1 / 42

Chapter 14: Chemical Equilibrium

Chapter 14: Chemical Equilibrium. Renee Y. Becker Valencia Community College. Introduction. How far does a reaction proceed toward completion before it reaches a state of chemical equilibrium? 2. Chemical equilibrium

ping
Download Presentation

Chapter 14: Chemical Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 14: Chemical Equilibrium Renee Y. Becker Valencia Community College

  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 5. 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. 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

  10. 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!

  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. 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)

  13. 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

  14. 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

  15. 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?

  16. 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)

  17. 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?

  18. 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

  19. Example 7: For each of the following reactions, write the equilibrium constant expression for Kc a)   2 Fe(s) + 3 H2O(g) Fe2O3(s) + 3 H2(g) b) 2 H2O(l) 2 H2(g) + O2(g) c) SiCl4(g) + 2 H2(g) Si(s) + 4 HCl(g) d) Hg22+(aq) + 2 Cl-(aq) Hg2Cl2(s)

  20. Example 8: Which of the following has a Heterogeneous equilibria? 1. 2 SO2(g) + O2(g) 2 SO3(g) • CO(g) + H2O(g) CO2(g) + H2(g) • SiCl4(g) + 2 H2(g) Si(s) + 4 HCl(g)

  21. 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

  22. 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

  23. 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

  24. Example 9: The equilibrium constant for the reaction 2 NO(g) + O2(g) 2 NO2(g) is 6.9 x 105 @ 500 K. A 5.0 L reaction vessel at this temperature was filled with 0.060 mol of NO, 1.0 mol O2, and 0.80 mol NO2. a) Is the reaction mixture at equilibrium? If not, in which direction does the net reaction proceed? b) What is the direction of the net reaction if the initial amounts are 5.0 x 10-3 mol of NO, 0.20 mol of O2 and 4.0 mol of NO2?

  25. 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

  26. 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

  27. 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

  28. Example 10: Consider the equilibrium for the water-gas shift reaction: CO(g) + H2O(g) CO2(g) + H2(g) Use Le Chatelier’s principle to predict how the concentration of H2 will change and what direction the reaction will flow when the equilibrium is disturbed by: 1.      Adding CO 2.      Adding CO2 3.      Removing H2O 4.      Removing CO2

  29. Example 11: In the following reaction, if I take away CO, which direction will the reaction proceed to equilibrium? CO2(g) + H2(g) CO(g) + H2O(g) • Products  • Reactants 

  30. 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.

  31. Example 12: Which direction will the reaction flow if the following equilibria is subjected to an increase in pressure by decreasing the volume? 1.      CO(g) + H2O(g) CO2(g) + H2(g) 2.      2 CO(g) C(s) + CO2(g) 3.      N2O4(g) 2 NO2(g)

  32. Example 13: If I increase the pressure by decreasing the volume, which direction will the reaction flow to reach equilibrium? C(s) + CO2(g)  2 CO(g) • Products  • Reactants 

  33. 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

  34. 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)

  35. Example 14: When air is heated at very high temperatures in an automobile engine, the air pollutant nitric oxide is produced by the reaction N2(g) + O2(g) 2 NO(g)H = 180.5 kJ 1. How does the equilibrium amount of NO vary with an increase in temperature? 2. What direction is the net reaction flowing?

  36. 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

  37. The Effect of a Catalyst on Equilibrium

  38. Example 15: A platinum catalyst is used in automobile catalytic converters to hasten the oxidation of carbon monoxide: 2 CO(g) + O2(g) 2 CO2(g)H = -566 kJ Suppose that you have a reaction vessel containing an equilibrium mixture. Will the amount of CO increase, decrease, or remain the same when: • A platinum catalyst is added • The temperature is increased • The pressure is increased by decreasing the volume • The pressure is increased by adding argon gas • The pressure is increased by adding O2 gas

  39. 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]

  40. 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

  41. Example 16: Nitric oxide emitted from the engines of supersonic transport planes can contribute to the destruction of stratospheric ozone: NO(g) + O3(g) NO2(g) + O2(g) This reaction is highly exothermic (E = -200 kJ), and its equilibrium constant Kc is 3.4 x 1034 at 300 K • Which rate constant is larger, kf or kr? • The value of kf at 300 K is 8.5 x 106 M-1 s-1. What is the value of kr at the same temperature? • A typical temperature in the stratosphere is 230 K. Do the values of kf, kr, and Kc increase or decrease when the temperature is lowered from 300 K to 230 K?

  42. Example 17: If I increase the temperature of reaction which way will the reaction flow to equilibrium? NO2(g) + O2(g)  NO(g) + O3(g)H = 200 kJ • Products  • Reactants 

More Related