Consider the following equilibrium in a closed container
`N_(2)O_(4)(g) hArr 2NO_(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 the degree of dissociation `(alpha)`?

Options

A. neither `K_(p)` nor `alpha` changes
B. both `K_(p)` and `alpha` change
C. `K_(p)` changes, but `alpha` does not change
D. `K_(p)` does not change, but `alpha` changes.

Answer

D

Explanation

At fixed temperature , the volume of the reaction container is decreased. This implies that pressure is increased. An increase in pressure shifts the equilibrium in the backward direction i.e., the direction in which number of moles decreases. Thus degree of dissociation `alpha` changes. As temperature remains constant , there is no change in equilibrium constant `K`.