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Lecture Notes

CHEM 142

Week 1 - Course Introduction

Week 2 - Quantum Mechanics

Week 3 - Quantum Mechanics and Chemical Bonding

Week 4 - Chemical Models

Week 5 - Midterm Review and Stoichiometry

Week 6 - Stoichiometry

Week 7 - Reaction Stoichiometry

Week 8 - Gas Stoichiometry

Week 10 - Kinetics

Week 11 - Kinetics and Review

Week 1 Wednesday Lecture

What is Chemistry?

  • Study of matter and how it changes
    • Study of transformation
    • Study of interactions between atoms and molecules
    • Designing and manipulating properties
    • Learning theories of underlying properties
  • Modern chemistry
    • iPhone and QLED TV
    • How can the phone emit light from electricity?
    • How can molecules emit different color lights?
    • Asking questions about modern chemistry in labs.
    • Design molecules and materials with properties.
    • Quantum computing - utilizing quantum nautre of molecules for computation.
    • Class is framed in real-world applications
    • Tokyo Narita Airport, Moscone Convention Center - photovoltaics

What Do Chemists Do?

  • Industry - developing new fuels, products, food additives, mediicnes, packaging, purification
  • Academia - research on new chemicals, new reactions, new instruments, no ideas
  • Other - patent law, clinical medicine, schooling
  • Many job companies - what skill set do you wnat on your resume?

Resources to help Navigate the General Chem Series

  • Content help with lectures, exams, quizzes, and worksheets - TA< instructor, website/syllabus
  • Missing quiz or exam - Chem absence reporting Google form
  • ALEKS - TA, no extensions for missed deadlines; register and begin working on problems.
  • Registration questions and entry codes - Chemistry Undergraduate Services (
  • Content help with prelab and lab report - TA
  • Canvas for prelabs – TA
  • Gradescope for Lab Reports - TA, read instructions on lab website. No extensions for missed deadlines, 0 for improperly linked or uploaded reports.
  • Missing a lab - UW CHem Absence Reporting Google Form
  • You are in the middle - learn how to manage it. Go to TA office hours, discussion sections, quiz section questions


  • Lectures - Live in KNE 130 - previously recorded videos on Canvas site, no demos.
    • Two demonstrations per lecture
  • Course Instructor - Prof. Xiaosong Li, email for coursework
  • Lab Instructor - Prof. Andrea Carroll, email for lab work
  • Both distinguished teaching faculty
  • Discussion Section and Office Hours - TAs.will lead


  • Textbook by Zumdahl - 7th UW custom edition. CHEM 142 version contains Chapters 2, 12, 13, 3, 4, 5, 15.
  • Student Solutions Manual
  • Lab notebook
  • Lab coat and goggles
  • Scientific calculator
  • ALEKS access
  • Course Components

  • Two midterm exams - 35% (17.5% on each), 45 minute exams.
  • Final Exam (Saturday, December 11th, 8:30 AM to 10:20 AM) - 30%
  • Participation (worksheets, week 2 - week 11) - 2%
  • ALEKS Pie - 18%
  • Laboratory - 15%
  • Failing early on is good for you - do not be afraid of dropping the class if you do not see it fit.
  • No set office hours; meet with professors individually and will be matched with prof. schedule.
  • Don’t just talk about grades, look at study habits, success, etc.
  • Reach out to TA for course material.

Course Policies

  • UW email address only; email your TA first
  • Absence Reporting Google Form
  • No extension for any ALEKS Objective
  • No makeup exam
  • Course grade of 0 for >1 unexcused lab absence or absence from final exam.
  • No makeup exams.

Start of Qurter Checklist

  • No lab, discussion section in week 1 (Sept. 27 to Oct. 1)
  • Setup ALEKS online homework account.
    • Reach/watch/register ALEKS tutorials
    • Finish Initial Knowledge Check, due Thursday, Sept. 30 midnight
    • First objective due Sunday Oct. 3 @ 11:59 PM. Start early, will take several hours.
    • Unique learning path for each person.
  • Read Syllabus and Course policies, finish sylabus and course policy quiz
  • Finish working through ALEKS tutorials and score 100% on the syllabus quiz to unlock lecture notes (includes worksheets, notes, etc.).
  • Learn how to use PollEverywhere.
  • First PollEverywhere quiz is on Oct. 1, Friday. Work on problems together; if majority of students make mistakes, will work on problem. Not graded.

Textbook Coverage

  • Ch. 12: Quantum Mechanics and Atomic Theory
  • Ch. 13: Bonding - General Concepts
  • Ch. 3: Stochiometry
  • Ch. 4: Chemical Reactions and Solution Stoichiometry
  • Ch. 5: Gases and the Kinetic-Molecular Theory
  • Ch. 15: Chemical Kinetics (needed for starting 152)
  • Chapters 3-5 are ‘boring’; there are being taught at a higher level. Putting everything together.
  • Material will be fun; don’t be too confident.

What is ALEKS?

  • ALEKS is an online learning enviornment for general chemistry.
  • ALEKS adapts to learning needs of each student.
  • Two components - working problems to develop skills and learn concepts, periodic assessment of learning using knowledge checks.
  • Based on your work in KCs, ALEKS determines the most effective pathway through the material to develop your knowledge.
  • Do not wait until last few hours.

Week 1 Friday Lecture


  • Next week - lab safety training.

Review Content

  • Mass number - number of protons and neutrons (excluding electrons). Have the same number of protons but different number of neutrons - different isotopes of the same element.
  • If there is no + or - in upper right, is neutral.
  • How many neutrons does a \(^{60}Co\) atom have (atomic number 27) - \(60-27=33\).
  • The chemical name for \(H_3PO_4\) is phosphoric acid - three hydrogens.
  • \[HClO_2\]
    • Hydrochloric Acid - \(HCl\), strong
    • Perchloric Acid - \(HClO_4\), strong
    • Chloric Acid - \(HClO_3\)
    • Chlorous Acid - \(HClO_2\), weak
    • Hypochlorous Acid - \(HClO\), weak

Live Experiment

View video of experiment:

Week 2 Monday Lecture


  • Required Chapters: 12.1-12.5, 12.8-12.3, 12.15-12.16.
    • A few sections will not be covered.
  • Quantum mechanics will be taught from an energy transformation perspective rather than a mathematical perspective.
  • This is a very important but chaotic chapter.
  • How to read and understand different concepts?

Energy Sources, Conversions, Use

  • All research deals with the transformation of different types of energy.
  • One source of energy - sunlight, is an energy form. Used to understand atomic structure, the transofrmations, etc.
  • Defining light as an energy form won Einstein a Nobel Prize.
    • Foundation of modern quantum emchanics.
  • Chemical energy, heat, nuclear, mehcanical work, electrochemical, electricity, etc.
  • Converting from one form of energy to another.
  • Two important concepts:
    • Energy (what forms, how they transfer from one form to another), 12.1 to 12.4.
    • \(\psi\) Wave Function, 12.5 to 12.10. Shape of any quantum mechanical property.
      • in the quantum mechanical world, you are looking at the wave function shape.
    • Builds into periodic table - remainder of chapter.
  • Schrodinger Equation
  • Need to understand how different shapes of electrons and forms of energy operate.
  • Chemical energy and light.

Chemical Internal Energy

  • Internal energy of a chemical system includes kinetic and potential energy of molecules.
  • Particles move in space; kinetic energy is \(\frac{1}{2}mv^2\).
    • \(m\) is particle (atom, electron, neutron, etc.) mass
    • \(v\) is the particle velocity.
  • Potential energy arises from electrostatic intermolecular attractions and chemical bonds.
    • If you have one electron and another, they will form electrostatic repulsion.
    • Potential energy only arises from interactions between more than 1 particle.
    • There is no potential energy in just one particle (only kinetic energy).
  • \[E = E_K + E_P\]

Light is Electromagnetic Radiation

  • Light is a wave; it has a diffraction property.
  • Microwave (long), infared (shorter), sunlight (quite narrow).
  • Visible light is a small part of the entire electromagnetic radiation spectrum.
  • Adding all the colors together makes white. If every pixel emits all colors of light, you will see pure white. The fact you see different colors is because each pixel can only emit a certain color of light.
  • Quantization: only certain colors can be emitted, not all.
  • Intensity of light is proportional to $$E^244.
  • Every wave obeys a certain rule.
  • Our goal is to discover these rules.

Characterizing Light

  • Wavelength (\(\lambda\)): distance between two consecutive peaks in the wave.
    • Standard SI unit is meters.
    • You will deal with units all over the place; convert to SI units so you don’t make a mistake.
    • Sometimes nanometers or picometers are used; learn how to perform conversion.
  • Frequency (\(\nu\)): number of waves that pass a given point in space per second.
    • The unit of frequency is \(s^{-1}\) (hertz).
  • Energy (\(E\)): increases with \(\nu\), decreases with \(\lambda\).
  • $$\lambda \times \nu = c = 3.0 \times 10^8 \text{ms}^{-1}
    • Must be memorized.
  • Light travels through space like a wave.

