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

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Presentation on theme: "Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop."— Presentation transcript:

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

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

3 CHAPTER 15 Chemical Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Lecture Road Map: ① Dynamic Equilibrium ② Equilibrium Laws ③ Equilibrium Constant ④ Le Chatelier’s Principle ⑤ Calculating Equilibrium

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

5 Le Chatelier Definition Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5 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)

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

7 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 Le Chatelier Change in Concentration Cu(H 2 O) 4 2+ (aq) + 4Cl – (aq) CuCl 4 2– (aq) + 4H 2 O blue yellow Equilibrium mixture is blue-green Add excess Cl – (conc. HCl) –Equilibrium shifts to products –Makes more yellow CuCl 4 2– –Solution becomes green

8 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8 Le Chatelier Change in Concentration Cu(H 2 O) 4 2+ (aq) + 4Cl – (aq) CuCl 4 2– (aq) + 4H 2 O blue yellow Add Ag + –Removes Cl – : Ag + (aq) + Cl – (aq)  AgCl(s) –Equilibrium shifts to reactants –Makes more blue Cu(H 2 O) 4 2+ –Solution becomes increasingly more blue Add H 2 O?

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

10 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

11 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 3H 2 (g) + N 2 (g) 2NH 3 (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)

12 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 H 2 (g) + I 2 (g) 2HI(g) If pressure is increased, what is the effect on equilibrium? –n reactant = = 2 –n product = 2 –Predict no change or shift in equilibrium

13 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 Le Chatelier Change in Pressure or Volume 2NaHSO 3 (s) NaSO 3 (s) + H 2 O(g) + SO 2 (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

14 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

15 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 Le Chatelier Change in Temperature Ice water Boiling water Cu(H 2 O) 4 2+ (aq) + 4Cl – (aq) CuCl 4 2– (aq) + 4H 2 O blue yellow –Reaction endothermic –Adding heat shifts equilibrium toward products –Cooling shifts equilibrium toward reactants

16 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Le Chatelier Change in Temperature  H f °=+6 kJ (at 0 °C) –Energy + H 2 O(s) H 2 O(l ) –Energy is reactant –Add heat energy, shift reaction right 3H 2 (g) + N 2 (g) 2NH 3 (g)  H rxn = –47.19 kJ – 3 H 2 (g) + N 2 (g) 2 NH 3 (g) + energy –Energy is product –Add heat, shift reaction left H 2 O(s) H 2 O(l)

17 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

18 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

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

20 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 N 2 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

21 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 Le Chatelier How To Use Le Chatelier’s Principle 1.Write mass action expression for reaction 2.Examine relationship between affected concentration and Q (direct or indirect) 3.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

22 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Group Problem Consider: H 3 PO 4 (aq) + 3OH – (aq) 3H 2 O(l) + PO 4 3– (aq) What will happen if PO 4 3– is removed?  Q is proportional to [PO 4 3– ]  Decrease [PO 4 3– ], decrease in Q  Q < K equilibrium shifts to right

23 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 Group Problem The reaction H 3 PO 4 (aq) + 3OH – (aq) 3H 2 O(aq) + PO 4 3– (aq) is exothermic. What will happen if system is cooled?  Since reaction is exothermic, heat is product  Heat is directly proportional to Q  Decrease in T, decrease in Q  Q < K equilibrium shifts to right heat

24 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(H 2 O) 6 ] 2+ (aq) + 4Cl – (aq) [Co(Cl) 4 ] 2– (aq) + 6H 2 O pink blue It is noted that heating a pink sample causes it to turn violet. The reaction is: A. endothermic B. exothermic C. cannot tell from the given information

25 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, K a. Which reaction proceeds the furthest to products? A. K a = 2.2 × 10 –3 B. K a = 1.8 × 10 –5 C. K a = 4.0 × 10 –10 D. K a = 6.3 × 10 –3


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