`K_p` of the reaction, `N_2O_4(g)hArr2NO(g)` at 773K is 640 mm of Hg. If the equilibrium pressure is 160 mm of Hg, then what percent of `N_2O_4` will be dissociated? At what pressure will its degree of dissociation be 50%?

Explanation

Let initial number of moles of `N_O_2`=1 mol and the degree of dissociation of `N_2O_4` at equilibrium`=prop`
`N_2O_4(g) Number of moles at equilibrium: `1-prop" "2prop`
Total number of moles at equilibrium
`=1-prop-2prop=1+prop`
At equilibrium
`P_(N_2O_4)=((1-prop)/(1+prop))xx160` and `P_(NO_2)=((2prop)/(1+prop))xx160`
`K_p=(p_(NO_2)^2)/(p_(NO_2O_4))=(4prop^2xx160)/(1-prop^2)` or `640=(4prop^2xx160)/(1-prop^2)`
`:.prop=0.707`
`:.` Percent dissociation of `N_2O_4=70.7%`
Let, at a pressure of p mm Hg, the degree of dissociation of `N_2O_4` is 0.5
`:p_(N_2O_4)=((1-0.5)/(1+0.5))p=p/3mmHg`
and `p_(NO_2)=((2xx0.5)/(1+0.5))p=(2p)/3 mm Hg`
`K_p=(p_(NO_2)^2)/(p_(N_2O_4))=((2p)/(3)^2)/(p/3)=(4p)/3` or `(4p)/3=640`
`:.` p=480 atm