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Chapter 15 Principles of Chemical Equilibrium

Chapter 15 Principles of Chemical Equilibrium. Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php. The equilibrium state.

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Chapter 15 Principles of Chemical Equilibrium

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  1. Chapter 15Principles of Chemical Equilibrium Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php

  2. The equilibrium state • Chemical equilibriumis the state reached when theconcentrations of the products and reactants remain constant over time. The mixture of reactants and productsin the equilibrium state is theequilibrium mixture.

  3. N2O4 (g)  2 NO2 (g) • We have used two directional arrows () to show that this reaction does not go to completion. • “The reaction occurs both ways.”

  4. Pitfall • The terms reactants and products are arbitrary. We must always refer to a balanced equation to be completely understood. • N2O4 (g)  2 NO2 (g) • reactant product • 2 NO2 (g)  N2O4 (g) • reactant product

  5. Each reaction occurs atits own rate as defined by its rate law. One reaction will initially have a faster rate than the other, and will initially dominate the system. The other reaction can be considered to be dominated in the system

  6. We saw in Kinetics that the rate of a reaction decreases as time proceeds because the concentrations of the “reactants” decrease. • This is what we will see in the dominant reaction.

  7. What will happen to the dominated reaction where the “reactant” concentrations increase? • Its reaction rate will increase!

  8. At some point in time the rate of the forward reaction is THE SAME as the rate of the reverse reaction. • This is a more correct means of defining equilibrium.

  9. Initially dominant reaction “slows down” Initially dominated reaction “speeds up”

  10. Equilibrium is a dynamic process At equilibrium the rates of the forward and reverse reactions are the same, BUT arenot equal to zero. • While no visible changeis occurring, • individual molecular events are still occurring.

  11. The equilibrium constant expression The final concentrations will not always be the same because we started with different numbers of N atoms and O atoms in the last three experiments. But ratio of [NO2]2 / [N2O4] is always the same!

  12. Experiments 1 and 2 Start first experiment with no NO2 and 0.04 M N2O4 Start first experiment with 0.08 M NO2 and no N2O4

  13. a A + b B  c C + d D • The concentrations of all species in an equilibrium mixture are related to each other through the equilibrium constant equation in terms of concentration. • This equilibrium equation applies only to this specific balanced equationand thevalueof theequilibrium constant, Kc must always be statedat aspecific temperature!

  14. N2O4 (g)  2 NO2 (g) • If we change the temperature then the equilibrium mixture will (most likely) change. This means the equilibrium constant will change. For this reaction • Kc = 1.53 at 127 C. • Notice there are no units for Kc!

  15. Be careful! • Equilibrium constant equations (and therefore the value of K) depend on the balanced equation referenced. • A + B  C + D • C + D  A + B • K’cDOES NOT EQUAL Kc, but rather • K’c = 1/Kc

  16. Be careful! • Equilibrium constant equations (and therefore the value of K) depend on the balanced equation referenced. • A + B  C + D • 2 A + 2 B  2 C + 2 D

  17. Thermodynamic equilibrium constant Keq • The thermodynamic equilibrium constant(Keq) equation takes the same mathematical form as the equilibrium constant equation in terms of concentrations (Kc). However, the composition of the equilibrium mixture is expressed in terms of activities.

  18. Thermodynamic equilibrium constant Keq • Activities relate effective properties(like concentration or pressure) of real substancesat given conditions in comparison to the same substance acting ideallyat standard conditions. • Since activities make comparisons of the same property typein ratio form, the property units cancel out and all activities are unitless.

  19. Thermodynamic equilibrium constant Keq a A + b B  c C + d D where ax = [X] / c0 (c0 is a standard concentration of 1 M) or ax = Px / P0 (P0 is a standard pressure of 1 bar) Note that ax = 1 for pure solids and liquids

  20. Problem • For the reaction • CO (g) + 2 H2 (g)  CH3OH (g) • the equilibrium concentrations of CH3OH and CO are found to be equal at 483 K. If Kc = 14.5 at 483 K, what is the equilibrium concentration of H2? Answer: [H2] = 0.263 M

  21. Problem • For the reaction • N2 (g) + 3 H2 (g)  2 NH3 (g) • Kc = 1.8 x 104 at a certain temperature. What is the equilibrium concentration of H2 if the equilibrium concentrations of N2 and NH3 are 0.015 M and 2.00 M respectively? Answer: [H2] = 0.25 M

  22. K for combined equilibria • If we can describe an overall equilibrium reaction as the sum of two or more other equilibrium processes, then the equilibrium constant for the overall reaction in the equilibrium constants of the processes multiplied together. • Krxn = K1 x K2 x K3 …

  23. Complex ions and solubility AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)

  24. Complex ions and solubility AgCl(s)  Ag+ (aq) + Cl- (aq) Ag+ (aq) + NH3 (aq)  [Ag(NH3)]+ (aq) [Ag(NH3)]+ (aq) + NH3 (aq)  [Ag(NH3)2]+ (aq) K for the first reaction is 1.8 x 10-10 K for the second reaction is 2.0 x 103 K for the third reaction is 7.9 x 103

  25. Complex ions and solubility The sum of the three reactions is AgCl(s) + 2 NH3(aq)  [Ag(NH3)2]+(aq) + Cl-(aq) which will have an equilibrium constant that is Krxn= K1 x K2 x K3 Krxn= 1.8 x 10-10 x 2.0 x 103 x 7.9 x 103 Krxn = 2.8 x 10-3

  26. The equilibrium constant Kp • Gas phase equilibrium constants are often expressed in terms of partial pressures because they are generally very easy to measure as a function of the total pressure of the system.

