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Quick Review of Skills by Week

Week 1

Week 2

Solution Stoichiometry

Balancing Redox Reaction Equations

The skill to balance redox reaction equation is rather complicated, but it can be broken down into some steps.

The first step is the skill to identify Oxidation states of any element in a chemical formula. This link helps you to acquire that skill, if you have not already done so.

The second step is to Balance Half Reaction Equations. a skill to acquire from this link.

As a review, here are the guidelines for balance reaction equations:

  1. Identify the elements that are oxidized and reduced by examining their Oxidation states
  2. Write the oxidation and reduction half-reactions and Balance Half Reaction Equations. In an acid solution, use H+ and H2O to balance the charges and other atoms. In a basic solution, use OH- and H2O to balance the charges and other atoms.
  3. Add the two half-reactions algebraically such that the electrons in the two half-reaction equations cancel completely. Cancel other species such as H+, OH-, and H2O common to the two sides, if necessary.
  4. Check your equation and make certain that numbers of atoms and charge are equal on both sides.

Week 4: The Gaseous State

Week 5: The Gaseous State, continue

Ideal gas law and Dalton's law of partial pressures

From the Ideal Gas law to Dalton's law for a system containing gases 1, 2, 3, ... P V = n R T
      = n
Total R T
      = (n1 + n2 + n3 + ...) R T
      = (P1 + P2 + P3 + ...)

Mole fraction of gas 1, X1

          n1
X1 = -----
          nTotal
Example: A 1.0-L cylinder contains 0.5 g each of N2, O2 and CO2 at 273 K. Calculate the total pressure, partial pressure, and mole fractions.
Hint

See the table of results

Gas Amount /mol P = n R T /atm Mole fraction
N2 0.5/28 = 0.179 P = 0.40 0.40
O2 0.5/32 = 0.156 P = 0.35 0.35
CO2 0.5/44 = 0.114 P = 0.2540.25
Total 0.449 1.004 1.00

Collecting Gases over Water

The vapour pressure of water is a function of temperature, and a plot is shown.
|Vapor Pressure
|
|     Illustrated during the lecture
|
|___________Temperature

The vapor-pressure vs. temperature curve is part of the Phase Diagram of water, and the vapor pressure is related to humidity and dew points. Example: On July 1, the atmosphere pressure was 745 mmHg kPa at 30 C and the air is saturated with water vapor, 31.8 mmHg. Estimate the mole fraction of water vapour.
Hint

X = 31.8 / 745 = 0.043 (4.3 %)

The partial pressure of dry air would have been 745 - 31.8 = 713 mmHg.

 

Kinetic Theory of Gases

Postulates: For 1.0 mole of ideal gas, N = Avogadro's No., V = molar volume.

P V = N m u2/2
      = (3/2) R T

u = (3 R T / N M)1/2
      = (3 k T / M)1/2

Comparing 2 gases

        M1 u12  =  M2 u22.
   Thus,

        M1     u2
        -- = (---)2
        M2     u1

  The Graham's law of effusion is

                              k             k '
rate of effusion = ------ = ------
                              d1/2       M1/2

Applications:

 

Real Gases

van Der Waals Equation (P + (n/V)2 a) (V - nb) = n R T

nb - Volume of n moles molecules.

(n/V)2 a - Pressure correction term.

Week 6: Thermochemistry

Explain what energy is and expand the discussion on the following:

Thermochemistry (Chemical energy)

Work on some examples and problems.

Week 7: Hess's Law and Introduction to Quantum Theory

Hess's law

Hess's law states that energy changes are state functions. C + 1/2 O2 -> CO,         dH° = -110 kJ/mol
CO + 1/2 O2 -> CO2,     dH° = -283 kJ/mol.
+ ___________
C + O2 -> CO2,         dH° = (-110)+(-283) = -393 kJ/mol.
From which, we write

       0 kJ ------------ C(graphite) + O2
              |       |
      -110 kJ |       |
              V       |
 CO + /2 O2 -----     |
              |       | -393 kJ
              |       |
      -283 kJ |       |
              |       |
              V       V
            ------------ CO2

Example 1

The enthalpy of combustion for H2, C(graphite) and CH4 are -285.8, -393.5, and -890.4 kJ/mol respectively. Calculate the standard enthalpy of formation dHf for CH4.

Solution:
Lets interpret the information about enthalpy of formation by writing out the equations as shown here.
dH°f
/(kJ/mol)
(1) H2(g) + 0.5 O2(g) -> H2O(l) -285.8
(2) C(graphite) + O2(g) -> CO2(g) -293.5
(3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) -890.4
From the above equations, derive
C + 2H2 -> CH4
Answer: C + 2H2 -> CH4 -74.7
Hint: 2*(1) + (2) - (3), Thus,
dHf = 2 * (-285.8) + (-393.5) - (-890.4) = ?

