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Unit 4-Systems and Equilibrium Dynamic Equilibrium

Unit 4-Systems and Equilibrium Dynamic Equilibrium. What Does “Equilibrium” Mean?. Equilibrium means achieving a state of balance. Ecological Equilibrium. Inner Ear Equilibrium. Economic Equilibrium. Types of Equilibrium. Static Equilibrium A state of balance without movement.

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Unit 4-Systems and Equilibrium Dynamic Equilibrium

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  1. Unit 4-Systems and EquilibriumDynamic Equilibrium

  2. What Does “Equilibrium” Mean? • Equilibrium means achieving a state of balance Ecological Equilibrium Inner Ear Equilibrium Economic Equilibrium

  3. Types of Equilibrium • Static Equilibrium • A state of balance without movement • Dynamic Equilibrium • A state of balance with movement • The water is moving out of the faucet and down the drain at the same rate so the level of water remains constant

  4. What Does Equilibrium Have To Do With Chemical Reactions? • Unlike what was covered in grade 11, not all chemical reactions go to completion • Most chemical reactions are reversible • 2H2(g) +O2(g) • Forward reaction: • 2H2(g) +O2(g)  2H2O (l) • Reverse reaction: • 2H2O(l) 2H2(g) + O2(g) 2H2O(l) Double arrow They can achieve an equilibrium between forming product and forming reactant!

  5. When is a System in Chemical Equilibrium? • When it is a closed system energy, but not mass, can be exchanged with surroundings • And… • The reactions are reversible • And… • The rate of product being formed is equal to the rate of reactant being formed • Rate of AB is equal to the rate of BA

  6. Types of Equilibrium systems H2O (g) 1. Phase Equilibria H2O(l) 2. Chemical Equilibria N2O4(g) 2NO2 (g)

  7. Types of Equilibrium systems 3. Solubility Equilibria AgCl(s) Ag+(aq) + Cl-(aq) Solubility Equilibrium Animation 4. Acid-Base Equilibria HNO2(aq)+H2O(l) H3O+(aq)+NO2−(aq)

  8. Types of Equilibrium systems • Heterogeneous equilibrium applies to reactions in which reactants and products are in different phases. A(g) +B(l) AB(s) • Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. A(g) +B(g) AB(g) What is Phase Equilibria an example of? Solubility? Chemical?

  9. Characteristics of Chemical EquilibriaClass Activity“Paper Wads” If a system is at equilibrium are the concentrations of products and reactants equal?

  10. The Addition of More Reactant to The System at equilibrium = Run 3

  11. Questions • Did the system always come to equilibrium and how do you know? Yes the concentrations of products and reactants became constant • Does it matter if you start on the reactants side or product side? Explain why or why not. No because the reactions are reversible, no matter where we start, the system will establish equilibrium • Was the "paper wad" concentration always the same on the reactant side and the product side once equilibrium was reached? If not, explain how this could still be equilibrium. No it still reaches equilibrium because the rate of the forward reaction equals the rate of the reverse reaction, NOT CONCENTRATION • If we are modelling equilibrium, why can we not stop the transfer or throwing of "paper wads“ Because at equilibrium reactions are still occurring, just at the same rate so the overall concentrations stay constant • During the first run (reaction), what happens to the rate of the forward reaction? The reverse reaction? Why? Forward reaction slows down, Reverse reaction starts at zero and increases as more product is formed until the 2 rates become equal • When we added 15 more “paper wads” to our last system that was in equilibrium, what happened to the equilibrium of the system? Our system re-established equilibrium and we ended up with LESS reactants than before and MORE product

  12. Factors Affecting EquilibriumIntro to Le Chatelier’s Principle • What happened when we added more “reactant” to the system that had reached equilibrium? • The system responded by generating more product • The reaction “shifted” to the product side = “shifted to the right” Equilibrium “shifts” to the right to get COUNTERACT the addition of more reactant Paper Wadsreactant Paper Wadsproduct Add more Paper Wads What would happen if we added more product?

  13. Factors Affecting EquilibriumIntro to Le Chatelier’s Principle • What would happen if instead of adding more reactant when the “system” was at equilibrium, we removed reactants instead? • How would the system respond? Equilibrium “shifts” to the left to COUNTERACT the removal of the reactant and make more Paper Wadsreactant Paper Wadsproduct remove some paper wads What would happen if removed some product instead?