Energy Quantization

  • Planck found that in order to correctly model blackbody radiation, one must assume energy transfer between light and matter is quantized.
  • Experiment from Friday: glowing piece of metal. Aluminum displaced iron oxide; at the end we have a piece of iron. Reaction was very exothermic and matter possessed high energy; energy was released in terms of light. Showcased energy transformation from energy transfer to light.
    • Won Max Planck the Nobel Prize.
    • Planck can change the temperature of the matter and measure the frequency of the light emitted from the metal; to build a connection.
    • Planck attempted to use a classical theory to explain this.
    • To explain the blackbody radiation experiment, a new hypothesis was derived.
  • \[\Delta E = nh\vu\]
    • \(\Delta E\) is energy change/released
    • \(n\) is an integer
    • \(h\) is Planck’s constant
    • \(\nu\) is the wavew frequency
  • Blackbody must release energy; energy is released as a small packet (many packets of energy). Packet of energy is defined as \(h\nu\).
    • \(\nu\) has units of \(s^{-1]\) and \(h\) is the Planck constant in Joules of seconds; yields the smallest packet of energy in Joules.
    • No packet of energy can be released less than \(h\nu\).
    • The blackbody can only release energy in multiples of \(h\nu\).
    • Before Planck, many believed energy release was a continuous spectrum.
  • Quantum mechanics was not derived mathematically, but from hypotheses that were verified via experiment (beginning from Planck).
  • While Planck introduced quantized chemical energy, Einstein introduced quantized light.

The Photoelectric Effect

  • We can shine light on a certain frequency of a metal surface, electrons will be emitted. According to classical physics, the energy of a wave is related to its amplitude. We should observe electron injection at any frequency if the amplitude is large enough.
  • Performed the experiment and came up with a crazy idea to explain what was observed.
  • View live experiment here
  • Concept of a photon: a packet of energy. Light is in a wave, but it is in a particle and has energy. Each particle has energy packet of \(h\nu\).
    • This energy is absorbed by electrons in the metal plate.
    • The electrons have kinetic energy and transform this packet of photon energy in kinetic energy.
    • \[h\nu \rightarrow \frac{1}{2}Mv^2\]
    • Leads to the development of quantum mechanics.
  • Why does long wavelength light not work but short wavelength light works?
    • Long wavelength light - because of \(\lambda\nu=c\), the frequency decreases and the energy is smaller.
    • Electron is bonded to the metal plate; it is trapped and cannot escape. For the electron to escape, it must pay a penalty in terms of energy (e.g. deposit for an apartment). If the energy is large enough, the electron can pay for it.
    • \(\phi\) - the penalty the electron must pay to escape; can be considered to be heat, bonding.
    • The electron absorbs \(h\nu - \phi\). Photon energy is \(h\nu\). This is what is left for the energy. The remaining energy is converted into electronic kinetic energy.
    • Completes Einstein’s photoelectric equation.
\[\frac{1}{2}m\nu^2 = h\nu - \phi\]
  • If \(h\nu < \phi\), the energy packet is not enough to pay the ‘penalty’ for escape.
  • If \(h\nu > \phi\), the energy packet is enough to pay the ‘penalty’ for escape.
  • This equation is very important because it has two energy forms: \(=\) implies transfer from one form to another, chemical and light.
  • The calculation of energy for particles cannot be used to define the energy of a photon.

Week 2 Wednesday Lecture

Energy Quantization

  • 1918 Nobel Prize winning equation: \(\Delta E = nh\nu\).
  • \(h\nu\) is a photon of energy. A system can absorb or emit \(n\) photons of energy \(h\nu\).
  • Energy transfer between light and matter is quantized.

The Photoelectric Effect

  • If you shine light of a certain frequency on a metal surface, electrons will be emitted.
  • We observe electron ejection at any frequency if the amplitude is large enougoh.
  • \[E_\text{photon} = h\nu\]
  • Electron - particle with mass. Photon - particle w/out mass.
  • Photon - packet of energy in the form of light. Propagates in space like waves.
  • Packets of energy come to the metal plate.
  • Electrons with mass reside in the metal plate; a particle w/out mass (photon) comes in. The photon exchanges energy (100%) to the electron; the electron absorbs the energy and the photon disappears.
  • The electron now possesses \(h\nu\) energy; pays a penalty (work function). Potassium has a work function of 2.0 eV.
  • Remaining energy - \(h\nu - \phi\) - will be converted into kinetic energy.
  • \(h\nu - \phi = \frac{1}{2}m_ev^2\).
  • Light is a packet of energy that can be absorbed.
  • Energy exchange exists; the energy can be re-emitted by certain species.
  • Solar energy conversion equation.
  • Concept of intensity - the energy of light is identical during bright and cloudy days; the frequency is the same.
  • The number of particles per second that you absorb is much more during sunny than cloudy days, though; it is quantity, not per-unit energy, that is the cause of variation.
  • If you increase the intensity of the light, you will see more photons ejected.
\[KE = \frac{1}{2}m_ev^2 = h\nu_\text{photon}-\phi\]

Workfunction \(\phi = h\nu_0\).

Photons and Energy

Example problem: What is the energy of a 500 nm photon?

Use \(c = \lambda \nu\) to find frequency \(\nu\) in \(s^{-1}\). Use \(\nu\) in \(E = h\nu\) to find the energy of a photon with that frequency. Note: If electron volt (eV) is used, conversion to Joules will be provided. Is a Level 2 problem - not a problem that should be missed.

Example problem: Electrons are ejected when sodium metal is irradiated with 400 nm light. What is the veloicity of the electrons? \(\phi_{Na} = 4.4 \times 10^{-19} J\).

We need to find \(\nu\) using \(c = \lambda\nu\) and use the photoelectric equation \(h\nu - \phi = \frac{1}{2} M_eV^2\).

Example problem: What is the momentum of an electron emitted under the same conditions as the previous problem?

Need to know the concept of moomentum - \(p_e = mv\). Solve for velocity. Another possible mathematical simplification - \(2M_e(h\nu - \phi) = M_e^2v^2 \implies \sqrt{2m_e(h\nu -\phi)} = p_e\). Not necessarily advised but a good exercise.

Notes for exams:

Exam is multiple choice. Circle things you need to pay attention to. Solve easy problems first - they are weighted the same as hard problems.

Wave Particle Duality

  • Einstein introduced the concept of a particle - light still passes through space like a wave.
  • Everything in this universe has a wave-particle duality - we are wave particles too, we have a wave.
  • After Einstein’s photon concept, light is also a stream of photons - a packet of energy.
  • Diffraction patterns are a consequence of the wave-like nature of light.
  • Quantized energy change is the consequence of the particle-like nature of light.
  • Matter must be quantized as well.

Atomic Emission and Absorption

  • Quantization of matter.
  • Atoms can emit EM energy only at specific, discrete wavelengths.
  • Why emit only certain types of light?

Bohr’s Quantum Model of the Hydrogen Atom

  • The electron circles the nucleus in only certain allowed circular orbits.
  • By convention, \(E=0\) at \(n=\infty\), because there is no interaction between the nucleus and the electron.
  • Orbit energies are quantized and described according to \(E = (-2.178]times10^{-18}J)\left(\frac{Z^2}{n^2}\right)\)
  • Successful explained discrete atomic emission.
  • Atom was originally stable; electrons jump to higher energy levels, which is not stable; electrons jump back to lower levels.
    • When the electron jumps to a lower energy level, energy is released \(\Dela E = h\nu\).
    • The jump is quantized (the electron cannot jump anywhere), meaning the energy is quantized.
    • Bohr’s equation can be applicable to Helium, Lithium, and other atoms.
  • Ionization at \(n = \infty\).
  • Transform photon into other forms of energy.
  • Understanding quantized energy is very important.
  • However, Bohr’s model works only for one-electron atoms.
  • *Problems like this will appear on the exam.8
  • This does not work for molecules; supercomputers must be used to solve quantum mechanical expressions.
  • Einstein introduced quantization of light, Bohr introduced quantization of matter, Planck introduced quantization of light to chemical energy.
  • Bohr’s model can be drawn as a series of circular orbits, which is wrong.
  • Energy levels inside atoms is quantized - you cannot have energy levels in between \(n\) as consecutive integers. As \(n\) indcreases, the gap between consecutive energy levels decreases.
\[\Delta E = \frac{hc}{\lambda_n}\]
  • The electron is not stable in an excited state; energy must be released and exchanged in the form of photons.
    • Photons - a packet of energy.
    • The release in energy must be equal to the energy gap - everything must be onserved.

Week 2 Friday Lecture


  • Began with Planck’s hypotheses of black body radiation.
  • Energy in matter must be quantized; when energy is released by a photon it also must be quantized.
  • Einstein introduced the concept of a photon to quantize light energy.
    • Every light particle carries a packet of energy.
  • Different route now - quantized nature of particles with mass.

Bohr’s Quantum Model of the H Atom

  • The electron circles the nucleus in only certain allowed circular orbits.
  • Electronic orbits are quantized; orbits are quantum numbers.
  • Derived Bohr’s expression for atomic levels: \(E = -2.178 \times 10^{-18} J \left(\frac{Z^2}{n^2}\right)\).
    • When \(n=\infty\), \(E = 0\).
    • Model successfully explained electron emission spectrum observed by other scientists.
  • All elements want to revert to the low-energy ground state.
  • When the electron makes the hop, the energy goes from high to low.
    • Energy must be conserved; therefore it is released in a packet - a photon.
  • \(E = h\nu\); \(c = \lambda \nu\); \(E = \frac{hc}{\lambda}\); \(\Delta E = E_\text{final} - E_\texts{initial}\)
  • Low level to high level - absorption. High levell to low level by emitting a photon - emission.
  • Bohr made a mistake by assuming electrons orbit around.

Example Problem


Use the Bohr model to predict what \(\Delta E\) is for any two energy levels.

\[E = E_\text{final} - E_\text{initial} = -2.178 \times 10^{-18} J \left(\frac{Z^2}{n^2_\text{final}} - \frac{Z^2}{n^2_\text{initial}}\]

For Hydrogen, \(Z=1\); for Helium, \(Z=2\). Must be used with one-electron atoms (e.g. Hydrogen, Helium+).

