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 whatpercent of N_2O_4 will be dissociated? At what pressure will its degree of dissociation be 50%?

K_p of the reaction, N_2O_4(g)hArr2NO(g) at 773K is 640 mm of Hg. If t

1.	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 whatpercent of N_2O_4 will be dissociated? At what pressure will its degree of dissociation be 50%?
Answer» Solution :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