For N2O4(g) 2NO2(g), K_p = 0.50 atm at a given temperature. Starting from pure N2O4, the degree of dissociation alpha at a total equilibrium pressure of 4.0 atm is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce046

For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{p} = 0.50$ atm at a given temperature. Starting from pure $N X_{2} O X_{4}$ , the degree of dissociation $α$ at a total equilibrium pressure of 4.0 atm is

With 1 mol at the start,

K_{p} = 4 α^{2} P / (1 - α^{2})

, so

α^{2} = K_{p} / (K_{p} + 4 P)

. Then

α = 0.50/ (0.50 + 16.00) = 0.50/16.50 = 0.174

. At 1 atm it would be 0.333, so compression lowers

α

, but not in simple proportion to

P