Use the Bohr model to predict the properties of a photon emitted by an energy jump from \(n_1\) to \(n_2\).

Level 2

At what wavelength will emission from \(n=4\) to \(n=1\) for the H atom to be observed?

\[\Delta E = -2.178 \times 10^{-18} J \left(\frac{1}{1} - \frac{1}{16}\right) \approx -2.04*10^{-18} J\]

In chemistry, \(-\) and \(+\) refers to exothermic or endothermic - energy is released in this case (exothermic) in terms of photons (a packet of energy).

\[\left|\Delta E \right| = hv = h\frac{c}{\lambda}\]

Level 3

At what wavelength will emission from \(n=4\) to \(n=1\) for the \(He^+\) atom to be observed?

Use \(Z=2\).

Level 3

*What is the longest wavelength of light that will result in tqhe removal of the electron from H in its ground state?

\[\Delta E = -2.178 \times 10^{-18} J \left(\frac{1}{\infty} - \frac{1}{1}\right)\]

Corresponds to ionization; the ionization condition is \(n_f = \infty\).

Atomic Emission

  • Lyman series, Balmer series, Paschen series; observe levels of emissions.
  • Electrons can make hops between any two levels - from \(n = 6\) to \(n = 4\), etc.
  • Different colors of light are emitted depending on the hop.
  • Bohr studied everyone else’s contributions and united them into a simple equation that explained everything; won the Nobel Prize in 1922.

Bohr Model Mistakes

  • Successfully desrcribed the line spectrum of hydrgoen and other one-electron systems, but could not describe the spectra of multi-electron atoms. It assumed a deterministic path for the electron -t he orbit. This assumes the electron is a particle.
  • The particle must collapse into the nucleus according to classical physics, since it has a negative charge and the nucleus has a positive charge.
  • Came up with a true model with a mathematical solution.

Those who are not shocked when they first come across quantum physics cannot possibly have understood it.

  • Niels Bohr, 1958.

Wave Interference

  • Light can produce diffraction patterns.
  • Diffraction pattern: two waves are propagating in space and time and form interference (destructive and constructive).
  • Diffraction experiment: waves go into a slit; the slits will propagate independently and form an interference pattern. Use diffraction nature to calculate distance between slits; study acoustic patterns, etc.

Diffraction of Light

  • Light is shined through a crystal, and the light is scattered as it passes by the ions in the crystal.
  • We can bomb X-rays through a sodium chloride crystal that forms a diffraction pattern.
  • The diffraction pattern can be used to calculate the crystal structure.

Diffraction of Particles

  • de Broglie was inspired by Einstein’s particle-wave duality.
  • Maybe everything has a wave particle duality.
\[\lambda = \frac{h}{mv}\]

For particles with mass, our wavelength can be calculated with the de Broglie equation.

  • No one believed de Broglie’s thesis; Einstein read it and convinced the committee to accept the thesis.
  • Quantum mechanics is based on many postulations and hypotheses proven by experiments and supported by mathematics. Was not derived from mathemaitcs.
  • How do we prove or disprove this?
  • If particles also behave like waves, we can bomb particles with mass through a crystal and observe diffraction patterns.

de Broglie Wavelength

What is the de Broglie wavelength of an electron traveling at 300 m/s? Note that \(m_e = 9.12*10^{-31} kg\).

\[lambda = \frac{6.63\times10^{-34}m^2 \cdot kg \cdot s^{-1}}{9.12\times10^{-31}kg \times 3.00m \cdot s^{-1} = 2.42\times10^{-4} m\]

Can only be used for particles with mass, not for light (use \(c = \nu\lambda\)).

  • de Broglie’s hypotheses is correct for electrons, neutrons, etc.
  • We cannot prove that humans are not waves because we cannot measure it, but we cannot disprove it either!
  • Why is this important? There is a nonzero probability that you can be of Mars.
  • Quantum entanglement - once our waves form a constructive and destructive pattern, they form entanglement and can be used for quantum information and computing.
  • However, we do not observe the quantized nature - the wavelength is so small that it cannot be observed.

Quantized vs Continuous Spectra

  • The energy levels available to electrons in atoms are quantized.
  • When atoms form molecules or extended solids, their atomic energy levles mix with each toher, forming molecular-scale energy levels that are also quantized.
  • As the number of interacting atoms increases, the number of molecular energy levles increases.

Wave Mechanics

  • Schrodinger and Louis de Broglie exploited wave-particle duality to describe matter and energy in terms of wave mechanics.
  • It is the electron shape that changes, not the orbital distance.
  • An electron can adopt only certain standing wave patterns of motion when subjecting to a constraining potential.

Week 3 Monday Lecture


  • Second part of chapter 12

Diffraction of Particles

  • Photon - particle without mass. Energy is calculated as \(h\nu = h\frac{c}{\lambda}\). Ultraviolet, infared, visible light, etc. are photons.
  • Shining EM radiateeion througha crystal yields a characteristic diffraction pattern, due to the wave-like nature of light. We can get a similar pattern with high-energy electrons.
  • Matter also exhibits the wave-particle duality.
  • Louis de Broglie described the wavelength of a particle travelling at a velocity \(v\): \(\lambda = \frac{h}{mv}\).
    • Everything, including things with mass, have a wavelength.
    • You can calculate your own wavelength (which is very tiny).
    • One is for particles, one is for photons.
    • One of the most revolutionary developments.
  • If we shine photons through a crystal, we will obtain a diffraction pattern.
    • We should be able to observe diffraction patterns if we shine electrons, neutrons, etc. - they are also waves, because only waves can generate diffraction patterns.
  • Our de Broglie wavelength is too small - no apparatus can detect this small change, it cannot be observed.
  • We can observe diffraction patterns for electrons and neutrons, which have longer wavelengths.
  • There is a nonzero probability of being anywhere.

The Schrodinger Equation

  • de Broglie introduced a beautiful expression - Schrodinger looked at the expression and attempted to find a mathematical way to calculate and quantify the wave nature of particles.
  • Inserted the de Broglie equation into classical wave equations and results in a beautiful Schrodinger equation.
\[\hat{H}\psi = E\psi; \hat{H} = \hat{E}_\text{kinetic} + \hat{E}_\text{potential}\]
  • \(\hat{H}\) is a collection of mathematical functions.
  • The solution comes as a pair; energy and psi (the wave function).
  • We can calculate the energy at the same time that we can calculate shape.

Wave Mechanics

  • If we solve the Schrodinger equation for one dimension, the shape of the wave function looks like a ‘string’; it is one particle, the shape of the one-particle. The solution is not unique; the equation can give rise to many solutions that are quantized.
    • A quantum particle can onlwy have a certain wave shape; the shape is quantized.
    • Transition in wave is associated with change in energy \(\Delta E\) - quantized shapes and orbitals.

2D Standing Waves

  • In two dimensions, we observe similar behavior - only certain wave functions are valid.
  • The whole wave is one quantum particle.
  • When electrons absorb a photon, they jump in their state from one quantized state to another.

Solutions to the Schrodinger Equation for H

  • Solutions come in pairs; we understand not only the energy but also the wave shape.
  • Once you solve the Schrodinger equation for the hydrogen atom, you obtain an energy constant identical to that of Bohr; the same energy levels that are computed using the Schrodinger equation.
  • Bohr’s model is correct, but Bohr made a mistake by assuming that electrons orbit around the nucleus - this is incorrect.
  • Electrons can take on many different shapes.
  • The probability amplitude of a wave function is referred to as an orbital.
  • A picture of an orbital represents a surface that contains 90% of the electron density.
  • Other solutions to the Schrodinger equation are increasingly complex.
  • Even in an orbital, shells can have different orientations; the \(m\) label is used to capture this.
  • Within the same level \(n\), there are different ‘sections’ \(l\); within each ‘section’, there are ‘seats’ \(m_l\).’=
  • In level \(1\), only \(s\) orbitals can be backed. Level \(2\) can pack \(s\) and \(p\) orbitals, etc.; we can pack more electrons as needed.

Electron Orbital Shapes

Must memorize electron orbital shapes.

  • How do we differentiate \(s\) orbitals?
    • Wave functions for \(s\) have only radial dependence - \(s\)-orbitals are spherical.
    • As \(n\) increases, orbitals demonstrate \(n-1\) nodes.
    • node - an area in a orbital with zero electron density.

Level 4 Question: Electrons in a hydrogen atom undergo a transition from the \(3s\) to the \(1s\) orbital. What is the wavelength of the emitted photon?

\[\Delta E = -2.178 \times 10^{-18} \left(\frac{Z^2}{n_f^2} - \frac{Z^2}{n_i^2}\right)\]

What is the initial and final quantum number? - \(n_i = 3\) and $$n_f = 1$.

Use derived energy to obtain the frequency using \(\Delta E = h\nu\).

  • \(p\) orbitals begin with level 2; they have radial and angular dependence, so they are not spherical but lobed; one node at the origin.
  • \(3p\) has more nodes as compared to \(2p\), as expected.
  • You can derive the principal quantum number using the number of nodes.

Level 5 Question: An electron in a hydrogen atom undergoes a transition from the 3p to the 2s orbital. What is the wavelength of the emitted photon?

\(n_i = 3\), \(n_f = 2\). They come from two different orbitals, but the only thing that affects the energy is the principal quantum number.

  • \(d\) orbitals begin with level 3; lowest possible energy level.
  • Five different possible orientations (not needed to be memorized).
  • There is a node that exists between successive quantum numbers of electrons.
  • \(f\) and \(p\) are beyond the focus of the class.

Week 3 Wednesday Lecture

  • Quantization and quantum phenomenon - discussed in four lectures.

