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Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson , Brady, & Hyslop. CHAPTER 15 Chemical Equilibrium. Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K)

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Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

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  1. Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

  2. CHAPTER 15 Chemical Equilibrium • Learning Objectives: • Reversible Reactions and Equilibrium • Writing Equilibrium Expressions and the Equilibrium Constant (K) • Reaction Quotient (Q) • KcvsKp • ICE Tables • Quadratic Formula vs Simplifying Assumptions • LeChatelier’s Principle • van’t Hoff Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  3. CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  4. CHAPTER 15 Chemical Equilibrium Le Chatelier’s Principle Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  5. Le Chatelier Definition • Equilibrium positions • Combination of concentrations that allow Q = K • Infinite number of possible equilibrium positions • Le Châtelier’s principle • System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress • System said to “shift to right” when forward reaction is dominant (Q < K) • System said to “shift to left” when reverse direction is dominant (Q > K) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  6. Le Chatelier Q & K Relationships • Q = K reaction at equilibrium • Q < K reactants go to products • Too many reactants • Must convert some reactant to product to move reaction toward equilibrium • Q > K products go to reactants • Too many products • Must convert some product to reactant to move reaction toward equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  7. Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blueyellow • Equilibrium mixture is blue-green • Add excess Cl– (conc. HCl) • Equilibrium shifts to products • Makes more yellow CuCl42– • Solution becomes green Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  8. Le Chatelier Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blueyellow • Add Ag+ • Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s) • Equilibrium shifts to reactants • Makes more blue Cu(H2O)42+ • Solution becomes increasingly more blue • Add H2O? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  9. Le Chatelier Change in Concentration: Example For the reaction 2SO2(g) + O2(g) 2SO3(g) Kc = 2.4 × 10–3 at 700 °C Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture? • Towards the products • Towards the reactants • No change will occur Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  10. Le Chatelier Change in Concentration • When changing concentrations of reactants or products • Equilibrium shifts to remove reactants or products that have been added • Equilibrium shifts to replace reactants or products that have been removed Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  11. Le Chatelier Change in Pressure or Volume • Consider gaseous system at constant T and n 3H2(g) + N2(g) 2NH3(g) • If volume is reduced • Expect pressure to increase • To reduce pressure, look at each side of reaction • Which has less moles of gas • Reactants = 3 mol + 1 mol = 4 mol gas • Products = 2 mol gas • Reaction favors products (shifts to right) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  12. Le Chatelier Change in Pressure or Volume Consider gaseous system at constant T and n H2(g) + I2(g) 2HI(g) • If pressure is increased, what is the effect on equilibrium? • nreactant = 1 + 1 = 2 • nproduct = 2 • Predict no change or shift in equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  13. Le Chatelier Change in Pressure or Volume 2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g) • If you decrease volume of reaction, what is the effect on equilibrium? • Reactants: All solids, no moles gas • Products: 2 moles gas • Decrease in V, causes an increase in P • Reaction shifts to left (reactants), as this has fewer moles of gas Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  14. Le Chatelier Change in Pressure or Volume • Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can • Increasing pressure • Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids • Substances are already almost incompressible • Changes in V, P and [X ] effect position of equilibrium (Q), but not K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  15. Le Chatelier Change in Temperature Boiling water Ice water Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blueyellow • Reaction endothermic • Adding heat shifts equilibrium toward products • Cooling shifts equilibrium toward reactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  16. Le Chatelier Change in Temperature Hf°=+6 kJ (at 0 °C) • Energy + H2O(s) H2O(l ) • Energy is reactant • Add heat energy, shift reaction right 3H2(g) + N2(g) 2NH3(g) Hrxn= –47.19 kJ • 3 H2(g) + N2(g) 2 NH3(g) + energy • Energy is product • Add heat, shift reaction left H2O(s) H2O(l) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  17. Le Chatelier Change in Temperature • Increase in temperature shifts reaction in direction that produces endothermic (heat absorbing) change • Decrease in temperature shifts reaction in direction that produces exothermic (heat releasing) change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  18. Le Chatelier Change in Temperature • Changes in T change value of mass action expression at equilibrium, so K changed • K depends on T • Increase in temperature of exothermic reaction makes K smaller • More heat (product) forces equilibrium to reactants • Increase in temperature of endothermic reaction makes K larger • More heat (reactant) forces equilibrium to products Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  19. Le Chatelier Change with Catalyst • Catalyst lowers Ea for both forward and reverse reaction • Change in Ea affects rates k r and k f equally • Catalysts have no effect on equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  20. Le Chatelier Addition of Inert Gas at Constant Volume Inert gas • One that does not react with components of reaction e.g. argon, helium, neon, usually N2 • Adding inert gas to reaction at fixed V (n and T), increase P of all reactants and products • Since it doesn’t react with anything • No change in concentrations of reactants or products • No net effect on reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  21. Le Chatelier How To Use Le Chatelier’s Principle • Write mass action expression for reaction • Examine relationship between affected concentration and Q (direct or indirect) • Compare Q to K • If change makes Q > K, shifts left • If change makes Q < K, shifts right • If change has no effect on Q, no shift expected Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  22. Group Problem Consider: H3PO4(aq) + 3OH–(aq) 3H2O(l) + PO43–(aq) What will happen if PO43– is removed? • Q is proportional to [PO43–] • Decrease [PO43–], decrease in Q • Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  23. Group Problem The reaction H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic. What will happen if system is cooled? heat • Since reaction is exothermic, heat is product • Heat is directly proportional to Q • Decrease in T, decrease inQ • Q < K equilibrium shifts to right Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  24. Group Problem The equilibrium between aqueous cobalt ion and the chlorine ion is shown: [Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O pinkblue It is noted that heating a pink sample causes it to turn violet. The reaction is: • endothermic • exothermic • cannot tell from the given information Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  25. Group Problem The following are equilibrium constants for the reaction of acids in water, Ka. Which reaction proceeds the furthest to products? • Ka = 2.2 × 10–3 • Ka = 1.8 × 10–5 • Ka = 4.0 × 10–10 • Ka = 6.3 × 10–3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

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