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Chapter 17: Equilibrium

Chapter 17: Equilibrium. 17.1: How chemical reactions occur 17.2: Conditions that affect Reaction Rates 17.3 Heterogeneous Reaction. 17.1 How Chemical Reactions Occur. Objectives: To understand the collision model of how reactions occur. 17.1 How Chemical Reactions Occur.

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Chapter 17: Equilibrium

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  1. Chapter 17: Equilibrium 17.1: How chemical reactions occur 17.2: Conditions that affect Reaction Rates 17.3 Heterogeneous Reaction

  2. 17.1 How Chemical Reactions Occur Objectives: To understand the collision model of how reactions occur.

  3. 17.1 How Chemical Reactions Occur How does a chemical reaction occur???? Atoms bump into each other hard enough to break bonds allowing the reactants to rearrange to form the products. Collision model: says that reactions occur during molecular collisions. A reaction proceeds faster -higher concentrations -higher temperatures

  4. Figure 17.2: (a)Two BrNO molecules approach each other at high speeds. (b) The collision occurs. (c) The energy of the collision causes Br-N bonds to break and Br-Br bonds to form. (d) The products: one Br2 and two NO molecules.

  5. 17.2 Conditions that Affect Reaction Rates Objectives: To understand activation energy. To understand how a catalyst speeds up a reaction.

  6. 17.2 Conditions that Affect Reaction Rates WHY DO REACTION RATES increase w/ Increasing Temperature? Not all collisions have enough energy to break bonds Activation Energy: the minimum amount of energy necessary to break bonds. Higher TempHigher speedsMore high-energyMore broken bondsFASTER collisions

  7. 17.2 Conditions that Affect Reaction Rates Is it possible to speed up a reaction w/o changing the temperature or the reactant concentration? YES! Catalyst: a substance that speeds up a reaction without being consumed. Enzymes: Biological catalysts Catalysts work by lowering the activation energy of a reaction.

  8. 17.2 Conditions that Affect Reaction Rates IMPT. EXAMPLES CO2(g) + H2O(l) H+(aq) + HCO3-(aq) Catalyzed by carbonic anhydrase: prevents the accumulation of CO2 in your blood Cl2+O3 ClO + O2 O + ClO Cl + O2 O + O3 2O2 Freons such as CF2Cl2 ; Banned in 1996.

  9. 17.3 Heterogeneous Reactions Objectives: To consider reactions with reactants or products in different phases.

  10. 17.3 Heterogeneous Reactions Homogeneous Reaction: reactions involving only one phase. (gas, liquid or solid) Heterogeneous Reaction: reactants in 2 phases.

  11. 17.3 Heterogeneous Reactions Factors that Affect Reaction Rates Nature of reactants: substances vary greatly in their tendency to react depending on their bond strengths and structures. Concentration(Pressure): Increase/Increase Temperature: Increase/Increase Surface Area: Heterogeneous rxns increase w/ increased surface area.

  12. 17.4 The Equilibrium Condition Objectives: To learn how equilibrium is established.

  13. REMEMBER THIS????? Figure 17.8: (a) Net transfer of molecules from the liquid state to the vapor state.(b) The amount of the substance in the vapor state becomes constant (c) The equilibrium state.

  14. 17.4 The Equilibrium Condition Equilibrium: exact balancing of 2 processes, one of which is the opposite of the other. Chemical equilibrium: a dynamic state where the concentrations of all reactants and products remain constant. Forward 2NO2(g) N2O4 (g) Reverse

  15. 17.5 Chemical Equilibrium: A Dynamic Condition Objective: To learn about the characteristics of chemical equilibrium.

  16. 17.5 Chemical Equilibrium: A Dynamic Condition Objective: To learn about the characteristics of chemical equilibrium.

  17. Figure 17.9: (a) Equal numbers of moles of H2O and CO are mixed in a closed container. (b) The reaction begins to occur. (c) The reaction continues, and more reactants are changed to products. (d) No further changes are seen as time continues to pass.

