Return Exam 3 Review for final exam: kinetics, equilibrium, acid-base

Transcription

1 Chem 106 Thurs Return Exam 3 Review for final exam: kinetics, equilibrium, acid-base Hour Ex 3; Ave=64, Hi=94 5/5/2011 1

2 ACS Final exam question types Topic # Calcul n Qualitative Intermol forces Solids Kinetics Equilibrium Acid-base Aqueous solution Free Ener/Entropy Electrochemistry Nuclear Other /5/2011 2

3 Kinetics 1) Obtaining rate law from initial rate data 2) Using integrated rate equations (these are given) 3) Using the Arrhenius equation (this is given) 4) including definitions of activation energy 5) Chemical mechanisms, intermediates, catalysts 5/5/2011 3

4 Concentration (mol/liter) R P The rate of the reaction (moles per liter per min) rate r Change in product concentration per unit time P t x Δ[P] Δt x Δ[P] Rate can be defined as (1) Average rate over some time interval Δt or (2) Instantaneous rate at some time = slope of a tangent 2/22/ Δt Time This one is the initial rate, which is the slope right at time = 0. 4

5 There are many known rate laws. We will study only THREE: zero order rate = k*r+ 0 =k first order rate = k[r] 1 second order rate = k*r+ 2 With more complex stoichiometry, any of these might apply, given the weird things that can happen in chemistry! For example,for: A + B C + D These are known for different A,B,C,D: Rate = k Rate = k[b] 1 +1 = 2 Rate = k[b] 2 Rate = k[a][b] Rate =.. Rate =. The kinetic order = SUM of exponents of concentration factors in the Rate Law. 2/22/2011 5

6 Rate = k [NO 2 ] 1 [F 2 ] e Homework: Determining the Rate Equation - Initial Rates Rate = k [NO 2 ] x [F 2 ] y x1 x 2 x 2 x 2 x1 x 2 rate k [ NO rate k [ NO ][ F ] k k x 10 4 ][ F 4 2 ] x 10 M s M M M s 1 2/22/2011 6

7 Summary of kinetic orders for Chem 106 Order Rate = -d[r]/dt = Integrated rate law (linear form) y = mx + b Alternate form Zero k [R] = -kt + [R] o First k[r] ln[r] = -kt + ln[r] o Second k[r] 2 2/24/2011 7

8 Intermediate. Product of 1 st step Reactant in 2 nd step Cancels out when you add 1st and 2nd steps (example C + reaction) Catalyst. Reactant in 1 st step Product of 2 nd step Cancels out when add 1 st and 2 nd steps (example H + reaction) 2/24/2011 8

9 Equilibrium -Is reaction at equilibrium? ( Q vs K) -Properly treat liquids and solids -K p vs K c -Calculate concentrations after equilibrium reached (ICE table) -Use Le Chatelier s principle to predict shifts in equilibrium with adding or removing reactants/products, changing pressure, or temperature. 5/5/2011 9

10 CH 2 Cl 2 (g) CH 4 (g) + CCl 4 (g) Initial (M) Change (M) Equilibrium (M) -2x +x +x x +x +x K c K c products reactants [ CH 4 ][ CCl [ CH Cl ] x ( x ) 2 2 ] 2 2 x ( x ) x x x /3/

11 Consider the following system at equilibrium, 2 NO(g) + Br 2 (g) 2 NOBr(g) If the volume of the system was suddenly increased, how would the system respond to re-establish equilibrium? 1. React forward, increasing [Br 2 ] 2. React forward, decreasing [Br 2 ] 3. React reverse direction, increasing [Br 2 ] 4. React reverse direction, decreasing [Br 2 ] /3/

12 Rule of thumb For an endothermic reaction, if T increases, remove heat by increasing K (makes more product) if T decreases, add heat by decreasing K For an exothermic reaction, if T increases, remove heat by decreasing K if T decreases, add heat by increasing K. 3/3/

13 Acid-Base -Ion product of water Kw -define ph and poh -Ionization constants K a for acid; K b for base (hydrolysis) -Remember the strong acids -Calculate ph from weak acid or base using ICE table 5/5/

14 AUTOIONIZATION = pure water reacting with itself to a small extent. H 2 O(l) + H 2 O(l) H 3 O + (aq) + HO - (aq) [H 3 O + ] = [HO - ] = 1.00 x 10-7 M (experimentally determined) To preserve electric neutrality, these must be equal concentrations. The equilibrium constant for autoionization is K w, also called the ION PRODUCT of water. products [H 3 O ][OH ] K w [H 3 O reactants 1 K w = (1.0 x 10-7 ) 2 = 1.00 x ][OH ] [H 2 O] (l) = 1 because it is the bulk solvent phase. [H 2 O] is essentially constant at 55 M. 3/8/

