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Effective Equilibrium Constant

Consider the ...

Question

Consider the following equilibrium in a closed container
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant $(K_{p})$ and degree of dissociation $(α)$ ?

Neither

K_{p}

nor '

α

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Both

K_{p}

and '

α

' changes

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K_{p}

changes but '

α

' does not

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K_{p}

does not change but '

α

' changes

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Solution

The correct option is A $K_{p}$ does not change but ' $α$ ' changes
Consider the following equilibrium in a closed container
$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
At a fixed temperature, the volume of the reaction container is halved.
For this change, the equilibrium constant $K_{p}$ does not change but degree of dissociation ' $α$ ' changes
The value of the equilibrium constant depends on temperature and is independent of pressure, concentration, volume, presence of catalyst etc.
When the volume is decreased, pressure increases and the equilibrium shifts to lesser number of moles of gaseous moles. In this case, the equilibrium shifts to backward direction. Hence, the degree of dissociation ' $α$ ' decreases.

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Consider the following equilibrium in a closed containerN2O4(g)⇌2NO2(g)At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant (Kp) and degree of dissociation (α)?