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

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Presentation on theme: "Homogeneous Gaseous Equilibria Many industrial processes involve gaseous systems eg. Haber Process It is more convenient to use pressure measurements to."— Presentation transcript:

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, K p, in terms of the partial pressures of the reactant and products 3H 2 (g) + N 2 (g) 2NH 3 (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 = p A + p b + p c Partial pressure, p A Partial pressure, p B Partial pressure, p C Total = 20 Mole fraction = 8/20Mole fraction = 7/20 Mole fraction = 5/20 Mole fraction, x = moles of component total moles of all components

4 Total pressure P = p A + p b + p c Partial pressure, p A Partial pressure, p B Partial pressure, p C Total = 20 Mole fraction = 8/20Mole fraction = 7/20 Mole fraction = 5/20 Equilibrium mixture is at 3000 kPa Partial pressure, p = mole fraction x total pressure p A = 8/20 x 3000 = 1200 kPa p C = 5/20 x 3000 = 750 kPa p B = 7/20 x 3000 = 1050 kPa

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

6 Total pressure P = p A + p b + p c Total = 20 Equilibrium mixture is at 3000 kPa p A = 8/20 x 3000 = 1200 kPa p C = 5/20 x 3000 = 750 kPa p B = 7/20 x 3000 = 1050 kPa K p = p C 2 p A x p B 3 A (g) + 3B (g) 2C (g) = x = 4.25 x kPa -2

7 K p Expressions & Units K p = (pNH 3 ) 2 (pN 2 ) x (pH 2 ) 3 N 2(g) + 3H 2(g) 2NH 3(g) K p = kPa 2 kPa x kPa 3 K p = 1 kPa 2 = kPa -2

8 Calculating K p values X (g) + Y (g) 2Z (g) 1. Calculate the mole fractions of each gas 2. Calculate partial pressure of each gas 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. Mole fraction, x = moles of component total moles of all components MolesMole fraction, xPartial pressure, p X77/10 = 0.7p x = 0.7 x 100 = 70kPa Y22/10 = 0.2p y = 0.2 x 100 = 20kPa Z11/10 = 0.1p z = 0.1 x 100 = 10kPa Total10 Partial pressure, p = mole fraction x total pressure

9 Calculating K p values X (g) + Y (g) 2Z (g) 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. MolesMole fraction, xPartial pressure, p X77/10 = 0.7p x = 0.7 x 100 = 70kPa Y22/10 = 0.2p y = 0.2 x 100 = 20kPa Z11/10 = 0.1p z = 0.1 x 100 = 10kPa Total10 K p = (pZ) 2 (pX) x (pY) 3. Calculate equilibrium constant Kp, and work out units = (10) 2 20 x 70 (kPa) 2 (kPa) x (kPa) = 0.071

10 Calculating K p values PCl 5 (g) PCl 3 (g) + Cl 2(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 PCl 5 is 80kPa. Calculate Kp. Partial pressure, p PCl 5 p PCL5 = 80kPa PCl 3 p PCl3 = Cl 2 p Cl2 = Total120kPa P tot = p PCl5 + p PCl3 + p Cl2 120 = 80+ p PCl3 + p Cl2 p PCl3 = p Cl2 (both 1 mole) p PCl3 = 20kPa p Cl2 = 20kPa

11 Calculating K p values PCl 5 (g) PCl 3 (g) + Cl 2(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 PCl 5 is 80kPa. Calculate Kp and state its units. Partial pressure, p PCl 5 p PCL5 = 80kPa PCl 3 p PCl3 = 20kPa Cl 2 p Cl2 = 20kPa Total120kPa K p = (pPCl 3 )(pCl 2 ) (pPCl 5 ) = 20 x K p = 5 (kPa)(kPa) (kPa) K p = 5 kPa

12 Calculating K p values 2SO 2 (g) + O 2 (g) 2SO 3 (g) Initially a vessel contained 12 moles of SO 2 and 6 moles of O 2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO 2 had reacted to form SO 3. Calculate Kp for this reaction, and state its units. Initial molesMoles at equlibMole fractionPartial pressure SO 2 12 O2O2 6 SO 3 0 Total200 kPa

13 Calculating K p values 2SO 2 (g) + O 2 (g) 2SO 3 (g) Initial molesMoles at equlib SO 2 12 O2O2 6 SO 3 0 Total Initially a vessel contained 12 moles of SO 2 and 6 moles of O 2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO 2 had reacted to form SO 3. Calculate Kp for this reaction, and state its units. 10% SO 2 remaining 10% x 12 = 1.2 moles SO 2 SO 2 reacts with O 2 so % O 2 remaining 10% x 6 = 0.6 moles O 2 90% SO 2 reacted to form SO 3 90% x 12 = 10.8 moles SO 3

14 Calculating K p values 2SO 2 (g) + O 2 (g) 2SO 3 (g) Initially a vessel contained 12 moles of SO 2 and 6 moles of O 2. At equilibrium, at a total pressure of 200kPa it was found that 90% of the SO 2 had reacted to form SO 3. Calculate Kp for this reaction, and state its units. At equilibrium Mole fraction = moles total moles Initial moles Moles at equlib Mole fraction SO /12.6 O2O /12.6 SO /12.6 Total12.6

15 Calculating K p values 2SO 2 (g) + O 2 (g) 2SO 3 (g) Initial molesMoles at equlibMole fractionPartial pressure SO /12.6 O2O /12.6 SO /12.6 Total kPa Partial pressure, p = mole fraction x total pressure p SO2 = 1.2/12.6 x 200 kPa = kPa p O2 = 0.6/12.6 x 200 kPa = 9.52 kPa p SO2 = 10.8/12.6 x 200 kPa = kPa

16 Calculating K p values 2SO 2 (g) + O 2 (g) 2SO 3 (g) Initial molesMoles at equlibMole fractionPartial pressure SO / O2O / SO / Total kPa K p = (pSO 3 ) 2. (pSO 2 ) 2 (pO 2 ) = (19.05) 2 x 9.52 = 8.51 K p = kPa 2. (kPa) 2 (kPa) K p = 8.51 kPa -1.


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