  27. The equilibrium constant Kp • Recall, for an ideal gas A • PAV = nART • PA = (nART) / V • PA = (nA/V) RT • What is n / V? It is moles over volume, which is concentration. So nA/V is [A] and • PA = [A]RT

  28. N2O4 (g)  2 NO2 (g) • We can express an equilibrium constant in terms of partial pressures because they are related to concentrations! • Again, the equilibrium constant is unitless.

  29. Kc and Kp are related • a A + b B  c C + d D (all are GASES!)

  30. Kc and Kp are related • Some of the RT terms in • will cancel each other out. In fact

  31. Kc and Kp are related Use R = 0.08206 (Latm)(Kmol)-1 because it relates molarity to pressure at a given temperature.

  32. N2O4 (g)  2 NO2 (g) • Dn = (2-1) = 1 so • At 25 C Kc = 4.64 x 10-3 • Kp = Kc(RT) • Kp = (4.64 x 10-3)(0.08206)(298) • note the lack of units and T is expressed in Kelvin! • Kp = 0.113 at 25 C

  33. Problem • In the industrial synthesis of hydrogen often the water-gas shift reaction is used • CO (g) + H2O (g)  CO2 (g) + H2 (g) • What is the value of Kpat 700 K if the partial pressures in an equilibrium mixture at 700 K are • 1.31 atm of CO • 10.0 atm of water • 6.12 atm of carbon dioxide, and • 20.3 atm of hydrogen gas?

  34. Problem answer • Kp = 9.48

  35. Problem • In the industrial synthesis of nitric acid: • 2 NO (g) + O2 (g)  2 NO2 (g) • If Kc = 6.9 x 105 at 227 C, • what is the value of Kp at this temperature? • If Kp = 1.3 x 10-2 at 1000 K, • what is the value of Kc at this temperature? Answers: Kp at 227C = 1.7 x 104 and Kc at 1000 K = 1.1

  36. Heterogeneous equilibria • Homogeneous equilibria occur in systems where all compounds in the equilibrium mixture are in the same state. • Heterogeneous equilibria occur in systems where some of the chemicals of the equilibrium mixture are in different states.

  37. CaCO3 (s)  CaO (s) + CO2 (g) • Since one of the products is a gas, while the other two compounds are solids, this is a heterogeneous equilibrium. • Now, if we were to express the equilibrium constant for this reaction, we would probably say • What is the concentration of a solid, though?

  38. What is the concentration of a solid or liquid? • Concentration is moles per unit volume. Also, density is mass divided by volume, and molar mass is mass per number of moles. So, for a pure substance • (mass / volume) / (mass / moles) = moles / volume • density / molar mass =(concentration)

  39. What is the concentration of a solid or liquid? • density / molar mass =(concentration) • Since both thedensityand molar mass of a pure solid or liquid substance are constant, the CONCENTRATION IS CONSTANT, and does not change in a reaction as long as some of the solid or liquid exists at all times. • This helps explain why the activitiesof solids and liquids are equal to one!

  40. CaCO3 (s)  CaO (s) + CO2 (g) • We choose not to include the concentrations of solids and liquids in the calculation of Kc! • The concentrations of the solids are “hidden” inside the equilibrium constant. • If we look at the reaction in terms of pressure, then • Kp = (PCO2)

  41. Thermodynamic equilibrium constant Keq • The activity of all pure solids and liquids is one, and so solids and liquids have no effect on the value of Keq

  42. CaCO3 (s)  CaO (s) + CO2 (g)

  43. Problem • For each of the following reactions, write the equilibrium constant expression for Kc. Where appropriate, do the same for Kp and give the relationship between Kc and Kp. • 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)  Hg2Cl (s)

  44. Using the equilibrium constant • Judging the extent of a reaction: The magnitude (size) of the constant K gives an idea of the extent to which reactants are converted to products. • We can makegeneralstatements about the“completeness”of a given equilibrium reaction based on the size of the value of the equilibrium constant.

  45. If the equilibrium constant is very large (>1000 for instance), then the forward reaction is initially very dominant and the reaction as written in the balanced equationproceeds nearly to completion before equilibrium is reached. • 2 H2 (g) + O2 (g)  2 H2O (g)

  46. If the equilibrium constant is very small (<10-3 for instance), then the reverse reaction is initially very dominant and the reaction as written in the balanced equation barely proceeds at all before equilibrium is reached. • 2 H2O (g)  2 H2 (g) + O2 (g)

  47. If the equilibrium constant is between 10-3 and 103, this means that the dominant reactionis not overpoweringthe other reaction and we reach equilibrium somewhere “in between” a state of “no reaction” and “completeness”. Appreciable concentrationsofallspecies are present in the equilibrium mixture. • H2 (g) + I2 (g)  2 HI (g)

  48. Predicting the direction of a reaction • If you put known concentrations of products and reactantsinto the equilibrium constant equation when you know the system is NOT at equilibriumyou would get a value thatdoes not equalthe equilibrium constant. • Can we use this value to tell which reaction is dominant in this non-equilibrium system?

  49. We define the reaction quotient Qc (or Qp or Qeq) in exactly the same way we define the equilibrium constant Kc (or Kp or Keq). • When the system is not at equilibrium, then • Qc Kc

  50. If Qc > Kc the reaction needs to create more reactants (and use up products)to get to equilibrium, so the reaction will be going from right to left. • If Qc < Kc the reaction needs to create more products (and use up reactants)to get to equilibrium, so the reaction will be going from left to right.

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