Solutions to the Schrodinger Equation for H

  • Successfully predicted the hydrogen energy levels.
  • Energy levels of electrons are quantized by the principal quantum number \(n\).
  • Bohr’s mistake - assigning an ‘orbital’ to each energy level.
  • For \(n=1\), there is one spherically-shaped orbital.
  • Nature of electrons - there exist many orbitals, rather than only one.

Electron Orbital Shapes

  • \(s\) orbitals - spherical shape; at different levels, they appear slightly different.
  • By looking at the number of nodes, we can identify the principal quantum number \(n\) and calculate energy.
  • \(m_l\) - methods to determine orientation.
  • Number of nodes is \(n - l - 1\), where \(n\) is the principal quantum number and \(l\) is the angular momentum number (counting from \(s=0\)).
  • Energy increases as \(\frac{1}{n^2}\) orbitals of same \(n\), but different \(l\) are considered to be of equal energy (degenerate). The ground or lowest energy orbital is \(1s\).
  • A hydrogen atom undergoes a transition from \(3s\) to \(1s\) orbital. Another hydrogen atom undergoes a transition from the \(3d\) to the \(1s\) orbital. Which emits a photon with higher energy? A: the photon frequency is identical because the two initial orbitals are at the same initial level.
  • Most quantum mechanics is conceptual rather than mathematics.

Electron Spin

  • There is one more quantum number - Schrodinger also made a mistake.
  • Some particles (electrons in particular) demonstrated inherent angular momentum.
  • Arised from experiments - a beam of electrons are shot through a magnetic field. If any particle rotates, the magnetic field will rotate it either upwards or downwards. 50% of electrons are deflected upwards; the other 50% are downwards. Electrons are spinning.
  • Electrons can spin clockwise or counterclockwise; spin upwards or downwards. This is spin quantum number \(m_s \in \left[-\frac{1}{2}, \frac{1}{2}\right]\).

Pauli Exclusion Principle

No two electrons may occupy the same quantum state simultaneously.

  • An atomic quantum state is not completely specified by \(n\), \(l\), and \(m_l\). We also need \(m_s\).
  • Only two electrons may occupy a given orbital, and they must have an opposite spin.
  • Every electron in an atom will have a unique set of quantum numbers - \(n\), \(l\), \(m_l\), and \(m_s\).
  • To visualize, convert the four quantum numbers into shapes, accounting for spin and rotation.

Polyelectronic Atoms

  • Atoms with more than one electron.
  • H has only one electron, so all the sublevels have the same energy within a principal level - they are degenerate.
  • When there is more than one electron, the energies are not well-defined; there is no direct solution to the Schrodinger equation.
  • In Hydrogen atoms, 2s and 2p orbitals are degenerate; when you have more than one electron, electrons can attract with one another - Coloumbic repulsion, entanglement, correlation, etc. They are no longer the same energy level - 2p is at a higher level.
  • Energy must be calculated using the Schrodinger equation with a supercomputer.
  • At level 3, there is a bizarre behavior - 4s is lower than 3d in energy. Afbau principle.
  • Hydrogen does not only need to be in the 1s orbital state; it can also occupy other quantum states (although 1s is the lowest energy state) in an excited state.

Filling Orbitals

  • Orbitals are filled starting from the lowest energy, and the two electrons in an orbital must have opposite spin.
  • Helium can be excited into higher-energy quantum states; \(1s^12s^1\), for instance. Is a valid solution, but is in higher-energy excited state.
  • Many electrons have have more or less an unlimited number of states. We can manipulate electrons into different states.
  • Electrons can be put in different states to emit different wavelenghts of light.
  • Hund’s rule - the lowest energy configuration is the one in which the maximum number of unpaired electrons with parallel spin are distributed amongst a set of degenerate orbitals.
  • Valence shell - the \(s\) and \(p\) orbitals in the highest occupied \(n\) levels.

Week 3 Friday Lecture

  • Will post exam information sheet soon - will include conversion units, equations, and constants.
  • Pay attention for announcements on the Canvas website.
  • Chapter 13: Read 13.1-13.3, 13.9-13.13.

Orbital Energies in Polyelectronic Atoms

  • Root of derivation of periodic table.
  • Wasn’t originally derived from orbital energies, but it is still useful.
  • Many ways to organize elements; Mendeleev’s is the most efficient known.
  • For polyelectronic systems, because of electron interactions creating repulsions, orbitals with different quantum angular momentum numbers but the same principal quantum number do not necessarily have the same energy.

Filling Orbitals

  • Orbitals are filled starting from the lowest energy.
  • The two electrons in an orbital must have opposite spin.
  • Hund’s rule - the lowest energy configuration is the one in which the maximum number of unpaired electrons with parallel spin are distributed amongst a set of degenerate orbitals.
  • An atom or ion with a full valence shell is a particularly stable species.
  • Which element has the most/least number of unpaired electrons in an orbital?
  • Short noble gas base notation.
  • Naming groups - group number, A category/B category (main groups/transition metals).
    • Transition metals: highest energy orbitals occupying \(d\) and \(f\) orbitals.

The Aufbau Principle

  • Orbitals are filled in order of increasing energy, not increasing \(n\).
  • Similar to sodium, we begin the next row of the periodic table by adding electrons to the \(4s\) orbital.
  • \(4s\) allows for closer approach and is energetically preferred (i.e. lower energy); thus it is filled before the \(3d\) orbital.
  • Aufbau principle applies well for neutral atoms.
  • When we remove electrons to create a cation, \(4s\) energy is higher than \(3d\) because there is one less electron; the electron-electron interaction changes.
  • Building the periodic table
    • Elements \(Z=19\) and \(Z=20\) behave as expected
    • \(Z \in [21, 30]\) have occupied \(d\) orbitals. Transition metals - highest energy orbital is \(d\). Transition metals are used everywhere - ion, manganese, etc. Electrons can easily be moved in the \(d\) orbital.
      • Exceptions: Chromium and Copper; making a complete half-filled \([Ar] 4s^13d^5\) or a complete shell \([Ar] 4s^13d^10\) is more stable.
    • After Lanthanum, we start filling \(4f\); highest energy occupies the \(f\) block. \(4f\) - Lanthanide series, \(5f\) - Actinide series. These are important elements
  • Derived from quantum mechanics.
  • We will be given the electron configurations for the Periodic Table in the exam.

The Periodic Table

  • Valence refers only to the \(s\) and \(p\) electrons in the highest occupied energy level.
  • Atoms in the same group have the same valence shells, so they tend to have similar chemical properties.
  • Column name has the number of valence electrons in it.
  • Properties generally correlate down a group of elements.
  • Electrons occupy energy levels further from the nucleus, atomic radii increases. Ionization potential is smaller - easier to remove.
  • Only three periodic trends need to be known: electron affinity, ionization potential, and atomic radii.
    • Straightforward concepts.

Example exam question: Which element has the most number of unpaired electrons from \(Z \in [19, 31]\)? Chromium - the \(4s\) orbital is unpaired as one of the electrons is donated to the \(3d\) orbital.

Week 4 Monday Lecture


  • 20 to 22 questions in 45 minutes.
  • There are some easy problems, some difficult problems; you should not spend more than 2 minutes per question.
  • Photoelectric problem - can you solve it iwthin 2 to 4 minutes instead of 10 minutes?
  • When you work on ALEKS problems, utilize the opportunity.
  • VSEPR structures are given; once you know the Lewis dot structure, you can identify the molecular structure.
  • Identify dipole moments - does a molecule have a dipole moment?

Covalent Bonds

  • \(H_2\) - mutual attraction of electrons to the protons in the nucleus form bonds.
  • Bonds - something that holds atoms together and is very difficult to break.
  • Two hydrogens with \(1s^1\) electron configuration - when they meet together, they can share electrons and achieve noble gas state.
    • Fulfill the duet rule; ‘become’ helium.
    • Homogenous distribution - the electron pulling identities are identical.
    • Nonpolar bond; no dipole moment.
    • Diatomic molecules are nonpolar.


  • Not required on the exams.
  • Electronegativity - relative ability of an atom to attract electrons in a bond.
  • In general, electronegativity increases left to right and down to up across the table. Fluorine has the strongest electronegativity and Francium has the smallestt.
  • The greater the EN difference, the more ionic they are.
  • Elements in the 1A group tend to lose electrons; elements in the 7A group tend to gain them to fulfill noble gas status.


  • Electrons are more likely to be found in the atom with higher electronegativity.
  • Dipole Moment: is mathematically defined.
  • No experiment can measure EN, but dipole moment can be measured.
  • As long as two atoms are identical, there is no polarity. The strength is identical and you cannot separate it.
  • To identify dipole moment, we have to think about molecule structure - what happens to polyatomic molecules?
  • Think of each dipole moment as a vector (magnitude and direction). Vectors can cancel or add.
  • Identify dipole moment per bond to begin with.
  • Linear molecules - \(CO_2\), nonpolar and no dipole moment.

Valence Electrons

  • The total number of \(s\) and \(p\) electorns in the highest occupied energy level form an atom’s valence electrons.
  • Elements in the same group in the periodic table have the same number of valence electrons.
  • \(d\) orbitals are not used in Lewis dot structures.

Lewis Dot Symbols

  • Lewis dot symbols indicate the number of valence electrons in an atom.
  • Electron structure = Lewis structure.
  • Does not tell you about molecular shape - just about the arrangement of electrons.
  • Begin by writing the name of the element. The number of electrons on the outside should be equivalent to the number of valence electrons. Each direction can have only 2 electrons; a full ring contains 8 electrons (octet rule).
  • How do we consider ions? Remove or add electrons accordingly and indicate charge in the diagram.
  • The most stable arrangement of electrons is one in which all atoms have a noble gas configuration.
  • Lone pair - does not form a bond.

