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 hold 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` change
C. `K_(p)` changes but `alpha` does not change
D. `K_(p)` does not change, but `alpha` changes

Answer

D

Explanation

For the equlibria `:N_(2)O_(4)(g)hArr2NO_(2)(g),K_(p)=K_(C)`because here `Deltan=1[K_(p)=K_(C)xx(RT)^(Deltan)].` will remain constant. Further since volume is halved, the pressure will be double so `alpha` will decrease so as to maintain the constancy of `K_(C) or K_(p).`
`N_(2)O_(4)hArr2NO_(2)`
`{:(a,0),(a-x,2x):}`
Let total pressure =P
`thereforep_(NO_(2))=(2x)/(a+x)xxP,p_(N_(2)O_(4))=(a-x)/(a+x)xxP`
`impliesK_(p)=((P_(NO_(2)))^(2))/(P_(N_(2)O_(4)))=(4x^(2)P^(2))/((a+x)^(2))xx((a+x))/(P(a-x))=(4x^(2)P)/((x^(2)-x^(2)))`
Since `K_(p)=` constants, so `x prop(1)/(sqrt(P)).` So when volume is halved, pressure gets doubled and thus x `( or alpha)` will decrease.