Chem 106 Thurs. 5-5-2011 Return Exam 3 Review for final exam: kinetics, equilibrium, acid-base

5/5/2011 1

Hour Ex 3; Ave=64, Hi=94

ACS Final exam question types Topic # Calcul’n Qualitative

Intermol forces 4 0 4

Solids 1 0 1

Kinetics 9 4 5

Equilibrium 9 2 7

Acid-base 12 4 8

Aqueous solution 5 2 3

Free Ener/Entropy 11 2 9

Electrochemistry 12 2 10

Nuclear 3 1 2

Other 4

70

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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

Time

Concentr

ation (

mol/lit

er)

0

R P

The rate of the reaction (moles per liter per min)

t

PtimeunitperionconcentratproductinChangerrate

Rate can be defined as (1) Average rate over some time interval Δt

Δt

Δ[P]

Δt

Δ[P]

x

x

(2) Instantaneous rate at some time = slope of a tangent

This one is the initial rate, which is the slope right at time = 0.

or

t

PtimeunitperionconcentratproductinChangerrate

2/22/2011 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] Rate = k[B]2

Rate = k[A][B] Rate = .. Rate = ….

+1 = 2 1

The kinetic order = SUM of exponents of concentration factors in the Rate Law.

2/22/2011

6

15.3e Homework: Determining the Rate Equation - Initial Rates

k[NO_2_][F_2_]

1.16e-4

114

14

22

22

1016.1

64.297.1

1003.6

]][[

]][[

sMxk

MM

sMxk

FNO

ratek

FNOkrate

1 1 4

1 4

2 2

2 2

10 16 . 1

64 . 2 97 . 1

10 03 . 6

] ][ [

] ][ [

s M x k

M M

s M x k

F NO

rate k

F NO k rate

x1 x 2 x 2

x1 x 2 x 2

Rate = k [NO2]x [F2]y

Rate = k [NO2]x [F2]1

Rate = k [NO2]1 [F2]1

2/22/2011

2/24/2011 7

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

Summary of kinetic orders for Chem 106

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

2/24/2011 8

5/5/2011 9

Equilibrium -Is reaction at equilibrium? ( Q vs K) -Properly treat liquids and solids -Kp vs Kc -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.

2 CH2Cl2(g) CH4(g) + CCl4(g)

Initial (M) 0.215 0 0

Change (M)

Equilibrium (M)

+x -2x +x

0.215 - 2x +x +x

2

22

44

][

]][[

reactants

products

ClCH

CClCHKc

2

2

)2215.0(5.10

x

xKc

240.35.10)2215.0( 2

2

x

x240.3

2215.0

x

x

0931.0x

0.0931

0.0931

0.0288

2

2

) 2 215 . 0 ( 5 . 10

x

x K c

240 . 3 5 . 10 ) 2 215 . 0 ( 2

2

x

x

240 . 3 2 215 . 0

x

x 0931 . 0 x

2 2 2

4 4

] [

] ][ [

reactants

products

Cl CH

CCl CH K c

3/3/2011 10

Consider the following system at equilibrium, 2 NO(g) + Br2(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 [Br2]

2. React forward, decreasing [Br2]

3. React reverse direction, increasing [Br2]

4. React reverse direction, decreasing [Br2]

3/3/2011 11

0

5

10

15

20

25

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.

3/3/2011 12

5/5/2011 13

Acid-Base -Ion product of water Kw -define pH and pOH -Ionization constants Ka for acid; Kb for base (hydrolysis) -Remember the strong acids -Calculate pH from weak acid or base using ICE table

H2O(l) + H2O(l) H3O+(aq) + HO-(aq)

AUTOIONIZATION = pure water reacting with itself to a small extent.

[H3O+] = [HO-] = 1.00 x 10-7 M (experimentally determined) To preserve electric neutrality, these must be equal concentrations.

The equilibrium constant for autoionization is Kw , also called the ION PRODUCT of water.

]][OHO[H1

]][OHO[H

reactants

productsK 3

3w

Kw = (1.0 x 10-7)2 = 1.00 x 10-14

] ][OH O [H 1

] ][OH O [H

reactants

products K 3

3 w

[H2O] (l) = 1 because it is the bulk solvent phase. [H2O] is essentially constant at 55 M.