  18. RATE of the FORWARD REACTION EQUALS the RATE of the REVERSE REACTION

  19. 17.6 The Equilibrium Constant: An Introduction Objective: To understand the law of chemical equilibrium and to learn how to calculate values for the equilibrium constant.

  20. 17.6 The Equilibrium Constant: An Introduction 1864: Cato Maximilian Guldberg and Peter Waage LAW of Chemical Equilibrium (Law of Mass Action) aA + bB cC + dD A, B, C, D: reactants and products a, b, c, d: coefficients [ ] mol/L Equilibrium Expression: equilibrium constant K=[C]c[D]d [A]a[B]b

  21. 17.6 The Equilibrium Constant: An Introduction Equilibrium Expression: equilibrium constant K=[C]c[D]d [A]a[B]b 2O3(g) 3O2(g) K=[O2]3 [O3]2 [ ] mol/L The equilibrium constant means that for a given reaction at a given temperature the ratio of the concentrations of the products to reactants defined by the equilibrium expression will always be equal to the same number.

  22. 17.6 The Equilibrium Constant: An Introduction Write the equilibrium constant for the following CH3OH(g) CH2O(g) + H2(g) 4NH3(g) + 5O2(g) 4NO(g) + 6 H2O(g) [ ] mol/L

  23. 17.6 The Equilibrium Constant: An Introduction N2(g) + 3 H2(g) 2NH3 K=6.02 x 10-2 If [N2]=0.921 M, [H2]=0.763M and [NH3]=0.157M Each set of equilibrium concentrations is called an equilibrium position THERE IS only ONE EQUILIBRIUM CONSTANT, but INFINITE EQUILIBRIUM POSITIONS

  24. 17.7 Heterogeneous Equilibria Homogeneous equilibria: all substances are in the same state. Heterogenous equilibria: more than one state CaCO3(s) CaO(s) + CO2(g) Lime Does not depend on the amounts of pure solids or liquids present…... SO…… K=[CO2]

  25. 17.7 Heterogeneous Equilibria Write the expression for K for the following processes. CO2(g) + H2O(l) H2CO3(l) ZrI4(s) Zr(s) + 2I2(g)

  26. 17.8 Le Chatelier’s Principle Objective: To learn to predict the changes that occur when a system at equilibrium is disturbed.

  27. 17.8 Le Chatelier’s Principle How do we understand the factors that control the position of a chemical equilibrium? LeChatelier’s Principle: a change is imposed on a system at equilibrium, the position of the equilibrium, shifts in a direction that tends to reduce the effect of that change.

  28. 17.8 Le Chatelier’s Principle The Effect of a Change in Concentration N2(g) + 3 H2(g) 2NH3 Add 1 mol of N2 What do you think will happen to equilibrium? 2) What about the equilibrium constant?

  29. 17.8 Le Chatelier’s Principle The Effect of a Change in Concentration N2(g) + 3 H2(g) 2NH3 Equilibrium Position I Equilibrium Position II [N2]= 0.399 M [N2]= 1.348M [H2]= 1.197 M [H2]= 1.044 M [NH3]= 0.203 M [NH3]=0.304 M Position I: [NH3]2 = (0.203)2 [N2][H2]3 (0.399)(1.197)3 Position II: ? LeChatelier’s says: when a reactant or product is added to a system at equilibrium, the system shifts away from the added component.

  30. 17.8 Le Chatelier’s Principle A real-life example: In the mountains, if you feel dizzy why? Lower air pressure, lower supply of oxygen Hb(aq) + 4O2(g) Hb(O2)4(aq)

  31. 17.8 Le Chatelier’s Principle 2SO2(g) + O2(g) 2SO3(g) What would happen if SO2(g) is added to the system? The SO3(g) present is liquefied and removed from the system? Some of the O2(g) is removed from the system.

  32. 17.8 Le Chatelier’s Principle The Effect of a Change in Volume CaCO3(s) CaO(s) + CO2(g) Shift to the left When the volume of a gas is DECREASED, the PRESSURE INCREASES. By Le Chatelier’s principle, the system will shift in the direction that reduces the pressure. When the volume of a gas is decreased, the system shifts in the direction that gives the smaller number of gas molecules.

