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Chapter 14: Chemical Equilibrium

Vanessa N. Prasad-Permaul Valencia Community College CHM 1046. Chapter 14: Chemical Equilibrium. Introduction. How far does a reaction proceed toward completion before it reaches a state of chemical equilibrium? 2. Chemical equilibrium

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Chapter 14: Chemical Equilibrium

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  1. Vanessa N. Prasad-Permaul Valencia Community College CHM 1046 Chapter 14: Chemical Equilibrium

  2. Introduction • How far does a reaction proceed toward completion before it reaches a state of chemical equilibrium? 2. Chemical equilibrium a) The state reached when the concentrations of reactants and products remain constant over time b) A state in which the concentration of reactants and products no longer change (net) 3. Equilibrium mixture A mixture of reactants and products in the equilibrium state

  3. Introduction • What are we interested in? a) What is the relationship between the concentration of reactants and products in an equilibrium mixture? b) How can we determine equilibrium concentrations from initial concentrations? c) What factors can be exploited to alter the composition of an equilibrium mixture?

  4. The Equilibrium State • In previous chapters we have generally assumed that chemical reactions result in complete conversion of reactants to products Many reactions do not go to completion!! Example1:

  5. The Equilibrium State

  6. The Equilibrium State The two experiments demonstrate that the interconversion of N2O4 and NO2 is reversibleand that the same equilibrium state is reached starting from either substance. 1. This is why we use a  instead of  2. Since both NO2 and N2O5 are products and reactants we will call the chemical on the left reactants and on the right products. 3. All chemical reactions are reversible     

  7. The Equilibrium State • We call a reaction irreversible when it proceed nearly to completion a. Equilibrium mixture contains almost all products and almost no reactants b. Reverse reaction is too slow to be detected • In an equilibrium state the reaction does not stop at particular concentrations of reactants and products, the rates of the forward and reverse reactions become equal. Important: reaction does not stop

  8. The Equilibrium State 6. Chemical equilibrium is a dynamic state in which forward and reverse reactions continue at equal rates so that there is no net conversion of reactants to products

  9. APPLYING STOICHIOMETRY TO AN EQUILIBRIUM MIXTURE EXAMPLE 14.1: CO(g) + 3H2(g) CH4(g) + H2O(g) When 1.000 mol CO & 3.000 mol H2 are placed in a 10.00 L vessel @ 927oC and allowed to come to equilibrium; the mixture is found to contain 0.387 mol H2O. What is the molar composition of the equilibrium mixture? (How many moles of each substance are present?) CO(g) + 3H2(g) CH4(g) + H2O(g) Starting 1.000 3.000 0 0 Change -x -3x +x +x Equilibrium 1.000 - x 3.000 - 3xxx = 0.387 CO = 1.000 – x = 1.000 – 0.387 = 0.613 mol H2 = 3.000 – x = 3.000 – 3(0.387) = 1.839 mol CH4 = x = 0.387 mol

  10. APPLYING STOICHIOMETRY TO AN EQUILIBRIUM MIXTURE EXERCISE 14.1: CO(g) + H2O(g) CO2(g) + H2(g) Suppose there is a mixture containing 1.00 mol CO and 1.00 mol H2O. When equilibrium is reached at 1000oC, the mixture contains 0.43 mol H2. What is the molar composition of the equilibrium mixture?

  11. The Equilibrium Constant, Kc a A + bB  cC + dD Kc = [C]c [D]d [A]a [B]b If we write the equation in the reverse direction cC + dD  aA + bB K’c = [A]a [B]b = 1 [C]c [D]d kc

  12. The Equilibrium Constant, Kc General equation: aA + bB  cC + dD Equilibrium equation: Kc = [C]c [D]d products [A]a [B]b reactants The substances in the equilibrium equation must be gases or molecules and ions in solution: NO SOLIDS! NO PURE LIQUIDS! Kc units are omitted but you must say at what temperature!

  13. WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXAMPLE 14.2: Write the equilibrium-constant expression Kc for catalytic methylation CO(g) + 3H2(g) CH4(g) + H2O(g) Kc = [CH4][H2O] [CO][H2]3 Write the equilibrium-constant expression Kc for the reverse reaction CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = [CO][H2]3 [CH4][H2O]

  14. WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXAMPLE 14.2: Write the equilibrium-constant expression Kc the synthesis of NH3 N2(g) + 3H2(g) 2NH3(g) Kc = [NH3]2 [N2][H2]3 Write the equilibrium-constant expression Kc for the following rxn. 1/2N2(g) + 3/2H2(g) NH3(g) Kc=[NH3] [N2]1/2 [H2]3/2

