The Behavior of Solutes:
EIT Review
S2007
Dr. J.A. Mack
Strong Electrolytes: complete dissociation into ions
1 M Na3PO4(aq) Æ 3M Na+ (a) + 1M PO43–
4M overall in ions
Non–Electrolytes:
www.csus.edu/indiv/m/mackj/
no dissociation into ions
1M CH3OH (aq)
methanol
1M overall in
molecules
Weak Electrolytes: partial dissociation into ions
Part 2
1M HC2H3O2(aq) Æ
H+ (aq) + C2H3O2–(aq)
between 1 & 2 M overall
1
Example: A solution of sodium phosphate is added to a solution of
aqueous barium nitrate. A white ppt is observed.
2
Solutions and Concentration
Unbalanced Equation:
Molarity:
Moles of solute per liter of solution.
Na3PO4 (aq) + Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + NaNO3 (aq)
Molecular:
moles of solute
2Na3PO4 (aq) + 3Ba(NO3)2 (aq) → Ba3(PO4)2 (s) + 6NaNO3 (aq)
Ionic:
/
{units: mol/L}
Molarity (M) =
L of Solution
/
6Na+ (aq) + 2PO43- (aq) + 3Ba2+ (aq) + 6NO3- (aq)
/
/
→ Ba3(PO4)2 (s) + 6Na+ (aq) + 6NO3- (aq)
Net Ionic:
molarity is a conversion factor that transforms units
of volume to mole and vise–versa
2PO43- (aq) + 3Ba2+ (aq) → Ba3(PO4)2 (s)
3
4
1
How many grams of sodium phosphate are in 35.0 mL of a
1.51 M Na3PO4 solution?
Dilutions:
old Molarity × old Volume
New Molarity =
mL solution Æ L Æ mols Na3PO4 Æ g Na3PO4
new Volume
use M as a
conversion factor
35.0mL ×
M1 × V1
M2 =
163.94g
L
1.51 mol Na 3 PO4
×
×
1 mol Na 3 PO4
103 mL
L
= 8.66g Na 3 PO4
Rearranging:
V2
M1 × V1 = M2 × V2
5
The pH Scale:
 1 
pH = log  + 
 H  
 
6
What is the pH of the 0.0515 M HCl solution that Jane made?
pH = -log[H+]
[H+]
= Molarity of
H+
(or anything for that matter…)
1
Since log   = − log(x)
Strong electrolyte!
pH = -log[H+]
x
HCl (aq) → H+(aq) + Cl−(aq)
Therefore [H+] = [HCl]
In a neutral solution,
[H+]
=
[OH-]
= 1.00 x
10-7
M at 25 oC
pH = - log [H+] = -log (1.00 x 10-7) = - [0 + ((-7)]
= 7
pH = − log[H + ] = − log(0.0515)
= 1.29
7
8
2
Combining Avagodro’s Law with the general gas law…
This in known as the “Ideal Gas Law”
Diffusion is the process of gas migration due to the random motions and
collisions of gas particles. It is diffusion that results in a gas completely
filling its container
P×V
= constant
n×T
P×V = n× R ×T
R = "gas constant" = 0.08206
Remove the partition and the
particles begin to mix.
L ⋅ atm
mol ⋅ K
PV = nRT
After sufficient time the mixture
becomes homogeneous.
9
Diffusion is the process of gas migration due to the random
motions and collisions of gas particles. It is diffusion that results in a
gas completely filling its container
gases separated
10
Effusion is the escape of a gas through an orifice or “pin hole”.
gases begin to mix
Effusion is governed by Graham’s Law:
The rate of effusion of a gas is proportional to its uRMS.
Rate ∝ u RMS ∝
over time, the mixture
becomes homogeneous.
1
M
Where M is the molar mass of a substance.
This implies that heavier gases will effuse slower than lighter gases.
11
12
3
Effusion is the escape of a gas through an orifice or “pin hole”.
Light is characterized by a wavelength and frequency:
The wavelength is
given the symbol
λ
λ
( lambda )
λ can be measured from
crest to crest or trough to trough.
u RMS
since uRMS is a function of
molar mass, the lighter
particles escape faster!
a node is a point of zero amplitude
3RT
=
M
The amplitude
measures the crest
height of a wave.
ν
The frequency of light (nu) measures the number
of crests that pass through a point in space per
second.
13
Wavelength has units of length:
What wavelengths correspond to FM radio (93.5 MHz) signals?
m… cm… µm… nm… pm
Frequency has units of inverse time:
14
λ×υ = c
c
λ=
υ
s −1 or Hz (hertz)
λ (m) × ν (s–1) = c (m s–1)
c”, the speed of electromagnetic radiation (light) moving through a
“
vacuum is 2.99792458 × 108 m/s
ν=
c
λ
3.00 × 108 m
λ=
93.5 MHz ×
λ=
c
ν
s
106 Hz 1s −1
×
1 MHz Hz
= 3.21 m
15
16
4
Orbital Shapes
p Orbital Shapes: The p orbitals are dumbbell–shaped.
s – orbital shapes: s orbitals are spherical.
