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Science NATIONALMATH + SCIENCEINITIATIVE

AP CHEMISTRY

Gas Laws

Presenter Name: ______________________________Please write down the name of the presenter for this session.You will need the title of the session and presenter's name forevery session you attend to complete the survey you will bereceiving by email later today.If you complete the survey by Tuesday morning, you will beentered for a chance to win a $25 Amazon gift card!

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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

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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

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ide

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ies:

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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

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e of

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ents

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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 mercuryg = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s)atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hνc = 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

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 =

=

 

AP*Chemistry: Gases

These terms may be used in a question testing your understanding of GASESpressure; volume; moles; molecular mass; temperature; total pressure; diffusion; effusion; ratio of gases; any non standard temperature/pressure conditions involving gases; molecular motion; kinetic energy; collected over water; O2; N2: CO2; dry gas; root mean square

Key Formulas or RelationshipsPV = nRT (this is the most commonly used…try it first)

L atmR 0.0821mol K

= or for ENERGY related calculations JR =8.31mol K

dRTMM=P

1 1 2 2

1 2

P V P V=

T T Solve for P, V, or T as needed; omit any variable that is constant

Ptotal = P1 + P2 … or Pgas = Ptotal – Pwater when a gas is collected over water Also in a mixture, the partial pressure of gas1 = mole fraction of gas1 × total pressure

STP for gases is 0°C or 273 K and 1 atm or 760 mmHg

Key ConceptsRemember to relate molecular mass to molecular motion (smaller = faster) especially when talking about diffusion or movement of gases.

Molar volume 1 mol = 22.4 L at STP only. Don’t use 22.4 if not at STP!

Understand conceptually that variables are either direct (P/T and V/T) and inverse (P/V) relationships

Van der Waals equation – adjusts ideal gas equation for IMF (a factor) and particlevolume (b factor) Gases will deviate more from ideal behavior at high pressure andlow temperature. Also, more electrons = greater IMFs and particle volume and thus more deviation from ideal behavior [key phrase to use for points!].

Connections to Other ChaptersStoichiometry ThermodynamicsEquilibrium Bonding

Potential Pitfalls and Unit WarningsKELVINS, KELVINS, KELVINS Make sure units match with constants“Collected over water” -- adjust P by subtracting vapor pressure of water at the given temperature before continuing with calculations

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Science NATIONALMATH + SCIENCEINITIATIVE

Figures or Graphs you may need to interpret

o The data Boyle collected isgraphed on (a) above.

o PV = k

o Therefore,P

kPkV 1==

which is the equation for a straight line of the type

o y = mx + b wherem = slope, and b is the y-intercept

o In this case, y = V, x = 1/Pand b = 0. Check out theplot on the right (b).

o P1V1 = P2V2 is the easiestform of Boyle’s law tomemorize

o Boyle’s Law has been testedfor over three centuries. Itholds true only at lowpressures.

As temperature increases the average speed of molecules increases but the variation in molecule speeds decreases

An ideal gas is expected to have a constant value of PV, as shown by the dotted line. CO2shows the largest change in PV, and this change is actually quite small: PV changes from about 22.39 L·atm at 0.25 atm to 22.26 L·atm at 1.00 atm. Thus Boyle’s Law is a good approximation at these relatively low pressures.

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Science

Jacques Charles was a French physicist and the first person to fill a hot “air” balloon with hydrogen gas and made the first solo balloon flight! o V ∝ T plot = straight lineo V1T2 = V2T1

o Temperature ∝ Volume at constant pressureo This figure shows the plots of V vs. T (Celsius) for several gases. The solid lines represent

experimental measurements on gases. The dashed lines represent extrapolation of the data intoregions where these gases would become liquids or solids. Note that the samples of the variousgases contain different numbers of moles.

o What is the temperature when the Volume extrapolates to zero?

These graphs are also classics and make great multiple choice questions on the AP exam.

When PV/nRT = 1.0, the gas is ideal This graph is just for nitrogen gas. All of these are at 200K. Note that although nonideal behavior Note the P’s where the curves is evident at each temperature, the cross the dashed line [ideality]. deviations are smaller at the higher Ts.

Don’t underestimate the power of understanding these graphs. We love to ask question comparing the behavior of ideal and real gases. It’s not likely you’ll be asked an entire free-response gas problem on the real exam in May. Gas Laws are tested extensively in the multiple choice since it’s easy to write questions involving them! You will most likely see PV = nRT as one part of a problem in the free-response, just not a whole problem!

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Science

2004B

Answer the following questions related to hydrocarbons.

(a) Determine the empirical formula of a hydrocarbon that contains 85.7 percent carbon by mass.

(b) The density of the hydrocarbon in part (a) is 2.0 g L–1 at 50°C and 0.948 atm.(i) Calculate the molar mass of the hydrocarbon.

(ii) Determine the molecular formula of the hydrocarbon.

(c) Two flasks are connected by a stopcock as shown below. The 5.0 L flask contains CH4 at a pressure of 3.0 atm, andthe 1.0 L flask contains C2H6 at a pressure of 0.55 atm. Calculate the total pressure of the system after the stopcock isopened. Assume that the temperature remains constant.

(d) Octane, C8H18(l), has a density of 0.703 g mL–1 at 20°C. A 255 mL sample of C8H18(l) measured at 20°C reactscompletely with excess oxygen as represented by the equation below.

2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

Calculate the total number of moles of gaseous products formed.

7

Science

2002B

A rigid 8.20 L flask contains a mixture of 2.50 moles of H2, 0.500 mole of O2, and sufficient Ar so that the partial pressure of Ar in the flask is 2.00 atm. The temperature is 127ºC.

(a) Calculate the total pressure in the flask.

(b) Calculate the mole fraction of H2 in the flask.

(c) Calculate the density (in g L−1) of the mixture in the flask.

The mixture in the flask is ignited by a spark, and the reaction represented below occurs until one of the reactants is entirely consumed.

2 H2(g) + O2(g) → 2H2O(g)

(d) Give the mole fraction of all species present in the flask at the end of the reaction.

8

Science

2009B

2 H2O2(aq) → 2 H2O(l) + O2(g)

The mass of an aqueous solution of H2O2 is 6.951 g. The H2O2 in the solution decomposes completely according to the reaction represented above. The O2(g) produced is collected in an inverted graduated tube over water at 23.4°C and has a volume of 182.4 mL when the water levels inside and outside of the tube are the same. The atmospheric pressure in the lab is 762.6 torr, and the equilibrium vapor pressure of water at 23.4°C is 21.6 torr.

(a) Calculate the partial pressure, in torr, of O2(g) in the gas-collection tube.

(b) Calculate the number of moles of O2(g) produced in the reaction.

(c) Calculate the mass, in grams, of H2O2 that decomposed.

(d) Calculate the percent of H2O2, by mass, in the original 6.951 g aqueous sample.

(e) Write the oxidation number of the oxygen atoms in H2O2 and the oxidation number of the oxygen atomsin O2 in the appropriate cells in the table below.

(f) Write the balanced oxidation half-reaction for the reaction.

9

Science