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Chemical equilibrium

First Year 2011-2012. Chemical equilibrium. By Dr. Hisham Ezzat Abdellatef Professor of Pharmaceutical Analytical Chemistry. Classification of reaction. Homogeneous only one phase. Heterogeneous mixture is not uniform. e.g: In the gas phase: H 2(g) + I 2 → 2 HI (g)

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Chemical equilibrium

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  1. First Year 2011-2012 Chemical equilibrium By Dr. HishamEzzatAbdellatefProfessor of Pharmaceutical Analytical Chemistry

  2. Classification of reaction Homogeneous only one phase Heterogeneous mixture is not uniform e.g: In the gas phase: H2(g) + I2 → 2 HI(g) In the liquid phase: CH3COCl + CH3OH → CH3COOCH3 + HCIAcetyl Chloride methyl alcohol methylacetate

  3. → irreversible reaction or complete reaction reversible reaction N2(g) + 3 H2(g) 2 NH3(g) A + B → AB at equilibrium ----the rate of forward reaction equal the rate of backward reaction

  4. Factors influencing equilibrium: • forward and reverse rates • partial pressures • concentrations • temperature

  5. Law of mass action A2(g) + B2(g) 2 AB(g) rate of the forward reaction = Kf. [A2] [B2] rate of the reverse reaction is = Kr [AB]2 At equilibrium ratef = rater Kf [A2] [B2] = Kr [AB] 2 K which is called the equilibrium constant

  6. In general, for any reversible reaction aA + bBeE + fFeE +fFaA + bB

  7. Mechanisms of more than one step 1 2 NO2CI(g)  2 NO2(g) + CI2(g) mechanism consisting of two steps: 1- NO2CI  NO2 + Cl 2. NO2CI  NO2 + CI2

  8. For reactions involving gases P α C using partial pressures instead of concentration. N2(g) + 3 H2(g)2 NH3(g) (Kc ≠ Kp)

  9. The Relationship Between Kp and KC: aA + bBeE + fF Assuming ideal gas PV= nRT concentration of a gas X Px is its partial pressure • Px = [X] RT Kp = Kc. (RT)n(g)

  10. Example 1: For the reaction N2O4(g) 2 NO2(g) The concentrations of the substances present in an equilibrium mixture at 25°C are [N2O4] = 4.27 x 10-2mol/L [NO2] = 1.41 x 10-2mol/L what is the value of Kcfor this temperature.

  11. Example 2: At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At equilibrium the ONCI(g) is 9.0% dissociated: 2 ONCI(g) 2 NO(g) + CI2(g) Calculate the value of Kc for equilibrium at 500 K. Solution: [ONCI(g) ] = 1 mol/L since 9.0% dissociated, Number of moles dissociated = at equilibrium, [ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/L amounts of CI2 : 2 ONCI  2NO + CI2 2 x x 0.09 mol 0.045 mol

  12. Example 3: For the reaction 2 SO3(g) 2 SO2(g) + O2(g) at 1100 K Kc is 0.0271 mol/L. what is Kp at same temperature. Solution: n = 3-2 =1 Kp = Kc (RT)+1 = 0.0271 mol/L x (0.0821 L. atm / K. mol) (1100 K) = 2.45 atm

  13. Example 4: What is Kc for the reaction? N2(g) + 3 H2(g) 2 NH3(g) At 500°C if Kp is 1.5 x 10-5 / atm-2 at this temperature. Solution: Kp = Kc (RT)n Kp = Kc (RT)-2 Kc = Kp (RT)2 = x [0.0821.atm/K.mol x 773K]2 = (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2) = 6.04 x 10-2 L2 / mol2 n = 2 - 4 = -2 T = 273 + 500 = 773 K

  14. Try At 127oC, K = 2.6 x 10-5 mol2/L2 for the reaction 2NH3(g)  N2(g) + 2H2(g) Calculate Kp at this temperature

  15. Reaction quotient (Q).

  16. Predicting the Direction of a Relation: • For the reaction PCl2(g) PCl3(g) + Cl2(g)at 250oC • Suppose that a mixture of 1.00 mol of PCI5(g)/ 0.05 mol of PCI3(g)/ and 0.03 mol of CI2(g) is placed in 1.0 L container. Is this an equilibrium system, or will a net reaction occur in one direction or the other?

  17. 1- Q < Kc from left to right (the forward direction) to approach equilibrium. 2- Q = KcThe system is in equilibrium. 3- Q > Kc from right to left (the reversible direction) to approach equilibrium.

