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 given statements holds true regarding the equilibrium constant ( $K_{p}$ ) and degree of dissociation ( $α$ )?

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 given statements holds true regarding the equilibrium constant ( $K_{p}$ ) and degree of dissociation ( $α$ )?

TestHub · Accepted Answer

For the reaction, N₂O₄(g) ⇌ 2NO₂(g). K c ​ = ( a − a α ) . V 4 a 2 α 2 ​ … ( i ) where 'a' is the initial mole of N₂O₄ present in V L, and α is its degree of dissociation. Also, K p ​ = K c ​ ( RT ) Δ n ​ . On reducing the volume of the container to V /2 L, the initial concentration of N₂O₄ becomes 2 a / V . An increase in concentration leads to more dissociation of N₂O₄ in order to have K c ​ constant, a characteristic constant for a given reaction at a temperature.

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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