Thermodynamics

802 | ThermodynamicsAssumptions 1 The equilibrium composition consists of CO 2 , CO, and O 2 .2 The constituents of the mixture are ideal gases.Analysis The stoichiometric and actual reactions in this case are as follows:Stoichiometric:Actual:C balance:CO 1 2 O 2 ∆ CO 2 1thus n CO2 1, n CO 1, and n O2 1 2 22CO 3O 2 ¡ xCO 2 yCO zO 2123 1552553products reactants(leftover)2 x y or y 2 xO balance:8 2x y 2zor z 3 x 2Total number of moles:Pressure:N total x y z 5 x 2P 304 kPa 3.0 atmThe closest reaction listed in Table A–28 is CO 2 ∆ CO 1 – 2 O 2 , for whichln K P 2.801 at 2600 K. The reaction we have is the inverse of this, andthus ln K P 2.801, or K P 16.461 in our case.Assuming ideal-gas behavior for all components, the equilibrium constantrelation (Eq. 16–15) becomesSubstituting, we getn CO2K P N CO 2n O2a P n CO2 n CO n O2bN totalnN COCO N O216.461 1>2x12 x213 x>22 a 31>2 5 x>2 bSolving for x yieldsThenx 1.906y 2 x 0.094z 3 x 2 2.047Therefore, the equilibrium composition of the mixture at 2600 K and 304kPa is1.906CO 2 0.094CO 2.074O 2Discussion In solving this problem, we disregarded the dissociation of O 2into O according to the reaction O 2 → 2O, which is a real possibility at hightemperatures. This is because ln K P 7.521 at 2600 K for this reaction,which indicates that the amount of O 2 that dissociates into O is negligible.(Besides, we have not learned how to deal with simultaneous reactions yet.We will do so in the next section.)