  15. WRITING EQUILIBRIUM-CONSTANT EXPRESSIONS EXERCISE 14.2: Write the equilibrium-constant expression Kc for the following reaction: 2NO2(g) + 7H2(g) 2NH3(g) + 4H2O(g) Write the equilibrium-constant expression Kc for the following reaction: NO2(g) + 7/2H2(g) NH3(g) + 2H2O(g)

  16. The Equilibrium Constant Kp Kp = equilibrium constant with respect to partial pressures of reactants and products a A + bB  cC + dD Kp = (PC)c (PD)d (PA)a (PB)b Relationship between Kc and Kp Kp = Kc(RT)n

  17. Heterogeneous Equilibria Introduction 1. So far we have been talking about homogeneous equilibria, in which all reactants and products are in a single phase (gas or solution) 2. Heterogeneous equilibria are those in which reactants and products are present in more than one phase

  18. Using the Equilibrium Constant Introduction Knowing the value of the equilibrium constant for a chemical reaction lets us: 1. Judge the extent of the reaction 2. Predict the direction of the reaction 3. Calculate the equilibrium concentrations from any initial concentrations

  19. Using the Equilibrium Constant The numerical value of the equilibrium constant for a reaction indicates the extent to which reactants are converted to products • Large value for Kc > 103 reaction proceeds essentially to 100% (mostly products) • Small value for Kc < 10-3 reaction proceeds hardly at all before equilibrium is reached (mostly reactants) • If a reaction has an intermediate value of Kc = 103 to 10-3 a.  Appreciable concentrations of both reactants and products are present in the equilibrium mixture

  20. OBTAINING AN EQUILIBRIUM CONSTANT FROM REACTION COMPOSITION EXAMPLE 14.3: What is the value of Kc for the decomposition of HI at room temp.? 2HI(g) H2(g) + I2(g) CONC Starting 0.800 M 0 0 4.00mol/5.00L = 0.800M Change -2x +x +x 0.442mol/5.00L = 0.0884M Equilibrium 4.00 mol – 2x x x(0.800 – 2(0.0884))M = 0.623M Kc = [H2] [I2] [HI]2 = [0.0884M][0.0884M] [0.623M]2 = 0.0201

  21. OBTAINING AN EQUILIBRIUM CONSTANT FROM REACTION COMPOSITION EXERCISE 14.3: Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating: 2H2S(g) 2H2(g) + S2(g) When 0.100 mol H2S was put into a 10.0L vessel and heated to 1132oC, the equilibrium mixture contained 0.0285 mol H2. What is the value of Kc at this temperature?

  22. WRITING Kc FOR A REACTION WITH PURE SOLIDS OR LIQUIDS EXAMPLE 14.4: Quicklime (calcium oxide, CaO), is prepared by heating a source of calcium carbonate CaCO3 such as limestone or seashells. CaCO3(s) CaO(s) + CO2(g) Write the expression for Kc. Kc = [CO2] An equilibrium constant can also be written for a physical equilibrium. H2O(l) H2O(g) Write the expression for Kc for the vaporization of water. Kc = [H2O(g)]

  23. WRITING Kc FOR A REACTION WITH PURE SOLIDS OR LIQUIDS EXERCISE 14.4: The Mond process for purifying nickel involves the formation of nickel tetracarbonyl Ni(CO)4, a volatile liquid, from nickel metal and carbon monoxide. Carbon monoxide is passed over impure nickel to form nickel carbonyl vapor, which, when heated, decomposes and deposits pure nickel. Ni(s) + 4CO(g) Ni(CO)4(g) Write the expression for Kc for this reaction

  24. Predicting the direction of Reaction Reaction Quotient = Qc 1. Not necessarily equilibrium concentrations, at some time, t, snapshot of reaction       2. As time passes, Qc changes toward the value of Kc       3. When the equilibrium state is reached Qc = Kc       4. Qc allows us to predict the direction of reaction by comparing the values of Kc and Qc        a) If Qc< Kc, net reaction goes from left to right, (reactant to products)       b) If Qc > Kc, net reaction goes from right to left, (products to reactants)       c) If Qc = Kc, no net reaction occurs

  25. USING THE REACTANT QUOTIENT EXAMPLE 14.5: A 50.0 L reaction vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500 mol NH3. Will more ammonia be formed or will it dissociate when a mixture goes to equilibrium @ 400oC? Kc is 0.500 @ 400oC. N2(g) + 3H2(g) 2NH3(g) Qc = [NH3]2 [N2][H2]3 = [0.500mol/50.0L]2 = [0.0100]2 [1.00mol/50.0L][3.00mol/50.0L]3 [0.0200][0.0600]3 Qc = 23.1 which is greater than Kc = 0.500  the reaction will go to the left or ammonia will dissociate.

