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Chemical equilibrium

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  • 1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 17 Equilibrium: The Extent of Chemical Reactions17-1
  • 2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.1 Reaching equilibrium on the macroscopic and molecular levels. N2O4(g) 2NO2(g)17-2
  • 3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibrium - the condition in which the concentrations of all the reactants and products in a closed system cease to change with time At equilibrium: rateforward = ratereverse no further net change is observed because changes in one direction are balanced by changes in the other but it doesn’t mean that the reaction had stopped. The amount of reactants and products are constant but they are not necessarily equal.17-3
  • 4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. If rateforward = ratereverse then kforward[reactants]m = kreverse[products]n kforward [products]n = = K the equilibrium constant kreverse [reactants]m The values of m and n are those of the coefficients in the balanced chemical equation. The rates of the forward and reverse reactions are equal, NOT the concentrations of reactants and products. K is dependent only of the temperature. Note: The terms for pure solids or pure liquids do not appear in the equilibrium constant expression17-4
  • 5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibrium constant expression for N2O4(g) 2NO2(g) 2 k fwd [NO ] 2 eq K= = k rev [N 2O 4 ]eq17-5
  • 6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Practice Problem Write the Equilibrium Constant for the combustion of Propane gas C3H8(g) + O2(g)  CO2(g) + H2O(g) 1. Balance the Equation C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) [CO 2 ]3 [H 2O]4 Kc = [C3 H 8 ]1[O 2 ]5 The subscript “c” in Kc indicates the equilibrium constant is based on reactant and product concentrations The value of “K” is usually shown as a unitless number, BUT IT ACTUALLY DOES HAVE A UNIT EXPRESSION17-603/12/12 6
  • 7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Equilibrium Constants in Terms of Partial Pressures • For reactions involving gases, concentrations are generally reported as partial pressures, and the equilibrium expression is often written: ( PP ) ( PQ ) p q Kp = ( PA ) ( PB ) a b • where the partial pressure of each reactant and product are given as Px in units of atmospheres (atm), and Kp is the equilibrium constant when concentration is given in partial pressures.17-7
  • 8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.2 The range of equilibrium constants small K large K intermediate K17-8
  • 9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. N 2( g ) + O2( g ) ← 2 NO( g ) → K c = 1.0 x10−30 Thus, a mixture of N2 and O2 will react to a very small extent to produce NO at equilibrium. N 2( g ) + 3H 2( g ) ← 2 NH 3( g ) K c = 5.0 x108 → Thus, a mixture of N2 and H2 will almost completely be converted to NH3 at equilibrium.17-9
  • 10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Relationsip between Kc and Kp Kp = Kc (RT)∆n(gas) Δn = (no. of moles of gaseous products) – (no. of moles of gaseous reactants)17-10
  • 11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.4 Converting Between Kc and Kp PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO2 from the gas and forms gypsum. Find Kc for the following reaction, if CO2 pressure is in atmospheres. CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K) PLAN: We know Kp and can calculate Kc after finding ∆ngas. R = 0.0821 L*atm/mol*K. SOLUTION: ∆ngas = 1 - 0 since there is only a gaseous product and no gaseous reactants. Kp = Kc(RT)∆n Kc = Kp/(RT)∆n = (2.1x10-4) / (0.0821 x 1000)1 = 2.6x10-617-11
  • 12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Q - The Reaction Quotient The reaction quotient, Q, is defined in the same way as the equilibrium constant Kc except that the concentrations in the equilibrium constant expression are not necessarily equilibrium We use the molar concentrations of the substances in the reaction. This is symbolized by using square brackets - [ ]. For a general reaction aA + bB cC + dD where a, b, c, and d are the numerical coefficients in the balanced equation, Q (and K) can be calculated as [C]c[D]d Q= [A]a[B]b17-12
  • 13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.3 The change in Q during the N2O4-NO2 reaction.17-13
  • 14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 17.5 Reaction direction and the relative sizes of Q and K. Reaction Reaction Progress Progress reactants products Equilibrium: reactants products no net change forward backward17-14
  • 15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.1 Writing the Reaction Quotient from the Balanced Equation Write the reaction quotient expression, Qc, for each of the following PROBLEM: reactions: (a) The decomposition of dinitrogen pentoxide, N2O5(g) NO2(g) + O2(g) (b) The combustion of propane gas, C3H8(g) + O2(g) CO2(g) + H2O(g) PLAN: Be sure to balance the equations before writing the Qc expression. SOLUTION: [NO2]4[O2] (a) 2 N2O5(g) 4NO2(g) + O2(g) Qc = [N2O5]2 [CO2]3[H2O]4 (b) C3H8(g) + 5O2(g) 3CO2(g) + 4 2O(g) H Qc = [C3H8][O2]517-15
