CHM102-Final Exam Review

CHM102-Final Exam Review - fl Properties of Solutions -...

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Unformatted text preview: fl Properties of Solutions - Recognize solution. solute, solvent - Concentration Units: — Know mass percent. parts per million, parts per billion. mole fraction. molarity, molality - Be able to convert from one unit to another - Principles of Solubility — "Like dissolves like" rule, ionic compounds - Be able to determine polarinonpolar based on formula - Gas solubility: fl as T i; i as P ) — Saturated. unsaturated. supersaturated Colilgative Properties — depend only on amount. not kind of solute — Be able to solve for van't Hoff factor. molar mass using coiligative properties nonelectrolyte: exist as molecules in solution electrolyte: exist as ions In solution van‘t Hoff tactor: i _ it ions after dissociation - it compounds before dissociation Osmotic Pressure: n=iMRT — Pressure needed to block osmosis, movement of solvent molecules through semipermeable membrane - Know isotonic, hypertonic. hypotonlc Boiling Point Elevation: ATb a lKh-m Freezing Point Depression: AT, = lKg-m Chemical Kinetics chemical kinetics: study of reaction rates or speeds reaction rate: a positive quantity expressing change of concentration with time Determining Rate from 1. Graphical data of concentration versus time 2. Experimental data of concentrations and time Recognize terms: rate law , rate constant (k) Reaction Order: ~ zero-order. first-order. second-order — overall order for more than one reactant Determining rate law and rate constant from 1. Experimental data of concentrations and rates 2. Graphical data of concentrations versus time — Know characteristics of plots for zero-order. first— order and second-order reactions Chemical Kinetics (Continued) Solve various problems Involving first-order reactions — Amount of sample at time. t — How old a sample is given half-life or k Collision Model — reactant molecules must collide to react Three Factors Affecting Fate 1. Concentration 2. Orientation of molecules 3. Kinetic Energy — molecules must have activation energy Activation Energy (El): - minimum energy needed for chemical reaction Reaction Rate and Tern rsture — As Til. reaction rate since more molecules have activation energy. -— Be able to translate distribution curves of molecules and energy at different temperatures Transition State Model Activation Energy Diagrams — Indicate transition state (activated complex). activation energy for reactants and products. effect of catalyst — Determine if reactions is endothermic or exothermic. Ctalyst: substance that lowers the activation energy, increasing reaction rate. without being consumed in reaction Arrhenius Equation: it I A e-EJRT where A=frequency factor. Ea=activation energy. R=8.314 JImol-K, T=temperature in K Be able to soive for any unknown given other variables or graphical data and variables. Two-Point Equation: - Solve for rates, temperatures. or activation energy k E ln_1=_..._I _.1__l kI R T2 T1 Reaction Mechanisms — sequence of steps by which a reaction occuni at molecular level molecularity of a reaction ~ unimolecular. bimoiecular. termolecutar — corresponding rate laws for each The slowest step in a mechanism is the rate- determlning step. intermediate: species produced in a step and consumed in later step of mechanism Chemical Kinetics chemical kinetics: study of the factors that influence reaction rates reaction rate: a positive quantity expressing change of concentration with time Collision Mode: — reactant molecules rnust collide to react Determine reaction rate — Given experimental data of concentrations and time. Recognize terms: rate law. rate constant (k) Reaction Order: _ — zero—order. first-order, second-order - Determine overall order for a reaction given rate law. Detennlne rate law using Initial rates method - Requires experimental data of concentrations and rates. . — Cannot be determined given only the balanced chemical equation. half-tife (tug): the time required for the concentration of a reactant to decrease by half Do calculations given integrated rate laws for zero-order, first-order, and second-order reactions. - Solve for concentration at a given time given initial concentration and rate constant. - Carry out naturai log (In) calculations for 1"-order reactions—review your algebra! - Solve for the time required for the concentration to decrease to a given amount. — Solve for half-life given rate constant. k. or vice versa. Summary of Zero-Order, First-Order, and Second-Order Reactions Rate Expression integrated Rate Law Half-Life 2 2 J— : kt 1 + Chemical Equilibrium equilibrium: state where the fonvard 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 K, for homogeneous and heterogeneous reactions. — include only gases; omit pure liquids and solids. For general reaction. jA + k3 '"-="- it: +1110 1 m P I P m K 2 LEI—[EL or : —C__D_ ° [Ali [81* pg pair Extent of reaction -— For large values of K, or K, (>103). the reaction essentially goes to completion. —> The equilibrium mixture consists mostly of products (product favored) a The equilibrium lies to the right. — For small values of Kc or K, (<1 0“), the reaction does not occurto any significant degree. —> The equilibrium mixture consists mostly of reactants (reactant favored) —> The equilibrium lies to the left. — For intermediate values (104 < Kc or K,<10’), the equilibrium mixture contains appreciable amounts of both reactants and products. —> For Kcor K,> 1. equilibrium lies to the right. Equilibrium positions: Set of equilibrium concentrations or partial pressures of reactants and products for a system. — While K, or K, for a reaction are constant for a given temperature, various equilibrium positions are possible for that reaction dependln on the initial concentrations of reactants and products. Relating K, and Kc: K,=K,(RT)”' - Be able to solve for on for any homogeneous or heterogeneous reaction. — Know that K,=Kc for An=0. Know litc or K, are unitlessl 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 equitibrlum Determination of K, or K, — Be able to solve for either using a variety of experimental data. - Given equilibrium concentrations or partial P’s. - Given initial 8. changes in cone. or partial P's. — Given total pressure at equilibrium. Equilibrium Problems Solving for K, or K.. 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 cone. or pressures into equilibrium expression for K, or Kc. 5. Solve for it. using quadratic method if necessary. 6. Substitute value for x into equilibrium conc. or pressures to solve for K, or K,. Le Chatelier‘s Principle — A system at equilibrium will shift (if possible) to minimize any stress (change in concentration, pressme. volume, or temperature) — Predict shins in equiiibrium given specific changes. — Only changes in temperature (1') effect K, or K... — For endothermic reactions, K, and Kc T as T T. - For exothermic reactions, K, and K. 1‘ as T 1.. Characteristics of Equilibrium Expression — For reverse reaction. equilibrium expression is reciprocal of that for forward reaction -- Multiplying coefficients by factor, n, raises equilibrium constant to n" 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. Acids and Bases Recognize hydronium ion. H,O* = H’ 1- H20 Know the strang acids: HCi. HBr, HI. HNOa. HCIOg. HCIO4, H2804. Know the common strong bases: LiOH. NaOH, KOH. Ca(OH);, Sr(OH}2, Ba(0H)2. Recognize strong acids and strong bases dissociate or ionize (break up) completely. —; Equilibrium lies far to the right. HN03(8Q) -i “76(1) + Nos—(5Q) Ca(OH)2(aq) —> Ca‘zitaq) + 2 OHTaq) anti for stOfi stodaq) —> HYan + HSOffaq) Recognize weak acids dissociate or ionize (break up) only to a small degree. —> Equilibrium lies far to the left. HF(a0) ==—" H’raq) + F‘raq) Write balanced equations and equilibrium expressions for the dissociation of any acid. - Omit pure liquids and solids. HMan = H’raq) + Nari) = [Hum-1 {HA} The strength of an acid is inversely related to strength of its conjugate base. -—) Strong acids have conjugate bases that are weaker than H20. -—> Weak acids have conjugate bases that are stronger than H20. —> Write the net ionic equation for the conjugate base reacting with H20 to form the conjugate acid and OH“. —> Be able to rank acids or bases in terms of increasing strength given K... Kb, or information on relative strength. K: Know water is amphoterlc, so it can act as an acid or a base. ~ Autoionization of water: 2 H100) =2 Hao'faq) + onraq) Kw = [H‘][0H']=1.0xio'“ Kw = water's ion product constant Acidic, Basic, and Neutral Solutions — When [H’] = [OH'], the solution is neutral. - When fH'] 3- [Oi-i1, the solution is acidic. — When [H‘] 4: [OH‘]. the solution is basic. — Be able to solve for [H‘] or [CH] givan the other then classify the substance as acidic. basic. neutral. pH scale: pH = 7: neutral & pH < 7: acidic & pH > 7: basic (or alkaline) [H‘i =1o‘t“ pH [OH‘J =10” pH = — loo [H’] 90H = — los [OH'] pH + pOH a 14.00 Know # of sig figs In [H‘] or [OH'] determines the # of decimal places in pH and pOH. Calculate pH of Strong Acids and Strong Bases — Strong acids and bases ionize completely. - Accounting for all the H’ and OH‘ ions, —+ [H‘] = original molarity of strong acid s [OH‘] = molarity of base x {# of OH‘ in base) pH Calculations for a Weak Acids — Weak acids remain mostly undissociated. — Write the equation for the dissociation of acid. - Setup lCE table with [HA] given or calculated. — Calculate pH of weak acid solution. — Use percent ionization of solution to get K... — Use [HX] - X = [HX] approx. fan [9% <5% - Use Quadratic Method or method of successive . . X a roxrmations for: —— 20.059 PD [HA] 0 + .. . Percent ionization = Mm—im x100% page I of? Acids and Bases (Continued) Convert K. and Kb: Kw = K.