Semester 2 Midterm Review (Bluffer s Guide) *Adapted from Paul Groves [South Pasadena HS]

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1 Semester 2 Midterm Review (Bluffer s Guide) *Adapted from Paul Groves [South Pasadena HS] Unit 5: Kinetics How to talk about Reaction Rate rate = Δ[chemical]/Δtime - Common Units: M/s, mol L - s rate of disappearance of reactant or rate of appearance of product use coefficients to change one rate to another Reaction: 2A + 3B 4C Δ[A] Δ[B] Δ[C] = 2 = 3 4 Δt Δt Δt watch your signs (Δ[React.] = -Δ[Prod.]) Rate law can be determined from initial rates. See Example 5.3 and Exercise 5.3 Rate Law matches the Molecularity of the Rate Determining Step in the Mechanism Examples for: 2A + 3B C (fill in from lecture) Rate Determining Step Rate Law in the mechanism Rate = k [A][B] A + B X (slow) How to Speed Up a Reaction [Use Collision Theory, Kinetic Molecular Theory] increase the concentration of reactants - increase molarity of solutions - increase partial pressure of gases [collision model: more collisions] more surface area between unlike phases [collision model: more collisions] increase the temperature [collision model: more & harder collisions] add a catalyst [collision model: alternate mechanism that requires lower energy collision or ensures that correct particles collide] Rate = k [A] 2 Rate = k [A] 2 [B] Rate = k A + A X (slow) A + A D X (fast) B + X Y (slow) Each step is usually bimolecular. A third order overall reaction often comes from a fast equilibrium before a slow step. This could be a mechanism that depends on a catalyst only. The concentrations would not matter. Because Rate depends on Concentration Rate Laws - what they mean - how to determine them - how they relate to the rate determining step - how they help you choose a mechanism General Form: Equation: A + B C Rate = k [A] x [B] y k is the specific rate constant Use experimental data to determine x, y, and k. order of rxn - first and second order reactions - what these look like graphically - how you can graphically tell the order of a reaction order straight-line plot Slope 0 [R] t vs. t -k ln[r] t vs. t -k 2 /[R] t vs. t k - how this relates to the rate law The Rate Law CANNOT be determined from the overall reaction. It MUST be determined experimentally because the rate law reflects only the rate determining step.

2 Two Important Diagrams PE energy profile of a reaction PE a b c reaction coordinate H of the reaction relates reactant and product PE s / exo- or endothermic/ downhill, -ΔH, or uphill, +ΔH activation energy (E a ) = energy barrier activated complex (at the peak) whether a reaction is fast or slow depends on the activation energy in the PE profile PE profile does not change with change in temperature of the reactants? adding a catalyst lowers the E a d e Reaction mechanisms - step-by-step...two particles at a time - example overall: 4 HBr + O 2 2 Br H 2 O mechanism: HBr + O 2 HOOBr HOOBr + HBr 2 HOBr HOBr + HBr Br 2 + H 2 O HOBr + HBr Br 2 + H 2 O [note: HOOBr and HOBr are not in the overall reaction because they are neither reactants nor products, they are reactive intermediates ] - overall reaction is sum of steps - slowest step is rate-determining step half-life - relationship to radioactivity (a first order reaction) - the equation [ A] ln o = kt [ A] t - the special case of half-life ln(2) = = kt ½ 2

3 Unit 6: Equilibrium aa +bb +... rr +ss +... r s [R] [S] K c = a b [A] [B] and for gases: r (PR ) (PS ) K p = a (P ) (P ) A 2. K > products favored K < reactants favored B s b 3. Excluded: solids; pure liquids; water (in aqueous solutions) because their [ ] s do not change. 5. Typical question: Given K c and the starting concentrations of reactants, find concentrations of products at equilibrium. Example: K c for acetic acid =.8 x 0-5. What is the equilibrium concentration of [H + ] in a 0.00 M solution of the acid? 6. Equilibrium constant for a reverse reaction = the value of the forward reaction. K 7. Equilibrium constant for a doubled reaction = K When using Hess s Law: K overall = K x K 2. Problem solving: Set up problems using the ICE box C = change or Δ. Example: A B + C A B C I 5.0 M 0 M 0 M C E E row only follows the stoichiometry of the equation. Learn when to make an approximation (needed for multiple choice questions!) 5% rule usually works when value of K is 0 3 smaller than value of known concentrations. Example: A B + C K = 3.0 x 0-6 if [A] = 5.0M initially; find [C] at equilibrium. If greater than 5% use the quadratic equation: (not usual on the AP exam) ax 2 + bx + c = 0 x = b ± b 2a 2 4ac Another easy to solve situation is the perfect squares situation. Example: H 2 + I 2 2HI K = 3.5 x 0 2 Calculate [HI] when [H 2 ] = [I 2 ] = 0.0 M 9. Le Châtelier s Principle: effect of changes in concentration, pressure, & temperature. Equilibrium always shifts away from what you add. Stress means too much or too little: chemical, heat, or room. 0. If out of equilibrium: Calculate the reaction quotient (Q) similar to the way an equilibrium constant would be found. If: Q < K forward reaction occurs to reach equilibrium Q > K reverse reaction occurs to reach equilibrium 3

