Transcript File

PCAT Prep
Chemistry Section
Laura Blue
Angie Proctor
Introduction to the Chemistry Section
Details of the Chemistry Section
• Format: 48 Multiple Choice Questions
• Time: 30 minutes
• Time/Question…37.5 seconds
Test Taking Strategies
• Eliminate and Simplify
– Since you are unable to use a calculator, try to eliminate choices by estimating a
probable order of magnitude for the answer or simplifying the math involved
•
Put it on Paper
– Although you are limited on time, do not be afraid to write down equations,
periodic trends, etc at the start of the section that may help you work through
some of the problems
•
Know the Trends
– Chemistry is all about knowing a little and applying it to a lot of applications.
•
Don’t know the Answer-GUESS
– There is NO penalty for wrong answers
•
•
Stay Calm and Confident
Keep Track of Time
Atomic Structure
Atomic Structure
Mass Number
A
X
Z
Element
Symbol
23
11
Na
Atomic Number
Mass Number = # of protons + # of neutrons
Atomic Number = # of protons
# electrons = Atomic Number ± charge
# neutrons = Mass Number - Atomic Number
Atomic Structure and Ions
23
11
35
Na
17
Cl
Neutral Atom
Neutral Atom
# protons = 11
# protons = 17
# electrons = 11
# electrons = 17
# neutrons = 12
# neutrons = 18
Na  Na+ + e-
Cl + e-  Cl-
Charged Atom-Cation
Charged Atom-Anion
# protons = 11
# protons = 17
# electrons = 10
(Minus 1 electron)
# neutrons =12
# electrons = 18
Na+
Cl-
(Plus 1 electron)
# neutrons =18
Isotopes
Isotopes - atoms of the same element with the same number of
protons but a different numbers of neutrons
23
Example:
11
Na
and
24
11
Na
Atomic Mass Unit (amu) – relative mass of element compared
to mass of the C12; each isotope has a different amu
Molecular weight (MW) – weighted average of relative
abundances of different isotopes of element
MW of Element Z = xA(amu A) + xB(amu B) +…
where:
x = fraction of composition or relative abundance
A,B, etc. = atomic mass of each isotope of Z
Bohr Model
An electron can exist only in certain fixed energy states
e- can be excited to a higher energy level
Excited state
return to ground state
n=5
n=4
n=3
n=2
n=1
Absorbs
Energy/
Photons
release energy
Emits
Energy/Photons
eˉ
Energy (hc/)
434.1nm (Violet)
656.3 nm (Red)
eˉ
486.1 nm
(Bluegreen)
Wavelength (nm)
Quantum Mechanical Model of Atoms
• Heisenberg Uncertainty Principle-Unable to calculate both the momentum
and location of an electron in an atom simultaneously
• Electrons travel in diffuse clouds or orbitals around the nucleus described
by probability distributions
• Quantum Numbers are used to describe the location of each electron in an
atom
–
–
–
–
n  Principle Quantum Number
ɭ  Angular Momentum Quantum Number
mɭ  Magnetic Quantum Number
ms  Spin Quantum Number
• Pauli exclusion principle – no 2 e- in a given atom can have the same
set of quantum numbers
Principal Quantum Number (n)
•Integer Value (n = 1, 2, 3….)
•Related to the size and energy of the orbital or shell
-Larger n: higher the energy of electron and larger radius of electron
•Maximum number of electrons in an energy level = 2n2
Energy
n=1
n=2
n=3
Angular Momentum Quantum Number (ɭ )
• Integral value from 0 to n-1 for each value of n
ɭ
Subshell
• Corresponds to the shape of the orbital
0
s
1
p
2
d
3
f
• Maximum number of electrons in subshell = 4 ɭ + 2
p-orbitals
s-orbitals
d-orbitals
f-orbitals
Magnetic Quantum Number (mɭ )
• Values range from – ɭ to + ɭ including 0
• Describes the orientation of an orbital in space
Subshell
ɭ
mɭ
s
0
0
p
1
-1, 0, +1
d
2
-2, -1, 0, 1, 2
f
3
-3, -2, -1,
0, 1, 2, 3
Spin Quantum Number (ms)
• Denoted as + ½ or - ½ electron spin
• Electron spin is based on the behavior of electrons as tiny
magnets.
