Download EIT Review S2012 Part 2 Dr. J. Mack CSUS Department of Chemistry

Solutions and Concentration
Molarity:
EIT Review S2012
Part 2
Dr. J. Mack
CSUS Department of Chemistry
Moles of solute per liter of solution.
moles of solute
{units: mol/L}
Molarity of X (cX) = L of Solution
cNaCl = [NaCl] = 1.00
http://www.csus.edu/indiv/m/mackj/
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Dr. Mack
×
Volume
1
mols
L
× L
=
= 1.00M
molarity is a conversion factor that transforms units of volume to mole and vise–versa
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2
A student adds 25.15 g of sodium sulfide into a 500.0mL
volumetric flask then fills the solution to the calibration mark with water. What is the molarity of this solution. if you know Molarity and volume, you know moles!
Molarity
mol NaCl
1L
moles
= moles
molarity (M) = moles of solute
L of solution
g Na2S
if you know mols and molarity, you know volume!
moles
×
mol ×
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1
M
=
mols Na2S
divide by
volume
M(Na2S)
Volume
1 mol Na 2S
78.05g Na 2S
L
500.0mL × 3
10 mL
25.15g Na2S ×
L
= L
mol
Dr. Mack
use
molar mass
M(Na2S) = 3
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Dr. Mack
= 0.6445M
Na2S
4
1
How many grams of sodium phosphate are in 35.0 mL of a 1.51 M Na3PO4 solution?
A student prepares a solution of NaOH(aq) by dilution.
She performs a standardization of the solution using KHP as the standard.
Solution:
Determine the [OH–] from the data collected:
mL solution → L → mols Na3PO4 → g Na3PO4
use M as a conversion factor
use molar mass as a conversion factor
mass KHP + flask:
95.3641 g
final buret reading:
30.12 mL
mass empty flask:
95.0422 g
initial buret reading:
1.56 mL
0.3219 g
vol. NaOH:
28.56 mL
mass KHP:
35.0mL ×
163.94g
L
1.51 mol Na3PO4
×
×
1 mol Na3PO4
103 mL
L
[OH − ] =
= 8.66g Na3PO4
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[OH–] = [NaOH] Strong Electrolyte!!!
5
Dr. Mack
HNO3(aq)
1mol KHP 1 mol OH −
1
103 mL
×
×
×
204.22 g
28.56 mL
1mol KHP
L
= 0.05519M OH
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mL KOH
[NaOH] = 0.05519 M
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KNO3(aq)
mols KOH
mols HNO3
+ H2O(l)
mL HNO3
L
0.105mol KOH 1mol HNO3
×
×
10 mL
1mol KOH
L
3
×
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+ KOH(aq)
−
25.1mL ×
remember… strong electrolyte
6
How many mL of 0.125M nitric acid are needed to completely neutralize 25.1mL of 0.105M potassium
hydroxide? from the stoichiometry, one mole of OH‐ reacts with 1 mol of KHP
0.3219g KHP ×
mol KHP
mols OH −
=
L of solution L titrated
[OH‐] = [NaOH]
7
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L
103 mL
×
0.125mol HNO3
1L
Dr. Mack
= 21.1mL
8
2
What is the concentration of 250.0mL of a solution that is initially 0.450M after 50.0 mL of water is added? Dilutions:
old Molarity × old Volume
New Molarity =
new Volume
M1 × V1
M1 × V1 = M2 × V2
M2 = let M1 and V1 = the old molarity and volume
V2
let M2 and V2 = the new molarity and volume
M1 × V1
M2 = M2 =
M1 × V1 = M2 × V2
V2
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The pH Scale: 1L
0.450 mols
×
3
10 ml
1L
1L
300.0mL × 3
10 ml
250.0mL ×
Rearranging:
9
0 to 14
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= 0.375 M Dr. Mack
10
Ch. 4
Example: The pH of a solution provides a way to express the acidity, or the concentration of H+ in solution:
pH = −log[H+]
A student is given a solution that is labeled pH = 4.72,
what is the molarity of H+ in this solution? pH = − log[H+]
[H+] =
10− pH
plugging in 10−4.72 into you calculators yields: 1.90546 × 10−5
but wait… how many sig. figs. are allowed?
