Question

Liquid N2O3 is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO2. At 25 degree C, a value of Kp = 1.91 has been established for this decomposition. If 0236 moles of N2O3 are placed in a 1.52-L vessel at 25 degree C, calculate the equilibrium partial pressures of N2O3(g), NO2(g) and NO(g).

Answer 1

N2O3(g) ----> NO2(g) + NO(g)

Kp = pNO*pNO2 / pN2O3

INITIAL PRESSURE OF n2o3 = nRT/V = 0.236*0.0821*298/1.52

   = 3.8 atm

at equilibrium

pN2O3 = 3.8-x

pNO = X

pNO2 = X

1.91 = X^2/(3.8-X)

X = 1.9

partial pressure of N2O3 = 3.8-1.9 = 1.9 atm

partial pressure of NO = 1.9 atm

partial pressure of NO2 = 1.9 atm