, and pharmacology that tests the ability of a larger substance to fragment reversibly into smaller components, such as when a complex breaks down into its component molecules or a salt splits apart into its component ions.
\[2HI \rightleftharpoons {H_2} + {I_2}\]
HIH₂I₂
At Initial Concentration100
At \[(1 - \dfrac{2}{9})\]concentration\[(1 - \dfrac{2}{9})\]\[\dfrac{1}{9}\]\[\dfrac{1}{9}\]


A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical mechanism after a period of time has passed in which its structure shows no discernible propensity to alter.
Equilibrium constants are calculated,
[HI] = \[\dfrac{{\dfrac{7}{9}}}{V}\]
$\Rightarrow[H_2]$ = \[\dfrac{{\dfrac{1}{9}}}{V}\]
$\Rightarrow[I_2] $= \[\dfrac{{\dfrac{1}{9}}}{V}\]
\[{K_c} = \dfrac{{[{H_2}][{I_2}]}}{{{{[HI]}^2}}}\]
$\Rightarrow$\[{K_c} = \dfrac{{[\dfrac{1}{{9V}}][\dfrac{1}{{9V}}]}}{{{{[\dfrac{7}{{9V}}]}^2}}}\]
\[{K_C} = \dfrac{1}{{49}}\]
Hence option D is correct.

Note:
A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical mechanism after a period of time has passed in which its structure shows no discernible propensity to alter.