10−4.72 = 10(0.28−5) = 100.28 ×10−5
low pH = high [H+]
high pH = low [H+]
acidic solution
basic solution
100.28 = 1.9 2 sig. figs.!
therefore the concentration should be reported as: a pH of 7 indicates that the solution is neutral
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1.9 ×10−5 M [H + ]
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3
Mass Percent
(a.k.a. Weight Percent)
What is the mass percent of a solution that is made by adding
2.45 g of potassium chloride to 75.4mL of water?
mass % =
mass of solute
How many grams of a 4.83% solution
of dextrose would it take to obtain 128g of dextrose. x 100
mass of solution
128g dextrose ×
4.83g dextrose
2.45g KCl
% KCl =
2.45g KCl
100g solution
+ 75.4 mL H2O ×
1.00g H2O
1.00mL H2O
= 2650g of solution
x 100
= 3.15 %
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V ∝
Boyle’s Law:
Pressure vs. Volume
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Unit:
Conversion to atm:
Atmosphere (atm)
V ∝ T
or
V
= Constant
T
1
760 mm Hg
= 1 atm
760 torr
= 1 atm
1 mm Hg = 1torr
14.696 lb/in2
V ∝ moles
or
Avogadro’s Law: Volume vs. moles
14
Pressure Measurement Units:
1
P
or
V × P = Constant
Charles’s Law: Volume vs. Temperature (absolute)
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= 1 atm
V
= Constant
n
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4
Combining Avagodro’s Law with the general gas law…
Gas: Temperature Scales
This in known as the “Ideal Gas Law”
For gas calculations, we need an absolute scale, one that does not take on negative values.
P× V
= constant n× T
The conversion between oC
and degrees K is:
PV = nRT
K = oC + 273.15
R = "gas constant" = 0.08206 When performing calculations with absolute temperatures, on must use the Kelvin scale.
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Example: What is the volume (in ml) of a gas initially at
30.0 °C, 775 torr and 3.25 L that is changed to 0.555 atm
and 25.0 °C?
P1×V1 P2 ×V2
=
T1
T2
Solution:
General Gas Law:
For a set number of moles:
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L ⋅ atm
mol ⋅ K
P×V
= n × R = a constant
T
So at two sets of conditions:
Solving for the new volume:
P1×V1
P ×V
= a constant = 2 2
T1
T2
V2 =
P1 × V1
× T2
P2 × T1
Substituting in the values given:
V2 =
103 ml × (25.0 + 273.15)K
1L
760 torr
× (30.0 + 273.15)K
0.555 atm ×
atm
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775 torr × 3.25 L ×
P1×V1 P2 ×V2
=
T1
T2
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19
= 5870 ml
20
5
Rules for Ideal Gas Law Calculations:
Standard Temperature and Pressure:
∙
Always convert the temperature to Kelvin (K = 273.15 + oC)
∙
Convert from grams to moles if necessary.
∙
Be sure to convert to the units of “R”
(L, atm, mol & K).
In order to provide a reference point for the comparison of gasses, standard temperature and pressure are set at:
1 atm (760 torr or mm Hg
0 oC (273.15K)
Problem: Calculate the volume occupied by 43.7 g of hydrogen at STP.
nRT
V=
PV = nRT
P
43.7 g H2 ×
V =
1mol H2 0.08206 L ⋅ atm
×
× 273.15K
2.02 g
mol ⋅ K
= 485 L
1.00 atm
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How many liters of hydrogen gas are required to react with 12.7 L of nitrogen to produce ammonia if the reaction occurs in a closed container at constant pressure?
3 H2(g) + N2(g) → 2 NH3(g)
Step 1: Write the reaction:
nRT
V=
P
Step 2: Racal from PV=nRT that...
so we have:
3 molsH2
1 molN2
12.7 L N2 ×
= EIT (2)
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated and the CO2 (g) collected in a 10.0 L flask had a pressure of 755 torr at 20.0° C. How much NaCl was in the original sample?
Solution: Recognize that the CO2 (g) produced is related to MgCO3 (s) by:
volume is proportional to moles!
Δ
MgCO3 (s) → MgO (s) + CO2 (g)
reaction stoichiometry
3 L H2
1 L N2
Volume CO2 → moles CO2 → mols MgCO3 → g MgCO3
3 L H2
= 38.1 L of H2
1 L N2
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Gas Stoichiometry
use PV = nRT
g Sample
Pressure works the same way!
