2014

Chemistry A

dvan

ced

Plac

emen

t

Integrated Rates

47.9

0

91.2

2

178.

49

(261

)

50.9

4

92.9

1

180.

95

(262

)

52.0

0

93.9

4

183.

85

(263

)

54.9

38

(98)

186.

21

(262

)

55.8

5

101.

1

190.

2

(265

)

58.9

3

102.

91

192.

2

(266

)

H Li Na K Rb

Cs Fr

He

Ne Ar

Kr

Xe

Rn

Ce

Th

Pr

Pa

Nd U

Pm Np

Sm Pu

Eu

Am

Gd

Cm

Tb Bk

Dy Cf

Ho

Es

Er

Fm

Tm Md

Yb

No

Lu Lr

1 3 11 19 37 55 87

2 10 18 36 54 86

Be

Mg

Ca Sr

Ba

Ra

B Al

Ga In Tl

C Si

Ge

Sn

Pb

N P As

Sb Bi

O S Se Te Po

F Cl

Br I At

Sc Y La Ac

Ti Zr Hf

Rf

V Nb Ta Db

Cr

Mo W Sg

Mn Tc Re

Bh

Fe Ru

Os

Hs

Co

Rh Ir Mt

Ni

Pd Pt §

Cu

Ag

Au §

Zn Cd

Hg §

1.00

79

6.94

1

22.9

9

39.1

0

85.4

7

132.

91

(223

)

4.00

26

20.1

79

39.9

48

83.8

0

131.

29

(222

)

140.

12

232.

04

140.

91

231.

04

144.

24

238.

03

(145

)

237.

05

150.

4

(244

)

151.

97

(243

)

157.

25

(247

)

158.

93

(247

)

162.

50

(251

)

164.

93

(252

)

167.

26

(257

)

168.

93

(258

)

173.

04

(259

)

174.

97

(260

)

9.01

2

24.3

0

40.0

8

87.6

2

137.

33

226.

02

10.8

11

26.9

8

69.7

2

114.

82

204.

38

12.0

11

28.0

9

72.5

9

118.

71

207.

2

14.0

07

30.9

74

74.9

2

121.

75

208.

98

16.0

0

32.0

6

78.9

6

127.

60

(209

)

19.0

0

35.4

53

79.9

0

126.

91

(210

)

44.9

6

88.9

1

138.

91

227.

03

58.6

9

106.

42

195.

08

(269

)

63.5

5

107.

87

196.

97

(272

)

65.3

9

112.

41

200.

59

(277

)

4 12 20 38 56 88

5 13 31 49 81

6 14 32 50 82

7 15 33 51 83

8 16 34 52 84

9 17 35 53 85

21 39 57 89

58 90

*Lan

than

ide

Ser

ies:

†Act

inid

e S

erie

s:

* †

59 91

60 92

61 93

62 94

63 95

64 96

65 97

66 98

67 99

68 100

69 101

70 102

71 103

22 40 72 104

23 41 73 105

24 42 74 106

25 43 75 107

26 44 76 108

27 45 77 109

28 46 78 110

29 47 79 111

30 48 80 112

Perio

dic

Tabl

e of

the

Elem

ents

§Not

yet

nam

ed

ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life

ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life

Page 143

GASES, LIQUIDS, AND SOLUTIONS

PV = nRT

PA = Ptotal × XA, where XA = moles A

total moles

Ptotal = PA + PB + PC + . . .

n = mM

K = °C + 273

D = m

V

KE per molecule = 12

mv2

Molarity, M = moles of solute per liter of solution

A = abc

1 1

pressurevolumetemperaturenumber of molesmassmolar massdensitykinetic energyvelocityabsorbancemolarabsorptivitypath lengthconcentration

Gas constant, 8.314 J mol K

0.08206

PVTnm

DKE

Aabc

R

Ã

- -

=============

==

M

1 1

1 1

L atm mol K

62.36 L torr mol K 1 atm 760 mm Hg

760 torr

STP 0.00 C and 1.000 atm

- -

- -====

THERMOCHEMISTRY/ ELECTROCHEMISTRY

products reactants

products reactants

products reactants

ln

ff

ff

q mc T

S S S

H H H

G G G

G H T S

RT K

n F E

qI

t

D

D

D D D

D D D

D D D

=

= -Â Â

= -Â Â

= -Â Â

= -

= -

= -

�

heatmassspecific heat capacitytemperature

standard entropy

standard enthalpy

standard free energynumber of moles

standard reduction potentialcurrent (amperes)charge (coulombs)t

qmcT

S

H

Gn

EIqt

============ ime (seconds)

Faraday’s constant , 96,485 coulombs per moleof electrons1 joule

1volt1coulomb

F =

=

 

Page 144

KINETICS Integrated Rates — Putting it All Together

What I Absolutely Have to Know to Survive the AP Exam

The following might indicate the question deals with kinetics: rate; reactant concentration; order; rate constant; mechanisms; rate determining step; intermediate; catalyst; half-life; instantaneous rate; relative rate; activation energy;

integrated rate law; rate expression; rate law

Instantaneous Rate Instantaneous rate is the rate at any one point in time during the experiment. To find instantaneous reaction rate you find the slope of the curve at the time in question (for those of you in calculus aka…derivative) i.e. the slope of the tangent line to that point in time.

