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Chapter 16 Principles of Chemical Reactivity: Equilibria

Chapter 16 Principles of Chemical Reactivity: Equilibria. Chemical Equilibrium: A Review. All chemical reactions are reversible, at least in principle. The concept of equilibrium is fundamental to chemistry.

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Chapter 16 Principles of Chemical Reactivity: Equilibria

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  1. Chapter 16Principles of Chemical Reactivity: Equilibria

  2. Chemical Equilibrium: A Review • All chemical reactions are reversible, at least in principle. • The concept of equilibrium is fundamental to chemistry. • The general concept of equilibrium was introduced in Chapter 3 to explain the limited dissociation of weak acids. • The goals of this and the following chapter will be to consider chemical equilibria in quantitative terms. • The extent to which that equilibrium lies (product favored verses reactant favored) will be discussed.

  3. Equilibrium • At some point in time during the progress of a reaction, if the concentration of the reactants and products remains constant, equilibrium is said to be achieved. • The concentrations are NOT equal.

  4. Equilibrium Moving towards equilibrium Equilibrium established

  5.    Properties of Chemical Equilibria Equilibrium systems are said to be: • Dynamic (in constant motion) • Reversible • Equilibrium can be approached from either direction. Pink to blue Co(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2

  6. Chemical EquilibriumFe3+(aq) + SCN (aq)  Fe(SCN)2+ (aq) • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained. • They are equal and opposite in rate.

  7. Examples of Chemical Equilibria Phase changes such as H2O(s) vs. H2O(liq)

  8. The Equilibrium Constant When a general chemical reaction is at equilibrium, the equilibrium constant is given by: If K > 0 then the reaction is said to be product favored. If K < 0 then the reaction is said to be reactant favored.

  9. The Equilibrium Constant Product-favored K > 1 Reactant-favored K < 1

  10. The Equilibrium Constant For the formation of HI(g) the equilibrium constant is given by:

  11. equilibrium equilibrium equilibrium N2O4(g) 2NO2(g) Start with NO2 Start with N2O4 Start with NO2 & N2O4

  12. Disturbing a Chemical Equilibrium The equilibrium between reactants and products may be disturbed in three ways: (1) by changing the temperature (2) by changing the concentration of a reactant (3) by changing the volume (for systems involving gases) A change in any of these factors will cause a system at equilibrium to shift back towards a state of equilibrium. This statement is often referred to as Le Chatelier’s principle.

  13. Disturbing a Chemical Equilibrium • Effect of the Addition or Removal of a Reactant or Product • If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually. • The new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K

  14. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. • Changes in Concentration

  15. Remove Remove Add Add aA + bB cC + dD Le Châtelier’s Principle • Changes in Concentration continued Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

  16. Disturbing a Chemical Equilibrium • Effect of Volume Changes on Gas-Phase Equilibria • For a reaction that involves gases, what happens to equilibrium concentrations or pressures if the size of the container is changed? (Such a change occurs, for example, when fuel and air are compressed in an automobile engine.) • To answer this question, recall that concentrations are in moles per liter. If the volume of a gas changes, its concentration therefore must also change, and the equilibrium composition can change.

  17. A (g) + B (g) C (g) Le Châtelier’s Principle • Changes in Volume and Pressure Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas • When the number of gas moles on either side is the same, there is no effect.

  18. Disturbing a Chemical Equilibrium Effect of Temperatue Changes on Gas-Phase Equilibria Consider the reaction of nitrogen and oxygen to form nitric oxide: As the temperature of the reaction is increased, the equilibrium constant increases. Why?

  19. Disturbing a Chemical Equilibrium Effect of Temperature Changes on Gas-Phase Equilibria Lets write the reaction in this manner: Notice that energy is included as a reactant!

  20. Disturbing a Chemical Equilibrium Effect of Temperature Changes on Gas-Phase Equilibria Lets write the reaction in this manner: As temperature (Energy) is increased, equilibrium shifts to the right, favoring products.

  21. Disturbing a Chemical Equilibrium Effect of Temperatue Changes on Gas-Phase Equilibria Lets write the reaction in this manner: As the concentration of products increases so does the value of the equilibrium constant.

  22. Disturbing a Chemical Equilibrium Effect of Temperatue Changes on Gas-Phase Equilibria This explains the increase in the equilibrium constant with increasing temperature.