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molar mass
– g MgCO3
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= g NaCl
24
6
Gas Stoichiometry
Solution continued:
Gas Stoichiometry
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated and the CO2 (g) collected in a 10.0 L flask had a pressure of 755 torr at 20.0° C. How much NaCl was in the original sample?
A 50.0 g sample of a mixture of MgCO3 and NaCl was heated and the CO2 (g) collected in a 10.0 L flask had a pressure of 755 torr at 20.0° C. How much NaCl was in the original sample?
moles CO2 → mols MgCO3 → g MgCO3
Volume CO2 → moles CO2 → mols MgCO3 → g MgCO3
PV
n =
RT
PV = nRT
Start with:
R = 0.08206 n =
L ⋅ atm
mol ⋅ K
755 torr ×
0.08206
moles CO2 → mols MgCO3 → g MgCO3
1 atm
× 10.0L
760 torr
L ⋅ atm
× 293K
mol ⋅ K
1mol MgCO3 84.3 g MgCO3
×
1 mol CO2
1mol MgCO3
= 34.8 g MgCO3
moles CO2
50.0 g sample
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×
PV
RT
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25
− 34.8g = EIT (2)
15.2 gNaCl
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The relationship between temperature (T) and velocity (u) (kinetic energy) can be found by the following:
Ideal gas law
P =
nRT
V
Setting the two equal:
solving:
P =
and
nRT
V
=
Nmu 2
3V
28.0 g/mol
Nmu 2
3V
3nRT
u =
Nm
2
=
u RMS =
32.0 g/mol
KMT
3RT
M
As molar mass (M) decreases, at a given temperature (T) the average speed of a gas particle increases:
18.0 g/mol
4.00 g/mol
3RT
M wt
J
mol ⋅ K
⎛ kg ⎞
is in ⎜
⎟
⎝ mol ⎠
R = 8.314
The “root mean square velocity” for a gas is:
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u RMS =
3RT
M wt
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M wt
27
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7
Effusion is the escape of a gas through an orifice or “pin hole”.
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
gases begin to mix
Effusion is governed by Graham’s Law:
The rate of effusion of a gas is proportional to its uRMS.
over time, the mixture
becomes homogeneous.
Rate ∝ uRMS ∝ 1
M
Where M is the molar mass of a substance.
This implies that heavier gases will effuse slower than lighter gases.
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Deviations from Ideal Gas Law
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Non‐Ideal Behavior: Real Gasses:
Under ideal conditions (moderate pressure, moderate temperature) gasses behave ideally: 1. The volume of the gas is << than the volume of the container
• Real molecules have 2. & the particles are not effected by “intermolecular forces”.
volume.
Under conditions of high pressure and low temperature, this is not the case. Corrections must be made for the volume and intermolecular forces of the gas particles and for the intermolecular forces • There are intermolecular forces.
van der Waal’s
Equation
– Otherwise a gas could not become a liquid.
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8
Phase Changes
The temperature dependence of vapor pressure goes as…
A plot of lnPvap vs. 1
yields a slope of…
T
∆vapH˚ is related to T and P by the Clausius‐
Clausius‐Clapeyron equation
slope :
−
ln Pvap = − Δ vap H + C
RT
o
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Δ vap H
R
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Energy Change Calculations:
or cooling)
=
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37
Determine the final temperature of a 25.0g block of metal that absorbs 255 cal of energy.
J
The initial temperature of The specific heat 2.72
the block was 17.0oC
capacity of the metal is:
g×K
Heating & Cooling:
q(heating
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m × C × ΔT
q = m × C × ΔT =m × C × ( Tf − Tin )
rearranging:
Heat absorbed or lost in a Phase change:
Tf =
q(phase change) = Δ(phase change)H× n
Tf
4.184 J
+255 cal ×
1 cal
J
25.0g × 2.72 o
g× C
=
q
+Tin
m× C
+ 17.0 oC =
Tf > Tin as expected
32.7 oC
(n = moles or grams) EIT (2)
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9
Adding energy to a substance increases its temperature until it reaches a phase phase change. During a phase change, the temperature remains constant
temperature remains constant. The energy change
added goes into overcoming the intermolecular forces
intermolecular forces. It does not increase the kinetic energy of the particles.