+½
e-
-½
+

+½ -½
e-
Electron Configuration-Key Concepts
• Aufbau Principle
– Electrons enter orbitals of lowest energy first
– s subshell is always the lowest in energy
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
Increasing Energy
• Hund’s Rule
– When electrons occupy orbitals of equal energy one electron enters each orbital until all the
orbitals contain one electron with spins parallel, then electrons are paired
2px 2py 2pz
Quantum Numbers-Practice Problem
What are the four quantum numbers for the 28th electron in Nickel?
1. Write the Electron Configuration
1s2 2s2 2p6 3s2 3p6 4s2 3d8
2. Determine Quantum Numbers
n =3
mɭ = 0
ɭ =2
ms = -1/2
S
0 0
d
-2 -1 0
1
2
p
-2
-1 0
1
2
Quantum Numbers - Practice Question
Which of the following combinations of quantum numbers is allowed
for a single electron in an element?
A. n = 2, ɭ = 2, mɭ = 1, ms = ½
Remember:
B. n = 3, ɭ = 1, mɭ = 0, ms = -½
ɭ = (0  n-1)
mɭ = (- ɭ  ɭ )
C. n = 5, ɭ = 1, mɭ = 2, ms = ½
D. n = 4, ɭ = -1, mɭ = 0, ms = ½
ms = +½ or -½
Quantum Numbers - Practice Question
The maximum number of electrons in a shell with the
principal quantum number equal to 4 is:
A. 2
B. 10
Remember:
Max # e- = 2n2
C. 16
D. 32
Electron Configuration - Practice Question
Within one principal quantum level of a many electron atom,
which orbital has the lowest energy?
A.
B.
C.
D.
s
p
d
f
1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6
Increasing Energy
What is the number of half-filled orbitals in one ground-state
atom of sulfur (atomic number 16)?
A.
B.
C.
D.
2
3
4
6
1s2 2s2 2p6 3s2 3p4
Sum Superscripts = 16 = # electrons
3px 3py 3pz
The Periodic Table
Atomic Properties and Electronic Structure
• Positively charged nucleus attracts
negatively charged electrons
• Negatively charged electrons repel other
electrons
Atomic Properties and Electronic Structure
Effective Nuclear Charge (Zeff)
Net positive charge experienced by an electron in a multielectron atom
Calculating zeff:
Zeff = Z – S
Z = Number of Protons
S = Number of Electrons in inner shells
Generally Zeff increases when…
-Atomic Radii Decreases
-Electronegativity Increases
-Ionization Energy Increases
Atomic Radius
Atomic Radius Increasing
Atomic Radius Increasing
Atomic Radius
Moving Across a Period:
• Increasing number of protons
• Same size electronic shell
• Greater Zeff
• Size gets smaller
Moving Down a Group:
• Larger n means a larger orbital
• More shielding of nuclear charge
• Overall size is larger
Trends in Ion Size
• Adding Electrons Forms Anions
– Negative ions are larger than the atoms from
which they are formed
– More electrons means increased repulsions
– Occupy a larger volume
ee-
e- e-
e-
Cl
e
e- e-
e-
+
e-
e
Cl
ee-
e-
e-
e-
*Electron Structure
Similar to Argon
Trends in Ion Size
• Losing Electrons Forms Cations
– Positive ions are smaller than the atoms from
which they are formed
– Electrons “feel” higher nuclear charge
– Remaining electrons are pulled closer to nucleus
e-
Na
e-
+ Na+
*Electron Structure
Similar to Neon
Ionization Energy
• Ionization Energy (IE)
– Energy required to remove an electron from:
– An isolated, gaseous atom
– An ion in its ground state
– Reflects how tightly the electron is held by the atom
– Successive ionization energies always increase
– Typically endothermic
Example: Magnesium
Mg(g)
738 kJ/mol
Mg+(g) + 1e-
Mg+(g)
1450 kJ/mol
Mg2+(g) + 1e-
Ionization Energy Increasing
Ionization Energy
Ionization Energy Increasing
Electron Affinity
• Electron Affinity (EA)
– Potential energy change associated with addition of
electron to:
– A gaseous atom
– An ion in its ground state
– Relative ease by which atoms gain electrons
– Typically exothermic
Example: Fluorine
F(g) + 1e-
F-(g) + 328 kJ
Electron Affinity Increasing
Electron Affinity
Electron Affinity Increasing
Electronegativity
• Electronegativity (EN)
– Attraction of electrons to atoms
of certain elements within a
compound
– Pauling scale from 1-4 used to
help predict bonding
– Noble gases omitted
Element
EN
F
4.0
O
3.5
Cl
3.0
N
3.0
C
2.5
H
2.1
Ag
1.9
Cs
0.79
Electronegativity Increasing
Electronegativity
Electronegativity Increasing
Periodic Trends - Practice Question
•
Arrange the following calcium species in terms of increasing size:
Ca, Ca+, Ca2+, Ca3+, Ca-, Ca2Positive Ions = Smaller Ionic Radius
Negative Ions = Larger Ionic Radius
Ca+3 < Ca+2 < Ca+ < Ca < Ca- < Ca2-
•
Elements in a given period have the same
A.