  26. USING THE REACTANT QUOTIENT EXAMPLE 14.5: A 10.0 L reaction vessel contains 0.0015 mol CO2 and 0.10 mol CO. If a small amount of carbon is added to this vessel and the temperature is raised to 1000oC, will more CO form? CO2(g) + C(s) 2CO(g) The value of Kc is 1.17 at 1000oC

  27. OBTAINING EQUILIBRIUM CONCENTRATION EXAMPLE 14.6: A gaseous mixture contains 0.30 mol CO, 0.10 mol H2 and 0.020 mol H2O plus an unknown amount of CH4, in each liter. The mixture is at equilibrium @ 1200K. What is the concentration of CH4 in this mixture? Kc = 3.92 CO(g) + 3H2(g) CH4(g) + H2O(g) Kc = [CH4][H2O] [CO][H2]3 3.92 = [CH4][0.020mol/1.0L] = [CH4][0.020M] [0.30mol/1.0L][0.10mol/1.0L]3 [0.30M][0.10M]3 3.92[3.00 x 10-4M] = [CH4] = 0.059M [0.020M]

  28. OBTAINING EQUILIBRIUM CONCENTRATION EXAMPLE 14.6: Phosphorus pentachloride gives an equilibrium mixture of PCl5, PCl3, and Cl2 when heated. PCl5(g) PCl3(g) + Cl2(s) A 1.00L vessel contains an unknown amount of PCl5 and 0.020 mol of each PCl3 and Cl2 at equilibrium at 250oC.How many moles of PCl5 are in the vessel if Kc for this reaction is 0.0415 @ 250oC?

  29. Altering an Equilibrium Mixture: Changes in Pressure and Volume In general Le Chatelier’s Principle predicts that: 1. An increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas 2. A decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas.

  30. Factors that Alter the Composition of an Equilibrium Mixture Introduction One of the principal goals of chemical synthesis is to maximize the conversion of reactants to products while minimizing the expenditure of energy. 1. Can be achieved if the reaction goes nearly to completion at mild temperatures and pressures. 2. If the equilibrium mixture is high in reactants and poor in products, the experimental conditions must be changed. 3. Several factors can be exploited to alter the composition of an equilibrium mixture. A.       The concentration of reactants or products      B. The pressure and volume C. The temperature

  31. Le Chatelier’s Principle Le Chatelier’s Principle If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress    1. Stress means a change in the concentration, pressure, volume, or temperature that disturbs the original equilibrium     2. Reaction then occurs to change the composition of the mixture until a new state of equilibrium is reached 3. The direction that the reaction takes (reactants to products or products to reactants) is the one that reduces the stress

  32. Altering an Equilibrium Mixture: Changes in Concentration In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Chatelier’s principle predicts that: 1.  The concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance 2. The concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance

  33. LE CHATLIER’S PRINCIPLE EXAMPLE 14.6: Predict the direction of reaction when H2 is removed from a mixture (lowering the concentration) in which the following equilibrium has been established: H2(g) + I2(g) 2HI(g) When H2 is removed from the reaction mixture, HI will dissociate to partially restore the H2 that was removed  The reaction will go to the left

  34. LE CHATLIER’S PRINCIPLE EXAMPLE 14.6: Consider each of the following equilibria, which are disturbed as indicated. Predict the direction of the reaction: The following equilibria is disturbed by increasing pressure of carbon dioxide: CaCO3(s) CaO(s) + CO2(g) The following equilibrium is disturbed by increasing the concentration of hydrogen: 2Fe(s) + 3H2O(g)Fe2O3(s) + 3H2(g)

  35. LE CHATLIER’S PRINCIPLE WITH ALTERED PRESSURE EXAMPLE 14.7: Will the products increase, decrease or have no effect if the pressure is increased: CO(g) + Cl2(g) COCl2(g) Reaction decreases the number of molecules of gas  an increase in pressure increases the amount of product and pushes the reaction to the right. 2H2S(g) 2H2(g) + S2(g) Reaction increases the number of molecules of gas  an increase of pressure decreases the amount of products and the reaction shifts to the left. C(s) + S2(g) CS2(g) Reaction does not change the number of molecules. Only look at gas volumes when decide the effect of pressure change on equilibrium composition. Pressure change has no effect.