How much energy is needed to melt 10.0g of ice at 0oC.
q2 = nice × Δ fusH
= 10.0g
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Enthalpy:
H = qp
H=E+W
the subscript “
the subscript “p” indicates constant pressure
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×
6.008kJ
mol
×
103 J
1kJ
= 3334J
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42
ΔH = Δ (E + PV )
= ΔE + PΔV
0
ΔH = ΔEp
Dr. Mack
Since Enthalpies are state functions, one must specify the conditions at which they are measured.
H(T,P): Enthalpy is a function of temperature and pressure.
“Ho” indicates that the Enthalpy is taken at Standard State
conditions.
H = E + PV
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1 mol
18.02g
Enthalpy Conditions:
Enthalpy, “H” is the heat transferred between the system and surroundings under conditions of constant pressure.
if no “
if no “PV”
PV” work is done by the system, ΔV = 0
×
Standard State Conditions are defined as:
the change in enthalpy is the change in internal energy at constant pressure 43
1 atm = 760 mm Hg or 760 torr & 298.15K or 25oC
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44
10
Enthalpies and Chemical Reactions: ΔrH
Enthalpies and Chemical Reactions
Reactants
All chemical processes are accompanied by energy changes:
ΔH = Hfinal − Hinitial
If a reaction gives off energy (heat), it is an exothermic reaction.
Products
Reactants
+
Enthalpy of reaction = ΔrH = Hproducts − Hreactants
Energy
Reactants
+
Energy
If a reaction absorbs energy (heat), it is an endothermic reaction.
Energy
Products
Energy can be treated as part of the reactants
reactants or products
products!
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H (Products)
H (Reactants)
ΔrH > 0
ΔrH < 0
H (Reactants)
H (Products)
Exothermic
Endothermic
45
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) Products
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How many hours would 6.42×103 kJ power a 100 W light bulb if one could use all of this energy?
ΔrH = −802 kJ
100 W =
g
→ mols
↑
molar mass
128.5g CH4 ×
EIT (2)
→J
↑
Reaction
enthalpy
46
exothermic reaction, positive energy out!
100J
s
3
1s
min
hr
×
×
=
6.42×103 kJ × 10 J ×
100J 60s 60 min
kJ
17.8 hrs
mol CH4
+802 kJ
×
= 6.42×103 kJ
1mol CH4
16.04 g
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11
Orbital Shapes
s – orbital shapes: s orbitals are spherical.
6.6
p ‐ Orbital Shapes: The p orbitals are dumbbell–shaped.
Each of the three p–orbitals is aligned on its own axis (x, y & z)
Each p‐orbital has a nodal plane. •A node is a point of zero probability.
•There are n − 1 nodal spheres for each s‐orbital
1s has zero, 2s has 1, 3s has 2 etc.
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6.6
The px orbital has a y‐z nodal plane and so on.
49
d and f ‐ Orbital Shapes: d and f orbitals have two different shapes.
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Electrons fill orbitals on the periodic table across then down”
And so on…
Each d–orbital has 2 nodal planes, the f’s have 4.
6.6
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12
Each element’s outermost electrons (valence) are related to the elements position on the periodic table.
Orbital Filling: The Aufbau Principle
Aufbau Principle: Lower energy orbitals fill first.
“s” orbitals can hold 2 electrons
“p” orbitals hold up to 6 electrons
“d” orbitals can hold up to 10 electrons
All of the subshells below the valence are full.
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53
One can find the electron configuration from the position on the
periodic table by counting backwards form the last electron to fill:
ch. 7
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1
Nitrogen:
2
Orbitals
↑
↑
3
1s
with one electron in the 3p orbital:
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3p1
N has an atomic number of 7, therefore 7 electrons
Electron Configuration
↑
2s
↑
↑
↑
1s2 2s2 2p3
2p
each electron fills an empty orbital until forced to pair (Hunds rule)
1s2 2s2 2p6 3s2 3p1
Dr. Mack
The electron configuration of an atom is the total sum of the electrons from lowest to highest shell.