B.
C.
D.
E.
Atomic weight
Maximum angular momentum quantum number
Maximum principal quantum number
Valence electron structure
Atomic number
C.
Bonding and
Chemical Reactions
Intermolecular Forces
Force
Ion-Induced Dipole
Strength
Players
(10-50 kJ/mol)
Ions and
Polar Solvents
Dipole-Dipole
(3-4 kJ/mol)
Polar Molecules
London Dispersion
(1-10 kJ/mol)
All Molecules
Hydrogen Bonding
(10-40 kJ/mol)
H with O,N,F
Figure
Bonding (Intramolecular Forces)
• Ionic Bond
– Transfer of electrons
– Often metals (cations) and nonmetals (anions)
– The potential energy or lattice energy of the
system is lowered (exothermic) when ion
complexes form
• Covalent Bond
– Sharing of electrons
– Shift of electron density
– Between two or more nonmetals
– Molecules are electrically neutral combinations
of atoms
Bonding (Intramolecular Forces)
EN Difference
Bond
> 1.7
Ionic
<1.7
Covalent
EN = Electronegativity
+
Na
e
+
e e
-
-
Cl
e
-
e e
-
EN
0.9
-
e
C + H
-
e
-
-
3.0
EN Difference = 2.1
Ionic Bond
EN
2.5
2.1
EN Difference = 0.4
Covalent Bond
Types of Covalent Bonding
• Polar Covalent
−Unbalanced electron density within a bond
−Each atom has partial charge
−Partial negative charge (d-)
−More electronegative atom
−Partial positive charge (d+)
−Less electronegative atom
d+
d-
O=C=O
H
F
No Net Dipole Moment
Types of Covalent Bonding
• Non-Polar Covalent
• Atoms with same electronegativity – equal sharing
• Example: Cl2
• Coordinate Covalent
• Shared electron pair comes from lone pair
of one of the atoms in the molecule
H H
I I
DEN
Bond
• Example: F B N H
> 1.7
Ionic
I
I
H H
1.7 < x < 0.5 Polar Covalent
< 0.5
Non-Polar
Covalent Bonding
• Octet Rule (General)
– Atoms gain or lose e- until 8 electrons are in outer shell
– Exceptions: Transition metals (>8e-), boron (stable
with 6e-), beryllium (stable with 4e-)
• Lewis Dot Structure Describes Bonding Between Atoms
1.Decide on Central Atom  Least Electronegative
2.Count Total Valence Electrons
3.Add Bonds to Central Atom
4.Complete Octets of More Electronegative Atoms
5.Place Remaining e- in Pairs on Central Atom
6.Consider Multiple Bonding
Covalent Bonding
• Formal Charge (FC) on Elements
FC = [# Valence e-] – [½ of bonding e-+ non-bonding e-]
− Electronegative elements will have 0 or negative FC
− Electropositive elements will have 0 or positive FC
• Multiple Bonding
− Higher bond order (double or triple bond)
− Shorter bond length
− Higher bond energy
-
..