  18. Q (0.015 mol/L) <kc (0.0415 mol/L). The system is not at equilibrium. The reaction will proceed from left to right.

  19. Example 4: • For the reaction 2 SO2(g) + O2(g) 2 SO3(g) at 827°C, kc is 36.9 L / mol. If 0.05 mol of SO2(g), 0.03 mol of O2(g), and 0.125 mol of SO3(g) are mixed in a 1.0 L container, in what direction will the reaction proceed? • Solution: Since Q (208 L/mol) > kc (36.9 L/mol), the reaction will proceed from right to left (SO3 will dissociate).

  20. Heterogeneous Equilibria: • The concentration of a pure solid or a pure liquidis constant and do not appear in the expression for the equilibrium constant. • For example CaCO3(S)CaO(S) + CO2(g) Kc = [CO2] 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)

  21. Example 5: Kc for the HI equilibrium at 425°C is 54.5: H2(g) + I2(g) 2 HI(g) A quantity of HI(g) is placed in a 1.01 container and allowed to come to equilibrium at 425°C What are the concentrations of H2(g) and I2(g) in equilibrium with 0.5 mol/L of HI(g)

  22. Solution: at equilibrium [H2] = [I2] [H2] = [I2] = x [HI] = 0.5 mol/L  x = 0.068 mol/L The equilibrium concentration is: [HI] = 0.5 mol/ L [H2] = [I2] = 0.068 mol/ L

  23. Example 6: For the reaction H2(g) + CO2(g) H2O(g) + CO(g) kc is 0.771 at 750°C. If 0.01 mol of H2 and 0.01 mol of CO2 are mixed in 1 liter container at 750°C, what are the concentrations of all substances present at equilibrium?

  24. Solution: If xmol of H2 reacts with xmol of CO2 out of the total amount supplied, xmol H2O and xmol CO will be produced. Hence

  25. square root X = 0.0878 – 0.878 X X= 0.00468 mol/l At equilibrium, therefore [H2] = [CO2] = 0.01 mol/L - 0.00468 mol/L = 0.0053 mol/L [H2O] = [CO] = 0.00468 mol/L

  26. Example 7: • For the reaction C(s) + CO2(g) 2 CO(g) Kp is 167.5 atm at 1000°C. What is the partial pressure of CO(g) in an equilibrium system in which the partial pressure ofCO2(g) is 0.1atm? • Solution: • PCO2= 16.8 • PCO= 4.10 atm

  27. Example 8: Kp for the equilibrium: FeO(s) + CO(g) Fe(s) + CO2(g) at 1000°C is 0.403. If CO(g) at a pressure of 1.0 atm, and excess FeO(s) are placed in a container at 1000°C, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?

  28. Solution: Let x equal the partial pressure of CO2 when equilibrium is attained X = PCO2 = 0.287 atm 1.0 – X = Pco = .713

  29. LeChatelier Principle • If a change is made to an equilibrium, the equilibrium shifts in the direction that consumes the change • Case 1: Changing the amounts of reactants / products. • Case 2: Changing the volume by changing pressure. • Case 3: Changing the temperature.

  30. 1- Concentration changes: • If the concentration of substance is increased, the equilibrium will shift in a way that will decrease the concentration of the substance that was added. e.g: H2(g) + I2(9) 2 HI(g) • Increase H2 or I2 → shift to to formation of HI • Removal of H2 or I2 ← Reaction shift to decomposition of HI.

  31. 2- Pressure changes: • Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas. 2 SO2(g) + O2(g) 2 SO3(g) 3 moles 2 moles • When the pressure on an equilibrium mixture is increased (or the volume of the system decreased), the position of equilibrium shifts to the right., and vice versa. • For reactions in which n = 0, pressure changes have no effect on the position equilibrium. e.g: N2(g) + O2(g) 2 NO(g)

  32. 3- Temperature changes: For the reaction N2(g) + 3 H2(g) 2 NH3(g)H = - 92.4 KJ Since H is -ve, the reaction to the right evolves heat N2(g) + 3 H2(g) 2 NH3(g) + 92.4 KJ The highest yields of NH3 will be obtained at the lowest temperatures and high pressures.

  33. Also consider the reaction CO2(g) + H2(g) CO(g) + H2O(g)H = + 41.2KJ Since H is + ve , we can write the equation 41.2 KJ + CO2(g) + H2(g) —— CO(g) + H2O(g) Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change.

  34. 4- Effect of catalyst: The addition of a catalyst causes a system to achieve equilibrium faster but does not alter the position of equilibrium. • 5- Addition of an inert gas: If an inert gas is introduced into a reaction vessel containing other gases at equilibrium, it will not affect the position of equilibrium because it will not alter the partial pressures or the concentrations of any of the substances already present.

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