Discussionn:


   ===C(graphite) + 2 H2(g) + 2 O2(g)===
- 74.7 kJ |              |
   == CH4 (g) + 2 O2(g)== |
          |              |
          |              |
          |              |
          |              |-965.1 kJ
-890.4 kJ |              | [(-2*285.8-393.5) kJ]
          |              |
          |              |
          |              |
          |              |
          V              V
   ==========CO2(g) + 2 H2O(l)==========

Example 3

The standard enthalpies of formation of SO2 and SO3 are -297 and -396 kJ/mol respectively. Calculate the standard enthalpy of reaction for the reaction: SO2 + 1/2 O2 -> SO3.

Solution:
In order to show how the chemical reactions take place, and for a better appreciation of the technique of problem solving, we write the equations according to the data given:

SO2(g) -> S(s) + O2(g)       dH = 297 kJ
S(s) + 3/2 O2 -> SO3           dH = -396 kJ
Add the two equations to give SO2(g) + 1/2 O2 -> SO3       dH = -99 kJ

Your turn to work:
Sketch an energy level diagram for the combinations of substances.

Calorimeter Measurements

Calorimetry the science of heat measurements, already mentioned. What is a Calorimeter?
Do you know how to apply the following: q = C dT = s m dT
C = q / dT

dH = dE + d(PV) = dE + P dV = dE + dn R T

Example 5 (see Calorimetry for other examples)

A table of thermodynamic data gives dHf = -285.8 kJ/mol for water. A bomb calorimeter measurement gives the heat of combustion for H2 as -282.0 kJ/mol. Estimate the error of the enthalpy measurement.

Solution
Reinterpret the problem, we have

H2(g) + 0.5 O2(g) = H2O(l),   dE = 282.0 kJ/mol.
Furthermore,
dn = -1.5
dH = dE + dn R T,
      = - 282.0 + (-1.5 mol * 8.314 J/(mol K) * 298 K)
      = (-282.0 - 3.72) kJ
      = -285.7 kJ
The error is (285.8-285.7)/285.8 = 0.03%

Discussion
More heat is giving of if the reaction is carried out at constant pressure, since the P-V work (1.5 R T) due to the compression of 1.5 moles of gases in the reactants would contribute to dH.

If 1.0 mole water is decomposed by electrolysis at constant pressure, we must supply an amount of energy equivalent to enthalpy change, dH, a little more than internal energy, dE. More energy must be supplied to perform the P-V work to be done by the products (H2 and O2).

Introduction to Quantum Theory

Stories told:

Here is an overallview of quantum theory:

Week 8

Review and examples. I hand-wrote the review lecture notes, which cannot be post here. Sorry!

Week 9

(Chapter 8) Electronic Configurations and Chemistry of the Periodic Table,
Midterm test on Friday for chapters 6, 7, and 8.

Review: Explain and apply the following terms

Week 10. Ionic and Covalent bonds (Chemical bonding)

Ionic Compounds Write out the Lewis Dot Symbols for the following

H

Li     Be     B     C     N     O     F     Ne

Na     Mg     Al     Si     P     S     Cl     Ar

Electronic configurations of ions

Main groupelements, ionization potentials (IP) of some elements

Element1st IP2nd IP3rd IP4th IP
Na 496 4,562 6,912 9,543
Mg 738 1,451 7,733 10,540
Al 578 1,817 2,754 11,577

Group IA, IIA, and IIIA - ions have pseudo-noble-gas electronic configurations

Groups IIIA to VA heavy metals: charge = (group number - 2)

Tl1+, Sn2+, Pb2+ and Bi3+

Anions of Groups VA to VIIA (pseudo-noble-gas electronic configurations)

N3-, O2-, Br1-

  Transition-Metal Ions

Ionic Radii

 

Born-Haber Cycle for the Calculation of Lattice Energy

Since the concepts of lattice energy and energy of crystallization are sound, and lattice energy is a useful quantity to judge the chemical properties of the salt, we can and usually calculate it by applying the law of conservation of energy, in a manner similar to the Hess's law. For the calculation of lattice energy, this process is called the Born-Haber cycle.

In 1919, M. Born and F. Haber separately devised a method to calculate the lattice energy from known thermodynamic data such as,

Hsub Enthalpy of sublimation,
Hmelt Enthalpy of melting,
Hvap Enthalpy of vaporization,
Hf Enthalpy of formation,
U Lattice energy,
Ecryst Energy of crystallization,
H Enthalpy of reaction,
D Bond (dissociation) energy
IP Ionization potential, or ionization energy, IE
EA Electron affinity,
For example, to calculate the lattice energy of NaBr, we start with Na metal and Br2 liquid. To make NaBr, each of the reactants need to go through a series of processes in order to calculate the lattice energy:

2 Na (s) --2 Hsub ® 2 Na (g) --2 IP ® 2 Na+(g)

Br2(l) --Hvap ® Br2 (g) --D ® 2 Br(g) --2EA ® 2 Br-(g)