-Kb = 1.0x10'“ Salts with an acidic cation and basic anion — Classify a given salt as acidic. basic. neutral. Acid-Base Properties of Salts — If K.>K.,, salt is acidic. —- salt= ionic compound — if K§>K., salt is basic. - Soluble salts dissociate into ions in water. — Account for different ion concentrations by — Classify a given salt as acidic. basic, neutral. multiplying initial concentration with K. or Kb. lens that produce acidic solutions pH calculations for 3 Weak Bases — NH4+ — Write equation for reaction of base with H20. - Highly Charged mata' ions-3 Ale”. 202'. em. — Write equilibrium expression for the weak base. except cation of strong base e The weaker the base. smaller the K... - Set up ICE table with [A‘] given or calculated. - Solve for x to calculate [OH] and pOH of weak base solution. ions that produce basic solutions — anions that are con}. bases of weak acids. - all anions except anions of strong acids — Note: 5042' + H20 —+ H305 + OH- Poiyprotlc Acids lens that produce neutral solutions ' mzrmiecsfigw'se diSSOCiaflo" Of any — do not react wrth H20 to make H or OH’ _ Recognize the [H1 is determined only by 1.1 . dissociation step for most poiyprotlc acids. — Solve for pH and equilibrium ion concentrations. Applications of Aqueous Equilibria 1 [HtiiOH-l - H” (from strong acids) will completely react with any base, A“. — OH' (from strong bases) will completely react with any acid. HA. — Be able to write neutralization reactions for H“ or OH’ added to any solution. Given Kg, = = 1.0x10“, Acid-Base Titrations — Distinguish between endpoint (when indicator changes color) and equivalence point (when equal amounts of H+ and OH‘ present) — Know general pH range for equivalence point of following: — strong acid-strong base titration (pH=7) — weak acid-strong base titration (pH>7) — strong acid-weak base titration (pl-k7) . — Recognize that pH=pK. at halfway to equivalence point for a weak acid-strong base titration. — Given titration data. calculate the pH an any point during an acid-base titration. — Account for amount of H‘ and OH' neutralized and new initial [HA] or [A']. - Setup ICE table and solve for x to get [H‘] or [0H‘]. — Calculate pH based on [H”] or [OH']. Applications of Aqueous Equilibria ~ - 1 u Gwen Kw - [H+]!0H_] _1.0x10 . — H” (from strong acids) will completely react with any base, A‘. - OH‘ (from strong bases) will completely react with any acid, HA. —- Be able to write neutralization reactions for H‘ or 0H‘ added to any solution. The Common ion Effect — The shift in equilibrium caused by the addition of a salt with an ion in common with the dissolved substances — Given a compound added to an acid or base at equilibrium, predict equilibrium shifts (to the left or right). if [H‘] i or i. if [OH’] 1 or i. and if pH T or Jr. — Calculate the pH of an acid or base solution when a common ion is added buffer: a solution of a weak acid or weak base and its conjugate that can resist large changes in pH upon addition of a small amount of strong acid or strong base Determining pH and [H‘] for any buffered solution - Calculate the number of moies of weak acidlbase and its conjugate present. - Know pH=pK. for [HA]=[A‘] (at halfway to the equivalence point for a titration) — Calculate the number of moles of strong acid (H‘) or strong base (OH') added to the buffer. _ Calculate the number of moles of weak acidlbase and its conjugate present after reaction with the strong acid (H") or strong base (OH‘). — Use Henderson-Hasselbaich equation: {1"} H = + io ~—~— P PK. 9 [HA] Know how to prepare a buffer system to maintain a given pH — Choose a weak acid with pK, near buffer pH range. — Use Henderson-Hasselbalch to determine the {fig ratio to get desired pH for a buffer system. — Be able to determine the amount of weak acid (volume and concentration) and the mass of a salt containing the conjugate base to prepare a buffer to maintain the desired pH. Acid-Base Indicators —- Know when color changes become apparent (at pKail). — Know effective range of indicators based on K.. — Given the pK. for different acid-base indicators. explain which indicators would be effective for specific acid-base titrations. Solubility Equilibria — Write the chemical equation for a saturated solution - Write equilibrium expressions for solubility product. Kat: — Express concentration in molarity - include only ions. omit pure liquids and solids Calculations involving Ksp — Solve for Ksp given concentration of ions m Solve for solubility (in glL) and molar solubility (in mollL) given Ksp Ion product. 