N
O
O
• Resonance
– Hybrid between two or more equivalent Lewis structures
– Lowers overall energy
– More stable
Covalent Bonding: Example
Draw Lewis Structure of NCO1. Central Atom
2. Count Total Valence Electrons
3. Add Bond to Central Atom
4. Complete Octets
5. Remaining e- in Pairs on Central
Atom
6. Consider Multiple Bonding and
Resonance
N-C-O
: :
1-
: :
:N-C-O:
: :
1-
: :
N=C=O
N=C=O
*NOTE: Atoms beyond the third period may be exceptions to the octet
rule since valence electrons can occupy the d orbitals
1-
: :
1-
: :
: N C O:
1-
: :
: :
1-
C=4; O=6; N=5
:N C O
:
•
EN C = 2.5
EN O = 3.5
EN N = 3.0
Covalent Bonding: Example
•
:
•
Dipole Moment
: :
:
1-
:
N=C=O
No Net Dipole
: :
:
Calculate Formal Charge
FC = [# Valence e-] – [½ of bonding e-+ non-bonding e-]
FCOxygen = 6 – [(½) 4 + 4] = 0
1FCCarbon = 4 – [(½) 8 + 0] = 0
N=C=O
FCNitrogen = 5 – [(½) 4 + 4) = -1
Total Formal Charge = -1
VSEPR Theory
• Valence Shell Electron Pair Repulsion Theory
– Electron pairs spread out to minimize repulsion
– Gives molecules its geometry
• Determining Geometry Using VSEPR Theory
– Draw the Lewis structure
– Arrange electron pairs around central atom so that
they are as far apart from each other as possible
– Lone pairs
– Bonding groups
VSEPR: Molecular Geometry
Name
Example
Hybrid
Lone Pair
Bond Angle
Linear Molecule
CO2
sp
0
180o
Planar Triangle
BCl3
sp2
0
120o
Tetrahedral
CH4
sp3
0
109.5o
NH3
sp3
1
107o
Bent
H2O
sp3
2
104.5o
Trigonal Bipyramidal
PCl5
sp3d
0
120o/90o
Unsymmetrical
Tetrahedron
SF4
sp3d
1
120o/90o
T-Shaped
ClF3
sp3d
2
90o
Linear
I3-
sp3d
3
180o
SF6
sp3d2
0
90o
BrF5
sp3d2
1
90o
XeF4
sp3d2
2
90o
Pyramidal
Structure
..
..
Octahedral
Square Pyramidal
Square Planar
..
..
..
Chemical Bonding - Practice Questions
•
Predict the geometry of ClO2
A.
B.
C.
D.
bent
linear
trigonal planar
pyramidal
+
1. Count total electrons
= 18
1Cl + 2O – 1e-
2. Draw Connectivity
3. Fill in electrons (octet rule)
4. Predict Geometry
•
Arrange the following compounds in order of increasing
boiling point
HINT: Consider Bonding Properties
A.
B.
C.
D.
C2H6
CH3OH
LiF
HCl
bp
C2H6
HCl
CH3OH
LiF
 Dispersion
 Dipole-Dipole
 Hydrogen Bonding
 Ionic
Compounds and
Stoichiometry
What is a Mole?
• Mole
– SI Unit for the amount of chemical substance
• Avogadro’s Number
– 1 mole = 6.022 x 1023 particles
– 1 mole Na atoms = 6.022 x 1023 Na atoms
– 1 mole He atoms = 6.022 x 1023 He atoms
– 1 mole cookies = 6.022 x 1023 cookies
The mass of 1 mole of a substance varies
depending on the substance!!
What is a Mole?
• Stochiometric coefficients
– Used to indicate the number of moles of a given
species involved in the reaction
– Two moles of H2 gas must be reacted with one
mole of O2 gas to form two moles of water
2 H2 (g) + O2 (g)  2 H2O (g)
MW (g/mol)
e.g. 1 mole Carbon =
12.011 g Carbon
Molar Heat of
Reaction
(energy/mol)
1 mole = 6.022 x 1023
particles
MOLE
1 mole = 22.4 L of a
gas at STP
Percent Composition
% Composition =
Mass of X in Formula
x 100
Molar Mass of Compound
Example:
Calculate percent composition of chromium in K2Cr2O7
MW: K = 39 g/mol, Cr = 52 g/mol, O = 16 g/mol
Molar Mass of K2Cr2O7 = 2(39g/mol) + 2(52g/mol) + 7(16g/mol)
Molar Mass of K2Cr2O7 = 294g/mol
%Cr = 2(52g/mol) x 100
294g/mol
%Cr = 35.4%
Molecular and Empirical Formula
• Molecular formula
• The exact number of atoms of each
element in the compound
C 4H 6O 2
• Empirical formula
• The simplest whole number ratio of the
elements in the compound
C 2H 3O 1
Percent Composition and Empirical Formulas
Determine the Empirical Formula According to the Following Percent
Compositions
2.1 % Hydrogen
29.8% Nitrogen
68.1% Oxygen
Step 1: Change Percent Composition to Grams Assuming a 100 g Sample
2.1 grams H
29.8 grams N
68.1 grams O
Step 2: Divide by the Molecular Weight of the Element
2.1 grams H
29.8 grams N
68.1 grams O
1.0 g/mol H
14.0 g/mol N
16.0 g/mol O
2.1 mol H
2.13 mol N
4.26 mol O
Step 3: Divide by the Smallest Value
