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Equilibrium

Equilibrium. Quantitative vs. Reversible. So far we have studied quantitative reactions , reactions which proceed until the limiting reactant is consumed However, many reactions are not quantitative, they are reversible meaning they can proceed in both forward and reverse directions

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Equilibrium

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  1. Equilibrium

  2. Quantitative vs. Reversible • So far we have studied quantitative reactions, reactions which proceed until the limiting reactant is consumed • However, many reactions are not quantitative, they are reversible meaning they can proceed in both forward and reverse directions • A + B C + D

  3. Dynamic Equilibrium • When the rate of the forward reaction and the rate of the reverse reaction are equal, the system is in dynamic equilibrium • This does NOT mean that nothing is happening!

  4. Properties of Equilibrium Systems • 1. Macroscopicproperties (e.g. colour, pressure, concentration, pH) are constant • I.e. It appears as if nothing is changing • 2. Can only be reached in a closed system • 3. Forward rate = Reverse rate • 4. Can be established from either direction At equilibrium, the concentrations of all reactants and products will remain constant

  5. Types of Equilibrium • Solubility Equilibrium: A dynamic equilibrium between a solute and solvent in a saturated solution in a closed system • Phase Equilibrium: a dynamic equilibrium between different physical states of a pure substance in a closed system • Ex: water/liquid in a sealed container • Chemical Reaction equilibrium: a dynamic equilibrium between reactants and products of a chemical reaction in a closed system

  6. Practice Sketch a graph of Concentration of Reactants and products vs. time for Experiment 1 and 2

  7. More Practice 1. Consider the reaction: • Hydrogen gas and hydrogen iodide gas are both clear colourless, while iodine gas is purple. • Explain why when hydrogen gas and iodine gas are added to a closed flask and allowed to react, the colour never fades to colourless. 2. Sketch a graph to illustrate the changes in concentration of each component over time as equilibrium is reached for Experiment 1

  8. 3. If experiment 1 did not reach equilibrium, what would the theoretical yield of HI be? • 4. What is the actual yield of HI in experiment 1?

  9. Solving Equilibrium Problems • Use ICE Tables • I: Initial Concentration (mol/L) • C: Change in concentration • +x if increasing in concentration • -x if decreasing in concentration • Multiply x by the # of moles • E: Equilibrium Concentration (mol/L)

  10. Practice #1 • In a 4.00L container, 2.50 mol of carbon dioxide gas is decomposed. At equilibrium, [CO2(g)]eq = 0.125 mol/L. Use an ICE table to find [O2(g)]eq and [CO(g)]eq

  11. Practice #2 • 0.500 mol of NOCl(g) is decomposed in a closed 2.00 L container. [NO(g)]eq = 0.040 mol/L. Find [Cl2(g)]eq and [NOCl(g)]eq.

  12. Practice #3 • When 2.00 mol of ethene gas and 1.50 mol of bromine vapour come to equilibrium in a closed 1.00 L container, the equilibrium concentration of bromine vapour is measured at 0.150 mol/L. Find [C2H4(g)]eq and [C2H4Br2(g)]eq.

  13. Practice #3 Continued • Graph the equilibrium reaction for the addition of bromine to ethene.

  14. At equilibrium, the rate of the forward and reverse reactions are equal Equilibrium

  15. The Equilibrium Constant, Keq For the reaction: aA + bB cC + dD At equilibrium: rfwd = rrev kfwd[A]a[B]b = krev[C]c[D]d kfwd = [C]c[D]d krev[A]a[B]b Keq = [C]c[D]d [A]a[B]b Sub in rate law equation Rearrange Sub in Keq for kfwd/krev Equilibrium constant

  16. Keq Conditions • Note: the equilibrium constant expression only works with reactions that occur in a single step • Keq will remain the same as long as the temperature is kept constant(changing the temperature changes the forward and reverse reactions by different amounts and therefore change the Keq)