One other reaction is required

2 Na + Br2 ® 2NaBr, 2Hf

The lattice energy corresponds to the reaction

2 NaBr(s) ® 2Na+(g) + 2 Br-(g)

To put all the information together, we may now draw the cycle:

          THE BORN-HABER CYCLE
-----------2Na+ + 2 Br(g)-------- ­ | |2IP + D |2 EA | ¯ 2Na(g) + Br2(g) 2 Na+(g) + 2Br-(g) ­ | |2Hsub+Hvap | | | 2Na(s) + Br2(l) | 2Ecryst | | |2Hf | ¯ ¯ ----------2 NaBr(s)---------------
Thus,
2Hf - (2Hsub + Hvap + 2IP + D) = 2EA + 2Ecryst

Note that we have used two moles of NaBr in the above diagram.

This scheme shows that we can calculate the lattice energy of NaBr from some known thermodynamic data. The same can be calculated from reaction equations and their associated energies. This is illustrated below

2 Na(s) + Br2(l) ® 2 NaBr(s) 2 Hf
2 Na(g) ® 2 Na(s)
or 2 Na(s) ¬ 2 Na(g)
- 2 Hsub
2 Na+(g) + 2e ® 2 Na(g)
or 2 Na(g) ¬ 2 Na+ + 2e
- 2 IP
Br2(g)®Br2(l)
or Br2(l) ¬ Br2(g)
- Hvap
2 Br(g) ® Br2(g)
or Br2(g) ¬ 2 Br(g)
- D
2 Br-(g) ® 2 Br(g) + 2 e
or 2 Br(g) + 2 e ¬ 2 Br-(g)
- 2 EA
Add all the above equations leading to
2 Na+(g) + 2 Br-(g) ® 2NaBr(s) 2 Ecryst
Thus, 2 Hf - 2Hsub - 2 IP - Hvap - D - 2 EA = 2 Ecryst
Ecryst = Hf - Hsub - IP - (Hvap + D) / 2 - EA
The same result as shown in the diagram

The H-H molecule

Describe the hydrogen molecule in light of the following:

Bondlength and bond energies

Week 11. Ionic and Covalent bonds (continue) and Molecular Geometry

Lewis Dot Structure

Valence-shell electron-pair repulsion models

  1. Identify the central atom in a molecule containing more than two atoms as a start.
  2. Identify the number of valence electrons of any element. This concept is important, because you need to know the number of valence electrons in order to write a Lewis dot structure for the molecule in question.
  3. Count the number of VSEPR pairs or steric number (SN) for the central atom in a molecule. You need this number in order to describe or predict the shape of the molecule in question.
  4. Determine the number of lone electron pairs that are not shared with other atoms. Often, a Lewis dot structure is useful to help you count this number.
  5. Predict the shape of molecules or inos as the key concept of VSEPR theory. From the shape and by applying the idea that lone electron pairs takes up more space, you can predict the bond angles withing 5% of the observed values.
  6. Predict the values of bond angles and describe the hybrid orbitals used by the central atoms in the molecules or ions.

Dipole moment and molecular geometry

Mostly hand-written notes, not available here.

Hybrid atomic orbitals valence bond theory and VSEPR model.

Week 12. Molecular Geometry

Valence bond theory and hybrid orbitals

Mostly contained in the website above.

Hybrid orbitals in chemical bonds of organic compounds

Describe the hybrid orbitals used in various organic compounds. The hybrid orbitals are described using hand-drawn diagrams and models. If you miss the lectures, it will take you a lot of time to catch up. Your ability to describe the bonding and shape of these molecules is very important. By applying the basic ideas provided here, you will be able to understand the complicated organic molecules in the future.
HCCH H2CCH2 H3CCH3 H3CNH2
H3COH C3CHO H3CCOOH H3CCCCH3
CO CN- NO SO2
C6H6 C6H12 NCS- CO2
H3C-CC-C-NH-CH(OH)-CH=C=O

Molecular orbital (MO) theory.
MO of Li2, Be2, B2, C2, N2, O2 and F2

Much more extensive than the coverage of the text book. Because hand drawing and explanation are required, no information is provided on the website. Final Examination Questions will require all concepts introduced during the lectures.

Week 13. (Monday only)

Review: stoichiometric calculations, chemical energy, Hess's law, Born-Habor cycle, etc.

Final Examination

Unknown yet

The final examination is divided into two parts.

Part A (30%) are questions that require you to provide written answers. In providing answers to them, you have to apply several concepts. Most of the exercises and CAcT questions are base on one concept. We have not finalized the exam paper yet, but we are currently considering letting you choose 3 out of 4 questions in Part A. Chemical bonding (Chapters 9 and 10) concepts are important for this part.

Part B (70%) consists of 35 multiple choice questions. You already know the style of these questions.

The average for the final examination is usually lower than those of term tests. The average for the written part is usually lower than the multiple choice part.

E-mail cchieh@uwaterloo.ca