1. CHAPTER 17 – CHEMICAL EQUILIBRIUM 8B-1 (of 37) COLLISION THEORY 1 - Molecules must collide to react 2 -Molecules must collide with sufficient energy to react (to break bonds) 3 - Molecules must collide in the proper orientation

2. 8B-2 (of 37)

3. Experimentally, rates of reactions are proportional to (1)Temperature (2)Concentration of reacting molecules REACTION RATE – The speed at which a chemical reaction takes place Measured by how fast a reactant reacts away, or how fast a product is produced 8B-3 (of 37)

4. Many chemical reactions run to completion 8B-4 (of 37) The forward reaction is the only reaction that can take place when a product escapes from the reaction vessel However, if the products stay in the reaction vessel, the reverse reaction may also take place

5. CHEMICAL EQUILBRIUM – When forward and reverse reactions are proceeding at equal rates in a system Concentration as a Function of Reaction Time 8B-5 (of 37) 2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g)

6. EQUILBRIUM CONSTANT (K eq ) – The ratio of product of the equilibrium product concentrations to the product of the equilibrium reactant concentrations EQUILBRIUM CONSTANT EXPRESSION 8B-6 (of 37) K eq = [HCl] 2 ___________ [H 2 ] [Cl 2 ] H 2 (g) + Cl 2 (g) ⇆ 2HCl (g) = [12] 2 _______ [3] [3] = 16 [ ] = concentrations, in mol/L

7. aA + bB ⇆ cC + dD K eq = [C] c [D] d ___________ [A] a [B] b Write the equilibrium constant expression for N 2 (g) + 3H 2 (g) ⇆ 2NH 3 (g) K eq = [NH 3 ] 2 ___________ [N 2 ] [H 2 ] 3 8B-7 (of 37)

8. A 2.0 L flask contains 2.0 moles hydrogen, 2.0 moles oxygen, and 6.0 moles water vapor at equilibrium. Find K eq for = 9.0 M -1 = (3.0 M) 2 ___________________ (1.0 M) 2 (1.0 M) [H 2 ] eq = [O 2 ] eq = [H 2 O] eq = 2.0 mol / 2.0 L 6.0 mol / 2.0 L 2H 2 (g) + O 2 (g) ⇆ 2H 2 O (g) = 1.0 M H 2 = 1.0 M O 2 = 3.0 M H 2 O K eq = [H 2 O] 2 ___________ [H 2 ] 2 [O 2 ] 8B-8 (of 37) K eq = 9.0 M -1 Notice the K eq > 1 What does this say about the container’s contents at equilibrium? CALCULATING AN EQUILIBRIUM CONSTANT

9. aA + bB ⇆ cC + dD K eq = [C] c [D] d ___________ [A] a [B] b If K eq > 1 There will be more products than reactants in the container when the reaction reaches equilibrium If K eq < 1 There will be more reactants than products in the container when the reaction reaches equilibrium 8B-9 (of 37)

10. C (s) + CO 2 (g) ⇆ 2CO (g) Gases (and dissolved solutes) have variable concentrations Therefore their concentrations appear in equilibrium constant expressions 8B-10 (of 37) EQUILIBRIA INVOLVING SOLIDS OR LIQUIDS

11. C (s) + CO 2 (g) ⇆ 2CO (g) Solids (and liquids) have constant concentrations Therefore they do not appear in equilibrium constant expressions K eq = [CO] 2 _________ [CO 2 ] 8B-11 (of 37) EQUILIBRIA INVOLVING SOLIDS OR LIQUIDS

12. CALCULATING EQUILIBRIUM CONCENTRATIONS 8B-12 (of 37)

13. H 2 (g) + Br 2 (g) ⇆ 2HBr (g) A tank is charged with HBr so that its concentration is 2.0 M. If K eq = 25, find the final concentrations of all three gases. H 2 (g) + Br 2 (g) ⇆ 2HBr (g) Initial M’s Change in M’s Equilibrium M’s 002.0 + x - 2x x2.0 - 2xx K eq = [HBr] 2 ___________ [H 2 ] [Br 2 ] 25= (2.0 – 2x) 2 _____________ x 2 8B-13 (of 37) CALCULATING EQUILIBRIUM CONCENTRATIONS