  17. Examples Write the equilibrium constant expression for the following two reactions

  18. Heterogeneous Equilibria • Equilibrium systems can involve all states of matter • However, the concentration of a pure solid or liquid cannot change • Therefore, equilibrium constant expressions will not include solids and liquids

  19. Keq = [C]c[D]d [A]a[B]b Magnitude of Keq Keq>> 1 At equilibrium there is more products than reactants. The reaction is product favoured Keq = 1 At equilibrium there is an equal amount of products and reactants Keq << 1 At equilibrium there is more reactants than products. The reaction is reactant favoured

  20. Example Calculate the value of Keq for the following system At Equilibrium: [CO2] = 0.0954 mol/L [H2] = 0.0454 mol/L [CO] = [H2O] = 0.00460 mol/L

  21. Practice For each reaction and their respective equilibrium constant, predict whether reactants or products are favoured. • N2(g) + O2(g)2NO(g)Keq = 4.7 x 10-31 • NO(g) + CO(g)2N2(g) + CO2(g)Keq = 2.2 x 1059

  22. Le Chatelier’s Principle • When a chemical system at equilibrium is disturbed by a stress, the system adjusts (shifts) to oppose the change • Stresses include: • Change in concentration • Change in pressure (or volume) • Change in temperature

  23. Change in Concentration A(g) + 3B(g) 2C(g) + heat • Increasing the concentration of the reactants OR • Decreasing the concentration of the products • Will favour the forward reaction, causing the equilibrium to shift to the RIGHT • Decreasing the concentration of the reactants OR • Increasing the concentration of the products • Will favour the reverse reaction, causing the equilibrium to shift to the LEFT • RECALL: Addition or removal of solid or liquids does not change the concentration. Therefore does not cause a shift. I.e. only applies to gases and aqueous solutions.

  24. Change in Concentration

  25. N2(g) + 3H2(g)2NH3

  26. Change in Pressure  volume  pressure  volume  pressure A(g) + 3B(g) 2C(g) + heat • Increasing the volume of the container OR Decreasing the pressure • Will cause a shift to the side with MORE gas molecules • In our example, it will shift left (4 molreactants > 2 molproducts) • Decreasing the volume of the container OR Increasing the pressure • Will cause a shift to the side with LESS gas molecules • In our example, it will shift right (4 molreactants > 2 molproducts)

  27. Change in Temperature In an exothermic reaction: • Increasing the temperature will cause a shift to the LEFT • Decreasing the temperature will cause a shift to the RIGHT

  28. Change in Temperature • In an endothermic reaction: • Increasing the temperature will cause a shift to the RIGHT • Decreasing the temperature will cause a shift to the LEFT

  29. Change in Temperature Recall: Keq is temperature dependent. Therefore, changes in temperature will also affect Keq Shift right =  products,  Keq Shift left =  reactants, Keq

  30. DEMONSTRATION

  31. Variables that do NOT Affect Equilibrium • Catalysts • Increases reaction rate by lowering activation energy (of BOTH the forward and the reverse reactions equally) • Decreases the time required to reach equilibrium but does not affect the final position of equilibrium • Inert Gases • Increases the pressure, which will increase reaction rate • Increases the probability of successful collisions for BOTH products and reactants equally • Decreases the time required to reach equilibrium but does not affect the final position of equilibrium

  32. Practice

  33. The Reaction Quotient (Q) • If a chemical system begins with reactants only, it is obvious that the reaction will shift right (to form products). • However, if BOTH reactants and products are present initially, how can we tell which direction the reaction will proceed? • Use a trial value called the reaction quotient, Q • When a reaction is NOT at equilibrium • Q=Keq the system is at equilibrium • Q > Keq the system shifts towards reactants to reach equilibrium • Q < Keqthe system shifts towards products to reach equilibrium

  34. Practice #1 (p. 464) In a container at 450°C, N2 and H2 react to produce NH3. K = 0.064. When the system is analysed, [N2] = 4.0 mol/L, [H2] = 2.0 X 10-2mol/L, and [NH3] = 2.2 X 10-4mol/L. Is the system at equilibrium, if not, predict the direction in which the reaction will proceed.