Example:
in the 3rd period
54
Electron Configurations: Orbital Box Notation
Electrons fill the orbitals from lowest to highest energy.
Aluminum is in the “p” block
therefore:
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↑
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paring requires opposite spin… (Pauli exclusion principle)
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13
Atomic radii of the elements:
ch. 7
There are 2 extreme forms of connecting or bonding atoms:
Ionic Bonding: complete transfer of 1 or more electrons from one atom to another (metals and non‐metals)
Covalent Bonding: valence electrons shared between atoms (non‐metals and non‐metals)
Most bonds are somewhere in between.
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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:
Metallic bond:
58
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As a reaction proceeds, the As a reaction proceeds, the concentration of the reactants concentration of the reactants decrease with time.
decrease with time.
electrostatic in nature, transfer of electrons from a metal to a nonmetal. (cation + anion)
attractive force holding pure metals together.
Cations in a “sea” of electrons.
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14
The concentrations of The concentrations of products increase with time
products increase with time
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Until at some point, they Until at some point, they remain constant, where remain constant, where equilibrium is established.
equilibrium is established.
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Properties of Chemical Equilibria
Moving towards equilibrium
Equilibrium established
Equilibrium systems are said to be:
•Dynamic (in constant motion)
•Reversible
•Equilibrium can be approached from either direction. Pink to blue
Co(H2O)6Cl2 → Co(H2O)4Cl2 + 2 H2O
forward
Blue to pink
→
Co(H2O)4Cl2 + 2 H2O Co(H
2O)6Cl2
reverse
→
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→
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15
Equilibrium From A Kinetic Perspective
forward reaction:
Equilibrium From A Kinetic Perspective
At equilibrium, the rates of the forward reaction is equal to the rate of the reverse:
aA + bB → cC +dD
Rate(f) = kf[A]a[B]b
reverse reaction:
kf[A]a[B]b = Rate(f) = Rate(r) = kr[C]c[D]d
The ratio of the rates of the forward to reverse reaction is given by:
cC +dD → aA + bB
Rate(r) = kr[C]c[D]d
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The Equilibrium Constant
c
d
Rate(f) [C] [D] k f
=
=
= Keq
Rate(r) [A]a[B]b kr
67
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Keq is known as the “equilibrium constant”
expression.
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The Equilibrium Constant
[C]c [D]d k f
Keq = a b =
[A] [B] kr
If kf > kr then Keq > 1
This indicates that the Ea for the forward reaction is lower than that of the reverse reaction. The reaction is said to be “product favored”.
If kf < kr then Keq < 1
Product‐favored
K > 1
This indicates that the Ea for the reverse reaction is lower than that of the forward reaction. The reaction is said to be “reactant favored”.
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Reactant‐favored
K < 1
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16
Writing Equilibrium Constant Expressions
The general form of an equilibrium constant expression is given by
aA + bB R cC + dD
products
[C]c [D]d
K=
[A]a [B]b
Values of K are dimensionless.
Write the expression for K for the reaction of ammonium ion with hydroxide ion.
NH4+(aq) + OH‐(aq) U NH3(aq) + H2O(l) (a) K =
[NH 3 ]
[NH +4 ][OH - ]
(b) K =
[NH +4 ][OH - ]
[NH 3 ]
(c ) K =
[NH 3 ]
[NH +4 ]
balancing coef’s
(d ) K =
reactants
[NH 3 ][H 2 O]
[NH +4 ][OH - ]
(e) K =
[NH +4 ][OH - ]
[NH 3 ][H 2 O]
The value of K is also temperature dependant.
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Writing Equilibrium Constant Expressions
Writing Equilibrium Constant Expressions
Reactions Involving Solids
• So long as a solid is present in the course of a reaction, its concentration is not included in the equilibrium constant expression.
Reactions in Solution
• If water is a participant in the chemical reaction, its concentration based on magnitude is considered to remain constant throughout.