  14. Factors Affecting EquilibriumIntro to Le Chatelier’s Principle ..\CHEM 3051 Fall 2012\Concept Presentation\reversible-reactions_en (5).jar

  15. Le Chatelier’s Principle “When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change”. There are 3 ways that a system can be “disturbed by a change in property”: 1. A change in concentration • adding/removing reactants and products 2. A change in temperature • Increasing or decreasing temperature 3. A change in pressure (or volume) • Only affects species in the gas phase e. g. NO2(g) When this happens, the system RESPONDS by moving or shifting to try to cancel out the change

  16. Remove Remove Add Add aA + bBcC+ dD Le Chatelier’s PrincipleA Change in Concentration • Change in concentration refers to adding or removing reactants or products Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right right Increase concentration of reactant(s) Decrease concentration of reactant(s) left

  17. N2 (g) + 3H2(g) 2NH3(g) Le Chatelier’s PrincipleA Change in Concentration • Example 1: Change Results in: Addition of NH3(g) a “shift” left and increase in N2(g) and H2(g) Removal of NH3(g) a “shift” right and increase in NH3(g) Addition of N2(g) a “shift” right and increase in NH3(g) Removal of H2(g) a “shift” left and increase in N2(g) and H2(g)

  18. Le Chatelier’s PrincipleA Change in Concentration 2SO2(g) + O2(g) 2SO3(g) Change Results in: Addition of SO3(g) a “shift” left and increase in SO2(g) and O2(g) Removal of SO3(g) a “shift” right and increase in SO3(g) Addition of O2(g) a “shift” right and increase in SO3(g) Removal of O2(g) a “shift” left and increase in SO2(g) and O2(g)

  19. CaO(s) + H2O(l) Ca(OH)2(aq) DH0 = -82kJ CaO(s) + H2O(l) Ca(OH)2(aq) heat Le Chatelier’s PrincipleA Change in Temperature • How would an increase in temperature affect the following system? • Step 1: Write “heat” as either a reactant or a product • Step 2: Treat the increase/decrease of heat the same as an addition/removal of concentration • Step 3: Determine the “shift” in equilibrium and the result on the equilibrium concentrations Increasing T is like increasing the concentration of heat”the equilibrium “shifts” left and the concentration of Ca(OH)2 (aq) decreases

  20. Le Chatelier’s PrincipleA Change in Temperature N2(g) + O2(g) 2NO(g) ∆H rxn = +180.6 kJ • Step 1: N2(g) + O2(g) +heat 2NO(g) • Step 2 and 3: Results in: Change “shift” right and an increase in NO2(g) Increase temperature “shift” left and an increase in N2(g) and O2(g) Decrease temperature

  21. Le Chatelier’s PrincipleA Change in Pressure/Volume • Since pressure and volume are inversely related(P 1/V), the result for one will be opposite to the same change in the other • i.e. An increase in pressure is the same as a decrease in volume and will affect the system in the same way • NOTE: This only applies for species in the gas phase

  22. A (g) + B (g) C (g) Le Chatelier’s PrincipleA Change in Pressure/Volume 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

  23. Le Chatelier’s PrincipleA Change in Pressure/Volume • Example 1: N2 (g) + 3H2(g) 2NH3(g) Results in: Change “shift” right and an increase in NH3(g) Increase pressure “shift” left and an increase in N2(g) and H2(g) Decrease pressure “shift” left and an increase in N2(g) and H2(g) Increase volume “shift” right and an increase in NH3(g) Decrease volume

  24. Le Chatelier’s Principle • What would happen if we added a catalyst to the reaction? • A catalyst speeds up BOTH the forward and reverse reactions equally • It has NO NET EFFECT on the equilibrium of the system With Catalyst

  25. Le Chatelier’s PrincipleClass Summary

  26. N2 (g) + 3H2(g) 2NH3(g) + heat Le Chatelier’s PrinciplePractice Problems • For the above system in equilibrium determine the resulting shift and change to relevant concentrations of the following disturbances: • An increase in volume = • The addition of H2 gas = • The removal of NH3 = • A decrease in pressure = • Applying heat to the system = • Cooling the system = shift left, increase in reactants shift right, increase in products shift right, increase in products shift left, increase in reactants shift left, increase in reactants shift right, increase in products

