CH 302 1st EditionExam 2 Study GuideChemical Equilibrium:- Equilibrium Constanto The ratio of activities of the products divided by the reactantso Always a constanto K > 1, favor the productso K < 1, favor the reactants - Kc and Kpo Kp = Kc (RT)^no n is the total number of moles in the reaction (products minus reactants)- Free Energy and Equilibriao G and K are related by the equation: G = -RT ln Ko Systems always want to move towards the lowest free energyo The free energy depends on both the compounds and the concentrations- Q vs Ko Q is also the ratio of activities of products divided by reactantso ONLY INVOLVES GASESo Q is the value we get from any point away from the equilibriumo Q depends on the concentrations right now, K is a fixed numbero Q > K then the equilibrium lies towards the reactantso Q < K then the equilibrium lies towards the productso Q = K then the system is in equilibrium- Le Chatelier’s Principleo Adding or removing reactant or product Adding reactant shifts reaction to the right Adding product shifts the reaction to the lefto Changing the pressure Increasing the pressure by decreasing the volume shifts the reaction towards the side with fewer moles of gaso Changing the temperature Exothermic reactions: increasing temperature causes the reaction to shift to the left Endothermic reactions: increasing the temperature causes the reaction toshift to the rightPractice Problems:1. Write the expression for Kc for the reaction below. Cu2+(aq) + 4 NH3(aq) ↔ Cu(NH3)42+(aq)2. Write the expression for K for the reaction below. Al2S3(s) ↔ 2 Al3+(aq) + 3 S2-(aq)3. Dinitrogen tetraoxide decomposes to nitrogen dioxide. N2O4(g) ↔ 2 NO2(g)Calculate the value of Kp, given that Kc = 5.88 × 10-3 at 273 K. (R = 0.08206 L·atm/mol·K)4. A 4.00 L flask is filled with 0.75 mol SO3, 2.50 mol SO2, and 1.30 mol O2, and allowed to reach equilibrium. Predict the effect on the concentrations of SO3 as equilibrium is achieved by using Q, the reaction quotient. Assume the temperature of the mixture is chosen so that Kc = 12. 2 SO3(g) ↔ 2 SO2(g) + O2(g)5. Excess Ca(IO3)2(s) is placed in 1.5 L of water. At equilibrium, the solution contains 0.011 M IO3-(aq). What is the equilibrium constant for the reaction below? Ca(IO3)2(s) ↔ Ca2+(aq) + 2 IO3-(aq)Answers:1. Kc = [Cu(NH3)4 2+]/ [Cu 2+][NH3]^42. K = [Al 3+]^2 [S 2-]^33. 0.1324. [SO3] will decrease because Q < K5. 6.7 x 10^-7Acid/Base Theory:- Strong Acidso HCl (hydrochloric acid)o HBr (hydrobromic acid)o HI (hydroiodic acid)o HClO4 (perchloric acid)o HClO3 (chloric acid)o H2SO4 (sulfuric acid)o HNO3 (nitric acid)- Strong Baseso LiOH (lithium hydroxide)o NaOH (sodium hydroxide)o KOH (potassium hydroxide)o RbOH (rubidium hydroxide)o CsOH (cesium hydroxide)o Ca(OH)2 (calcium hydroxide)o Sr(OH)2 (strontium hydroxide)o Ba(OH)2 (barium hydroxide)- Bronsted-Lowry Definitiono Acid: Proton donoro Base: Proton acceptor- Weak Acids and Weak Baseso Weak acids only partially dissociateo Weak bases only partially dissociateo Ka is used to describe the equilibrium constant for acids Larger Ka favors the dissociated side of the reaction Smaller Ka favors the less it favors the dissociated side of the reactiono Kb is used to describe the equilibrium constant for bases- Strong Acid Equilibriao Strong acids and bases go to 100% dissociationo Equilibrium constants are usually very large- Weak Acid Equilibriao Set up the equilibrium expression equal to the Ka or Kbo Used to solve for concentrations o Do not include liquids or solids in the expression- Acid/Base Strengtho The larger the Ka or Kb, the stronger the acid or baseo The stronger the acid or base, the more it dissociatesPractice Problems:1. The substance NH3 is considered:A. A weak acidB. A weak baseC. A strong acidD. A strong baseE. A neutral compound2. The substance (CH3CH2)2NH is considered:A. A weak acidB. A weak baseC. A strong acidD. A strong baseE. A neutral compound3. The substance HClO4 is considered:A. A weak acidB. A weak baseC. A strong acidD. A strong baseE. A neutral compound4. Which of the following is the strongest acid?A. CH3COOHB. HFC. H3PO4D. H2SO3E. HI5. Which of the following pairs has the stronger acid listed first?A. HBr, HIB. HClO2, HClO3C. H2SeO4, H2SeO3D. HNO2, HNO3E. HF, HCl6. Which of the following acids has the lowest pH?0.1 M HBO, pKa = 2.43 0.1 M HA, pKa = 4.55 0.1 M HMO, pKa = 8.23 0.1 M HST, pKa = 11.89 pure waterA. HAB. HSTC. HMOD. HBOE. Pure water7. Which of the following liquids contains the strongest acid?0.1 M HA, pH = 6.85 0.1 M HD, pH = 7.22 0.1 M HE, pH = 8.34 0.1 M HJ, pH = 11.88pure waterA. HEB. HAC. HJD. HDE. Pure water8. What is the pH of a 0.20 M HCl solution?A. < 0B. 0.70C. 1.61D. 12.39E. 13.309. What is the pH of a 0.75 M HNO3 solution?A. 0.12B. 0.29C. 0.63D. 0.82E. > 1.010. What is the pH of a 0.050 M LiOH solution?A. < 1.0B. 1.30C. 3.00D. 11.00E. 12.70Answers:1. B2. B3. C4. E5. C6. D7. B8. B9. A10. EWater and pH:- Autoionization of watero Kw = [H3O+][OH-]o Kw = 1.0 x 10^-14o Used to solve for the concentration of [H3O+] or [OH-]- Concentrations at Room Temperatureo At room temperature, Kw = 1.0 x 10^-14o Increase in temperature causes the equilibrium constant to increase tooo The higher the temperature, the higher the concentrations will increaseo The solution will still be neutral where [H3O+] = [OH-]- Neutral, Acidic, and Basico [H3O+] = [OH-] is neutralo [H3O+] > [OH-] is acidico [H3O+] < [OH-] is basic- pH and pOHo pH is the –log [H3O+]o pOH is the –log [OH-]o pOH + pH = 14Practice Problems:1. Which of the following is true in pure water at any temperature?A. pH = 7.0B. Kw decreases with increasing temperature.C. [H3O+] = [OH−]D. pH = 7.0 or greater than 7.0 E. [H3O+][OH−] = 1.0 × 10−142. What is [OH−] in a 0.0050 M HCl solution?3. Hydroxylamine is a weak molecular base with Kb = 6.6 × 10−9. What is the pH of a 0.0500 M solution of hydroxylamine?4. Which acid is weaker?A. HBrOB. HBrO3C. They have the same strength5. Which of the following would be expected to act as a Lewis acid?A. OH−B. NH3C. H3O+D. NH+4E. BF36. What is the conjugate base of ammonia?A. NH2OHB. NH2 –C. NH4+D. NH3E. OH-Answers:1. C2. 2.0 x 10^ -123. 9.264. A5. E6. BNeutralizations/Buffers:- Saltso Strong acid + strong base = neutral salto Strong acid + weak base = acidic salto Strong base + weak acid = basic salt- Identifying acids, bases, and saltso Begin by memorizing strong acids
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