3/8/2011 14

A common way to express acidity (and

basicity) is with “pH”

Define: pH = - log [H3O+]

(= - log10 [H3O+])

In a neutral solution,

[H3O+] = 1.00 x 10-7 at 25 oC

pH = -log (1.00 x 10-7)

= - (-7.000) = 7.000

3/8/2011 15

Other p Scales In general pX = -log10 X

and so pOH = - log [OH-]

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

Take the -log of both sides

-log (10-14) = - log [H3O+] + (-log [OH-])

pKw = 14 = pH + pOH

3/8/2011 16

Acid dissociation constant Ka

acetic acid, CH3CO2H (HOAc)

HOAc(aq) + H2O(l) H3O+(aq) + OAc-(aq)

Acid Conj. Base

(acetate ion)

K a =

[ H 3 O + ] [ O A c - ]

[ H O A c ] = 1 . 8 x 1 0 - 5

3/8/2011 17

Acids Conjugate Bases

Increase acid strength

Ka

3/8/2011 18

Increase base strength

3/8/2011 19

Strong acids you should know

HCl, HBr, HI

HNO3

HClO4

H2SO4

Strong acids ionize completely in water: HCl(aq) + H2O (l) H3O+(aq) + Cl-(aq)

Weak acids ionize partially in water: HF(aq) + H2O (l) H3O+(aq) + Cl-(aq)

3/8/2011 20

Strong bases ionize completely in water: NaOH(aq) Na+(aq) + HO-(aq)

Weak bases ionize partially in water: F-(aq) + H2O (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

3/10/2011 21

Types of Acid/Base Reactions

1. Strong Acid + Strong Base ----> Water + salt .. pH = Neutral

Example: HCl + NaOH ----> H2O + NaCl

Net Ionic Equation: H3O+ + OH- ----> 2 H2O

--------------------------------------------------------------------------------

2. Weak Acid + Strong Base ----> Water + weak base ... pH = Basic

Example: HF + NaOH ----> H2O + NaF

Net Ionic Equation: HF + OH- ----> H2O + F-

(Conjugate base of a weak acid = Basic.)

--------------------------------------------------------------------------------

3. Strong Acid + Weak Base ----> Water + Weak Acid ... pH = Acidic

Example: HCl + F- ----> HF

Net Ionic Equation: H3O+ + F- ----> HF + H2O

(Conjugate acid of a weak base = Acidic.)

--------------------------------------------------------------------------------

4. Weak Acid + Weak Base ----> Weak Acid + Weak Base ..

Net Ionic Equ: HCN + F- ----> HF + CN-

( pH depends on Ka’s ) -------------------------------------------------------------------------------

3/10/2011 22

A 0.015 M solution of hydrogen cyanate HOCN has a pH = 2.67. What is Ka for HOCN?

HOCN + H2O H3O+ + OCN-

I 0.015 0 0 C E

[H3O+] = 10-2.67 = 0.00214 M

0.00214 M 0.00214 M

+0.00214 M +0.00214 M -0.00214 M

0.01286 M

= 3.56 x 10-4

5/5/2011 23

Other Aqueous Equilibria - Henderson-Hasselbalch equation and buffers - Solubility Product, Ksp

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 + H2O A- + H3O+

B + H2O BH+ + OH-

(Notice that either equilibrium can absorb H+ or OH- ions)

3/22/2011

3/22/2011 25

Logic of the H-H equation

Say [A-] = [HA] Then log(1) = 0

pH = pKa + 0 = pKa

or [H3O+] = Ka

Say 2/3 [A-] and 1/3 [HA] Then log(2/1) = 0.30 pH = pKa+ 0.30

or [H3O+] < Ka

Say 1/5 [A-], 4/5 [HA] Then log(1/4) = -0.60 pH = pKa - 0.60

or [H3O+] > Ka

][

][log

HA

ApKpH

a

Please remember this equation and know how to use it.

3/22/2011 26

Say you prepared a buffer using equal moles of sodium nitrite NaNO2 and nitrous acid HNO2 (Ka = 3.2 x 10-4). What is the pH of the resulting

solution? (Try using pencil and paper, and not a calculator…)

1. 3.0

2. 3.5

3. 4.2

4. 4.5

pH = pKa + log(A-/HA) = pKa

pH = -log(3.2 x 10-4) pH = -[log3.2 + log(10-4)] pH = -(~0.5 - 4) pH = -(-3.5) = 3.5

0

2

4

6

8

10

12

14

0.00 0.50 1.00 1.50

pH

Mol OH-/mol HA

Acid 2

Titrating a weak acid (1.0 M, Ka = 1 x 10-6) with strong base NaOH

“pH of weak acid”

“buffer of weak acid & its conjugate base”

“pH of conjugate base”

“pH of NaOH solutions”

“*weak acid+=*conjugate base+ pH = pKa” 3/22/2011 27

titration equivalence pt

HA + OH- A- + H2O