  27. Le Chatelier’s Principle • Illustrating the effects of increasing the amount of Carbon Dioxide in the air

  28. aA + bBcC + dD Equilibrium Law and the Equilibrium Constant • We can describe a chemical system in equilibrium in 2 ways: • Qualitatively Le Chatelier’s Principle • Quantitatively = MATHEMATICALLY • The “Law of Mass Action” and the equilibrium constant Keq • For the reaction • The ratio of products over reactants at equilibrium is constant…

  29. Equilibrium Law and the Equilibrium Constant aA + bBcC + dD [C]c[D]d CONSTANT K = [A]a[B]b • Definition of Equilibrium Law (The Law of Mass Action): • If the system is at equilibrium at a given temperature, then the ratio of the concentration of the product to the concentration of the reactant is a constant. • In other words, for the reaction: WHY?

  30. aA + bBcC + dD MATH • What do we know about a system at equilibrium? 1. RATEforward = RATEreverse • In other words, the speed of making products is equal to the speed of making reactants • What does the rate/speed of reaction depend on? 2.  RATEforward [A]a[B]b RATEforward= kf[A]a[B]b

  31. aA + bBcC + dD [C]c[D]d = Keq= kf [A]a[B]b kr MORE MATH 3.  RATEreverse [C]c[D]d RATEreverse= kr[C]c[D]d 4.  RATEforward = RATEreverse kf[A]a[B]b = kr[C]c[D]d

  32. [C]c[D]d [C]c[D]d [NH3]2 = Keq= kf [A]a[B]b [A]a[B]b [N2][H2]3 kr Equilibrium Law and the Equilibrium Constant • …means that if we know the equilibrium constant, we can figure out equilibrium concentrations and vice versa • Example 1: N2 (g) + 3H2(g) 2NH3(g) Keq= Keq=

  33. [C]c[D]d [CH3OH] [0.00261] [A]a[B]b [H2]2[CO] [0.250]2[0.105] Equilibrium Law and the Equilibrium Constant-Calculating Keq CO(g) + 2H2(g)  CH3OH(g) • Calculate Keq for the above reaction if the equilibrium concentration of the reactants and products respectively are [CH3OH]= 0.00261M, [CO]= 0.105 M and [H2] = 0.250 M Keq= Keq= Keq= Keq = 0.398

  34. [C]c[D]d [CH3OH] [x] [A]a[B]b [H2]2[CO] [0.100]2[0.210] Equilibrium Law and the Equilibrium Constant-Calculating Equilibrium Concentrations CO(g) + 2H2(g)  CH3OH(g) • What is the equilibrium concentration of methanol if Keq = 0.500 and the equilibrium concentration of CO is 0.210M and H2 is 0.100M? x = 0.00105M = 1.05 * 10-3 M Keq= Keq= 0.500 =

  35. Calculating Equilibrium Concentrations From Initial concentrations N2 (g) + 3H2(g) 2NH3(g) • A reaction vessel containing 1.00mol/L of N2(g) and 1.00mol/L H2(g) was heated to 5000C. The equilibrium concentration of NH3(g) was found to be 0.25mol/L • Determine the equilibrium concentrations of N2(g) and H2(g) • Determine the equilibrium constant, Keq for the formation of NH3(g) • Determine the equilibrium constant for the decomposition of NH3(g) • What is the relationship between the equilibrium constant for the forward and reverse reactions?

  36. Initial-Change-Equilibrium Calculations (I.C.E. Calculations) Question 3a-Determine the concentration for all species at equilibrium N2 (g) + 3H2(g) 2NH3(g) “The equilibrium concentration of NH3(g) was found to be 0.250mol/L” ∴ 0.250 = 2x x = 0.125 mol/L At Equilibrium: [N2(g)] = 1.00 – 0.125 = 0.875 mol/L [H2(g)] = 1.00 – 3(0.125) = 0.625 mol/L [NH3(g)] = 0 + 2(0.125) = 0.250 mol/L

  37. [0.250]2 [NH3]2 [N2][H2]3 [0.875][0.625]3 KeqCalculations Question 3b-Determine Keq for the formation of NH3(g) N2 (g) + 3H2(g) 2NH3(g) At Equilibrium: [N2(g)] = 0.875 mol/L [H2(g)] = 0.625 mol/L [NH3(g)] = 0.250 mol/L Keq = 0.293 Keq= Keq=

  38. [0.875][0.625]3 [0.250]2 Keq Calculations Question 3c-Determine Keq for the decomposition of NH3(g) N2 (g) + 3H2(g) 2NH3(g) At Equilibrium: [N2(g)] = 0.875 mol/L [H2(g)] = 0.625 mol/L [NH3(g)] = 0.250 mol/L [N2][H2]3 [NH3]2 Keq =3.42 Keq= Keq= Is there a faster way we could have done this question?

