Question

One mole each of CO 2 , O 2 , and N 2 are fed to a batch reactor and heated to 3000 K and 5.0 atm.

Two reactions as shown proceed to equilibrium

CO 2<-----> CO + ½ O 2 K p1 = 0.3272 atm 1/2

½ N 2 + ½ O 2 <-------> NO, K p2 = 0.1222

Calculate the equilibrium composition of the reactor content. [CO 2 = 25.74 mole%; CO = 4.50

mole%; O 2 = 33.55 mole%; N 2 = 30.30 mole%; NO = 3.90 mole%]

Answer 1

The reactions and equilibrium constants are CO2-àCO+1/2O2., Kp1= 0.3272

Partial pressures : Mole fraction* total pressure

Partial pressures : PCO2= PN2= PO2= 5/3= 1.66 atm

Let P =drop in pressure of CO2 to reach equilibrium

Hence at equilibrium, PCO2= 1.66-P, O2= 1.66+P/2, PCO= P

Kp1= [PCO] Sqrt(PO2)/ PCO2.

= P*(1.66+P/2)^0.5/ (1.66-P)= 0.3272

When solved, P=0.52 atm

P= 0.52 atm, PCO2= 1.66-0.52 = 1.4 atm, PO2= 1.66+0.52/2 =1.92 atm, PCO= 0.52 atm

From the second reaction 0.5N2+0.5O2-àNO

Kp = PNO/ (sqrt(PN2*PO2)

Let P= partial pressure of NO

So at equilibrium   PN2= 1.66- P/2 and PO2= 1.92-P/2

0.1222= P/ {( sqrt(1.66-P/2)* (1.92-P/2)}

P= 0.21atm, PCO= 0.21atm, PN2= 1.66-0.221/2= 1.555, PO2= 1.92-0.21/2= 1.815 atm

So at eq. PCO= 0.21 atm, PN2= 1.555 atm , PO2= 1.815 atm, PCO2= 1.4 atm

Total pressure= 0.21+1.555+1.815+1.4= 4.98 atm

Mole fractions : CO= 0.21/4.98= 0.042, N2= 1.555/4.98= 0.311, O2= 1.815/4.98= 0.365, CO2= 1.4/4.98= 0.2811