One Ibmol of carbon reacts with 2 lbmol of oxygen (O₂) to form an equilibrium mixture of CO2, CO, and O₂ at 5240°F, 1 atm.For the equilibrium mixture, determine the number of moles of:(a) COnco(b) CO2nco₂=(c) 0₂no₂=i 0.393i 0.2091.238molesmolesmoles

Given that- Carbon (C)= 1 lb.mol Oxygen (O2)=2 lb.mol Temperature, T=52400F Pressure, P= 1 atm The reaction forms equilibrium mixture of CO, CO2 and O2 so as the reaction is given as follows- C+2O2→xCO + yCO2+ zO2 Comparing coefficient of elements- For(C), 1=x+yy=1-xFor(O2), 4=x+2y+2z4=x+2(1-x)+2z=x+2-2x+2z2=2z-xz=(2+x)2 Hence, the reaction becomes, C+2O2→xCO + (1-x)CO2+ 2+x2O2 Now, in 1 lbmol, the amount of mixture is- n=x+(1-x)+2+x2n=4+x2 Further, at equilibrium- CO2↔CO+12O2 Now, using the relationship between equilibrium constant (k), stoichiometric coefficients of reactants and products, number of moles of each component species and the total number of moles of product mixture is given by- k=ncνc×nOνOnAνA×nBνB×P/Prefnνc+νO-νA-νBNow, from the given equilibrium reaction, νACO2↔νCCO+νOO2 Comparing, the stoichiometric coefficients are obtained as- (Since component B is not involved so ignoring B) νA=νC=1, νO=1/2nA(moles of CO2) =1-xnC(moles of CO)=xnO(moles of O2)=2+x2n (total moles of mixture=4+x2 Hence, at given pressure, P= 1 atm, and reference pressure Pref= 1 atm, the equilibrium constant is obtained as- k=x1×2+x21/2(1-x)1×1/14+x21+1/2-1=x1-x2+x21/224+x1/2k=x(1-x)[(2+x)(4+x)]1/2 Now, at 52400F=3166.38K, for reaction- CO2↔CO+12O2 The corresponding value of equilibrium constant using interpolation of values is- logk=-0.170298k=10-0.170298k=0.67562 Now, substituting this value of (k) in the obtained equation- k=x1-x2+x4+x1/20.675622=x1-x22+x4+x=x21+x2-2x2+x4+x=x3+2x24+4x2-8x+x+x3-2x20.4565=x3+2x2x3+2x2-7x+40.4565x3+0.913x2-3.1955x+1.826=x3+2x20.5435x3+1.087x2+3.1955x-1.826=0 Solving the cubic equation, x=0.476 Hence, for the equilibrium mixture, (a) Number of moles of CO is- nCO=x=0.476 mol (a) Number of moles of CO2 is- nCO2=(1-x)=(1-0.476)=0.524 mol (a) Number of moles of O2 is- nO2=2+x2=2+0.4762=1.238 mol