2.10 mol H
2.13 mol N
2.10 mol
2.10 mol
1H
:
1N
4.26 mol O
2.10 mol
:
Empirical Formula: HNO2
2O
Types of Chemical Reactions
1. Combination Reaction
– When 2 or more reactants combine to form one
product
– Exothermic reaction
S (s) + O2 (g)  SO2 (g)
2. Decomposition Reaction
– When a reactant decomposes into two or more
products (at least one gas)
– Endothermic reaction
2 H2O (g)  2 H2 (g) + O2 (g)
Types of Chemical Reactions
3. Single Displacement Reaction
– When a single atom (or ion) of one reactant is
displaced by an atom of another element based
on activity to form a product
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
4. Double Displacement Reaction
– When elements from two different reactants
displace each other to form two new products
– Exothermic reaction
CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2NaCl(aq)
**Always removes one compound from solution as a precipitate (solid) or a gas**
Types of Chemical Reactions
5. Combustion Reaction
– When a hydrocarbon in the presence of O2
oxidizes or burns to form a gas and H2O
– Exothermic reactions
2C4H10 + 13O2(g)  18CO2(g) + 10H2O(aq)
6. Neutralization Reaction
− An acid and base react to form a salt in water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
**More about these in the Acid/Base section**
Writing and Balancing Chemical Equations
Solubility Rules
•
•
•
SOLUBLE
Group 1 Metals, NH4+, NO3-, ClO3-, ClO4-, C2H3O2Cl-, Br-, I- ; Except when combined with Pb2+, Ag+, Hg22+
SO42- ; Except when combined with Pb2+, Ag+, Hg22+, Ca2+, Sr2+, Ba2+
•
•
INSOLUBLE
Metal Oxides and Hydroxides Insoluble Except Group 1, NH4+, Ca2+, Ba2+, Sr2+
PO43-, CO32-, SO32-, S2- Insoluble Except Group 1 and NH4+
GASES
•
•
•
•
•
•
•
H+ + Sulfide  H2S (g)
H+ + CN HCN (g)
H+ + HCO3-  CO2 (g) + H2O
H+ + CO3-2  CO2 (g) + H2O
H+ + HSO3  SO2 (g) + H2O
H+ + SO3-  SO2 (g) + H2O
OH- + NH4+  NH3 (g) + H2O
Writing and Balancing Chemical Equations
Balance the following reaction:
C4H10 (l) + O2 (g)  CO2 (g) + H2O (l)
Combustion Reaction
1. Balance the carbon in reactants and products
C4H10 (l) + O2 (g)  4 CO2 (g) + H2O (l)
2.Balance the hydrogen in reactants and products
C4H10 (l) + O2 (g)  4 CO2 (g) + 5 H2O (l)
3. Balance the oxygen in reactants and products
C4H10 (l) + 6.5 O2 (g)  4 CO2 (g) + 5 H2O (l)
4. Make stoichiometric coefficients whole numbers
2 C4H10 (l) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l)
Coefficients, not subscripts, are used to balance chemical equations
Smallest whole-number coefficients used to write balanced equations
Applications of Stoichiometry
How many grams of calcium chloride are needed to prepare
72 g of silver chloride according to the following reaction:
CaCl2 (aq) + 2 AgNO3 (aq)  Ca(NO3)2 (aq) + 2 AgCl (s)
Calculate Molar Mass of the two compounds:
•CaCl2 = 110 g/mol
•AgCl = 144 g/mol
72 g AgCl x 1 mol AgCl x 1 mol CaCl2 x 110g CaCl2
144 g AgCl 2 mol AgCl
1 mol CaCl2
= 27.5 g CaCl2
Chemical Kinetics and
Equilibrium
Reaction Mechanisms
A2 + 2 B  2 AB
Overall Reaction:
rate = -1 D[A2] =
Dt
-1 D[B] =
2 Dt
1 D[AB]
2
Dt
• B consumed twice as fast as A2
• AB produced twice as fast as A2 is consumed
• AB produced at same rate as B is consumed
Two-step reaction:
Step 1:
A2 + B  A2B
Step 2:
A2B + B  2 AB (Fast)
(Slow)
A2B is an intermediate
rate-determining step - The slowest step (step 1) in a proposed mechanism
Kinetics
• Definition of Rate:
decrease in [reactants]
increase in [products]
=
rate =
time
time
aA + bB
rate = -1 D[A]
a Dt
=
cC + dD
-1 D[B]
b
Dt
1 D[C]
=
c
Dt
=
1
D[D]
d
Dt
rate = k [A]x [B]y
rate constant
reaction order
*Note: Do not assume
reaction order = coefficients
in chemical equation
Determining the Exponents of a Rate Law
aA + bB
cC + dD
rate = k[A]x [B]y
Experimental determination of x and y:
Trial
[A]initial(M)
[B]initial(M)
rinitial(M/sec)
1
1.0
1.0
2.0
2
1.0
2.0
8.1
3
2.0
2.0
16
Find the rate law for the given reaction at 300K.