  • 16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.5 Comparing Q and K to Determine Reaction Direction PROBLEM: For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 1000C. At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M. Is the reaction at equilibrium. If not, in which direction is it progressing? PLAN: Write an expression for Qc, substitute with the values given, and compare the Qc with the given Kc. SOLUTION: [NO2]2 (0.55)2 Qc = = = 2.5 [N2O4] (0.12) Qc is > Kc, therefore the reaction is not at equilibrium and will proceed from right to left, from products to reactants, until Qc = Kc.17-16
  • 17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.10 Predicting Reaction Direction and Calculating Equilibrium Concentrations PROBLEM: Given the reaction: CH4(g) + 2H2S(g) CS2(g) + 4H2(g) In one experiment, 4.0 M of CH4, 4.0 M of CS2, 8.0 M of H2S, and 8.0Mof H2 are initially mixed in a vessel at 9600C. At this temperature, Kc = 0.036 In which direction will the reaction proceed to reach equilibrium? SOLUTION: [CH4]initial = 4.0M [CS2]initial = 4.0M [H2S]initial = 8.0M [H2]initial = 8.0M17-17
  • 18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.10 Predicting Reaction Direction and Calculating Equilibrium Concentrations continued [CS2][H2]4 [4.0][8.0]4 Qc = = = 64 [CH4][H2S]2 [4.0][8.0]2 A Qc of 64 is >> than Kc = 0.036 The reaction will progress to the left.17-18
  • 19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. LeChatelier’s Principle When a chemical system at equilibrium is subjected to a stress, the system will return to equilibrium by shifting to reduce the stress. If the concentration increases, the system reacts to consume some of it. If the concentration decreases, the system reacts to produce some of it.17-19
  • 20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Factors Affecting Equilibrium • Concentration Changes • Pressure Changes • Temperature Changes • Addition of a Catalyst17-20
  • 21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. I. Concentration Changes If the conc. of a substance is increased, the equilibrium will shift in a way that will decrease the conc. of the substance that was added. H 2( g ) + I 2( g ) ← 2 HI ( g ) → Where will the reaction shift? Decrease HI - forward Decrease I2 - backward H2 HI Increase H2 - forward I2 Increase HI - backward17-21
  • 22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Note: adding/removing a solid/liquid to an equilibrium system will not cause any shift in the position of equilibrium. addition of an inert gas such as He, Ar, Kr, etc. at constant volume, pressure and temperature does not affect the equilibrium. CaCO3( s ) ← CaO( s ) + CO2( g ) → Increase CO2 : - backward Increase CaO - No effect Decrease CO2 - forward - No effect Adding Kr17-22
  • 23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Practice Exercise: Predicting the Effect of a Change in Concentration on the Equilibrium Position PROBLEM: To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2; 2H2S(g) + O2(g) 2S(s) + 2H2O(g) In what direction will the rection shift if (a) O2 is added? (b) O2 is added? (c) H2S is removed? (d) sulfur is added?17-23
  • 24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. I. Pressure Changes + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas17-24
  • 25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The effect of pressure (volume) on an equilibrium system. An increase in pressure (decrease in volume) shifts the position of the equilibrium in such a way as to decrease the number of moles of gaseous component. When the volume is increased (pressure decreased), a net reaction occurs in the direction that produces more moles of gaseous component N 2( g ) + 3H 2( g ) ← 2 NH 3( g ) → Decrease in pressure : backward Decrease in volume: forward17-25
  • 26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.12 Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position PROBLEM: For the following reactions, predict the direction of the reaction if the pressure is increased: (a) CaCO3(s) CaO(s) + CO2(g) (b) S(s) + 3F2(g) SF6(g) (c) Cl2(g) + I2(g) 2ICl(g) (a) CO2 is the only gas present. The equilibrium will shift toSOLUTION: the direction with less moles of gas. Answer: backward (a) There are more moles of gaseous reactants than products. Answer: forward (c) There are an equal number of moles of gases on both sides of the reaction. Answer: no effect17-26
  • 27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Effect of a Change in Temperature on an Equilibrium Consider heat as a product or a reactant. In an exothermic reaction, heat is a product, ∆H0rxn = negative In an endothermic reaction, heat is a reactant, ∆H0rxn = positive CaCO3( s ) ← CaO( s ) + CO2( g ) → ∆H = −92.4kJ CaCO3( s ) ← CaO( s ) + CO2( g ) + 92.4kJ → Increase temperature: backward CO2 92.4 kJ17-27
  • 28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 17.13 Predicting the Effect of a Change in Temperature on the Equilibrium Position PROBLEM: In what direction will the reaction shift if there is a decrease in temperature (a) CaO(s) + H2O(l) Ca(OH)2(aq) ∆H0 = -82kJ (b) CaCO3(s) CaO(s) + CO2(g) ∆H0 = 178kJ (c) SO2(g) S(s) + O2(g) ∆H0 = 297kJ SOLUTION: (a) CaO(s) + H2O(l) Ca(OH)2(aq) + heat A decrease in temperature will shift the reaction to the right (b) CaCO3(s) + heat CaO(s) + CO2(g) The reaction will shift to the left (c) SO2(g) + heat S(s) + O2(g) The reaction will shift to the left17-28
  • 29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Addition of a Catalyst The presence of a catalyst has no effect on the position of the chemical equilibrium, since a catalyst affects the rates of the forward and reverse reactions equally.17-29
  • 30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.17-30

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