Each of the three p–orbitals is aligned on its own axis (x, y & z)
A node is a point of zero probability
17
18
In a multi–electron atom (more than
electron) the orbitals of a subshell
are shifted by electron–electron
interactions.
d and f Orbital Shapes: d and f orbitals have two different
shapes.
Increasing Energy
Core
[He]
[Ne]
[Ar]
[Kr]
[Xe]
[Rn]
1s
2s
3s
4s
5s
6s
7s
2p
3p
4p
5p
6p
7p
3d
4d 4f
5d 5f
6d
Each “s” holds 2–electrons
Each d–orbital has 2 nodal planes, the f’s have 4.
Each “p” holds 6–electrons
Each “d” holds 10–electrons
19
20
5
Electron Configurations cont…
Each element’s outermost electrons (valence) are related to the
elements position on the periodic table.
Orbital box notation:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
1s
2s
2p
3s
↑
3p
↑
3p
↑↓
Aluminum: Al (13 electrons)
Electron Configuration
3s
↑↓
↑↓
↑↓
↑↓
2p
2s
1s2 2s2 2p6 3s2 3p1
All of the subshells below the valence are full.
↑↓
1s
21
Chemical bond:
attractive force holding two or more atoms
together
Covalent bond:
a sharing electrons between the atoms.
non–metal / non–metal bonds.
Ionic bond:
22
Electronegativity:
The measure of the tendency for a given atom to polarize the
electrons in a covalent bond is called the Electronegativity (EN)
of an atom.
EN increases
EN
decreases
electrostatic in nature, transfer of electrons from a
metal to a nonmetal. (cation + anion)
Periodic Table
Metallic bond: attractive force holding pure metals together.
Cations in a “sea” of electrons.
23
24
6
When there exists a difference in the EN between the two bonding
atoms, (∆EN) the bond is said to be polar.
ionic
N.
+
→
: :
: :
polar-covalent
3.1
O:
O=O
“double bond”
→
:N≡N:
“triple bond”
N⋅
:
0.9
NaF
+
:
HCl
O:
non-polar
:
0
:
Br2
: :
Bond
The oxygen and nitrogen that makes up the bulk of the atmosphere also
exhibits covalent bonding in forming diatomic molecules.
: :
∆EN
Example:
Multiple bonds in covalent molecules:
Polyatomic Molecules (More than two atoms)
: :
: :
Carbon dioxide: CO2
O=C=O
25
26
Bond Strength and Bond Properties:
Covalent bond strength increases with increasing ∆EN
example:
HCl bond is stronger than the HBr bond
Covalent bond strength increases with increasing bond order
example:
O=O bond in stronger than O–O bond
triple > double > single
Bond Order:
3
2
1
H–bonding in water brings about a network of interactions which explain
phenomena such as:
Bond length decreases with increasing bond order (Strength)
example:
O=O bond is shorter than O–O bond
capillary action
27
surface tension
why ice floats
28
7
Kinetics: Rate Law & Reaction Order
Reaction Orders:
The reaction rate law expression relates the rate of a reaction to
the concentrations of the reactants.
Each concentration is expressed with an order (exponent).
The rate constant converts the concentration expression into the
correct units of rate (Ms−1). (It also has deeper significance,
which will be discussed later)
For the general reaction:
A reaction order can be zero, or positive integer and fractional number.
Order
0
1
2
0.5
1.5
0.667
aA + bB → cC + dD
Rate = k [A]x [B]y
x and y are the reactant orders
determined from experiment.
x and y are NOT the
stoichiometric coefficients.
Name
zeroth
first
second
one-half
three-half
two-thirds
29
Recognizing a first order process:
AÆproducts
30
Determining the Rate constant for a first order process
Taking the log of the integrated rate law for a first order process we find:
Whenever the conc. of a
reactant falls off
exponentially, the kinetics
follow first order.
ln ([A] = [A]o e − kt )
ln[A] = ln[A]o − k × t
A plot of ln[A] versus time (t)
is a straight line with slope -k
and intercept ln[A]o
[A] = [A]o e − kt
31
32
8
A certain reaction proceeds through t first order kinetics.
The half-life of the reaction is 180 s.
What percent of the initial concentration remains after 900s?
A certain reaction proceeds through t first order kinetics.
The half-life of the reaction is 180 s.
What percent of the initial concentration remains after 900s?
