Question

34. Consider the reaction 2NO(g) + O2(g) ⇆ 2NO2(g) At a certain temperature, an equilibrium mixture consists of 0.0230 mole O2, 0.0470 mole NO, and 0.919 mole of NO2. Calculate KP for the reaction, given that the total pressure is 0.200 atm. KP =

Answer 1

Mole fraction of O2 = 0.0230 / (0.0230 + 0.0470 + 0.919) = 0.0232

Mole fraction of NO = 0.0470 / (0.0230 + 0.0470 + 0.919) = 0.0475

Mole fraction of NO2 = 0.919 / (0.0230 + 0.0470 + 0.919) = 0.929

According to Raoult's law,

Partial pressure of each gas in the mixture = total pressure * mole fraction of that gas,

So,

P_O2 = Ptotal * X_O2 = 0.200 * 0.0232 = 0.00464 atm

P_NO = Ptotal * X_NO = 0.200 * 0.0475 = 0.00950 atm

P_NO2 = Ptotal * X_NO2 = 0.200 * 0.929 = 0.186 atm

Now, equilibrium constant expression intems of pressure can be written as,

Kp = P_NO2² / (P_NO² * P_O2)

Kp = (0.186)² / [(0.00950)²*(0.00464)]

Kp = 8.26 * 10⁴