N2O4(g) (molar mass 92 g/mol) partly dissociates as N2O4(g) 2NO2(g). At a total pressure of 1.0 atm, the equilibrium mixture has a vapour density (relative to H2) of 36. Take the degree of dissociation as alpha = (D - d)/d, where D is the vapour density without dissociation and d is the observed value. The value of K_p (in atm) is

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$N X_{2} O X_{4} (g)$ (molar mass 92 g/mol) partly dissociates as $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ . At a total pressure of 1.0 atm, the equilibrium mixture has a vapour density (relative to $H X_{2}$ ) of 36. Take the degree of dissociation as $α = (D - d) / d$ , where $D$ is the vapour density without dissociation and $d$ is the observed value. The value of $K_{p}$ (in atm) is

Without dissociation,

D = 92/2 = 46

. So

α = (46 - 36) /36 = 0.2778

. For

N X_{2} O X_{4} 2 NO X_{2}

with 1 mol start,

K_{p} = 4 α^{2} P / (1 - α^{2}) = 4 (0.0772) (1.0) /0.9228 = 0.334

atm.