S(s) + O2 (g) R SO2 (g)
NH3 (aq) + H2O(l) R NH4+ (aq) + OH− (aq)
• Equilibrium constant:
• Equilibrium constant:
K=
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[SO2 ]
[O2 ]
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K=
74
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[NH+4 ][OH− ]
[NH3 ]
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17
Writing Equilibrium Constant Expressions
Writing Equilibrium Constant Expressions
Reactions Involving Gases: Kc and Kp
• Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems. • In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in Kc. • For gases, however, equilibrium constant expressions can be written in another way: in terms of partial pressures of reactants and products, Kp. Reactions Involving Gases: Kp
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The Reaction Quotient
Kp =
PHI2
PH2 × PI2
• Notice that the basic form of the equilibrium constant expression is the same as for Kc. In some cases, the numerical values of Kc and Kp are the same. They are different when the numbers of moles of gaseous reactants and products are different.
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The Reaction Quotient
In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q
REACTION QUOTIENT, Q.
If Q < K then the system is heading towards equilibrium:
There are more reactants than products as expected at equilibrium. The reaction is said to be headed to the “right”.
aA + bB R cC + dD
Reaction Quotient = Q =
If Q > K the system has gone past equilibrium:
There are more products than reactants as expected at equilibrium. The reaction is said to be headed to the “left”.
[C]c [D]d
[A]a [B]b
If Q = K, then system is at equilibrium.
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H2 (g) + I2 (g) R 2HI(g)
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If Q = K then the system is at equilibrium.
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Oxidation/Reduction Reactions Electrochemistry
Applications:
• Batteries
• Corrosion
• Industrial production of chemicals such as Cl2, NaOH, F2 and Al
• Biological electron transfer reactions
Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s)
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Electrochemical Cells
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When the two half reactions are separated, but connected electrically, the flow of electrons can do work!
If a RedOx reaction is spontaneous, the electron flow will have a positive potential which can be used to do work.
When the individual processes are separated by a salt bridge, the electron flow can be completed by a wire.
A group of such cells is
called a “Battery”
This configuration is called
a “Galvanic” or
“Voltaic” cell.
In the cell, oxidation (loss of electrons) occurs at the Anode
reduction (gain of electrons) occurs at the cathode
Electrons flow from the anode to the cathode
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Cathode
(reduction)
Anode
(oxidation)
Electrochemical Cells
1. Oxidation occurs at the anode (electrons are generated).
[cations] build up
[cations] reduce
2. Electrons flow from the anode to the cathode by means of a wire where reduction occurs (electrons are consumed)
NO3−
e−
3. The generation of cations at the anode is balanced by the flow of anions across the salt bridge to the anode.
K+
M → Mn+ + ne−
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ne−
+ Xn+
4. The loss of cations at the cathode is balanced by the flow of cations across the salt bridge to the cathode.
→X
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Electrochemical Cell Notation:
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Calculating Eocell:
A Galvanic or Voltaic cell is diagramed in the following manner.
Anode
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Eocell = Eocathode − Eoanode
•Where the cell potentials are the individual reduction potentials at standard state conditions.
Catode
Fe(s)|Fe2+(aq) || Cu2+(aq)|Cu(s)
•The oxidation half reaction represents the anode.
•The reduction half reaction represents the cathode.
Oxidation
reduced form
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salt bridge
Reduction
The single vertical lines represent a phase boundary.
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reduced form
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•If Eocell is > 0 (positive) then the reaction is spontaneous as written.
•If Eocell is < 0 (negative) then the reaction is spontaneous in the reverse direction.
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Eocell = Eocathode − Eoanode
Consider the reaction of Ag+(aq) and Cu(s):
When copper metal is dropped in to a solution of silver ions, the silver plates out.
= 0.799 V − 0.337 V = 0.462 V Cu(s) +2Ag+(aq) → Cu2+(aq) +2Ag(s)
What is Eocell? Cu(s) +2Ag+(aq) → Cu2+(aq) +2Ag(s)
(oxidation)
Cu(s) → Cu2+(aq) + 2e−
(reduction)
2Ag+(aq) + 2e−
(anode)
→ 2Ag(s) (cathode)
Eocell = Eocathode − Eoanode
= 0.799 V − 0.337 V = 0.462 V EIT (2)
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Eocell
+0.337 V
+0.799 V
The reaction is spontaneous as written!
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Colligative solution properties are properties that depend directly on the concentration of solute particles in the solution. Experiments demonstrate that the vapor pressure of water (solvent) above a solution is lower than the vapor pressure of pure water at a given temperature.