  35. Practice #2 In a container, carbon monoxide and water vapour are producing carbon dioxide and hydrogen at 900oC. CO(g) + H2O(g) H2(g) + CO2(g)Keq= 4.00 at 900oC If the concentrations at one point in the reaction are: [CO(g)] = 4.00 mol/L, [H2O(g)] = 2.00 mol/L, [CO2(g)] = 4.00 mol/L, and [H2(g)] = 2.00 mol/L. Determine whether the reaction has reached equilibrium, and, if not, in which direction it will proceed to establish equilibrium.

  36. Practice #2 Answer products

  37. Practice #3Calculating Equilibrium Concentrations from Initial Concentrations • Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900oC, Keq is 4.200. Calculate the concentrations of all entities at equilibrium if 4.000 mol of each entity are initially place in a 1.00L closed container.

  38. Practice #4Calculating Equilibrium Concentrations Involving a Quadratic Equation • If 0.50 mol of N2O4is placed in a 1.0L closed container at 150oC, what will be the concentrations of N2O4and NO2 at equilibrium? (Keq = 4.50)

  39. Practice #5 Simplifying Assumption: 100 rule (for small K values) If: [initial reactant] > 100, you can simplify the Keqexpression K Ex: 2CO2(g) 2CO(g) + O2(g) If K = 6.40 x 10-7, determine the concentrations of all substances at equilibrium if it starts with [CO2] = 0.250 mol/L

  40. Solubility Equilibrium • Not all ionic compounds are equally soluble • Ionic compounds dissociate into individual ions in aqueous solutions • This can be a reversible system • Example: CaCl2(s) Ca2+(aq) + 2Cl-(aq) • Equilibrium can be reached between the solid substance and its dissolved ions (saturation point) • The solution is saturated at equilibrium (no more ions can dissociate)

  41. Solubility Product Constant (Ksp) • An equilibrium equation can be written for solubility reactions • Ex: AgCl(s) Ag+ (aq) + Cl- (aq) Recall: Since AgCl is a solid, the concentration is not changing, so it is “built in” to the K value: • The new constant is the solubility product constant (Ksp)

  42. Example • Eg: Lead (II) chloride has a molar solubility of 1.62x102 mol/L at 25oC. What is the Ksp of this salt? PbCl2 Pb2+ + 2Cl- Ksp = [Pb2+][Cl-]2 [Pb2+] = [PbCl2] = 1.62x10-2mol/L [Cl-] = 2[PbCl2] = 2(1.62x10-2mol/L) = 3.24 x 10-2mol/L Ksp = [1.62x10-2mol/L][3.24 x 10-2mol/L]2 = 1.7x10-5

  43. Example 2 • The Ksp of silver chloride at 25oC is 1.8x10-10. What is the molar solubility of AgCl? AgCl Ag+ + Cl- Ksp = [Ag+][Cl-] 1.8x10-10 = [X][X] 1.8x10-10 = X2 X = 1.34x10-5M

  44. The size of Ksp depends on the solubility of the salt. • Large Ksp: [ions] at equilibrium is high, salt is very soluble • Small Ksp: [ions] at equilibrium is low, salt has low solubility • To determine whether a precipitate will form during a reaction, a trial solubility product constant can be determine which is denoted by the symbol Qsp. Qsp < Ksp: Shifts right to equilibrium – all solid dissolving Qsp > Ksp : Shifts left to equilibrium – precipitate forms Qsp = Ksp : Equilibrium (saturated) – no precipitate

  45. Example 1 • Will a ppt form if a solution of CaSO4 is made such that [Ca2+] = 0.0104 mol/L and [SO42-]=0.0082 mol/L? • Ksp =7.1x10-5 • (Ans: Q = 8.5 X 10-5, ppt forms)

  46. Example 2 • Will a precipitate form when a solution is made by dissolving PbCl2 into water. • [Pb2+]=0.0001mol/L. [Cl-]=0.05mol/L. Ksp = 1.2x10-5

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