Using the integrated rate law, substituting in the value of k and 900s we find:
Step 1: Determine the magnitude of the rate constant, k.
k = 0.00385 s-1
t 1 = ln 2 = 0.693
2
k
k
k=
[A]
= e − kt
[A]o
ln 2
ln 2
=
= 0.00385s −1
t1
180s
−1
[A]
= e −0.00385 s × 900 s
[A]o
= 0.0312
Since the ratio of [A] to [A]0 represents the fraction of [A] that remains, the
% is given by:
2
100 × 0.0312 = 3.12%
33
Chemical Equilibrium:
34
Chemical Equilibrium:
Reactions that form only products may be written:
Initially, only the reactants
are present:
Reactants → Products
Consider the reaction:
H 2 (g) + I 2 (g) U 2HI(g)
[HI(g)]o = 0
Reactions that do not form products may be written:
tEQ
As the reaction proceeds
the reactant concentrations
decrease and the product
concentration increases.
Reactants ← Products
Reactions that are governed by equilibrium are written:
[HI(g)] > 0
[I2(g)]<[I2(g)]o
[H2(g)]<[H2(g)]o
Reactants U Products
reaction time
At some point in time, the concentration of all species remain constant:
Where the symbol U indicates that the reaction proceeds in both
directions, forward and reverse.
When this occurs, equilibrium has been established.
35
36
9
Chemical Equilibrium:
Chemical Equilibrium:
H 2 (g) + I 2 (g) U 2HI(g)
Lets think about equilibrium form a Kinetic perspective:
H 2 (g) + I 2 (g) U 2HI(g)
Initially, the forward reaction dominates the process:
Rate forward =rate reverse
Rateforward = k f [I 2 ][H 2 ]
tEQ
The rate of the reverse reaction is zero because the initial concentration
of HI is zero.
Rate reverse = k r [HI]2 = 0
therefore:
0
As time goes by, the concentration of products increases, increasing the
rate of the reverse reaction.
Rate reverse = k r [HI]2 ≠ 0
At time t = tEQ the rate of the forward reaction equals the rate of the
reverse reaction.
k f [I 2 ][H 2 ] = k r [HI]2
reaction time
At time t = tEQ the rate of the forward reaction equals the rate of the
reverse reaction.
37
38
The Equilibrium Constant:
The Equilibrium Constant:
At equilibrium the rates of the forward and reverse reactions are
equal.
rate forward = rate reverse
In general for a given reaction that exhibits equilibrium, the
equilibrium constant expression is given by:
aA + bB U cC + dD
kf × [reactants] = kr × [products]
c
[C]
Comparing the forward and reverse rate constants:
K=
kf
[products]
=
kr
[reactants]
since the rate constants are “constants”, the ratio is also a constant…
kf
[products]
=
=K
k r [reactants]
a
[A]
d
×
[D]
×
[B]
b
Where the concentrations of the product and reactant species are
raised to the power of their respective stoichiometric coefficients.
the “equilibrium”
constant
39
40
10
The Equilibrium Constant:
The Equilibrium Constant:
The magnitude of the equilibrium constant K, tells chemists which side
of the chemical equation is favored at equilibrium.
low concentration of reactants
at equilibrium
high concentration of reactants
at equilibrium.
Reactants U Products
K=
[products]
[reactants]
If K >> 1,
1 then the numerator (products) concentration dominates at Eq.
The reaction is said to be “product favored”
If K << 1,
1 then the denominator (reactants) concentration dominates at
Eq.
The reaction is said to be “reactant favored”
If K ≅ 1 then neither dominates
ProductProduct-favored
K >> 1
ReactantReactant-favored
K << 1
41
Determining the Equilibrium Constant:
The Reaction Quotient: “Q”
Consider the reaction:
When a reaction is not at a state of equilibrium, the ratio of the
products and reactants concentrations yields information as to the
reaction direction.
aA + bB U cC + dD
c
[C]
Q=
a
[A]
42
2NOCl(g) U 2NO(g) + Cl 2 (g)
Initially 2.00 mol of NOCl(g) is added to a 1.00 L flask.
At equilibrium you find the concentration of NO(g) to be 0.66 M.
Calculate K for the reaction.
d
×
[D]
×
[B]
b
if Q < K the reaction is heading towards the formation of products
(not yet to equilibrium)
if Q = K the reaction is at equilibrium
Solution: From the information given, we know that 0.66 mols of
NO(g) are present at equilibrium.
0.66M NO(g) =
if Q > K the reaction is heading towards the formation of reactants
(gone past equilibrium)
43
0.66 mols NO(g)
× 1.00 L = 0.66 mols NO(g)
1L
44
11
Determining the Equilibrium Constant:
Determining the Equilibrium Constant:
Consider the reaction:
Consider the reaction:
2NOCl(g) U 2NO(g) + Cl 2 (g)
2NOCl(g) U 2NO(g) + Cl 2 (g)
Initially 2.00 mol of NOCl(g) is added to a 1.00 L flask.