When a solute is added to a solvent, the boiling point increase and the freezing point decreases. Also, when a pure solvent is separated from a solution by a semi‐permeable membrane, solvent molecules flow across the membrane towards the solvent side. This phenomenon is know as osmosis.
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(oxidation)
Cu(s) →
(reduction)
2Ag+(aq) + 2e− → 2Ag(s) (cathode)
(anode)
Eocell
+0.337 V
+0.799 V
Note!
•Even though the oxidation reaction is the reverse of the reaction in the reduction table, do not reverse the sign of Eo.
•Even though there are 2 mols of Ag+, do not double the reduction potential!
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Boiling Point Elevation:
COLLIGATIVE PROPERTIES OF SOLUTIONS
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Cu2+(aq) + 2e−
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The boiling point of a solution is always higher than the boiling point of the pure solvent of the solution. tb (solution) > tb(solvent) The difference in boiling point between pure solvent and solution depends on the concentration of solute particles, and is calculated using the following equation: ∆tb = nKbM
∆tb = tb(solution) − tb(solvent) M = the molarity of the solution
Kb = bp constant (depends on the substance)
n = total number of particles in solution
n = 1 for molecular compounds
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n = 2 for NaCl (Na+ and Cl−)
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Calculate the freezing point of a solution made up by adding and
completely dissolving 4.52g sodium phosphate to 100.0 mL of water.
Calculate the freezing point of a solution made up by adding and
completely dissolving 4.52g sodium phosphate to 100.0 mL of water.
∆tf = nKfM
∆tf = nKfM
K f = −1.86
o
C
M
step 1: Calculate the molarity of the solution
step 2: Recognize that n = 4 K f = −1.86
n = 4
o
C
M
o
1molNa 3 PO 4
⎛
C⋅L ⎞
1
103 mL
×
∆tf = 4 × ⎜ −1.86
×
× 4.52g Na3PO4 ×
⎟
163.94g
mol ⎠
100.0mL 1L
⎝
3−
Na3PO4(aq) → 3Na+(aq) + PO 4 (aq)
∆tf = −2.05 oC
n = 3 + 1= 4
tf = 0.00 oC + ∆tf
= 0.00 oC − 2.05 oC =
−2.05 oC
step 3: enter the values into the equaton
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Reaction Orders:
For the general reaction:
A reaction order can be zero, or positive integer and fractional number.
aA + bB → cC + dD
Order
Name
0
zeroth
Each concentration is expressed with an order (exponent).
1
first
The rate constant converts the concentration expression into the
correct units of rate (Ms−1). 2
second
0.5
one‐half
1.5
three‐half
0.667
two‐thirds
Rate = k [A]x [B]y
x and y are the reactant orders determined from experiment.
x and y are NOT the stoichiometric coefficients.
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Rate law Expression:
x
The Overall Order of a reaction is the sum of the individual orders:
y
Rate = k [A] [B]
1
Rate (Ms −1 ) = k[A][B] 2 [C]2
If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one…
one… 1
1 + ½ + 2 = 3.5 = 7/2 Overall order:
or If the rate remains constant as [B] doubles and [A] is held constant, the order for [A] is? zero…
zero… 0
seven−halves order
note: when the order of a reaction is 1 (first order) no exponent is written.
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EXAMPLE: The reaction,
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The reaction of nitric oxide with hydrogen at 1280°C is: third order in NO
is experimentally found to be first order
first order in H2 and third order
a) Write the rate law.
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
the “
the “1” in first order is omitted
1
Rate(Ms‐1) = k [H2] [NO]
3
From the following experimental data, determine the rate law and
rate constant.
b) What is the overall order of the reaction?
“4th order”
order”
1 + 3 = 4
c) What are the units of the rate constant?