Lewis Dot Structure Mechanics

  • Replace bonding pairs shared between two species with a bar, known as a single bond. One bond represents an electron pair.
  • \(4A > 5A > 6A > 7A\) when determining central atom, can form more bonds.
  • In a polyatomic molecule, the atom that can make the most bonds is typically in the center. This is also typically the least electronegative atom.
  • \(H\) can only form one bond, so it goes on the outside of the molecule - \(H\) is a terminal atom.
  • If \(O\) and \(H\) both appear in a chemical formula, they are often bonded to each other.
  • When several \(C\) atoms appear together in a formula, they are probably bonded to each other in a chain.

Multiple Bonds

  • Sometimes atoms have to share more than one pair of electrons to fulfill the octet rule, like \(O_2\).
  • Single bond and double bond are not the same concepts.
  • Sometimes there are more than one solution for a Lewis dot structure.

Week 4 Wednesday Lecture


  • Get Scantron sheet from bookstore.
  • Monday - clicker session, make sure to write down questions - reflective of questions on the exam.

Lewis Dot Structure Mechanics

  • Analyzing molecular structure to dipole moment from electron configuration.
  • Look at Periodic Table and identify the name of each column to obtain the numbe rof valence electrons in the valence shell.
  • Rule: \(4A > 5A > 7A > 7A\) when determining atoms to put in the middle.
  • Must satisfy octet rule
  • We have assumed that there is only one correct Lewis structure for a given chemical formula.
  • There are systems for which more than one Lewis structure is possible:
    • Resonance structures: same atomic linkages, but different bonding (single vs double, etc.) - real structure is an ‘average’ of the available resonance structures. Fix bonds but move electrons around.
    • Structural isomers: different atomic linkages. “Formal Charge” is used to determine most likely structure.

Resonance Structures

  • When you see a double bond - can you move it around?
    • Not with single bonds - you may eliminate the bond altogether.
  • Arise from moving electrons around.
  • Conceptually, we think of the bonding as an average of these two structures. Electrons are delocalized between the oxygens such that on average the bond strength is equivalent to 1.5 \(O-O\) bonds.
  • Some resonance structures are not equivalent.
    • \(N_3^-\) - 16 total valence electrons.
      • Can form double-double, triple-single resonance structures. These are nonequivalent Lewis Structures, but there are still 3.
      • Nonequivalent: one ‘unique’ design is lower-energy than the other.

Structural Isomers

  • Different sets of atomic linkages can be used to construct correct Lewis Dot Structures.
  • \(Cl_2 O\) can be arranged at \(Cl - O - Cl\) or \(Cl - Cl - O\), etc.
  • Structural isomers can be derived from moving different atoms around.
  • Structural isomers are not equivalent because the structures are different.

Formal Charge

  • Understanding how energy is distributed across an atom.
  • A method to identify which Lewis Dot Structures are more ‘correct’ or ‘stable’.
  • Formal charge is the different between the number of valence electrons and the number of valence electrons owned.
  • The structure with the minimal FC on all atoms is more correct; the negative formla charges on the more electronegative atoms tend to be more correct.
  • Example question - what is the formal charge on \(X\) atom in \(Y\) molecule?
  • Some questions will require 2-5 minutes to finish, but many questions are also easier. Make sure you know which questions you need to skip.
    • Get the easy points - it’s very very hard to fail, but also very very hard to get full points.

Beyond the Octet Rule

  • There are numerous exceptions to the octet rule.
  • Two categories of violation:
    • Sub-octet systems (atom has less than 8 electrons and is still OK) - beryllium, boron, and aluminum.
    • Valence shell expansion (more than 8 electrons) - phosphorus (5 bonds), sulfur (6 bonds), etc.
  • Sub-octet rule
    • Beryllium is happy with two bonds - beryllium difluoride, for instance. We can use formal charge to analyze this.
    • Boron is happy with three bonds - boron trifluoride, for instance.
      • If you force boron to have a double bond, the formal charge outcome will be worse.
    • Be, B, Al can form sub-octet rule.
  • Valence shell expansion
    • Third-row elements and higher - energetic proximity of \(d\) orbitals allow for the participation of these orbitals in bonding.
    • \(d\) orbital is not too ‘far’; can access \(d\) orbitals and can hold more than eight electrons - but only if they are forced.
    • Hydrogen through Neon cannot expand their valence shell, because they have no access to the \(d\) orbitals.
    • Larger halogens like Cl, Br, and I, in addition to some noble gases like Argon and Xenon, can demonstrate valence shell expansion because of their access to \(d\) orbitals.

Week 4 Friday Lecture


  • Will post exam seating chart grouped by sections. Bring your own scantron form.
  • Exam will be next Friday.
  • Write down information (5 point deducation if missing)
  • Monday - clicker question

Beyond the Octet Rule

  • Numerous exceptions to the octet rule - sub-octet systems and valence shell expansion.
  • Only three atoms can exhibit sub-octet systems - beryllium, boron, and aluminum (can form 2, 3, and 3 bonds, respectively).
  • Valence shell expansion - when can it be used? Phosphorus, sulfur, etc. - have access to \(3d\) orbitals.
    • Don’t force it - it is only used if it is needed.
    • Exam question - which molecule has zero lone pairs/valence shell expansion?
  • \(I_3^-\) - has three resonanc estructures resulting from the moving of a lone pair electron across each of the iodine atoms.
  • AXE/ABE notation.
  • In molecular structure, we do not consider lone pairs.


  • There are two types of electron pairs - bonding and lone.
  • Valence Shell Electron Pair Repulsion model - 3d structure is determined by minimizing repulsion of electron pairs.
  • Convert from Lewis dot structure to a three-dimensional geometry.

Arranging Electron Groups

  • Consider both bonding and lone pairs when minimizing repulsion.
  • \(CH_4 \implies AX_4\), forms tetrahedral shape.

Two Electron Groups

  • Linear (\(AB_2\)) - angle between bonds is \(180^\circ\).
  • Molecules with multiple bonds can also be linear.

Three Electron Groups

  • Common in organic compounds
  • Trigonal Planar - angle between bonds is 120 degrees.
  • Example: \(BF_3\), no lone pairs; \(CH_2O\) (formaldehyde)

Four Electron Groups

  • Tetrahedral \(AB_4\), angle between bonds is \(109.5^\circ\).
  • Bonding pair - constrained by two nuclear potentials; more localized in space. Lone pair is constrained by only one nuclear potential; less localized and thus ‘needs’ more room.
  • Trigonal pyramidal (\(AB_3E\)) - bond angles are less than \(109.5^\circ\) (about \(107^\circ\)), and the structure is nonplanar due to the repulsion of the lone pair. Dipole moment exists.
  • Tetrahedral angle shift from \(109.5^\circ\) - \(AB_2E_2\), bent structure (e.g. \(H_2O\)).
  • \(H_F\) (\(ABE_3\)) case, linear.
  • Be careful when looking at electron geometries; bent structures derived from \(AB_4\) are different from structures derived from \(AB_3\).

Beyond the Octet

  • Systems with expanded valence shells will have five or six electron pairs around a central atom.
  • Five electron pair groups - what is the optimum arrangement of electron pairs? See-saw structure for \(AB_4E\). \(AB_2E_3\) - linear group.
  • Six electron groups - maximum number of groups a molecule can have. Only a few molecular configurations exist; \(AB_4E_2\), geometry corresponds to two lone pairs opposite and perpendicular to a square plane, square planar.

Molecular Dipole Moments

  • We can use VSEPR to determine the polarity of a whole molecule.
  • Draw Lewis structures to determine 3D arrangement of atom. If one side of the molecular has more EN atoms than the other, the molecule has a net dipole.
  • Molecules capable of exhibiting molecular dipole moments: perfect linear, perfect trigonal planar perfect trigonal bipyramidal, perfect octahedral, perfect square planar, etc.
  • Completely symmetric molecules will not hav ea dipole regardless of the polarity of the bonds.
  • Consider the directionality of the dipole moments.

Week 5 Monday Lecture

  • First midterm is this Friday; the seating chart will be printed byt he end of the day. Look for where you will be seated.
  • Midterm will start on time - get seated as soon as possible. Will start no later than 1 minute after 9 thirty. Bring your scantron sheet; don’t worry about printing out the information sheet.
  • Can bring one double-sided cheat sheet.
  • If you miss 6/20 in this clicker section, you are in trouble.

  • Emission vs absorption sepctrum - should be the opposite. Energy levels go up and down; energy required is a same. One process absorbs photons; the other corresponds to emission - high to low energies.
  • What is the wavelength of an emitted electron? Use the de Broglie equation - particles with mass.
  • Does hydrogen have a \(3d orbital\)? Yes - even if the orbital is empty, it exists.
  • Excitation - moving electrons from one orbital to another. Electron affinity - adding another electron. Ionization energy - removing an electron from the orbital.

  • 25 questions: 5 from Ch.2, 11 from Ch. 12, and 9 from Ch. 13.

Week 5 Wednesday Lecture


  • Friday - midterm, covering Ch. 2, 12, and 13 in 45 minutes. Number of questions was reduced from 25 to 22.
  • Chapter 3, Stoichiometry: 3.1-3.10.
  • Some questions are intentionally difficult.
  • Learn how to manage time well throughout the test.
  • Exam packets will be distributed at 9:25 AM, starting as soon as possible.
  • Four questions from Ch.2, ten from Ch. 12, and eight from Ch. 13.
  • Do not miss writing Scantron information, bring a no. 2 pencil.
  • The exam seating chart was posted.
  • First part of Chapter 3 is probably the most boring part of Chem 142. The challenge is the last part, and the most challenging part of the second exam.
  • Chain reactions within limiting reactant.
  • Second exam will have 18 or so questions.