0“: instant state of system 0., < K»: Solution is supersaturated, so precipitate forms until equilibrium is established 0., > K»: Solution is unsaturated. so no precipitate forms Q... = Ksp: Solution is saturated, so no precipitate forms, but solution is at the point of precipitation Recognize Factors that Affect Solubility - Common ion effect — pH and solubility: How strong acids can be used to dissolved metal hydroxides and precipitates where the anion is conjugate base of a weak acid CHEMICAL THERMODYNAMICS Entropy, S: randomness factor - Ssolid < Sliquld < Sgas S = 0 only for a perfect crystalline solid at 0 K -— S > 0 for all other substances. even naturally occuring elements (3rd Law of Thermodynamics) Be able to calculate AS' given Sf' data G=Gibhs free energy - Be able to calculate AG' given Gf’ data Gibbs-Helmholtz Equation: AG = AH - T AS Under standard state conditions: AG' = AH' - T A3“ — Be able to calculate 156', AH', T in Kelvins, andior AS‘ given the other variables If AG<0 —> a spontaneous reaction l_ m If AG>D —> a nonspontaneous reaction 5+ 5 If AG=0 —; reaction is at equilibrium (3-; i Know conditions for DH, DS. and T (6.9. we or —ve AH or AS. high T or l0w T) to bring about a spontaneous reaction if possible Pressure and Concentration: — Use AGaoG'+RTInQ — Know how partial pressure of gases and concentrations of solutions affect AG At equilibrium, use: A6' = — RT In K ELECTROCHEMISTRY Electrochemlstry: the study of the relationship between electricity and chemical reactions Be able to determine oxidation numbers for all atoms in an oxidation-reduction (redox) reaction. — Determine what is oxidized. what is reduced. the oxidizing agent. and the reducing agent in a redox reaction. — Be able to batance the electrons transferred in a redox reaction by balancing atoms gainingliosing electrons cathode: where reduction occurs anode: where oxidation occurs Voltaic (or Galvanic) Cetle Know line notation (cell diagram) for an electrochemical cell: Zn lZn2+ || Cu" | Cu where 1. anode reaction (oxidation) is shown at left 2. The separation of the two half cells is indicated by the symbol [1. 3. cathode reaction (reduction) is shown at right 4. single vertical line "i" indicates a phase boundary (e.g. a solid electrode in an aqueous solution) For any voltaic cell. be able to identify: 1. The half-reaction at the cathode 2. The half~reaction at the anode 3. The electron flow in the cells and through the external circuit 4. The ion flow In both cells. including the salt bridge Calculation of 31.).“ from 3’; and 3nd“: 32m" = 8:: + gndo — The cell's overall voltage or potential (32.13) is the sum of the tvvo half reactions: & /s‘r-°, C £512; 2 gazw’ “Ml” General Guidelines for Voltaic Cells 1. Calculated cell voltages. gnu“. must be positive for reactions in a voltaic cell. 2. Because cell potentials are intensive properties (independent of amount). you never multiply by the coefficients in a half reaction to get the cell potential. Strength of Oxldlzing and Reducing Agents u The species reduced = oxidizing agent — The species oxidized = reducing agent The strength of an oxidizing agent is directly related to its aims. --> The more positive 8:..." —> the stronger the oxidizing agent (or tendency to be reduced) —> The more negative 89,...“ -> the stronger the reducing agent (or tendency to be oxidized) — Be able to rank different oxidizing and reducing agents given 3...,“ values. Spontaneity of Redox Reactions — if 82.1." > 0. the reaction is spontaneous. — if 8:...“ < 0, the reaction is nonspontaneous; the reverse reaction is spontaneous. Relations Between 3:...‘3 AG'. and K: AG =— nFfl." and AG°=-n F89“)? where n=# of moles of electrons transferred. F=faraday=96.485 J/mol-V "aa' H’I—r/ C’— -: CF5— ELECTROCHEMISTRY Nernst equation: goon = 519"" — "B; In Q At equilibrium at 25’C, 325"" = an or Emu" = logK n n Not aiequillbrlum at 25°C. 3;." =8T=au°— 922M Inc Or 829" =3’ceu°- M logo n Concentration Cells: - Recognize that electrons are transferred from the less concentrated to the more concentrated solution in an effort to equalize the concentration of ions in solution. —)- Oxidation occurs in the cell with the less concentrated solution. -—> Reduction occurs in the cell with the more concentrated solution. Electrolysis: the process of using electrical energy to cause a nonspontaneous redo): reaction Unite to know: faraday (F) = 1 mole of electrons = 96,485 coulombs -_: 6? Q (—0 0 . 1 coulomb = amperes - seconds or 1 C = 1 As 1 joule = volts - coulombs or 1 J = 1 V-C For an electrolytic cell, be able: 1. Identify what is oxidized (reducing agent) 2. Identify what is reduced (oxidizing agent) 3. Write half reactions Quantitative Relationships The mass of product formed (or reactant consumed) at an electrode is directly proportional to 1. the amount of electricity transferred at the electrode 2. the molar mass of the substance BE PREPARED T0 SOLVE PROBLEMS COMBINING CONCEPTS FROM VARIOUS CHAPTERS. ...
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This note was uploaded on 03/15/2011 for the course CHE 215 taught by Professor Aboyousef during the Spring '11 term at American University of Sharjah.

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CHM102-Final Exam Review - fl Properties of Solutions -...

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