The `K_(p)` for the reaction `N_(2)O_(4)hArr2NO_(2)` is `640 mm` at `775 K`. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of `160 mm`. At what pressure, the dissociation will be `50%`?

Explanation

Suppose intially `N_(2)O_(4)` taken = 1 mole and its degree of dissociation = `alpha`
` {:(,N_(2)O_(4),hArr,2NO_(2),),("Intial ",1 "mole",,,),("At eqm.",1-alpha,2alpha,"Total " =1-alpha + 2 alpha = 1+alpha,):}`
If P is the total pressure at equilibrium , then ` p_(N_(2)O_(4)) = (1-alpha)/(1+alpha) xxP and p_(NO_(2)) = (2 alpha)/(1+alpha) xxP `
Now, ` K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= ((2 alpha)/(1+alpha).P)^(2)/((1-alpha)/(1+alpha) *P)= (4 a^(2))/((1+alpha)(1-alpha)) = (4a^(2))/(1-alpha^(2)) xx P`
Putting ` K_(p) = 640 " mm (Given ) and equilibrium pressure , P+ 160 mm, we get "`
` 640 = (4a^(2))/(1-alpha)^(2) xx 160 or (alpha^(2))/(1-alpha)^(2) = or alpha^(2) = 1 - alpha^(2) or 2 alpha^(2) = 1 or alpha^(2) = 0*5 or alpha = 0* 707 = 70* 7 % `
For dissociation to be ` 50 % , alpha = 0*50 , K_(p) = 640` mm ( constant )
` :. 640 = (4 ( 0*5)^(2))/(1-(0*5)^(2)) xx P or 640 = 1/(1-1/4) P= 4/3 P or P =480 "mm"` .