The oxidation of nitric oxide NO + 1/2O_2 NO_2 takes place in an isothermal batc

Question

The oxidation of nitric oxide NO + 1/2O_2 NO_2 takes place in an isothermal batch reactor. The reactor is charged with a mixture containing 20.0 volume percent NO and the balance air at an initial pressure of 380 kPa (absolute). Assuming ideal gas behavior, determine the composition of the mixture (component mole fractions) and the final pressure (kPa) if the conversion of NO is 90%. Suppose the pressure in the reactor eventually equilibrates (levels out) at 360 kPa. What is the equilibrium percent conversion of NO? Calculate the reaction equilibrium constant at the prevailing temperature, k_p[(atm)^-0.5], defined as K_p = (Pno_2) / (P_no)(Po_2)^0.5

Explanation / Answer

Remember the Dalton law and gas law (p=nRT/V, i.e, p = const.n)

a.

The initial volume ratio is NO:air is 2:8

The same value for molar ratio , molar concentration ratio and partial pressure ratio.

The initial pressure of NO is 0.20x 380 kPa = 76 kPa

The initial pressure of air is 0.80x 380 kPa = 304 kPa (0.21x304 = 63.84 kPa for O2)

At equilibrium:

The partial pressure of NO is 0.10x76 kPa = 7.6 kPa (a variation of 7.6 -76 = - 68.4kPa)

The variation of the partial pressure of O2 (and also for air) is -68.4/2= - 34.2 kPa.

The partial pressure of the new formed NO2 is 68.4 kPa (a variation of 68.4-0=68.4 kPa) .

The total pressure variation is – 34.2 kPa. The final pressure will be

          380 -34.2 = 345.8 kPa.

The mol fraction of NO will be

7.6 kPa/ 345.8 kPa = 0.0220

The partial pressure of O2 t equilibrium is

63.84 – 34.2 = 29.6 kPa

The molar fraction of O2 is

29.6 kPa/345.8 kPa = 0.086

The molar fraction of NO2 is

68.4/345.8 = 0.197                        (the rest is molar fraction of N2)

b.

Assume no gas loss. Only the pressure is higher.

NO      +   0.5O2   +     = NO2

76 kPa          63.83 kPa       0                Total pressure 380 kPa,   initial values

-x            - 0.5 x               + x   (total variation - 0.5x at the new equilibrium, p=360kPa)                    

- 0.5 x = 360 kPa – 380 Pa

x = 40 kPa

pNO = 76-40 = 36 kPa

PO2 = 63.84 – 20 = 43.84 kPa

pNO2   = 40 kPa

Calculate Kp using these values.

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