  39. [0.875][0.625]3 [0.250]2 [NH3]2 [0.875][0.625]3 [N2][H2]3 [0.250]2 Keq Calculations Question 3d-What is the relationship between Kf and Kr? N2 (g) + 3H2(g) 2NH3(g) [N2][H2]3 Forward Reverse [NH3]2 Keq= Keq= Keq= Keq= Keq = 0.293 Keq = 3.42

  40. [NH3]2 1 Keq (forward) [N2][H2]3 Keq Calculations • Because…. [N2][H2]3 [NH3]2 1 = Keq (reverse)=

  41. [C] [C]2 [C] [A][B]2 [A]2[B]4 [A][B]2 Characteristics of Keq-More Math • Look at the 2 reactions below…what happens to Keq when we double the coefficients of the equation? Rxn 1 A + 2B C Keq= 8 Rxn 2 2A + 4B 2C Keq = ? 2 = 82 Keq = 8 = Keq = ? = = = 64

  42. [C] [C] [C]1/2 [A]1/2[B] [A][B]2 [A][B]2 Characteristics of Keq-More Math = (Keq)3 • If doubling the coefficients changed Keq by a POWER OF 2 then… • How would Keq change if we tripled the coefficients? • How would Keq change if we HALVED the coefficients? Rxn 1 A + 2B C Keq= 8 Rxn 2 1/2A + B 1/2C Keq = ? 1/2 Keq = ? = = Keq = 8 = = 81/2 = √8

  43. [SO3] [SO3] [SO3] [O2]1/2[NO] [O2]1/2[NO] [O2]1/2[SO2] [O2]1/2[SO2] [NO2] [NO2] [SO2] Characteristics of Keq-More Math Look at the following two reactions: • SO2(g) + 1/2O2(g) SO3(g)      Keq1 = 2.2 • NO2(g) NO(g) + ½ O2(g)      Keq2 = 4.0 What is the equilibrium constant for the reaction • SO2(g) + NO2(g) SO3(g) + NO(g) Keq3 = ? [NO] * = Keq3 = ? = Keq2 = 4.0 = Keq1 = 2.2 = [NO2]

  44. [SO3] [SO3] [O2]1/2[NO] [O2]1/2[SO2] [NO2] [SO2] Characteristics of Keq-More Math Keq1 Keq2 [NO] = [NO2] = Keq1 *Keq2 = 2.2 *4.0 = 8.8 Concentration versus Pressure We can write Keqin terms of concentrations (mol/L) or in terms of pressure, Kp if all species are in the gas phase Kp= Keq(RT)∆n Dn = moles of gaseous products – moles of gaseous reactants * Keq3 = ? =

  45. Keq • Playing with Keq-Computer Animation • For homework, please follow the instructions on the site..for each run, calculate the concentrations of A and B using the Keq value given in the plot

  46. Summary of Mathematics of Keq • The Keq for forward and reverse reactions are reciprocals (the inverse) of each other • If you multiply the stoichiometry of a reaction by n, Keq is changed by the power of n • If you add chemical equations together, you multiply their Keq’s together • Kp= Keq(RT)∆n (Kp is for gases only) • If K>>1, then the concentration of products is large compared to reactants (reaction heads towards completion) • If K<<1, then the concentration of reactants is large compared to products (very small amount of product formed)

  47. Characteristics of Keq- No More Math • The equilibrium constant is a dimensionless quantity (no units). • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. • The concentrations of pure solids, pure liquids and solvents (like H2O) do not appear in the equilibrium constant expressions... • Because the concentrations of pure solids, liquids, and solvents are constant • Concentrations are expressed as mol/L . For gases, concentrations can be expressed as either mol/L or atm

  48. Solving Equilibrium CalculationsUsing I.C.E Step 1: Look for a perfect square, if you see one, solve Step 2: See if you can use the “approximation rule” if there is no perfect square to avoid having to do the quadratic formula Step 3: If you cannot use the approximation rule, then solve using the quadratic equation • Remember to check both roots and keep the one that is valid under the circumstances!

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