Determining the Exponents of a Rate Law
aA + bB
cC + dD
rate = k[A]x [B]y
Experimental determination of x and y:
Trial
[A]initial(M)
[B]initial(M)
rinitial(M/sec)
1
1.0
1.0
2.0
2
1.0
2.0
8.1
3
2.0
2.0
16
Trial 1: r1 = k[A]x [B]y = k(1.00)x (1.00)y
Trial 2: r2 = k[A]x [B]y = k(1.00)x (2.00)y
r2
r1
k(1.00)x (2.00)y
8.1
=
2.0
=
k(1.00)x (1.00)y
= 4 = (2.00)y  y = 2
Determining the Exponents of a Rate Law
aA + bB
cC + dD
rate = k[A]x [B]y
Experimental determination of x and y:
Trial
[A]initial(M)
[B]initial(M)
rinitial(M/sec)
1
1.0
1.0
2.0
2
1.0
2.0
8.1
3
2.0
2.0
16
Trial 2: r2 = k[A]x [B]y = k(1.00)x (2.00)y
Trial 3: r3 = k[A]x [B]y = k(2.00)x (2.00)y
r2
r1
k(2.00)x (2.00)y
16
=
8.1
=
k(1.00)x (2.00)y
= 2 = (2.00)x  x = 1
Determining the Exponents of a Rate Law
aA + bB
cC + dD
rate = k[A]x [B]y
Experimental determination of x and y:
Trial
[A]initial(M)
1
1.0
1.0
2.0
2
1.0
2.0
8.1
3
2.0
2.0
16
x=1
y=2
rate = k[A] [B]2
[B]initial(M)
rinitial(M/sec)
Solving for k:
k=
Rate
[A]1 [B]2
=
2.0 M s-1
=
(1.0 M)1 (1.0 M)2
2.0 M-2 sec-1
Reaction Orders
Zero-Order – constant rate (M *sec-1); independent of reactants’
concentrations
rate = k
[A]
Differential Equation:
slope = -k
rate = -d[A]/dt = k
Integrate
t
[A] = [A]0 - kt
Half-life – time needed for the concentration of the substance to
decrease to one-half the initial value
t1/2 = 0.5 A0/k
Reaction Orders
First-Order – rate proportional to the concentration of one reactant (sec-1)
rate = k[A]
ln[A]
Differential Equation:
slope = -k
rate = -1 * d[A] = k[A]
a dt
ln[A] = - kt + ln[A]0
Half-life
t1/2 = 0.693/k
** Math Tip: ln x = (log x) (2.3) **
Reaction Orders
1 / [A]
Second-Order – rate proportional to the concentration of two reactants,
or the square of one reactant (M-1sec-1)
rate = k[A]2 or (k[A][B])
slope = k
t
Energy Diagram
H2 + Cl2
2 HCl
H
H
H
H
H
H
Cl
Cl
Cl
Cl
Cl
Cl
Energy
transition state
Ea forward
Reactants
Ea reverse
DHrxn
Products
Progress of Rxn (Reaction Coordinate)
Energy Diagram
Energy
Exothermic
DHreactants > DHproducts
DHrxn is negative
Temperaturesystem
Endothermic
DHreactants < DHproducts
DHrxn is positive
Temperaturesysem
Progress of Rxn (Reaction Coordinate)
Factors Affecting Reaction Rate
• Nature of reactants and differences in chemical
reactivity
• Ability of reactants to collide
• Concentration of reactants
• Temperature
• Activation energy (Ea)
• Presence of catalysts
Equilibrium Constant
aA+bB
cC+dD
c
d
[C] [D]
K 
eq
a
b
[A] [B]
Properties of Equilibrium Constant (Keq):
• Do not include pure solids and liquids in Keq expression
•
If Keq > 1, Products Favored
If Keq < 1, Reactants Favored
If Keq = 1, Products = Reactants
•
Keq is dependent on temperature
•
If reaction is more than one step, multiply together Keq from each step in
the reaction
•
When the system is not at equilibrium, use Q instead of Keq
Reaction Quotient (Q) – measures degree of reaction completion
Le Chatelier and Chemical Equilibria
Le Chateliers Principle - if an external stress if applied to the system that
is at equilibrium, the system will attempt to adjust itself to offset the stress
CH3OH (l) + H2 (g)
CH4 (g) + H2O (l) DH = -30 kcal
1.
