1. This is a continuation of our discussion of Chapter 14: Equilibria

2. Approximations to Simplify the Math When the equilibrium constant is very small, the position of equilibrium favors the reactants For relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium [A] eq = ([A] o  x)  [A] o The equilibrium concentration of reactant is assumed to be the same as the initial concentration.

3. Checking the Approximation and Refining as Necessary Compare the approximate value of x to the initial concentration If the approximate value of x is less than 5% of the initial concentration, the approximation is valid.

4. Sulfur trioxide decomposes to give sulfur dioxide and molecular oxygen. The equilibrium constant at 300˚C is 1.6 x 10 -10. Calculate the equilibrium concentrations if the initial concentration of sulfur trioxide is 0.100M.

5. For the isomerization reaction of butane to isobutane the equilibrium constant at 25˚C is 2.50. In a 10.0 L bulb at equilibrium the concentrations of butane and isobutane are 0.140 M and 0.350 M respectively. If 2.00 mol isobutane is added to the equilibrium mixture, what will be the new equilibrium concentrations?

6. Changes in the Position of Equilibrium Once at equilibrium, the concentrations of all the reactants and products remain the same If the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re- established The new concentrations will be different, but the equilibrium constant (K) will be the same –unless you change the temperature

7. Le Châtelier’s Principle: If a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance Predicts effects changes in conditions have on position of equilibrium. Look at results from the butane isomerization

8. The Effect of Concentration Changes on Equilibrium Adding a reactant: other reactants decrease and products increase until a new position of equilibrium is found Removing a product: other products increase and reactants decrease. –you can use this to drive a reaction to completion! K stays the same!

9. The Effect of Concentration Changes on Equilibrium Equilibrium shifts away from side with added chemicals or toward side with removed chemicals Important: adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

10. The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium Adding a gaseous reactant increases its partial pressure and concentration, causing a shift to the products Removing a gaseous reactant increases its partial pressure and concentration, causing a shift to the reactants Adding an inert gas to the mixture has no effect on the position of equilibrium

11. The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, some of it decomposes to make more NO 2 2 NO 2  N 2 O 4

12. Effect of Volume Change on Equilibrium Decreasing the size of the container increases partial pressure of all the gases  P increases Le Châtelier’s Principle: the equilibrium should shift to remove that pressure When the volume decreases, the equilibrium shifts to the side with fewer gas molecules Solids, liquids, or solutions: changing the size of the container has no effect on the position of equilibrium

13. The Effect of Temperature Change on Equilibrium When T changes, K changes Use Le Châtelier’s Principle to predict the effect of temperature changes

14. The Effect of Temperature Change on Equilibrium Exothermic reactions release energy –Write heat as a product in an exothermic reaction –As T increases, K decreases shifting to the reactants Endothermic reactions absorb energy –Write heat as a reactant in an endothermic reaction –As T increases, K increases shifting to the products

15. Catalysts Catalysts provide an alternative, more efficient mechanism Works for both forward and reverse reactions Affects the rate of the forward and reverse reactions by the same factor Catalysts do not affect the position of equilibrium

16. The reaction 2 SO 2 (g) + O 2 (g) Û 2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? 1.adding more O 2 to the container 2.condensing and removing SO 3 3.compressing the gases 4.cooling the container 5.doubling the volume of the container 6.warming the mixture 7.adding the inert gas helium to the container 8.adding a catalyst to the mixture

17. Chapter 15: Acids and Bases

18. Properties of Acids Sour taste React with “active” metals  Al, Zn, Fe, but not Cu, Ag, or Au 2 Al + 6 HCl ®  AlCl 3 + 3 H 2  Corrosive React with carbonates, producing CO 2  marble, baking soda, chalk, limestone CaCO 3 + 2 HCl ®  CaCl 2 + CO 2 + H 2 O Change color of vegetable dyes  blue litmus turns red React with bases to form ionic salts (neutralization)

19. Common Acids Chemical NameFormulaUsesStrength Nitric AcidHNO 3 explosive, fertilizer, dye, glue Strong Sulfuric AcidH 2 SO 4 explosive, fertilizer, dye, glue, batteries Strong Hydrochloric AcidHClmetal cleaning, food prep, ore refining, stomach acid Strong Phosphoric AcidH 3 PO 4 fertilizer, plastics & rubber, food preservation Moderate Acetic AcidHC 2 H 3 O 2 plastics & rubber, food preservation, Vinegar Weak Hydrofluoric AcidHFmetal cleaning, glass etching Weak Carbonic AcidH 2 CO 3 soda waterWeak Boric AcidH 3 BO 3 eye washWeak

20. Structures of Acids Binary acids have acid hydrogens attached to a nonmetal atom –HCl, HF

21. Structure of Acids oxy acids have acid hydrogens attached to an oxygen atom –H 2 SO 4, HNO 3

22. Structure of Acids Carboxylic acids have COOH group –HC 2 H 3 O 2, H 3 C 6 H 5 O 7 Only the first H in the formula is acidic –the H is on the COOH

23. Properties of Bases Also called alkalis Tastes bitter –alkaloids = plant product that is alkaline often poisonous Solutions feel slippery Changes color of vegetable dyes –red litmus turns blue React with acids to form ionic salts (neutralization)

24. Common Bases Chemical Name FormulaCommon Name UsesStrength sodium hydroxide NaOHlye, caustic soda soap, plastic, petrol refining Strong potassium hydroxide KOHcaustic potash soap, cotton, electroplating Strong calcium hydroxide Ca(OH) 2 slaked limecementStrong sodium bicarbonate NaHCO 3 baking sodacooking, antacidWeak magnesium hydroxide Mg(OH) 2 milk of magnesia antacidWeak ammonium hydroxide NH 4 OH, {NH 3 (aq)} ammonia water detergent, fertilizer, explosives, fibers Weak