At equilibrium you find the concentration of NO(g) to be 0.66 M.
Calculate K for the reaction.
Initially 2.00 mol of NOCl(g) is added to a 1.00 L flask.
At equilibrium you find the concentration of NO(g) to be 0.66 M.
Calculate K for the reaction.
Using stoichiometry, one can calculate the moles and concentrationd
of Cl2(g) formed at Eq. and the moles of NOCl(g) reacted to get to Eq.
1mol Cl (g)
2
0.66 mols NO(g) × 2mol NO(g)
0.66 mols NO(g) ×
1
= 0.33M
1.00L
Cl2(g)
1
=
= 0.66 mols NOCl(g) ×
1.00L
(reacted)
= 0.33 mols Cl2(g) ×
2mol NOCl(g)
2mol NO(g)
0.66M NOCl(g)
Now we tabulate the results of our calculations:
[NOCl]
[NO]
[Cl2]
initial concentration
2.00
0
change to achieve
equilibrium
− 0.66
+ 0.66
+ 0.33
equilibrium
concentration
1.34
0.66
0.33
0
45
[NOCl]
[NO]
[Cl2]
initial concentration
2.00
0
0
change to achieve
equilibrium
− 0.66
+ 0.66
+ 0.33
equilibrium
concentration
1.34
0.66
0.33
Heat Transfer:
Surroundings
2NOCl(g) U 2NO(g) + Cl 2 (g)
heat in
The equilibrium constant expression for the reaction is:
[ NO(g)]2 × [Cl2 (g)]
K=
[ NOCl(g)]2
=
(0.66)2
×
(1.34)2
K is always reported as a “unit-less” quantity.
0.33
46
heat out
System
= 0.080
q>0
(+)
∆E > 0
q<0
(–)
∆E < 0
(reactant favored)
47
48
12
Thermochemistry:
Energy in
E Final
Ef > Ei
Heat and the Specific Heat Capacity:
Energy out
Ef < Ei
— When heat is absorbed or lost by a body, the temperature
must change as long as the phase (s, g or l) remains constant.
E initial
work out
energy
work in
— The amount of heat (q) transfer is related to the mass and
temperature by:
q = m × C × ∆T
q = heat lost or gained (J)

E Final
E initial
∆Esystem > 0
(+)
∆Esystem < 0
(–)
∆T is the temperature change in degrees Celsius or Kelvin’s
50
Ideal Gas Law
How many kJ of energy are released when 128.5 g of methane
(CH4(g)) is combusted?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
P×V
= constant for a set number of gas moles
n×T
∆H = -802 kJ
802 kJ
mol CH 4
×
= 6.42×103 kJ
1mol CH 4
16.04 g
P×V = n × R ×T
How many hours would this power a 100 W light bulb if one could
use all of this energy?
100 W = 100 Js–1
6.42×103 kJ ×

49
Enthalpy: ∆H
128.5g CH 4 ×
J
C = the Specific Heat Capacity of a compound  o 
 g⋅ C 
q out
q in
m = mass of substance (g)
103 J × s × min × hr =
100J
60s 60 min
kJ
R = "gas constant" = 0.08206
L ⋅ atm
mol ⋅ K
PV = nRT
17.8 hrs
51
52
13
Problems: How many grams of krypton (Kr) will it take to exert a
pressure of 11.2 atm in an 18.5 L at 28.2 oC.
PV = nRT
n=
PV
=
RT
R = 0.082057
L ⋅ atm
mol ⋅ K
11.2 atm × 18.5 L
0.08206
L ⋅ atm
× ( 28.2 + 273.15) K
mol ⋅ K
8.3789 mols Kr ×
83.80 g
= 702 g Kr
mol
= 8.3789 mols Kr
(3sf)
53
54
We can use the periodic table to determine the electron configuration by
counting:
Each element’s outermost electrons (valence) are related to the
elements position on the periodic table.
N: 7 electrons
1
3
2
4
5
1: 1s1
All of the subshells below the valence are full.
55
2: 1s2
3: 2s1
4: 2s2
1s2
2s2
5: 2p1
6
6: 2p2
7
7: 2p3
2p3
56
14
Electron Configurations cont…
Periodic Trends
Orbital box notation:
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
1s
2s
2p
3s
↑
3p
↑
3p
↑↓
Aluminum: Al (13 electrons)
Electron Configuration
3s
↑↓
↑↓
↑↓
↑↓
2p
2s
1s2 2s2 2p6 3s2 3p1
↑↓
1s
57
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15