The Atomic Mass Unit

  • Defined as 1/12 the mass of a carbon-12 atom.
  • Masses of all other atoms are given relative to this standard.
  • The atomic masses on the periodic table are a weighted average of the masses of each isotope of that element.
  • amu is equivalent to grams/mol.
  • Avagadro’s number - a method to count. A mole - an Avagadro’s number quantity of particles.
  • Naturally occuring elements are often mixed with different isotopes.
  • Hard to build a quantum device because it’s hard to make the isotope free.
  • Once you have different isotopes, their nuclear spins mess up quantum interactions.
  • Want to make isotopes as pure as possible.

Atomic Mass and Mass of an Atom

  • There is no carbon isotope that weights 12.011 amu.
  • Atomic mass is not the mass of an atom - it is an average weighted by abundance.
  • Standard SI unit is kilograms; amu is g/mol. Be careful with measurements.

Interpreting the Mole

  • Mole ratio: number of particles per mole. Not mass ratio. Allows to count the number of particles.
  • Elemental analysis: can only get information about mass.
  • Molar mass - allows us to convert between moles and mass.

Percent Composition of Compounds

  • Helpful to know a compound’s composition int erms of the masses of elements.
  • Use reactions to identify the weight of each element.
  • Deduce mass frraction and mass percent to convert from mass to moles.

Empirical and Molecular Formulas

  • Empirical formula - simplest formula for a compound that agrees with elemental analysis.
  • Molecular formula - a multiple of the empirical formula reflecting the measured result.
  • Molecular structures formed from multiples of empirical formulas are different molecules in three-dimensional space; they are not simply repeating units.

Week 6 Monday Lecture

  • Grades will be released by Midterm.
  • You will be taught how to project your grade.
  • Most students do well wtih the new material.
  • For chapters 3-5 int he second exam, differences between people with a good and poor high school education are different.
  • Content is taught at a higher level.

Chemical Equations

  • Chemistry is the study of the rearrangmeent of matter due to the flow of energy.
  • In a chemical reaction, some bonds are broken and tohers are formed, resulting in a reorganization of atoms.
  • Law of Conservation of Mass.
  • Chapter 3 is straightforward.
  • Moreover, indicate the phase/state of the molecule.

Upon combustion with excess oxygen, a 6.49 mg sample yielded 9.74 mg carbon dioxide and 2.64 mg water. Calculate the empirical formula of ascorbic acid.

We have \(C_n H_m O_l + O_2 \to CO_2 + H_2O\). The only source of carbon and hydrogen are in the carbon dioxide and water reactants. Use molar mass ratio multiplied by the total mass to obtain the total mass of carbon, and the same to obtain the total mass of hydrogen. Then, obtain the amount of oxygen left over given the mass of the total sample. Convert mass to moles, then figure out the final result - \(C_3H_4O_3\).

Formulas of Elements and Compounds

  • Elemental forms of nonmetals are often found as molecules - hydrogne, nitrogen, oxygen, fluorine, chlorine, bromine iodine - BrINClHOF.

Interpreting Chemical Reactions

  • Tables are important - regardless how complex the problem.
  • Do not write down mass; mass must be always converted into moles.

Balancing Equations

*Write the balanced chemical equation for the combustion of hexane \(C_6H_{14} (l)\).

\[C_6H_{14} + O_2 (g) \to CO_2(g) + H_2O(g)\] \[C_6H_{14} + O_2(g) \to 6CO_2(g) + 14H_2O(g)\] \[C_6H_{14} + 9.5 O_2(g) \to 6CO_2(g) + 14H_2O(g)\] \[2C_6H_{14} + 19 O_2(g) \to 12CO_2(g) + 28 H_2O(g)\]
  • Mole ratio matters the most.

Mole Ratios

  • We can use a balanced chemical equation to predict the number of moles of products that a given number of moles of reactants will produce.
  • Mole ratios are the only thing that are constant throughout the reaction.
  • Table method:
\(2 Al\)\(3 I_2\)\(2 Al\)
\(\frac{35.0 g}{26.9 \frac{g}{mol}} = 1.3 mol\)x 

We can conclude that \(\frac{2}{3} = \frac{1.3}{x}\) and solve for \(x\).

Limiting Reagant

  • What if we have a mixture of reactants?
  • If we mix two reactants together, one of them becomes a limiting reagant - there is not enough of it to react.
  • Use the limiting reagant to perform calculations.

Week 6 Wednesday Lecture


  • Midterm 1. Median: 68%. Standard deviation: 16%.
  • ALEKS Pie Progress. Median: 67%. Standard deviation: 13%.
  • Top 15% - primed for a 4.0!

Percent Yield

  • Actual yield divided by theoretical yield.
  • A comparison of how much product actually was produced as opposed to how much product could theoretically be produced.


  • Solute - substance being dissolved, mixed, and diluted.
  • Solvent - substance doing the dissolving.
  • Solution - final combination of dissolution, mixing, and dilution.
  • We consider water to be a solvent, although it can also be a solute.

Water as a Solvent

  • Water has a dipole moment, meaning it has an unbalanced distribution of electrons.
  • Water can reorient itself to form a compound; it can pul charges apart.

Ionic Solutes

  • Polar water molecules dissolve ionic compounds (salts).
  • Hydration breaks ionic compounds into anions and cations, pulling them out of the lattice.
  • Water dissolves different ionic compoudns to different degrees.
  • Electrical conductivity - flow of electricity in a solution indicates the presence of ions in the solution.
  • Electrolyte - a substance that conducts a current when dissolved in water.
  • Ions become solvated/hydrated - surrounded by water molecules.
  • Ions are labelled as ‘aqueous’ - free to move throughout the solution and conduct electricity.

Electrolytes and Nonelectrolytes

  • If a solution conducts electricity, it contains ions.
  • A solution containing many ions is a strong electrolyte.
  • A solution that contains only a few ions is a weak electrolyte.
  • A solution that contains no ions is a nonelectrolyte.
  • Nonelectrolyte - all molecules can be fully separated by water.

Strong Electrolytes

  • Good conductors of electricity.
  • Substances break up to produce many ions in water.
\[NaCl(s) \to Na^+ (aq) + Cl^- (aq)\]
  • Strong acids and bases fully dissoasociate.

Weak Electrolytes

  • Poor conductors of electricity.
  • Substances remain mostly intact as compounds, producing few ions in water.


  • Subtances that cannot conduct electricity - molecules never break down into ions.
  • Always remain as nuclear molecules with no charge.

Dissolving Compounds in Water

  • When ionic compounds are dissolved in water -
    • Ions are hydrated, separated from the solid crystal, and become individual ions in the solution.
  • Ionic equation

Week 6 Friday Lecture


  • 45 minutes to solve roughly 18 problems. No luxury of taking time on problems.
  • Test how well you know the concept - do not spend more than 30 seconds.

Dissolving Compounds in Water

  • Learning more about wet chemistry - different types of reactions in solutions.
  • Determine moles of ions in aqeous solutions of ionic compounds.


  • Many chemical reactions take place “in solution”.
  • Still need to know amounts of reactants and products.
  • How do we make solutions of known concentrations?
  • We measure concentration in terms of moles per volume. Molarity is moles of solute per liters of solution.
  • Bracket notation - molarity notation. Equivalent to \(M\).
  • Do not subtract the mass of an electron to molar mass. The mass is so small it will not change the calculations.
  • Hints are given in units - density is mass per volume. Look at units to understand how to go from one unit to another.

Preparing Solutions

  • Dissolve 1 mole of \(NaCl\).
  • We assume when we divide solids into water that the volume of the solid is 0 - it is dissolved, it does not change the volume of the water.
  • When you have two solutions, a precipate will form - some try to remove the precipitate from the total volume. Do not worry about this. The volumes of two solutions are assumed to be additive.

Solution Dilution

  • Dilution - the process of making a solution with a certain molarity from a solution with a greater molarity.
  • Sometimes we want to change the concentration of solutions.
  • We have \(n\) moles of a solute; we only add water. The number of moles is the same at the end of the dilution.
  • Since the number of moles \(n\) stays the same, onlyt eh volume changes. \(M_1V_1 = M_2V_2 = n\); \(M\) and \(V\) represent sets of molarity and volume.

Types of Chemical Reactions

  • Precipitation reactions, acid-base neutralization reactions, redox reactions (oxidation-reduction).
  • Every reaction has a driving force. Each oft he categories of reactions has their own force.
  • Precipitation reactions - formation of a solid (precipitate) that does not dissolve.
  • Acid-Base Neutralization - formation of water from a hydroxide group and a proton.
  • Oxidation-Reduction reaction - many subcategories, including combination, decomposition, combustion, and single-replacement
    • Driving force: electron transfer between species.
    • When balancing reactions, we must keep in mind additional rules. Must think about charge conservation - cannot gain electrons from nowhere.
  • Precipitation reaction - widely used in environmental science.

Precipitation Reactions

  • Two solutions react to form a precipitate - an insoluble solid.
  • Driving force is the formation of a solid.
  • Know solubility rules for salts in water.
  • If a cation-anion pair is not soluble, then it will completely dissolve.

Week 7 Monday Lecture


  • Second midterm will cover chapters 3, 4, and 5
  • Covers stoichiometry problems. Make sure to balance equations properly. The most difficult part is balancing equations.
  • Review basic techniques to write equations. Use net ionic equations rather than the complete ionic equations.
  • Will be given solubility table; must be familiarized with solubility rules, though.