25. Structure of Bases Most ionic bases contain OH ions –NaOH, Ca(OH) 2 Some contain CO 3 2- ions –CaCO 3 NaHCO 3 Molecular bases contain structures that react with H + –NH 3, other amine groups

26. Arrhenius Theory Bases dissociate in water to produce OH - ions and cations NaOH(aq) → Na + (aq) + OH – (aq) Acids ionize in water to produce H + ions and anions HCl(aq) → H + (aq) + Cl – (aq) HC 2 H 3 O 2 (aq) → H + (aq) + C 2 H 3 O 2 – (aq)

27. Arrhenius Acid-Base Reactions H + from the acid combines with the OH - from the base to make a molecule of H 2 O The cation from the base combines with the anion from the acid to make a salt acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)

28. Problems with Arrhenius Theory Can’t explain molecular bases like NH 3 that don’t contain OH – Can’t explain why molecular substances, like CO 2, dissolve in water to form acidic solutions – even though they do not contain H + ions Does not include acid-base reactions that take place outside aqueous solution

29. Brønsted-Lowry Theory In a Brønsted-Lowry Acid-Base reaction, an H + is transferred -Does not have to take place in aqueous solution -Broader definition than Arrhenius Acid is H donor, base is H acceptor -Base must have an unshared pair of electrons In an acid-base reaction, the acid molecule gives an H + to the base molecule H–A + :B  :A – + H–B +

30. Hydronium Ion H + ions produced by the acid don’t exist alone in water They react with a water molecule(s) to produce complex ions, mainly hydronium ion, H 3 O + H + + H 2 O  H 3 O + –there are also minor amounts of H + with multiple water molecules, H(H 2 O) n +

31. Brønsted-Lowry Acids Acids are H + donors -Needs an H to be a Brønsted-Lowry acid HCl(aq) is acidic because HCl transfers an H + to H 2 O, forming H 3 O + ions -water acts as base, accepting H + HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) acidbase

32. Brønsted-Lowry Bases Brønsted-Lowry bases are H + acceptors -Needs lone pairs to act as a Brønsted-Lowry base NH 3 (aq) is basic because NH 3 accepts an H + from H 2 O, forming OH – (aq) -water acts as acid, donating H + NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) base acid

33. Amphoteric Substances amphoteric substances can act as either an acid or a base –have both transferable H and atom with lone pair Water acts as base, accepting H + from HCl HCl(aq) + H 2 O(l) → Cl – (aq) + H 3 O + (aq) Water acts as acid, donating H + to NH 3 NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq)

34. Polyprotic Acids polyprotic acids --Acid molecules with more than one ionizable H –HCl, monoprotic, –H 2 SO 4, diprotic, –H 3 PO 4, triprotic Ionize in steps, each ionizable H removed sequentially Removing of the first H automatically makes removal of the second H harder –H 2 SO 4 is a stronger acid than HSO 4 

35. Conjugate Pairs A Brønsted-Lowry acid becomes a Brønsted- Lowry base after it donates a proton. A Brønsted-Lowry base becomes a Brønsted- Lowry acid after it accepts a proton. Each reactant and the product it becomes is called a conjugate acid/base pair

36. Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reactions H 2 SO 4 + H 2 O  HSO 4 – +H 3 O + NH 3 + H 2 O  NH 4 + +OH -

37. Write the formula for the conjugate acid of the following Base H 2 O NH 3 CO 3 2− H 2 PO 4 1−

38. Write the formula for the conjugate base of the following Acid H 2 O NH 3 CO 3 2− H 2 PO 4 1−

39. Strength of Acids and Bases The strength of an acid is inversely related to the strength of its conjugate base --The stronger the acid, the weaker it’s conjugate base The strength of an acid is inversely related to the strength of its conjugate base --The stronger the base, the weaker it’s conjugate acid

40. Increasing Acidity Increasing Basicity

41. Acid Ionization Constant, K a For the reaction of an acid with water HA + H 2 O  A -1 + H 3 O +1 The equilibrium constant is called the acid ionization constant, K a –larger K a = stronger acid

43. Base Equilibrium Constant, K b For the reaction of a base with water B + H 2 O  OH - + BH + The equilibrium constant is called the base constant, K b –larger K b = stronger base

44. Autoionization of Water ~ 1 out of every 10 6 water molecules form ions through a process called autoionization H 2 O Û H + + OH – H 2 O + H 2 O Û H 3 O + + OH – All aqueous solutions contain both H 3 O + and OH – –H 3 O + and OH – are equal in water –[H 3 O + ] = [OH – ] = 10 -7 M @ 25°C

45. Ion Product of Water The equilibrium constant for the autoionization of water is called the ion product of water and has the symbol K w K w =[H 3 O + ] [OH – ] = 1 x 10 -14 at 25°C As [H 3 O + ] , [OH – ]  (so the product stays constant)

46. Acidic and Basic Solutions Neutral solutions : [H 3 O + ] = [OH – ] = 1 x 10 -7 Acidic solutions: [H 3 O + ] > 1 x 10 -7 [OH – ] < 1 x 10 -7 Basic solutions: [H 3 O + ] 1 x 10 -7

47. Calculate the [OH  ] at 25°C when the [H 3 O + ] = 1.5 x 10 -9 M, and determine if the solution is acidic, basic, or neutral

48. pH The acidity/basicity of a solution is often expressed as pH pH = -log[H 3 O + ] [H 3 O + ] = 10 -pH pH 7 is basic, pH = 7 is neutral

49. Sig. Figs. & Logs When you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 10 6 ) = log(10 6 ) + log(2.0) = 6 + 0.30303… = 6.30303... The part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log log(2.0 x 10 6 ) = 6.30