In the forward direction the reaction creates more

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Unformatted text preview: d an excess amount of PbO. Let us calculate the number of moles of CO and CO2 at equilibrium at 298 K under a total pressure of 1 bar. Assuming we are dealing with the red PbO (s) and not the yellow PbO (s), which is less stable at room temperature, we can write: ΔRG 298 K = ΔFG 298 K (CO2(g)) + ΔFG 298 K (Pb(s)) - ΔFG 298 K (CO(g)) - ∅ ∅ ∅ ∅ ΔFG 298 K (PbO(s)) ∅ ΔRG 298 K = (- 394.36 kJ) + (0 kJ) - (- 137.17 kJ) - (- 188.93 kJ) = - 68.26 kJ ∅ Therefore, ln Keq = - (- 68.26 * 103 / (8.314*298) = 27.55 and K = 9.22 1011 Furthermore, Keq is given by: K eq = ( ) ( ) = PCO / P∅ = PCO PCO a(CO[g])a(Pb[s]) PCO / P ∅ a CO 2 [g] a Pb[s] 2 2 = x CO2 P x COP = x CO2 x CO Note that the activities of PbO and Pb are each unity since these are pur...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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