15 A common way to express acidity (and basicity) is with ph Define: ph = - log [H 3 O + ] (= - log 10 [H 3 O + ]) In a neutral solution, [H 3 O + ] = 1.00 x 10-7 at 25 o C ph = -log (1.00 x 10-7 ) = - (-7.000) = /8/

16 In general Other p Scales px = -log 10 X and so poh = - log [OH - ] K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C Take the -log of both sides -log (10-14 ) = - log [H 3 O + ] + (-log [OH - ]) pk w = 14 = ph + poh 3/8/

17 Acid dissociation constant K a acetic acid, CH 3 CO 2 H (HOAc) HOAc(aq) + H 2 O(l) H 3 O + (aq) + OAc - (aq) Acid Conj. Base (acetate ion) K a = [ H 3 O + ] [ O A c - ] [ H O A c ] = 1. 8 x /8/

18 Acids K a Conjugate Bases Increase acid strength Increase base strength 3/8/

19 Strong acids ionize completely in water: HCl(aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) Weak acids ionize partially in water: HF(aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) Strong acids you should know HCl, HBr, HI HNO 3 HClO 4 H 2 SO 4 3/8/

20 Strong bases ionize completely in water: NaOH(aq) Na + (aq) + HO - (aq) Weak bases ionize partially in water: F - (aq) + H 2 O (l) HF(aq) + HO - (aq) Strong bases you should know Group I Group II LiOH NaOH KOH Ca(OH) 2 (s) slightly soluble Sr(OH) 2 Ba(OH) 2 3/8/

21 Types of Acid/Base Reactions 1. Strong Acid + Strong Base ----> Water + salt.. ph = Neutral Example: HCl + NaOH ----> H 2 O + NaCl Net Ionic Equation: H 3 O + + OH- ----> 2 H 2 O Weak Acid + Strong Base ----> Water + weak base... ph = Basic Example: HF + NaOH ----> H 2 O + NaF Net Ionic Equation: HF + OH > H 2 O + F - (Conjugate base of a weak acid = Basic.) Strong Acid + Weak Base ----> Water + Weak Acid... ph = Acidic Example: HCl + F > HF Net Ionic Equation: H 3 O + + F > HF + H 2 O (Conjugate acid of a weak base = Acidic.) Weak Acid + Weak Base ----> Weak Acid + Weak Base.. Net Ionic Equ: HCN + F > HF + CN - ( ph depends on K a s ) /10/

22 A M solution of hydrogen cyanate HOCN has a ph = What is K a for HOCN? [H 3 O + ] = = M HOCN + H 2 O H 3 O+ + OCN - I C M M M E M M M = 3.56 x /10/

23 Other Aqueous Equilibria - Henderson-Hasselbalch equation and buffers - Solubility Product, K sp 5/5/

24 Definition A buffer is an aqueous solution containing a mixture of a weak acid and its conjugate base, (or a weak base and its conjugate acid). The function of a buffer is to absorb H + or OH - ions, minimizing the change in ph that might otherwise occur. HA + H 2 O A - + H 3 O + B + H 2 O BH + + OH - (Notice that either equilibrium can absorb H + or OH - ions) 3/22/

25 Logic of the H-H equation ph pk a [ A ] log [ HA ] Please remember this equation and know how to use it. Say 1/5 [A - ], 4/5 [HA] Then log(1/4) = ph = pk a Say [A - ] = [HA] Then log(1) = 0 ph = pk a + 0 = pk a Say 2/3 [A - ] and 1/3 [HA] Then log(2/1) = 0.30 ph = pk a or [H 3 O + ] > K a or [H 3 O + ] = K a or [H 3 O + ] < K a 3/22/

26 Say you prepared a buffer using equal moles of sodium nitrite NaNO 2 and nitrous acid HNO 2 (K a = 3.2 x 10-4 ). What is the ph of the resulting solution? (Try using pencil and paper, and not a calculator ) ph = pk a + log(a-/ha) = pk a ph = -log(3.2 x 10-4 ) ph = -[log3.2 + log(10-4 )] ph = -(~0.5-4) ph = -(-3.5) = 3.5 3/22/

27 ph Titrating a weak acid (1.0 M, K a = 1 x 10-6 ) with strong base NaOH HA + OH - A - + H 2 O 14 Acid 2 ph of NaOH solutions buffer of weak acid & its conjugate base ph of conjugate base ph of weak acid *weak acid+=*conjugate base+ ph = pk a Mol OH-/mol HA titration equivalence pt 3/22/