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//(14.4) One kilomole of carbon monoxide reacts with the theoretical amount of air to form an equilibrium mixture of CO2, CO, O2, and N2 at 2500 K and 1 atm. Determine the equilibrium composition in terms of mole fractions, and compare with the result of Example 14.2.
//solution
//For a complete reaction of CO with the theoretical amount of air
//CO + .5 O2 + 1.88N2 -----> CO2 + 1.88N2
//Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is
//CO + .5O2 + 1.88N2 ---> zCO + z/2 O2 + (1-z)CO2 + 1.88N2
K = .0363 //equilibrium constant the solution to Example 14.2
p =1 //in atm
pref = 1 //in atm
//solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives
z = .175
yCO = 2*z/(5.76 + z)
yO2 = z/(5.76 + z)
yCO2 = 2*(1-z)/(5.76 + z)
yN2 = 3.76/(5.76 + z)
printf('the mole fraction of CO is: %f',yCO)
printf('\nthe mole fraction of O2 is: %f',yO2)
printf('\nthe mole fraction of CO2 is: %f',yCO2)
printf('\nthe mole fraction of N2 is: %f',yN2)
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