CHEM 162 Exam 1 Review – Winter2013 page 1 of 4 CHEM 162: Exam 1 Study Guide Chapter 16: Chemical Equilibrium equilibrium: state where the forward and reverse reactions or processes occur at the same rate – Know that concentrations are not changing at equilibrium, but they need not be equal to one another. – Be able to indicate when equilibrium is achieved given concentration vs. time plots Use the law of mass action to write equilibrium expressions for Kc or Kp for homogeneous and heterogeneous reactions. – Include only gases; omit pure liquids and solids. For general reaction, j A + k B l C + m D kjml[B] [A][D] [C]cK  or k j m l BADCpP PP PK  Extent of reaction – For large values of Kc or Kp (>103), the reaction essentially goes to completion.  The equilibrium mixture consists mostly of products (product favored)  The equilibrium lies to the right. – For small values of Kc or Kp (<10‐3), the reaction does not occur to any significant degree.  The equilibrium mixture consists mostly of reactants (reactant favored)  The equilibrium lies to the left. – For intermediate values (10‐3 < Kc or Kp < 103), the equilibrium mixture contains appreciable amounts of both reactants and products.  For Kc or Kp > 1, equilibrium lies to the right. Equilibrium positions: Set of equilibrium concentrations or partial pressures of reactants and products for a system. – While Kc or Kp for a reaction are constant for a given temperature, various equilibrium positions are possible for that reaction depending on the initial concentrations of reactants and products. Relating Kp and Kc: Kp=Kc(RT)n – Be able to solve for n for any homogeneous or heterogeneous reaction. – Know that Kp=Kc for n=0. Know Kc or Kp are unitless! Reaction Quotient (Q): instant state of system, not necessarily at equilibrium. – Also determined using law of mass action. – Q < K: too many reactants  System shifts right to make more products. – Q > K: too many products  System shifts left to make more reactants. – Q = K: system at equilibrium – To compare Q and K, Q must be determined in terms of partial pressure for Kp and in concentrations for Kc. Determination of Kp or Kc – Be able to solve for either using a variety of experimental data. – Given equilibrium concentrations or partial P’s. – Given initial & changes in conc. or partial P’s. – Given total pressure at equilibrium. Equilibrium Problems Solving for Kp or Kc. 1. Get balanced chemical equation 2. Write equilibrium expression 3. Set up equilibrium ICE table. – Let x=change in conc. or partial pressure. 4. Substitute equilibrium conc. or pressures into equilibrium expression for Kp or Kc. 5. Solve for x, using quadratic method if necessary. 6. Substitute value for x into equilibrium conc. or pressures to solve for Kp or Kc. Le Chatelier's Principle – A system at equilibrium will shift (if possible) to minimize any stress (change in concentration, pressure, volume, or temperature) – Predict shifts in equilibrium given specific changes. – Only changes in temperature (T) affect Kp or Kc. – For endothermic reactions, Kp and Kc  as T . – For exothermic reactions, Kp and Kc  as T . CHEM 162 Exam 1 Review – Winter2013 page 2 of 4 Chapter 16: Chemical Equilibrium (Continued) Characteristics of the Equilibrium Expression – For reverse reaction, equilibrium expression is reciprocal of that for forward reaction – Multiplying coefficients by factor, n, raises equilibrium constant to nth power – Multiple Equilibria – When a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is simply the product of the equilibrium constants for the individual reactions. – Be able to manipulate a series of reactions to determine the equilibrium constant for an overall reaction. Chapter 17: Acids And Bases • Know the properties of acids and bases. • Know Arrhenius definitions for acids and bases. – Know the general form of an Arrhenius acid‐base neutralization reaction: – acid + base  water + salt • Know terms monoprotic, polyprotic, etc. • Know Brønsted‐Lowry definitions for acids and bases. – Recognize conjugate acid‐base pairs. • Recognize hydronium ion, H3O+ = H+ + H2O • Know the strong acids: HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4. • Know the common strong bases: LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2. • Recognize strong acids and strong bases dissociate or ionize (break up) completely.  Equilibrium lies far to the right. HNO3(aq)  H+(aq) + NO3(aq) Ca(OH)2(aq)  Ca+2(aq) + 2 OH(aq) and for H2SO4: H2SO4(aq)  H+(aq) + HSO4(aq) • Recognize weak acids dissociate or ionize (break up) only to a small degree.  Equilibrium lies far to the left. HF(aq) H+(aq) + F(aq) • Write balanced equations and equilibrium expressions for the dissociation of any acid. – Omit pure liquids and solids. HA(aq) H+(aq) + A(aq) + ‐[H ] [A ]K=a[HA] • Recognize the strength of an acid is inversely related to the strength of its conjugate base.  Strong acids have conjugate bases that are weaker than H2O.  Weak acids have conjugate bases that are stronger than H2O.  Be able to write the net ionic equation for the conjugate base reacting with H2O to form the conjugate acid and OH. • Know Lewis definitions for acids and bases. – Be able to identify the Lewis acid or base in a given reaction. – Know why nitrogen‐containing compounds are generally Lewis bases. – Know highly charged cations (Al3+, Cu2+, etc.) can act as Lewis acids and why. • Recognize how structure, bond strength and bond polarity influence the properties of an acid. – Be able to explain: – Why HF is weak while HCl, HBr, HI are strong – Why CH4 does not display acidic properties – Why some ternary oxyacids are strong while others are weak (e.g. HNO3 vs HNO2) – Ka values for different acids based on structure CHEM 162 Exam 1 Review – Winter2013 page 3 of 4 Chapter 17: Acids And Bases (Continued) • Water is amphoteric; it can act as an acid or a base. • Autoionization of water: 2