Adding Removing a Reactant or Product
-Add Reactant Shift Right; Add Product Shift Left
-Remove Reactant Shift Left; Remove Product Shift Right
2.
Changing Volume of Gaseous Reactions
-Decrease volume of gas increases pressure. Shift towards lower # molecules
Ex: 1M H2 vs 1M CH4, since same number moles no effect on change in volume
3.
Change in Temperature
-Treat as reactant (endothermic) or product (exothermic); shift accordingly
4.
Effect of a Catalyst
-Affect forward and reverse reactions equally  No net change
Chemical Kinetics - Practice Question
•
Which would NOT result in a formation of more XeF6(g)
based on Le Chatelier’s Principle?
A.
Increasing the concentration of F2(g)
B.
Decreasing the concentration of XeF6(g)
C.
Decreasing the volume of the container
D.
Decreasing the pressure of the container
Thermochemistry
Some Basic Concepts of Thermochemistry
Thermodynamics
- energy changes determine whether a reaction can occur and how easily
reaction will proceed
Types of Systems:
• Isolated: Cannot exchange energy or matter with surroundings
• Closed: Can exchange energy but not matter with surroundings
• Open: Can exchange energy and matter with surroundings
Types of Processes:
• Isothermal: No change in temperature
• Adiabatic: No heat exchange
• Isobaric: No change in pressure
• Isochoric: No change in volume
Calorimetry
Heat Capacity (C = J g-1 K-1) - The amount of heat required to raise the temperature of 1 g
of a substance 1 oC
•General Equation
q = mCΔT
q = the heat absorbed or released (J = kg m2 s-2)
m = the mass (g)
C = the specific heat of the compound (J g-1 K-1)
ΔT = the change in temperature (K)
•Constant Pressure (Cp)
Measures directly the enthalpy change during the reaction
q = mCpDT
•Constant Volume (Cv)
Measures the internal energy change between reactants and products
q = mCvDT
•The difference between CP and CV is the capacity of the gas to expand and do work and is
equal to Cp = CV + R for an ideal gas where R is the universal gas constant
State Functions
State function – property whose magnitude depends only on
initial and final states of system (not on path)
1) Enthalpy (H)- reflects the heat exchange of a reaction
at constant pressure
2) Entropy (S) – is a measure of disorder, or
randomness of system
3) Gibbs Free Energy (G) – maximum energy released
by the process
Enthalpy
DHrxn = DHproducts – DHreactants
+DHrxn = endothermic process
-DHrxn = exothermic process
• Standard heat of formation (DHof)
DH if one mole of a compound were formed directly from its
elements in their standard state
• Standard conditions (o)
T = 25oC or 298 K, P = 1 atm
Hess’s Law
Hess’s law - Enthalpies of reaction are additive
Overall Reaction
(1) 2C (s) + 2O2  2CO2
DHrxn = -787 kJ
Multi-Step Reaction:
(2) 2C (s) + O2  2CO
DHrxn = ?
(3) 2CO + O2  2CO2
DHrxn = -566 kJ
What is the Heat of Reaction for one mole of CO?
