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Homogeneous Gaseous Equilibria

Homogeneous Gaseous Equilibria. Many industrial processes involve gaseous systems eg. Haber Process It is more convenient to use pressure measurements to calculate equilibrium constants

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Homogeneous Gaseous Equilibria

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  1. Homogeneous Gaseous Equilibria • Many industrial processes involve gaseous systems eg. Haber Process • It is more convenient to use pressure measurements to calculate equilibrium constants • This generates expressions for equilibrium constants, Kp, in terms of the partial pressures of the reactant and products 3H2(g) + N2(g) 2NH3(g)

  2. Partial Pressures • In an equilibrium mixture of gases, each component will contribute to the pressure • The total pressure is the sum of the pressures of all the gases • Each gas contributes a partial pressure, p, to the total pressure, P. • However, only total pressure can be measured

  3. Total pressure P = pA + pb + pc 8 7 Total = 20 5 Partial pressure, pA Partial pressure, pB Partial pressure, pC Mole fraction = 8/20 Mole fraction = 7/20 Mole fraction = 5/20 Mole fraction, x = moles of component total moles of all components

  4. Total pressure P = pA + pb + pc Equilibrium mixture is at 3000 kPa 8 7 Total = 20 5 Partial pressure, pA Partial pressure, pB Partial pressure, pC Mole fraction = 8/20 Mole fraction = 7/20 Mole fraction = 5/20 Partial pressure, p= mole fraction x total pressure pB = 7/20 x 3000 = 1050 kPa pC = 5/20 x 3000 = 750 kPa pA = 8/20 x 3000 = 1200 kPa

  5. The Equilibrium Constant Kp The general equation for any homogeneous gaseous reaction at equilibrium is… aA(g) + bB(g) cC(g) + dD(g) Product pressures Kp = pCc pDd pAa pBb Reactant pressures pA represents the partial pressure in kPa or Pa a,b,c & d are the numbers of moles of substances A, B, C & D

  6. Total pressure P = pA + pb + pc Equilibrium mixture is at 3000 kPa 8 A(g) + 3B(g) 2C(g) 7 Total = 20 5 pA = 8/20 x 3000 = 1200 kPa pB = 7/20 x 3000 = 1050 kPa pC = 5/20 x 3000 = 750 kPa Kp = pC2 pA x pB3 = 7502 1200 x 10503 = 4.25 x 10-4 kPa-2

  7. Kp Expressions & Units N2(g) + 3H2(g) 2NH3(g) Kp = (pNH3)2 (pN2)x(pH2)3 Kp = kPa2 kPa xkPa3 Kp = 1 kPa2 = kPa-2

  8. Calculating Kp values A gaseous mixture contains 7 moles of X, 2 moles of Y and 1 mole of Z at equilibrium at a total presssure of 100kPa. Calculate the value of Kp. X (g) + Y(g) 2Z (g) 1. Calculate the mole fractions of each gas Mole fraction, x = moles of component total moles of all components 2. Calculate partial pressure of each gas Partial pressure, p= mole fraction x total pressure

  9. Calculating Kp values A gaseous mixture contains 7 moles of X, 2 moles of Y and 1 mole of Z at equilibrium at a total presssure of 100kPa. Calculate the value of Kp. X (g) + Y(g) 2Z (g) 3. Calculate equilibrium constant Kp, and work out units = (10)2 20x70 (kPa)2 (kPa)x(kPa) Kp = (pZ)2 (pX)x(pY) = 0.071

  10. Calculating Kp values PCl5(g) PCl3(g) + Cl2(g) A sample of phosporus (V) chloride is introduced to the reaction vessel. The reaction is carried out at 120kPa. At equilibrium the partial pressure of PCl5 is 80kPa. Calculate Kp. Ptot = pPCl5+ pPCl3+ pCl2 120 = 80+ pPCl3+ pCl2 pPCl3= pCl2(both 1 mole) pPCl3 = 20kPa pCl2 = 20kPa

  11. Calculating Kp values PCl5(g) PCl3(g) + Cl2(g) A sample of phosporus (V) chloride is introduced to the reaction vessel. The reaction is carried out at 120kPa. At equilibrium the partial pressure of PCl5 is 80kPa. Calculate Kp and state its units. Kp = (pPCl3)(pCl2) (pPCl5) = 20 x 20 80 Kp = 5 (kPa)(kPa) (kPa) Kp = 5 kPa

  12. Calculating Kp values 2SO3(g) 2SO2(g) + O2(g) Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units.

  13. Calculating Kp values 2SO3(g) 2SO2(g) + O2(g) Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units. 90% SO2 reacted to form SO3 90% x 12 = 10.8 moles SO3 10% SO2 remaining 10% x 12 = 1.2 moles SO2 SO2 reacts with O2 so..... 10% O2 remaining 10% x 6 = 0.6 moles O2

  14. Calculating Kp values 2SO3(g) 2SO2(g) + O2(g) Initially a vessel contained 12 moles of SO2 and 6 moles of O2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO2 had reacted to form SO3. Calculate Kp for this reaction, and state its units. At equilibrium Mole fraction = moles total moles

  15. Calculating Kp values 2SO3(g) 2SO2(g) + O2(g) Partial pressure, p= mole fraction x total pressure pSO2 = 1.2/12.6 x 200 kPa = 19.05 kPa pO2 = 0.6/12.6 x 200 kPa = 9.52 kPa pSO2 = 10.8/12.6 x 200 kPa = 171.43 kPa

  16. Calculating Kp values 2SO3(g) 2SO2(g) + O2(g) = 8.51 Kp = (pSO3)2. (pSO2)2(pO2) = 171.43 2 (19.05)2 x 9.52 Kp = kPa2. (kPa)2(kPa) Kp = 8.51 kPa-1.

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