Rate =
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M
s
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Determining Reaction Order: The Method of Initial Rates
2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g)
Overall order =
Dr. Mack
= k × M × M3 = k × M 4
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k=
M
= M −3s −1
s × M4
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Run
[NO]o (M)
[H2]o (M)
Initial Rate (Mmin‐1)
1
0.0100
0.0100
0.00600
2
0.0200
0.0300
0.144
3
0.0100
0.0200
0.0120
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2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
y
y
0.0120M min −1 [0.0200]
⎡ 0.0200 ⎤
=
=
y
⎢⎣ 0.0100 ⎥⎦
0.00600M min −1 [0.0100]
The rate law for the reaction is given by:
Rate(M/min) = k [NO]x [H2]y
= [2.00]
y
2.00
Taking the ratio of the rates of runs 3 and 1 one finds: Take the log of the equation:
x
y
k [NO](3)
[H 2 ](3)
Rate (run 3)
= x
(1)
log
y
2 (1)
k [NO] [H ]
Rate (run 1)
= [ 2.00]
2.00
y
Plugging in the values from the data:
k [0.0100]x [0.0200]y
0.0120M / min
=
k [0.0100]x [0.0100]y
0.00600M / min
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=
[0.0200]y
[0.0100]y
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move the exponent out in front
log(2.00) =
⎛ 0.0200 ⎞
log ⎜
⎟
⎝ 0.0100 ⎠
= log [ 2.00]
log(2.00) EIT (2)
Similarly for x:
y
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Rate(M/min) = k [NO]x[H2]y
x
y
k [NO](1)
[H 2 ](1)
Rate(1)
=
x
y
Rate(2)
k [NO](2) [H 2 ](2)
y
0.00600 k [0.0100]x [0.0100]
=
0.144
k [0.0200]x [0.0300]
⎛ 0.0200 ⎞
log(2.00) = y × log ⎜
⎟ = y × log(2.00)
⎝ 0.0100 ⎠
1
=
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log(2.00)
y=
log(2.00)
x
⎛1⎞ ⎛1⎞
⎜ ⎟ ×⎜ ⎟
⎝ 2⎠ ⎝3⎠
x
⎛1⎞ 1
⎜ ⎟ =
⎝2⎠ 8
y = 1
Now that “y” is known, we may solve for x in a similar manner:
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Rate(M/min) = k [NO]x[H2]
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⎛1⎞
⎛1⎞
x log ⎜ ⎟ = log ⎜ ⎟
⎝2⎠
⎝8⎠
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x = 3
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The Rate Law is:
The Rate Law expression is:
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
Rate = k
Rate(M/min) = k [NO]3[H2]
To find the rate constant, choose one set of data and solve:
3
[NO]x [H 2 ]y
0.0120
The order for NO was 3
M
3
= k ( 0.0100M ) ( 0.0200M )
min
M
min
k=
3
( 0.0100M ) ( 0.0200M )
0.0120
The order for H2 was 1
The over all order is 3 + 1 =4
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rate law expression
M
min
=
3
( 0.0100 ) ( 0.0200 ) M 4
k = 6.00 × 105
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0.0120
M −3
min
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Homogeneous and Heterogeneous Catalysts
A homogeneous catalyst exists in the same phase as that of the reactants.
In the presence of the bromide ion, the decomposition of hydrogen peroxide occurs rapidly
2H2O2(aq) →
2H2O(l)
Step 1:
2Br−(aq)
aq) + H2O2(aq) + 2H+(aq) →
Step 2:
Br2(aq) + H2O2(aq)
Net:
Ea
+ O2(g)
Br2(aq) + 2H2O(l).
→ 2Br−(aq)
aq) + 2H+(aq) + O2(g).
2H2O2(aq) → 2H2O(l)
+ O2(g)
Br− acts as a catalyst since it regenerates at the end of the reaction.
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A heterogeneous catalyst exists in a different phase from the reactants.
In the gas phase, H2(g) + C2H4(g) → C2H6(g) is slow…
Ea
The catalyst allows the H2
molecule to dissociate at lower than normal temperature.
The molecules adhere to the catalyst surface….
Ea(cat)
platinum metal
The Ea for the catalyzed pathway is lower.
The reaction takes a “shortcut”, therefore the rate increases!
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Consider H2
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bond energy = 218 kJ/mol
On the surface of the metal, there are full orbitals that stick out into space:
When the H2 sits on the surface, the electrons from the metal can fill the anti‐bonding orbital
↑↓
σ*1s
1s
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↑↓
σ1s
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Since the energy of activation is lowered, the reaction proceeds with a faster rate!
Eact for the gas
phase reaction
Eact for the surface
reaction
20 electrons in an anti‐bonding MO
BO = 10
1s
2 electrons in a bonding MO
The bond order becomes zero!
Dissociation!
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