  36. LE CHATLIER’S PRINCIPLE WITH ALTERED PRESSURE EXERCISE 14.7: Can the amount of product be increased in each of the following reactions by increasing the pressure? explain: CO2(g) + H2(g) CO(g) + H2O(g) 4CuO(s) 2Cu2O(s) + O2(g) 2SO2(g) + O2(g) 2SO3(g)

  37. Altering the Equilibrium Mixture: Changes in Temperature In general, the temperature dependence of the equilibrium constant depends on the sign of H for the reaction 1.      The equilibrium constant for an exothermic reaction (negative H) decreases as the temperature increases 2.      The equilibrium constant for an endothermic reaction (positive H) increases as the temperature increases. 3.      H = standard enthalpy of reaction, enthalpy change measured under standard conditions 4. Standard conditions = most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25C; 1 M concentration for all substances

  38. Altering the Equilibrium Mixture: Changes in Temperature Le Chatelier’s Principle says that if heat is added to an equilibrium mixture (increasing the temperature) net reaction occurs in the direction that relieves the stress of the added heat. 1.     For an endothermic reaction heat is absorbed by reaction in the forward direction. The equilibrium shifts to the right at the higher temperatures, Kc increases with increasing temperature 2.     For an exothermic heat is absorbed by net reaction in the reverse direction, so Kc decreases with temperature, and the reaction would flow to the left (reactants)

  39. LE CHATLIER’S PRINCIPLE WITH ALTERED TEMPERATURE EXAMPLE 14.8: Carbon monoxide is formed when carbon dioxide reacts with solid carbon: CO2(g) + C(s) 2CO(g) ; DHo = 172.5kJ Is a high or low temperature more favorable to the formation of carbon monoxide? Increase products need to shift equilibrium to the right Endothermic The temperature needs to be raised High temperature is more favorable to the formation of carbon monoxide.

  40. LE CHATLIER’S PRINCIPLE WITH ALTERED TEMPERATURE EXAMPLE 14.8: Carbon monoxide is formed when carbon dioxide reacts with solid carbon: CO2(g) + H2(g) CO(g) + H2O(g) Is a high or a low temperature more favorable to the production of carbon monoxide? Explain.

  41. The Effect of a Catalyst on Equilibrium A catalyst increases the rate of a chemical reaction by making available a new, lower-energy pathway for conversion of reactants to products. 1.  Since the forward and reverse reaction pass through the same transition state, a catalyst lowers the activation energy for both 2.   The rates of the forward and reverse reactions increase by the same factor 3. Catalyst accelerates the rate at which equilibrium is reached 4. Catalyst does not affect the composition of the equilibrium mixture

  42. The Effect of a Catalyst on Equilibrium

  43. The Link Between Chemical Equilibrium and Chemical Kinetics A + B  C + D Assuming that the forward and reverse reactions occur in a single bimolecular step, elementary reactions, we can write the following rate laws Rate of forward reaction = kf [A] [B] Rate of reverse reaction = kr [C] [D] When t=0 [C] = [D] = 0 As A and B are converted to C and D the rate of the forward reaction decreases and the rate of the reverse reaction is increasing, until they are equal, chemical equilibrium kf [A] [B] = kr [C] [D]

  44. The Link Between Chemical Equilibrium and Chemical Kinetics kf = [C] [D] kr [A] [B] The right side of this equation is the equilibrium constant expression for the forward reaction, which equals the equilibrium constant Kc Kc = [C] [D] [A] [B] Therefore the equilibrium constant is simply the ratio of the rate constants for the forward and reverse reactions: Kc = kf kr

  45. Example 1: Which of the following is correct? • Some reactions are truly not reversible • All reactants go to all products in all reactions • All reactions are reversible to some extent • The rates of the forward and reverse reactions will never be equal

  46. Example 2:  Write the equilibrium equation for each of the following reaction: a)      N2(g) + 3 H2(g) 2 NH3(g) b) 2 NH3(g) N2(g) + 3 H2(g)

  47. Example 3: • The oxidation of sulfur dioxide to give sulfur trioxide is an important step in the industrial process for synthesis of sulfuric acid. Write the equilibrium equation for each of the following reactions: a)      2 SO2(g) + O2(g) 2 SO3(g) b) 2 SO3(g) 2 SO2(g) + O2(g) The following equilibrium concentrations were measured at 800 K: [SO2] = 3.0 x 10-3 M [O2] = 3.5 x 10-3 M [SO3] = 5.0 x 10-2 M Calculate the equilibrium constant at 800 K a and b

  48. Example 4: In the industrial synthesis of hydrogen, mixtures of CO and H2O are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is: CO(g) + H2O(g) CO2(g) + H2(g) What is the value of Kp at 700 K if the partial pressures in an equilibrium mixture at 700 K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm H2?

  49. Example 5: When will kc = kp ? 1. 2 SO2(g) + O2(g) 2 SO3(g) • CO(g) + H2O(g) CO2(g) + H2(g) 3. N2(g) + 3 H2(g) 2 NH3(g)

  50. Example 6: Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2 NO(g) + O2(g) 2 NO2(g) • If Kc = 6.9 x 105 @ 227C, what is the value of Kp @ 227C? b) If Kp = 1.3 x 10-2 @ 1000 K, what is the value of Kc @ 1000 K?

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