  33. 17.8 Le Chatelier’s Principle N2(g) + 3 H2(g) 2NH3 4 molecules 2 gaseous molecuels If we reduce the volume, what happens? Equilibrium shifts to the right

  34. 17.8 Le Chatelier’s Principle Predict the shift in equilibrium when volume is increased 4NH3(g) + 5 O2(g) 4NO(g) + 6H2O(g) Shift to the right NH4NO3(s) 4N2O(g) + 2H2O(g) Shift to the right

  35. 17.8 LeChatelier’s Principle The Effect of a Change in Temperature This affects the equilibrium constant In an endothermic reaction, heat can be considered a reactant. In an exothermic reaction, heat can be considered a product. In summary to use LeChatelier’s principle: treat heat as a reactant or product. N2(g) + 3 H2(g) 2NH3 + 92 kJ CaCO3(s) +556kJ CaO(s) + CO2(g)

  36. 17.8 Le Chatelier’s Principle The reaction C2H2(g) + 2Br2(g) C2H2Br4(g) This is exothermic…an increase in temperature shifts the equilibrium Shifts to the left ZrI4(s) Zr(s) + 2 I2(g) This is endothermic…an increase in temperature will shift equilibrium to… Shifts to the right

  37. 17.9 Applications Involving the Equilibrium Constant • Objective: To learn to calculate equilibrium concentrations from equilibrium constants.

  38. 17.9 Applications Involving the Equilibrium Constant K tells us: >1 that the equilibrium favors products <1 that the equilibrium favors reactants Also, we can calculate the equilibrium constant.

  39. 17.9 Applications Involving the Equilibrium Constant Assume the equilibrium constant for the reaction H2(g) + F2(g) 2HF(g) Has the value 2.1x103 at a particular temperature. When the system is analyzed at equilibrium at this temperature, the concentrations of both H2(g) and F2(g) are 0.0021 M. What is the concentration of HF(g) in the equilibrium system under these conditions? [HF(g)]= 9.6 x 10-2M

  40. 17.9 Applications Involving the Equilibrium Constant Assume the equilibrium constant for the reaction 2H2O(g) 2H2(g) + O2(g) Has the value 2.4x10-3 at a particular temperature. When the system is analyzed at equilibrium at this temperature, it is found that [H2O(g)]=1.1 x 10-1 M and [H2(g)]=1.9x10-2M. What is the concentration of O2(g) in the equilibrium system under these conditions? [O2(g)]= 8.0 x 10-2M

  41. 17.10 Solubility Equilibria Objective: To learn to calculate the solubility product of a salt given its solubility and vice versa.

  42. 17.10 Solubility Equilibria Consider the equilibria associated with dissolving solids in water to form aqueous solutions. CaF2(s)Ca 2+ (aq) + 2F-(aq) Ksp=[Ca 2+ ][F-]2 Ksp=solubility product constant or solubility product So where’s the solid….dissolving happens at the surface so if you increase the surface increase the rates in both directions.

  43. 17.10 Solubility Equilibria Write the balanced equation for dissolving each of the following solids in water. Also write the Ksp expression for each solid. Cr(OH)3 Ca3(PO4)2

  44. 17.10 Solubility Equilibria Zinc carbonate, ZnCO3(s), dissolves in water to give a solution that is 1.7 x 10-5M at 22oC. Calculate Ksp for ZnCO3(s) at this temperature (ignoring the fact that CO32- reacts with water). Ksp=2.9x10-10

  45. 17.10 Solubility Equilibria Copper(II) chromate, CuCrO4(s) dissolves in water to give a solution that contains 1.1 x 10-5 g CuCrO4 per liter at 23oC. Calculate Ksp for CuCrO4(s) at this temperature. Ksp=3.8x 10-15

  46. 17.10 Solubility Equilibria PbCrO4(s) Pb 2+ + CrO42- Calculate the solubility at 25oC if the Ksp value is 2.0 x 10-16

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