Selective Precipitation

  • Some ionic compounds are soluble, while others are not.
  • We can use precipitation behavior to remove species selectively.
  • Example: we have a solution of silver, barium, and iron. We want to measure the amount of a metal ion in a solution.
    • Design a precipitation experiment.
    • In the first step, add chloride. Only chloride and silver can form precipate.
    • Second step, add sulfate ion. Only barium and sulfate can form precipate.
    • Third step, add hydroxide ion. Only iron can precipate.
    • Modern laboratories use newer techniques.
  • Important reaction.
  • Always check for limiting reagant. Make sure that the amount of silver is enough to precipitate the chloride.

Acid Base Reactions

  • In an acid-base reaction, a hydrogen proton from the acid reacts with the hydroxide group from the base to form water.
  • The cation from the base combines with an anion from the acid to form a salt.
  • Bronsted-Lowry Theory: acid/base reactions are proton-transfer processes. Acids are proton-donors, bases are proton-acceptors.

Strong and Weak Acids

  • Strong acids undergo complete ionization.
  • Weak acids undergo incomplete ionization. A bit like insoluble salts; don’t like to dissociate much. Not a subject of CHEM 142.
  • In 142, anything that will be given will be a strong acid or a strong base.
  • Neutralization and precipitation reactions can occur at once.
  • Don’t consider new volumes formed as products in reactions - they are not as significant.

Acid Base Titrations

  • Titration is relatively straightforward and neat. A way to analyze the molarity of an unknown acid or base.
  • Consider reaction of hydrochloric and sodium hydroxide to form sodium chloride and water (neutralization reaction).
    • 20 mL or 2M hydrochloric acid is neeeded to neutralize 40 mL.
    • Hydrochloric acid is slowly added to the solution until the color changes; the moles of hydrochloric acid is equal to the moles of sodium hydroxide.
    • This point is the equivalence point.
    • Read the volume of the hydrochloric acid solution - now, you know the moles of the reactants.

Redox Reactions

  • Redox reactions are probably the most important reactions daily life.
  • Most reactions used for energy production are redox reactions.
  • Oxidation - loss of electrons. Reduction - gain of electrons.
  • OIL RIG: Oxidation Is Loss, Reduction Is Gain.
  • Charge conservation - we cannot get electrons from nowhere. The charges must be balanced.

Week 7 Wednesday Lecture

Redox Reactions

  • Make sure that the charges are balanced on both sides.
  • Redox reactions are important; involve electron transfers and significant energy. You can deposit energy into molecular compounds, or harvest it; etc.

Oxidation Numbers

  • Oxidation number represents the number of electrons required to produce the effective charge on a species.
  • Oxidation number means the charge of an atom in a reaction; atoms can have different charges, going from one state to another.
  • Oxidation numbers must sum to the total charge.
  • If you are not comfortable with naming for groups, make sure you remember it.
  • The same atom in a species can have different oxidation states.

Key Rules

  1. Elements in their elemental state have an oxidation state of 0
  2. Monatomic ion oxidation states are equivalent to the charge on the ion.
  3. The sum of all oxidation states in a compound equals the overall charge on the compound.

Specific Rules

  1. Group 1 ions are +1.
  2. Group 2 ions are +2.
  3. Hydrogen is +1 when with a non-metal and -1 when with metals and boron.
  4. Fluorine is -1.
  5. Oxygen is -1 in peroxides and -2 in all other compounds.
  6. Halogens are -1, except for O and halogens above.
  7. Figure out everything else using math.

Types of Redox Reactions

  • Combination, decomposition, replacement/displacement, combustio

Combination Reactions

  • A metal and non-metal react to form a solid ionic compound

Decomposition Reaction

  • The technical reverse of a combination reaction.
  • One reactant forms two or more products.

Combustion Reaction

  • Combustion of a carbohydrate in oxygen; the product must be carbon dioxide and water.
  • Limited working direction in the context of chem 142.
  • All combustion reactions are redox reactions.

Displacement Reactions

  • All displacement reactions are redox reactions.
  • Activity series: relative order of elemetns arranged by their ability to replace cations in aqueous solution. Their ease of oxidation (loss of electrons to form a cation).
  • Lithium is more reactive than Potassium, which is more reactive than Barium, etc. Metals with larger activity can displace metals with smaller activity.
  • To complete and balance displacement reactions, check the activity series to see which metal is more active.
  • For metal and acid solutions, compare metal with hydrogen in activity series.
  • Gold, silver are resistant to acid rain; smaller activity than hydrogen. Cannot displace.

Recognizing Oxidizing and Reducing Agents

  • How do we know if a reaction is an oxidation or a reduction?
  • Redox reaction involves electron transfer. If there is no change in charge state, there is no transfer.
  • Oxidation and reduction ‘half-reduction’.
  • Oxidizing and reducing agent: the reducing agent is the one that undergoes oxidation; the oxidizing agent is the one that undergoes reduction.
  • Sometimes, oxidation and reduction agent can be the same species.
\[A + BZ \to AZ + B\]

Week 7 Friday Lecture

Writing Half-Reactions

  • Separate electorn-transfer reactions between elements into half-reactions, where the electron loss or gain is explicitly stated.
  • Provides us a reliable method of balancing redox reactions in an aqueous solution.
  • Involve electrons in the calculation.
  • In energy research, everything is a half-reaction; want to figure out which steps are limiting steps, etc.
  • Analyzing half-reactions can be used to make reactions more efficient.
  • Electrochemistry - inject electricity into a solution to induce a reaction.

Half Reaction Method

  1. Identify oxidation numbers.
  2. Write unbalanced half-reactions.
  3. Balance non-hydrogen and non-oxygen atoms within each of the half-reactions.
  4. Add water to balance oxygen - aqueous solution, water is available.
  5. Add protons to balance hydrogen. Assume it is an acidic environment.
  6. If in a basic solution, add \(OH^-\) cations needed to offset the hydrogen protons.
  7. Balance charges. Add electrons to balance charges.
  8. Multiply both half-reactions by whatever respective coefficient necessary such that the number of electrons on both sides when combined is equal (i.e. conservation of charge).
  9. Remove redundant species (i.e. species that are the same as products and reactants).
  • Example question: how many protons are on the left side of the oxidation reaction?
  • Example exam question: \(Cl_2 \to Cl^- + ClO_3^-\).
    • This is a redox reaction; you need to balance the equation.

Oxidation Number Technique

  1. Identify oxidation numbers.
  2. Find how many electrons are lost or gained for the oxidation and reduction half-reactions.
  3. Multiply the corresponding reactant of a half-reaction such that the sum of electrons lost and gained is 0; conserved.
  4. Add water and \(H^+\)/\(OH^-\) groups, balance as appropriate.
  • Example exam question: how many hydroxide groups are on the left side of a redox reaction?


  • Only one equation covered in the exam - only cover 5.3 to 5.5.
  • Simple extension of high school chemistry, put in the context of gas-based stoichiometry.

Earth’s Atmosphere

  • Earth’s atmosphere is a layer of gases around 31 miles thick, consisting primarily of nitrogen, oxygen, and trace gases.
  • Gravity pulls gases in the atmosphere towards the surface, exerting pressure.

Atmospheric Pressure

  • Pressure must be exhibited by molecules.
  • Aqueous - dissolution of liquids and solids into water.
  • In gas, molecules are diffused and volume matters. Must consider volume, which is dependent on temperature.
  • Particle movement is random and independent.
  • An equation to quantify reactions is needed.


Pressure is force per unit area \(P = \frac{F}{A}\). Different units: pascals, torr, PSI, atm, mm-Hg etc. Challenge of unit conversion. Many of these are given in the exam cheat sheet.

  • One pascal is the force of 1 newton per square meter.


  • People ussed mercury to measure pressure.
  • Put a mercury column in the device; as you move it around, the mercury height changes depending on the pressure.
  • 1 atm is 760 mm-Hg.
  • \(p = hgd\) (\(h\) is height, \(g\) is gravity acceleration, \(d\) is density).

Week 8 Monday Lecture


  • Cover final new material on Chapter 15.
  • Midterm on Monday covers Chapters 3-5 - molarity, ideal gas equation, etc.
  • Expect class median around 70%


  • For ideal gas equation, convert everything to atm.


  • Manometers operate on the same principle as barometers, but measure the pressure of an isolated gas sample rather than the whole atmosphere.
  • \(P_{\text{gas}} + h = P_\text{atm}\) when gas pressure is less than atmospheric pressure; \(P_\text{gas} = P_\text{atm} + h\) when gas pressure is greater than atmospheric pressure.
  • Find a horizontal plane; same pressure at that point.

Ideal Gas Law

\[PV = nRT\]

\(P\) - pressure, \(V\) - volume, \(n\) - number of moles, \(R\) - gas constant, \(T\) - temperature in Kelvin. Common \(R\) values: \(0.082057 \frac{L \cdot atm}{\mol \cdot K}\), \(8.3145 \frac{J}{mol \cdot K}\). For the ideal gas law, use the first constant but pay attention to constants.

Celsius + 273.15 degrees is Kelvin.

Standard Temperature and Pressure

  • STP is 0 degrees Celsius (273 Kelvin), 1 atm (101.325 kPa, 1.01325 bar, 760 mmHg). This will be on the exam.
  • At STP, 1 mol of any ideal gas occupies 22.4 L.
  • Simple question - give 3 out of 4 unknowns, ask to calculate remaining.
  • Level II - give different units and calculate quantity (Kelvin, mmHg, etc.). Do not miss these kinds of questions.
  • All that matters in stoichiometry is moles.
  • If more than one product is a gas, the measured pressure is influenced by all and stoichiometry must involve all gas products.