14. [H 2 ] eq =x [Br 2 ] eq =x [HBr] eq =2.0 - 2x H 2 (g) + Br 2 (g) ⇆ 2HBr (g) A tank is charged with HBr so that its concentration is 2.0 M. If K eq = 25, find the final concentrations of all three gases. 25= (2.0 – 2x) 2 _____________ x 2 5.0= 2.0 – 2x __________ x 5.0x= 2.0 – 2x 7.0x= 2.0 x= 0.286 = 0.29 M = 2.0 M – 2(0.286 M) = 1.4 M 8B-14 (of 37) CALCULATING EQUILIBRIUM CONCENTRATIONS

15. 0.200 - x2x H 2 (g) + I 2 (g) ⇆ 2HI (g)K eq = 36.0 A tank is charged with 0.200 M H 2 and 0.200 M I 2. Find the equilibrium concentration of HI. H 2 (g) + I 2 (g) ⇆ 2HI (g) Initial M’s Change in M’s Equilibrium M’s 0.200 0 - x + 2x 0.200 - x K eq = p HI 2 ________ p H2 p I2 36.0= (2x) 2 _______________ (0.200 – x) 2 8B-15 (of 37)

16. H 2 (g) + I 2 (g) ⇆ 2HI (g)K eq = 36.0 A tank is charged with 0.200 M H 2 and 0.200 M I 2. Find the equilibrium concentration of HI. 36.0= (2x) 2 _______________ (0.200 – x) 2 6.00= 2x ___________ 0.200 - x 6.00 (0.200 – x)= 2x 1.20 – 6.00x= 2x 1.20= 8.00x [HI] eq =2x = 2(0.150 M) = 0.300 M 0.150= x 8B-16 (of 37)

17. x0.050 - 2x H 2 (g) + Cl 2 (g) ⇆ 2HCl (g)K eq = 75 If 0.10 moles of HCl are placed in a 2.0 L flask, what will be the concentration of the HCl at equilibrium? H 2 (g) + Cl 2 (g) ⇆ 2HCl (g) Initial M’s Change in M’s Equilibrium M’s 000.050 + x - 2x x = 0.050 M HCl[HCl] in =0.10 mol / 2.0 L K eq = [HCl] 2 ___________ [H 2 ] [Cl 2 ] 75= (0.050 – 2x) 2 ________________ x 2 x= 0.0047 8B-17 (of 37)

18. H 2 (g) + Cl 2 (g) ⇆ 2HCl (g)K eq = 75 If 0.10 moles of HCl are placed in a 2.0 L flask, what will be the concentration of the HCl at equilibrium? = 0.050 M – 2(0.00469 M) [HCl] eq =0.050 – 2x 8B-18 (of 37) 75= (0.050 – 2x) 2 ________________ x 2 8.66= 0.050 – 2x ______________ x 8.66x= 0.050 – 2x 10.66x= 0.050 x= 0.00469 = 0.041 M

19. AB ⇆ A + B Initial Molecules/L 10000 Equilibrium Molecules/L 8020 K eq = [A][B] _________ [AB] = (20) 2 _______ (80) = 5 Initial Molecules/L 0100 Equilibrium Molecules/L 8020 The same ratio of products to reactants (K eq ) is always achieved, and starting at equivalent stoichiometric points the numerical values of A, B, and AB will always be achieved 8B-19 (of 37)

20. AB ⇆ A + B Equilibrium Molecules/L 8020 Add AB Molecules/L13020 Equilibrium Molecules/L12525 K eq = [A][B] _________ [AB] = (25) 2 _______ (125) = 5 The same ratio of products to reactants (K eq ) is always achieved, although the numerical values of A, B, and AB can be different 8B-20 (of 37)

21. LE CHATELIER’S PRINCIPLE – If a system is changed so that it is no longer at equilibrium, either the forward or reverse reaction will become spontaneous until the system reaches equilibrium again If a stress is applied to a system at equilibrium, the equilibrium will shift to relieve the stress 8B-21 (of 37)

22. At Equilibrium: H 2 (g) + I 2 (g) ⇆ 2HI (g) Spontaneous ReactionShift Direction H 2 is addedForwardRight I 2 is addedForwardRight H 2 is removedReverseLeft HI is addedReverseLeft 8B-22 (of 37) He is addedNo Change

