We disregard the oxides of nitrogen in this example

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Unformatted text preview: t gas, the KP value of the reaction is the same as that used in Example 163. The stoichiometric and actual reactions in this case are FIGURE 1613 Schematic for Example 164. Stoichiometric: Actual: CO 3CO 1 2 O2 CO2 1thus nCO2 1, nCO 1, and nO2 1 22 2.5O2 8N2 xCO2 123 yCO zO2 152553 reactants (leftover) 8N2 123 inert products C balance: O balance: Total number of moles: 3 8 N total x 2x x y y y 2z z or y or z 8 13.5 3 2.5 x 2 x x 2 Assuming ideal-gas behavior for all components, the equilibrium constant relation (Eq. 1615) becomes KP Substituting, we get N nCO2 CO 2 N nCO N nO22 CO O a P nCO b N total 2 nCO nO2 16.461 Solving for x yields 13 x x2 12.5 x>22 1>2 a 5 13.5 x>2 b 1>2 x 2.754 804 | Thermodynamics Then y z 3 2.5 x x 2 0.246 1.123 Therefore, the equilibrium composition of the mixture at 2600 K and 5 atm is 2.754CO2 0.246CO 1.123O2 8N2 Discussion Note that the inert gases do not affect the KP value or the KP relation for a reaction, but they do affect the equilibrium com...
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