Charles’ Law

  • Quantifies the relationship between gas volume and temperature. Volume and temperature are proportional.

Boyle’s Law

  • Relationship between volume and pressure. Volume is inverse proportional to pressutr

Avagadro’s Law

  • Volume proprtional to number of gas molecules.

Dalton’s Law of Partial Pressure

  • You can calculate total pressure as a sum of partial pressures.
  • Gas molecules are non-interacting. Increasing the number of point particles increases the pressure by an amount that is proportional to the number of particles.

Week 8 Wednesday Lecture


  • Easier than first midterm but still expect around 70/100 class average.
  • 20 questions.

Dalton’s Law of Partial Pressures

  • According to the ideal gas law, gas molecules are non-interacting point particles.
  • Increasing the number of point particles increases the pressure by an amount proportional to the number of particles.
  • Ideal gas equation does not discriminate by type
  • Partial pressure - pressure exerted by one gas molecule.
  • Total pressure is a sum of these partial pressures.
  • Another way to calculate total pressure - calculate total number of moles and calculate total pressure.

Mole Fraction and Partial Pressure

  • Mole fraction: ratio of number of moles of a component in a mixture to the total number of moles in the mixture.
  • Partial pressure of a gas in a mixture of gases is equalt o the ratio of its number of moles to the total number of moles times the total pressure. Can get pressure of individual gas by multiplying mole fraction by total pressure.
  • Look at vapor pressure of water at different temperatures and correct for the pressure exerted by water.

Chapter 15: Chemical Kinetics

Read 15.1-15.8. About stoichiometry with time; how many moles are consumed within a given time?

  • Reactants are used up and products are formed. The reaction takes time. The final concentrations approach those on the right-hand side of the reaction equation. The rate of change of a molecule’s concentration changes with time.
  • There are some general trends and behaviors that are important to understand as they change across time.
  • Stoichiometric ratios must be constant.

Week 10 Monday Lecture


  • Second midterm was easier.
  • Class median - 75. Standard deviation - 20.
  • Total: median 62, standard deviation 15.
  • Class beat expectations, grading is not completely curved. Median can be raised.
  • Final is same level of difficulty.

What happens during a chemical reaction?

  • Reactants are used up and products are formed.
  • The reaction takes time.
  • Final concentrations approach those on the right-hand side of the reaction.
  • Reactions can go in the reverse direction as well.
  • Some reactions may have higher forward speed than reverse speed, and to some extent we can assume only forward motion.

Defining Rate

  • Rate is consumption/production of something over a period of time.
  • Calculated as $$\left\frac{\delta \text{reactant/product}}{\delta t}$$.
  • The rate changes over time.
  • As \(\delta t\) approaches 0, instantaneous rate becomes the tangent \(\frac{d \text{NO}_2}{dt}\) for example.
  • If \(2A \to 2B + C\), we have \(r_A = r_B = 2r_C\). The uniform rate for the whole reaction, on the other hand, is \(R = -\frac{dA}{2dt} = \frac{dB}{2dt} = \frac{dC}{dt} \implies \frac{r_A}{2} = \frac{r_B}{2} = r_C\).
    • Divide rate by stoichiometric coefficient.
  • Must balance equation.

Rate Laws

  • Equations that show how the rate depends on the concentration of the reactants.
  • Speed of reaction depends on the concentration; using different concentrations will make the reaction proceed differently.
\[r_A = -\frac{d[A]}{dt} = k[A]^n\]

Differential rate law

  • Rate expressed as a function of concentration \(A\) for \(A \to \text{product(s)}\)
  • \(k\) is rate constant
  • \(n\) is order of reaction wrt \([A]\)
  • \(k\) and \(n\) and determined experimentally
  • Assumes negligble reverse reaction

Integrated Rate Law

First order rate laws:

\[\ln [A] - \ln [A]_0 = -kt\] \[\ln \frac{[A]}{[A]_0} = -kt\] \[[A] = [A]_0 e^{-kt}\]

When plotting slope \(-k\) and intercept \(\ln[A]_0\), the graph should be lienar for first-order reactions.

Week 10 Wednesday Lecture

Final Exam

  • 110 minutes
  • 50 questions
  • Expected class median 70%
  • Will cover everything - the whole quarter.

Rate of the Reaction

  • Defined rate law in previous lecture - rate is the derivative of the concentration wrt time.
  • Rate Laws

  • Differential rate law
  • Gives relationship between concentration of reactants and rate.
  • If n is one, first-order rate law. Integrated rate law can be derived from integrating the differential rate law.
  • First order is most difficult.
  • Plotting only works for first or second order.

Half Life of a First Order Reaction

  • Half life of a reaction - \(t_\frac{1}[2}\)
  • Half life is independent of starting concentration for a first-order reaction.

Second Order Rate Laws

  • Second order rate law is simpler
  • Always plot the data
  • In the exam, will probably give you a plot and identify the rate law; calculate the rate constant.
  • Half life depends on initial concentration - \(t_\frac{1}{2} = \frac{1}{k[A]_0\)

Zero Order Rate Laws

  • A reaction can be completely independent on concentration.
  • Concentration is linearly dependent on time.

Rate Law for Reactions with 1+ Reactant

  • Every species has its own rate.
  • Rates are governed by stoichiometric coefficients; \(r_A/a = r_B/b = r_C/c = ... = r\).
  • Rate of reaction: only worry about the reactant.
  • Overall reaction order - sum of individual reactant orders
  • w.r.t. some reactant - only the order for that reactant.

Method of Initial Rates

  • Find the orders of individual reactants.
  • Keep all other reactants constant.
\[\frac{r_2}{r_1} = \left[\frac{[A]_2}{[A]_1}\right]^n\]

Week 10 Friday Lecture

  • Two more regular lectures; wednesday will be a clicker section to review Chapter 15, and no lecture on Friday.
  • Chapter 15 has two important concepts left - reaction mechanism and Arrhenius equation.
  • Most questions will come from previous lecture - reaction rate laws, determining rate law, etc.

Rate Laws

  • Consider different units for rate to produce final mol/s.
  • By looking at the units for rate, you can identify the reaction order.

Measuring Rate Laws

  • If the concentration is very large, we don’t need to worry about the concentration of others; they are considered essentially constant, any fluctuation will have a negligble effect. Buffer - create conditions such that other concentrations do not functionally change.

Elementary Step

  • An elemntary reaction or step is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.
  • A chemical reaction consists of one or more elementary steps.
  • Unlike the rate law for an overall reaction, the rate law for an elementary step can be obtained from the stoichiometry of the species reacting in that step.
  • Unimolecular, bimolecular, termolecular. Most basic step that triggers reaction.
  • Most reactions don’t obey elementary step simplicities because they involve multiple elementary steps.
  • All elementary steps in a reaction mechanism will always add up to give the balanced overall chemical equation.
  • If experiments do not match elementary rate predicition, then it is not an elementary reaction. How do we figure out reaction mechanisms?
  • You’ll be given a set of steps and asked to verify if the mechanism is correct.
  • A reaction can only be as fast as the slowest stop.

Week 11 Monday Lecture

Final ‘regular’ lecture. Clicker review section on Wednesday, no lecture on Friday.

Elementary Step

  • Simplest step in a single reaction step: all elementary steps in a reaction mechanism will always add up to give the balanced overall chemical equation.
  • Only involves one collision process. Some reactions with multiple collisions are not elementary steps.
  • Rate law is constant multiplied by product of reactant concentrations raised to the stoichiometric coefficient.
  • Which step is the rate-limiting step?
  • Reactions are as fast as the slowest step.
  • What if the second step is slow and the first step is fast?
  • Reaction intermediates cannot be part of the final rate law.
  • Equillibrium - if the second step is slow, the first step forward and reverse reaction rates will be equal.
  • Be comfortable with the process.
  • Practice Q #33: answer is wrong.

Steady State Approximation

  • Won’t be on this year’s exam.
  • We don’t know which reaction is faster or slower, but we can make an assumption: the concentration of intermediates is constant.

Arrhenius Equation

  • For molecules to react upon collision, they must become activated, and the parameter \(E_a\) is Activation Energy.
  • When temperature changes, \(k\) changes.
  • Allows us to change the reaction speed by altering \(k\) and temperature.
  • Arrhenius curve
  • Underlies all chemistry concepts.
  • For a reaction to take place, a reaction must pass an energy barrier.
  • Reverse and forward reactions may have different activation energies.
  • Change in energy tells you if the reaction is exothermic or endothermic.
  • \[k = Ae^{-E_a/RT}\]
  • \(k\) varies exponentially with \(1/T\).
    • Gas constant - make sure you use the right constant.
    • Ideal Gas Equation, use R in liter etc.
    • Arrhenius Equation, use R in joules etc.
  • If you take the logarithm of both sides of the Arrhenius equation, you can plot \(\ln k\) against \(\frac{1]{T}\), which sohld be straight with slope \(-\frac{E_a}{R}\).
  • We can calculate activation energy from two sets of data at different values of \(T\).
  • Solve for activation energy.
  • All temperature should be in units of Kelvin.


  • Will not be on exam.
  • Lower activation barrier to a reaction.
  • Energy barrier changes
  • Catalysts interact with the reaction to produce the desired result.

Final Exam Information

  • One question - fast followed by slow, another question - Arrhenius equation, two temperatures + two rate constants are given.
  • Final: 45 instead of 50 questions.
    • 11 from Ch. 15
    • 14 from Ch. 2, 12, 13
    • 20 from Ch. 3, 4, 5
  • Expected class median is 70/100.
  • Read the questions very carefully. Proton vs photon, etc.
  • One question is very difficult - photoelectric equation. Question #3. Come back and do it later.