4 Unit 6 Cont Acids and Bases. H 2 O H + + OH K w = [H + ][OH ] = 0 4 ph = -log[h + ] ph+poh = 4 [H+] =0 ph Convert between ph, poh, [H + ], & [OH ] 2. Acid Ionization Constant (K a ): HA + H 2 O H 3 O + + A - K a = [A - ][H 3 O + ]/[HA] Example: HF + H 2 O H 3 O + + F - K a = [F - ][H 3 O + ]/[HF] 0. Acid Strength-know the 6 strong acids: HCl, HBr, HI, HNO 3, HClO 4, and H 2 SO 4 (removal of the first H + only) (a) binary acids - acid strength increases with increasing size and electronegativity of the other element. ( NOTE: Size predominates over electronegativity in determining acid strength.) Examples: H 2 Te > H 2 O & HF > NH 3 3. Typical question: Given K a and the starting concentrations of acid, find concentrations (or ph) of [H + ] at equilibrium. Example: K a for acetic acid =.8 x 0-5. Find the ph of 0.00M acetic acid. 4. Polyprotic Acids: H 3 PO 4, H 2 SO 4, H 2 C 2 O 4, etc. The st dissociation is strong for H 2 SO 4. When using Hess s Law with a polyprotic acid: K overall = K a x K a2 Calculating ph, use K a 5. Bronsted-Lowry Definitions. Acids = H+ donors; Bases = H+ acceptors Conjugate acid-base pairs. 6. Base Ionization Constant (K b ): B + H 2 O BH + + OH - K b = [BH + ][OH - ]/[B] Example: F - + H 2 O HF + OH - K b = [HF][OH - ]/[F - ] 7. Salt solns can have ph s 7 (hydrolysis) ions from weak acids basic solutions C 2 H 3 O 2 + H 2 O HC 2 H 3 O 2 + OH ions from weak bases acidic solutions NH H 2 O NH 4 OH + H + 8. K a x K b = K w = 0-4 only applies for conjugate acids & bases! Example: K a HC 2 H 3 O 2 =.8 x 0-5 K b C 2 H 3 O 2 - = 0-4 /.8 x 0-5. Buffers: A solution that resists a change in ph when small amounts of acid or base are added. Buffers are a mixture of a weak acid & its conjugate base or a weak base & its conjugate acid. Examples: HC 2 H 3 O 2 & C 2 H 3 O 2 - or NH 3 & NH 4 + Your blood is a buffer. The equilibrium is the acid equilibrium: HA + H 2 O H 3 O + + A- in which both the acid and its conjugate base are available to counteract the stress of adding acid or base (Le Châtelier s). The equilibrium shifts, but is almost completely counteracted by the proton donor or acceptor. Similarly, for bases with conjugate acids: B + H 2 O HB+ + OH - The best buffer contains large, equal amounts of the proton donor and the proton acceptor. [HA] = [A - ] (Note: they cancel out of K a ) 4

5 Consider: HA H + + A [H ][A ] K a = [HA] The best buffer, K a = [H + ]; ph = pk a. The ph of a buffer can be adjusted by changing the ratio of acid and base. [H + [HA] ] = K a - [A ] [HA] ph = pk a - log - [A ] Similarly, for bases with conjugate acids: B + H 2 O HB + + OH - + [OH ][HB ] K b = [B] [OH - [B] ] = K b + [HB ] [B] poh = pk b - log + [HB ] Buffers can also be formed by changing a weak acid into its conjugate base by neutralizing some of the acid.. HA + OH - H 2 O + A - The same can be done with a weak base: B + H + HB + So, a weak acid and some strong base can form a buffer. A weak base and some strong acid can also form a buffer. 2. Titration: A carefully measured neutralization. Acid + Base H 2 O + Salt 3. Equivalence Point: The point in a titration when stoichiometric amounts of acid and base have reacted. Note that the salt solution that is formed may have a ph >, <, or = 7. (Remember hydrolysis.) Indicators for a titration are selected based on the ph at the equivalence point. acid base ph at eq. pt. STRONG STRONG ph = 7 STRONG weak ph < 7 weak STRONG ph > 7 weak weak it depends 4. Titration Curves: This graph shows how the ph changes as a titration occurs. (A) Strong acid/strong Base HCl + NaOH H 2 O + NaCl NOTE: Graph should have ph as the vertical axis and added base as the horizontal axis. The graph should be in an S shape. The middle of the S is the equivalence point (ph = 7). The top part of the S levels off at the ph of the base solution. Since volumes are measured, this is a volumetric analysis. HA + OH - H 2 O + A - The salt is the conjugate base of the weak acid or the conjugate acid of the weak base. A half-titration (neutralizing half the weak acid or weak base) forms a buffer. In that case the ph = pk a of the acid. (poh = pk b ) 5

6 (B) Weak acid/strong Base HA + OH - A - + H 2 O (D) Weak diprotic acid/strong base H 2 A + OH - HA - + H 2 O HA - + OH - A 2- + H 2 O NOTE: The middle of the lower part of the S indicates the point of maximum buffering where [HA]/[A-] =. The middle of the S is the equivalence point (above ph = 7) and [HA] = 0. The top part of the S levels off at the ph of the base solution. (C) Weak base/strong acid B + H 3 O + BH + + H 2 O NOTE: Graph should have ph as the vertical axis and added acid as the horizontal axis. The graph should be in a backwards S shape. The middle of the upper part of the backwards S indicates the point of maximum buffering where [B]/[HB + ] =. The middle of the backwards S is the equivalence point (below ph = 7) and [B] = 0. The bottom part of the backwards S levels off at the ph of the acid solution. NOTE: Graph should have ph as the vertical axis and added base as the horizontal axis. The graph should be in a double S shape. The middle of the lower part of the first S indicates the point of maximum buffering of the first buffering zone where [H 2 A]/[HA - ] =. The middle of the first S is the first equivalence point where [H 2 A] = 0. The top part of the first S (i.e. the lower part of the second S ) indicates the point of maximum buffering of the second buffering zone where [HA - ]/[A 2- ] =. The middle of the second S is the second equivalence point where [HA - ] = 0. The top part of the second S levels off at the ph of the base solution. 6