+
2 C (s) + 2 O2  2 CO2
DHrxn = -787 kJ
2 CO2  2 CO + O2
DHrxn = 566 kJ
DHrxn = -221 kJ
2 C (s) + O2  2 CO
The product has a coefficient of 2,
so 1 mole =
-221 kJ
2 mol
= -110 kJ/mol
Entropy
DSrxn = DSproducts – DSreactants
DS = qrev/T
+DSrxn = Increase in Disorder
-DSrxn = More Ordered
•
Predicting the Sign of DS
-Volume: For gases, entropy increases with increasing volume
-Temperature: Temperature = S
-Physical State: Ssolid < Sliquid < Sgas
Gibbs Free Energy
DG = DH – TDS
DG < 0 spontaneous reaction
DG > 0 non-spontaneous reaction
DG = 0 equilibrium (DH = TDS)
DH
DS
Outcome
-
+
Spontaneous at All Temperatures
+
-
Non-Spontaneous at All Temperatures
+
+
Spontaneous only at High Temperatures
-
-
Spontaneous only at Low Temperatures
Gibbs Free Energy
Equilibrium Constants from Thermodynamics
DG = DGo + RT ln Q
At Equilibrium… DGo = -RT ln K
R = Gas Constant (8.314 J mol-1 K-1)
T = Temperature (K)
Thermochemistry - Practice Question
•
Consider the exothermic reaction given above. Predict the
sign of DH, DS, and DG
A.
.DH = neg, DS = neg, DG = neg
B.
.DH = neg, DS = pos, DG = neg
C.
.DH = neg, DS = pos, DG = pos
D.
.DH = pos, DS = pos, DG = pos
Remember:
Exothermic = -DH
Disordered = +DS
DG = DH - TDS
Phase Changes and
the Gas Laws
The Phase Diagram
Pressure
Freezing
Critical
Point
Liquid
Melting
Condensation
Solid
Vaporization
Deposition
Triple
Point
Gas
Sublimation
Temperature
•Triple Point:
-T and P in which all three phases are in equilibrium
-For H2O, Triple Point = 0.01oC and 4.58 torr
•Critical Point:
-T and P in which no distinction between gas and liquid is possible
-Substance above critical point = supercritical fluid (e.g. CO2)
Kinetic Molecular Theory for Ideal Gases
Basic Assumptions
•
•
A gas consists of objects with a defined mass and zero volume
A gas consists of a large number of very tiny particles that are in
constant random motion.
KEavg  Temperature
•
•
•
Particles collide in perfectly elastic collisions (no energy gained or
lost)
Molecules travel in rapid, random, straight line motion without
attraction or repulsions
States of matter are determined by kinetic energy and attraction
The Gas Laws
Boyle’s Law: the volume of a fixed quantity of gas at constant temperature (isothermal)
is inversely proportional to the pressure
V1/P

P1V1 = P2V2
Charles’s Law: the volume of a fixed amount of gas at constant pressure (isobaric) is
directly proportional to its absolute temperature
VT

V1/T1 = V2/T2
Gay-Lussac’s Law: the pressure of a fixed amount of gas held at constant volume is
directly proportional to the absolute temperature
PT

P1/T1 = P2/T2
Avogadro’s Law: the volume of a gas at constant temperature and pressure is directly
proportional to the number of moles
Vn

n1/V1 = n2/V2
The Ideal Gas Law
Ideal gas law – represents hypothetical gas whose molecules have no
intermolecular attraction and occupy no volume
PV=nRT
STP: standard conditions of temperature and pressure
1 atm  101.3 kPa  29.96 inches Hg  760 torr
273.15 K (0oC)
R = 0.0821 L atm mol-1 K-1 or 8.314 J K-1 mol-1
At constant T and P, all gases have the same number of moles in
the same volume
1 mol gas = 22.4 L
Gas Laws - Practice Problem
A sample of argon occupies 50 L at standard temperature. Assuming
constant pressure, what volume will the gas occupy if the temperature
is doubled?
(A)
(B)
(C)
(D)
(E)
25 L
50 L
100 L
200 L
2500 L
Solution:
Step1:
V1= V2
T1 T2
Step 2:
50L
273K
=
“x” = 100 L
“x”
546K
Dalton’s Law of Partial Pressure
•
Dalton’s Law of Partial Pressures: the total pressure (Ptotal) of a mixture of
gases is equal to the sum of the partial pressures (Px) of the individual
components
Ptotal = PA+ PB + PC + …
PA = Ptotal xA
•
Mole Fraction:
XA =
nA
nTOTAL
Colligative Properties
Colligative properties – physical properties derived solely from the
number of particles present
1) Freezing Point Depression - change in freezing point due
to presence of another solute
DTf = Kfm
2) Boiling Point Elevation – change in boiling point due to
presence of another solute
DTb = Kbm
3) Osmotic Pressure – tendency of solute to move from a
region of higher concentration/pressure to a region of
lower concentration/pressure
 = MRT
molality (m) = mol solute/ kg solvent