  23. Disturbing a Chemical Equilibrium • Effect of Temperature Changes on Gas-Phase Equilibria • Conclusion: • Increasing the temperature of an endothermic reaction favors the products, equilibrium shifts to the right. • Increasing the temperature of an exothermic reaction favors the reactants, equilibrium shifts to the left. • Lowering temperature results in the reverse effects.

  24. Temperature Effects on Equilibrium Kc (273 K) = 0.00077 Kc (298 K) = 0.0059

  25. Le Châtelier’s Principle • Adding a Catalyst • does not change K • does not shift the position of an equilibrium system • system will reach equilibrium sooner Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

  26. Le Chatelier’s Principle Practice

  27. Change Equilibrium Constant Change Shift Equilibrium Le Châtelier’s Principle - Summary Concentration yes no Pressure yes* no Volume yes* no Temperature yes yes Catalyst no no *Dependent on relative moles of gaseous reactants and products

  28. Writing Equilibrium Constant Expressions • In an equilibrium constant expression, all concentrations are reported as equilibrium values. • Product concentrations appear in the numerator, and reactant concentrations appear in the denominator. • Each concentration is raised to the power of its stoichiometric balancing coefficient. • Values of K are dimensionless. • The value of the constant K is particular to the given reaction at a specific temperature.

  29. Writing Equilibrium Constant Expressions Reactions Involving Solids • So long as a solid is present in the course of a reaction, its concentration is not included in the equilibrium constant expression. • Equilibrium constant:

  30. Writing Equilibrium Constant Expressions Reactions in Solution • If water is a participant in the chemical reaction, its concentration based on magnitude is considered to remain constant throughout. • Equilibrium constant:

  31. Writing Equilibrium Constant Expressions Reactions Involving Gases: Kc and Kp • Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems. • In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in Kc. • For gases, however, equilibrium constant expressions can be written in another way: in terms of partial pressures of reactants and products.

  32. Writing Equilibrium Constant Expressions Reactions Involving Gases: Kp • Notice that the basic form of the equilibrium constant expression is the same as for Kc. In some cases, the numerical values of Kc and Kp are the same. They are different when the numbers of moles of gaseous reactants and products are different.

  33. R = the gas constant T = the absolute temperature n = (mols gas product)  mols gas reactant) Writing Equilibrium Constant Expressions Reactions Involving Gases: Kp& Kc • The general relationship between Kp and Kc is derived in chapter 26, pa726. • When the number of gas mole is equivalent on either side of the chemical equation, the two equilibrium constants are the same value.

  34. The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium.

  35. The Reaction Quotient, Q • If Q < K then the system is heading towards equilibrium: There are more reactants than products as expected at equilibrium. The reaction is said to be headed to the “right”. • If Q > K the system has gone past equilibrium. There are more products than reactants as expected at equilibrium. The reaction is said to be headed to the left. • If Q = K then the system is at equilibrium.

  36. aA + bB cC + dD [C]c[D]d K = [A]a[B]b Equilibrium Will K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants

  37. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2(g) COCl2(g) = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] = 220 Kc= Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp= 220 x (0.0821 x 347)-1 = 7.7 37

  38. CaCO3(s) CaO (s) + CO2(g) ′ ′ Kc = Kc x [CaO][CO2] [CaCO3] [CaCO3] [CaO] Kp = PCO 2 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. [CaCO3] = constant [CaO] = constant Kc= [CO2] = The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

  39. Determining the Equilibrium Constant • The value of a reaction’s equilibrium constant is determined by measuring the concentrations of the reactants and products when a system is at equilibrium. • The equilibrium constant can also determined by looking at the changes in concentrations as a system achieves equilibrium. • This is know as an “ICE” table.

  40. Determining the Equilibrium Constant • ICE tables: • I = Initial concentration • C = change in concentration • E = concentrations at equilibrium

  41. Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

  42. Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

  43. Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

  44. Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K? note the reaction stoichiometry

  45. Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

  46. Determining K

  47. Equilibrium Concentrations from K 1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?

  48. Equilibrium Concentrations from K 1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?

  49. Equilibrium Concentrations from K 1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?

  50. Equilibrium Concentrations from K 1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3? Where x is defined as amount of H2 and I2 consumed on approaching equilibrium in moles.

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