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Kinetics – 2

Integrated Rate – Concentration and Time

Where the differential rate law expresses rate as a function of reactant concentration(s) at an instant in time (hence instantaneous rate), integrated rates express the reactant concentrations as a function of time. To solve integrated rate problems, construct a graph with time on the x-axis and then make 3 plots where the y-axis is

§ Concentration of A [A] vs. t § Natural log of the concentration of A ln [A] vs. t

§ Reciprocal of [A]

1[A]

vs. t

LINEAR IS THE WINNER § Zero Order reaction are linear for [A] vs. t § First Order reactions are linear for ln [A] vs. t

§ Second Order reactions are linear for

1[A]

vs. t

Think y = mx + b It is imperative that you can determine reaction order simply by analyzing a graph. What is important?

§ What is plotted on each axis? § What does the slope of the line indicate?

If you know this, the order and rate constant can easily be determined.

Zero Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]0

When this relationship is integrated from t to tt then… [A]t = −kt + [A]0

y = mx + b

Where… § [A]t is the concentration at time t – [A]t represents what is plotted on the y-axis § [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant - which is the SLOPE of the graph! § Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant

If the reaction is zero order then the graph will be linear with a negative slope.

slope = −k

time

[A]

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Kinetics – 2

First Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]1

When this relationship is integrated from t to tt then… ln [A]t = −kt + ln [A]0

y = mx + b Where…

§ [A]t is the concentration at time t – ln [A]t represents what is plotted on the y-axis § [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant – which is the SLOPE of the graph! § Notice that in this equation k has a negative sign; thus the SLOPE of the graph is negative not the rate constant

If the reaction is first order then the graph will be linear with a negative slope.

slope = −k

time

ln[A

]

Half life… § the time required for the initial concentration, [A]0, to decrease to ½ of its value.

ln [A]t = −kt + ln [A]0 ln 1 = −kt + ln 2

ln 12= −kt

−0.693= −kt0.693

k= t

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Kinetics – 2

Second Order Reactions For the reaction…

A(g) → products

rate = –tA

ΔΔ ][

= k[A]2

When this relationship is integrated from t to tt then…

0

1 1[ ] [ ]

ktA A

= +

y = mx + b

Where…

§ [A]t is the concentration at time t –

1[A]

represents what is plotted on the y-axis

§ [A]0 is the initial concentration § t is the time – represent what is plotted on the x-axis § k is the rate constant – which is the SLOPE of the graph! § Notice that in this equation k is positive; thus the SLOPE of the graph is positive

If the reaction is 2nd order then the graph will be linear with a positive slope.

slope = +k

time

1[A]

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Kinetics – 2

Kinetics Cheat Sheet

Relationships Differential Rate Law (concentration vs. rate data): Rate = k [A]x[B]y Integrated Rate Laws (concentration vs. time data): Zero order [A] = (−)kt + [A]0 First order ln [A] = (−)kt + ln[A]0

Second order 0

1 1 = t + [A] [A]

k

1/20.693tk

= for first order reactions and all nuclear decay

Connections Stoichiometry − “using up” one component of the system might indicate a limiting reactant in effect

Electrochemistry − if reaction is redox in nature, rate problems could come into play

Thermochemistry − Ea and ΔH°rxn and reaction diagrams Potential Pitfalls

Units on k ! Make sure you can solve for units for k

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Kinetics – 2

NMSI SUPER PROBLEM

The decomposition of substance X was experimentally observed at 25°C and shown to be first order with respect to X. Data from the experiment are shown below.

[X] M Time (min)

0.100 0 0.088 2 0.069 6 0.054 10 0.043 14 0.030 ???

time

[X]

time

ln

[X]

(a) For each of the graphs above

i. Sketch the expected curve based on the labeled axes. You do not need to plot the exact data.

ii. Write the rate law for the decomposition of substance X.

iii. Explain how one of the two graphs above can be used to determine the rate constant, k. Be sure to specify which graph.

(b) Based on the above data

i. Calculate the rate constant for this reaction. Be sure to include units. ii. How many minutes will it take for [X] to become 0.030 M?

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Kinetics – 2

In a different experiment, the decomposition of substance Y at 50°C was determined to have the following rate law.

Rate = k [Y]2

time

__________

(c) On the axes above

i. Sketch the graph that is expected to provide a linear relationship when plotted against time. Be sure to label the y-axis.

ii. What does the slope of this line represent?

(d) The temperature of this reaction was increased from 50°C to 100°C. Predict the effect this would have on each of the following.

i. Rate of the reaction

ii. Rate constant, k

(e) Sketch the graph of the reaction at 100°C on the plot in part (c)

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