23. energy + AB ⇆ A + B Equilibrium Molecules 8020 Equilibrium at Higher T7030 K eq = [A][B] _________ [AB] = (30) 2 _______ (70) = 13 Energy can be treated as a reactant (for endothermic reactions) or a product (for exothermic reactions) to predict shifts in equilibrium Only changes in temperature can change the numerical value of the K eq ENDOTHERMIC – A process that absorbs energy EXOTHERMIC – A process that releases energy 8B-23 (of 37)

24. H 2 (g) + I 2 (g) ⇆ 2HI (g) + energy Spontaneous ReactionShift Direction T increasedReverseLeft T decreasedForwardRight For exothermic reactions: T ↑, K eq ↓ T ↓, K eq ↑ 8B-24 (of 37) Exothermic Reaction at Equilibrium: H 2 (g) + I 2 (g) ⇆ 2HI (g)

25. AB ⇆ A + B Equilibrium Molecules/L8020 Change Volume from 1 L to 0.5 L 16040 Equilibrium Molecules/L 17129 Q = [A][B] _________ [AB] = (40) 2 _______ (160) = 10K eq = [A][B] _________ [AB] = (29) 2 _______ (171) = 5 A decrease in volume will cause the reaction that produces the lesser number of gas molecules to be spontaneous An increase in volume will cause the reaction that produces the greater number of gas molecules to be spontaneous 8B-25 (of 37)

26. 4RbCl (s) + O 2 (g) ⇆ 2Cl 2 (g) + 2Rb 2 O (s) Spontaneous ReactionShift Direction Volume increasedForward Volume decreasedReverse 8B-26 (of 37) Right Left

27. THE FINAL REVIEW Metric Prefixes Significant Figures Reading Graduated Equipment Density Matter : Elements, Compounds, and Mixtures Elements : Metals, Nonmetals, Metalloids Physical and Chemical Changes Ion Charges Energy Calculations Using Specific Heat Capacities 8B-27 (of 37)

28. THE FINAL REVIEW Subatomic Particles Atomic Number Mass Number Periodic Trends Activity of Metals and Nonmetals Atomic Radii Ionization Energy Electron Affinity Electronegativity 8B-28 (of 37)

29. THE FINAL REVIEW Atomic Orbitals Organization of Atomic Orbitals Energy Levels Sublevels Orbital Notation Electron Configuration Notation Electron Dot Notation 8B-29 (of 37)

30. THE FINAL REVIEW Ionic Bonding Covalent Bonding Polar and Nonpolar Covalent Bonds Lewis Structures Molecular Shape Molecular Polarity 8B-30 (of 37)

31. THE FINAL REVIEW Formulas and Names Ionic Compounds Covalent Compounds Acids Conversions to Moles Percentage Composition by Mass Empirical and Molecular Formulas 8B-31 (of 37)

32. THE FINAL REVIEW Balancing Equations Oxidation and Reduction Predict Products of Chemical Reactions Composition, Decomposition, Combustion, Replacement, Precipitation, Acid/Base Chemical Reaction Calculations Involving Masses Involving Gas Volumes Involving Solution Data Limiting Reactant 8B-32 (of 37)

33. THE FINAL REVIEW Conversions to Moles Percentage Composition by Mass Empirical and Molecular Formulas Balancing Equations Oxidation and Reduction Predict Products of Chemical Reactions Composition, Decomposition, Combustion, Replacement, Precipitation, Acid/Base 8B-33 (of 37)

34. THE FINAL REVIEW Standard Temperature and Pressure Gas Volume Relationships with Pressure Temperature Quantity of Matter Ideal Gas Law Equation Water Vapor Pressure 8B-34 (of 37)

35. THE FINAL REVIEW Attractive Forces in Different Types of Matter Nonpolar Molecular, Polar Molecular, Metallic Network, Nonmetallic Network, Ionic Heating Curves Net Ionic Equations Likes Dissolve Likes Strong Electrolyte, Weak Electrolyte, Nonelectrolyte Molarity 8B-35 (of 37)

36. THE FINAL REVIEW Arrhenius Acids and Bases Brønsted Acids and Bases Conjugate Acid/Base Pairs Acid Strengths Ion-Product Constant for Water Hydronium and Hydroxide Ion Concentrations pH Buffers 8B-36 (of 37)

37. THE FINAL REVIEW Collision Theory Chemical Equilibrium Equilibrium Constant Equilibrium Constant Expression Significance of the Numerical Value of the Equilibrium Constant Calculating the Equilibrium Constant Calculating Equilibrium Concentrations Using an ICE Table LeChatelier’s Principle 8B-37 (of 37)