Consider the following equilibrium in a closed container:
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
At a fixed temperature, the volume of the reaction container is halved. For this change which of the following statements holds true regarding the equilibrium constant `(K_(p))` and degree of dissociation `(alpha)`?

Options

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

Explanation

For the reaction, `N_(2)O_(4)hArr2NO_(2(g))`
`K_(c)=(4alpha^(2)a^(2))/((a-aalpha)V)` …(`1`)
Where `a` is initial mole of `N_(2)O_(4)` in `V` litre vessel and `alpha` is degree of dissociation of `N_(2)O_(4)`.
If volume is reduced to `V//2`, the initial concentration of `N_(2)O_(4)` becomes `(a)/((V)/2)=(2a)/(V)`.
An increase in concentration of reactants leads to forward reaction i.e., the decomposition of `N_(2)O_(4)` to have constant value of `K_(p)`.