Question

a) One kmol of N2O4 dissociates at 25℃, 1 atm to form an equilibrium ideal gas mixture of N2O4 and NO2 in which the amount of N2O4 present is 0.8154 kmol. Determine the amount of N2O4 that would be present in an equilibrium mixture at 25℃, 0.5 atm.

Answer 1

Answer:

neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C

Explanation:

             ni      change    eq.

N2O4    1          1 - x       0.8154.....P = 1 atm; T = 25°C

NO2      0        0 + x          x

∴ x = neq = Peq.V / R.T.....ideal gas mix

if P = 0.5 atm, T = 25°C; assuming: V = 1 L

⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))

⇒ x = neq = 0.0205 mol

⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol