This page summarizes the contents of each lecture, with links to its slides and video recording. Please note that the slides and their contents are copyrighted by Dan Dill.

Use "Find" in your Web browser to find in this page those lectures where a particular topic is discussed.

For each lecture: there is a link to the PDF of the PowerPoint slides used in that lecture and a link to the lecture recording, showing the notations made to each slide during the lecture. Click on the desired lecture number:

**Lecture 1, Friday, January 19, 2018**

**Continue Mahaffy et al., chapter 10: Modeling bonding in molecules.**

PDF notes:
Bonding in diatomic molecules.

Which AOs form bonding and antibonding MOs? Inner-shell AOs overlap negligibly and so do not contribute to bonding.

CDF animation: Li_{2} 1s and 2s bonding electron clouds.

CDF animation: Li_{2} 1s and 2s antibonding electron clouds.

CDF animation: Li_{2} 2s bonding and antibonding electron clouds.

CDF animation: XY 2s correlation diagram.

CDF animation: XY 2s correlation diagram, bonding electron cloud and its dipole moment.

CDF animation: XY 2s correlation diagram, antibonding electron cloud and its dipole moment.

SOE: The role of AO symmetry, overlap, and relative energy.

PDF notes:
Questions on Symmetry, Overlap, Energy.

Only valence AOs form MOs; inner-shell AOs contribute negligibly to bonding or antibonding.

Lecture slides and
lecture recording.

**Lecture 2, Monday, January 22, 2018**

Covalent versus ionic character in bonding.
MO description of hydroxide, OH^{−}.

Lecture slides and
lecture recording (video only).

**Lecture 3, Wednesday, January 24, 2018**

Simple MO description of water, HOH, predicts a bond angle of 90^{o}.

PDF notes:
Hybrid AOs and Polyatomic MOs.

Hybrid AOs (sp, sp^{2}, and sp^{3}) account for molecular shape.

CDF animation:
Hybrid Orbitals in Organic Chemistry.

Hybrid orbitals have energies intermediate between those of the constituent AOs.
Calculating energy of hybrid orbitals.
Hybrid orbital water AO-MO correlation diagram.

Lecture slides and
lecture recording.

**Lecture 4, Friday, January 26, 2018**

PDF notes: Polyatomic MO recipe.

\(\sigma\) and \(\pi\) bonding in formaldehyde, H_{2}CO.

Formic acid, HC(O)OH (localized \(\pi\) bonds).

Formate, HC(O)O^{−} (delocalized \(\pi\) bonds).

Lecture slides and
lecture recording.

**Lecture 5, Monday, January 29, 2018**

Complete formate, HC(O)O^{−} (delocalized \(\pi\) bonds).
Postscript on recipe for polyatomic MO recipe.

**Begin Mahaffy et al., chapter 11: States of matter.**

Macroscopic versus microscopic understanding of the ideal gas law.

Lecture slides and
lecture recording.

**Lecture 6, Wednesday, January 31, 2018**

Review: Review: Localized and delocalized \(\pi\) orbitals versus localized and delocalized electrons.

PDF notes: Kinetic molecular theory

Derivation of pressure in terms of time rate of change of momentum per collision with a container wall.
Pressure is due to collisions of all \(N\) particles of gas.

Lecture slides and
lecture recording.

**Lecture 7, Friday, February 2, 2018**

Complete kinetic molecular theory.
Root mean squared speed, \(u_{rms}\) goes up with temperature and down with speed.
Particle picture of gases.
Understanding gas behavior in terms of motion and speed of individual particles.
Mixtures of gases.

Lecture slides and
lecture recording.

**Lecture 8, Monday, February 5, 2018**

Review: Practice with particle picture of gases.
Units of pressure: Pascal (Pa), bar, and atm.
Units of the gas constant, \(R\): J versus L atm.
Distribution of molecular speeds is due to collisions of gas particles with one another.
After not too many collisions the relative number of particles with a given speed is given by the Maxwell-Boltzmann distribution.
The higher the temperature, the broader but lower the distribution.
The rms speed, \(u_{rms}\), is slightly larger than the most probable speed,
\(u_{mp} = \sqrt{2/3} u_{rms}\),
because of the exponential shape of the speed distribution at hight speeds.
In calculating rms speed, it is crucial to carefully cancel units.

PDF article: Bonomo & Riggi, 1984, The evolution of the speed distribution for a two-dimensional ideal gas: A computer simulation.

Lecture slides and
lecture recording.

**Lecture 9, Wednesday, February 7, 2018**

Complete molecular speeds and their distribution.

CDF animation:
Maxwell-Boltzmann distribution of speeds.

Real gases:
Van der Waals \(a\) reflects intermolecular attractions present when gas particles encounter one another;
therefore, hydrogen bonding can affect the value of van der Waals a.
Because of the random orientation of close encounters,
dipole-dipole interaction is typically less strong than dispersion interaction.

Lecture slides and
lecture recording.

**Lecture 10, Friday, February 9, 2018**

Real gases: effect of molecular size (van der Waals \(b\)).
Van der Waals equation.
Conditions which require taking into account deviations of gas behavior from that described by the idea gas law,
\(PV=nRT\).
Phase diagram lines are combinations of pressure and temperature at which two different phases (solid-liquid, liquid-gas, and solid-gas) are simultaneously present and in equilibrium.
The triple point is when all three phases---solid, liquid, and gas---are in equilibrium.

YouTube: Triple point video, Tert Butyl Alcohol

The critical point is the temperature and pressure above which it is not meaningful to distinguish between liquid (too diffuse) and gas (too dense)---the supercritical region. By passing through the critical region, gas can be converted to liquid without any phase transition, that is, without gas and liquid being simultaneously present.

YouTube: Supercritical fluid video, Liquid Cl_{2}

Lecture slides and
lecture recording.

**Lecture 11, Wednesday, February 14, 2018**

**Begin Mahaffy et al., chapter 12: Solutions and their behavior.**

Ionic solids dissolve as a result of competition between
attraction of oppositely charged ions in the solid and
attraction of polar water molecules for the individual ions.
Some ionic solids release heat when dissolving, and some absorb heat when dissolving.

Lattice enthalpy is the enthalpy change required to separate one mole of ionic solid into its individual ions in the gas phase,
so that they are so far apart that they no longer interact with one another electrically.
Lattice enthalpy is always positive, since energy is required to separate oppositely charged ions from one another.

Enthalpy of aquation (enthalpy of hydration) is the enthalpy change when one mole of oppositely charged ion pairs,
initially in the gas phase, so that they are so far apart they no longer interact with one another electrically,
is place in liquid water.
Enthalpy of aquation is always negative, due to the attractive interaction of the polar water molecules with the individual ions.

Enthalpy change of solution is the enthalpy change when one mole of an ionic solid dissolves in water.
By Hess's law, enthalpy of solution is the sum of the lattice enthalpy and the enthalpy of aquation.
Since lattice enthalpy and the enthalpy of aquation are each large but opposite in sign,
the magnitude of their sum is much smaller and whether enthalpy of solution is positive (endothermic) or negative (exothermic)
**cannot** be predicted without further information.

Relative values of lattice enthalpy are determined by relative values of the Coulomb interaction energy between the oppositely charged ions.
The smaller the ions, the closer they can be and so the larger the lattice enthalpy.
The greater the charge on the ions, the larger the lattice enthalpy.

Lecture slides and
lecture recording.

**Lecture 12, Friday, February 16, 2018**

Enthalpy of aquation (enthalpy of hydration) is the enthalpy change when one mole of oppositely charged ion pairs, initially in the gas phase, so that they are so far apart they no longer interact with one another electrically, is place in liquid water. Enthalpy of aquation is always negative, due to the attractive interaction of the polar water molecules with the individual ions.

Crystal ionic radii values: https://en.wikipedia.org/wiki/Ionic_radius

Relative values of aquation (hydration) enthalpy are determined by relative values of the Coulomb interaction energy between ions and surrounding polar water molecules.
The smaller the ions, the closer the water molecules can be and so the larger the aquation (hydration) enthalpy.
The greater the charge on the ions, the greater the attraction between the ions and the water molecules and so the larger the aquation (hydration) enthalpy.
Enthalpy change of solution is the enthalpy change when one mole of an ionic solid dissolves in water.
By Hess's law, enthalpy of solution is the sum of the lattice enthalpy and the enthalpy of aquation.
Since lattice enthalpy and the enthalpy of aquation are each large but opposite in sign, the magnitude of their sum is much smaller and whether enthalpy of solution is positive (endothermic) or negative (exothermic) **cannot** be predicted without further information.
Colligative properties are due to the presence of nonvolatile solute particles.
There are four colligative properties: vapor pressure lowering, boiling point elevation, freezing point lowering, and osmotic pressure.
At this point you are responsible for calculating colligative properties using provided formulas, in lab and in discussion. Later, we will learn where these formulas come from, taking into account the affect of the solute on the entropy of the system.

Lecture slides and
lecture recording.

**Lecture 13, Tuesday, February 20, 2018**

Example osmotic pressure problem and answer.

**Begin Mahaffy et al., chapter 13: Dynamic and chemical equilibrium.**

If reactants are consumed to form products as time passes, a reaction is said to be spontaneous.
If products are consumed to form reactants as time passes, a reaction is said to be non-spontaneous.
If the amounts of reactants and products does not change with time, a reaction is said to be at equilibrium.
Reaction quotient Q tell where amounts of reactants and products are relative to equilibrium.
The value of the reaction quotient, Q, relative to its value at equilibrium, the equilibrium constant K, tell whether a reaction is spontaneous (Q < K), non-spontaneous (Q > K), or at equilibrium (Q = K).
Therefore, knowing Q and K, we know what will happen to the relative amounts of reactants and products as time passes.
Each chemical equation has its own equilibrium constant.
However, the numerical value of an equilibrium constant changes depending on how the chemical equation is written, according the three rules:
(1) Multiplying each stoichiometric coefficient by a constant a (which can be fractional) raises the equilibrium constant to that power.
(2) Reversing a chemical equation changes its equilibrium constant to its reciprocal.
(3) A chemical equation that is the sum of tow other chemical equations has an equilibrium constant equal to the product (not sum) of the equilibrium constant of the two other chemical equations.

Lecture slides and
lecture recording.

**Lecture 14, Wednesday, February 21, 2018**

For a given K, equilibrium concentrations can be different.
Equilibrium amounts (molarities, pressure, etc.) are different for different starting conditions.
If equilibrium is disturbed, by abruptly changing the amount of a reactant or product, once equilibrium is restored,
the new equilibrium amounts will only partially have offset the change (Le Chatelier's principle)

**Begin Mahaffy et al., chapter 14: Acid-base equilibria in aqueous solutions.**

Water reacts with itself (to a very small extent) to form equal amounts of H_{3}O^{+}(\(aq)\) and OH^{−}(\(aq)\).
The higher the temperature, the more H_{3}O^{+}(\(aq)\) and OH^{−}(\(aq)\) are formed.
This means the pH = − log[H_{3}O^{+}] of pure water is different at different temperatures.
At 25 ^{o}C, the pH of pure water is 7.0.
At 50 ^{o}C, the pH of pure water is 6.6.
This means the K_{w} = [H_{3}O^{+}] [OH^{−}] is
(10^{−7.0})^{2} at 25 ^{o}C and
(10^{−6.6})^{2} at 50 ^{o}C.
In both cases water is "neutral", because it contains equal amounts of H_{3}O^{+}(\(aq)\) and OH^{−}(\(aq)\).

Lecture slides and
lecture recording.

**Lecture 15, Friday, February 23, 2018**

In general
if K increases with temperature, the chemical equilibrium is endothermic, and
if K decreases with temperature, the chemical equilibrium is exothermic.
An acid in aqueous solution is anything that when added to water results in
more [H_{3}O^{+}] than [OH^{−}].
A strong acid reacts essentially 100%, and so a c_{a} molar solution results in [H_{3}O^{+}] = c_{a} M.
A weak acid reacts much less than 100%, and so a c_{a} molar solution results in [H_{3}O^{+}] << c_{a} M.

Lecture slides and
lecture recording.

**Lecture 16, Monday, February 26, 2018**

From [H_{3}O^{+}] alone, we cannot tell if an acid is weak or strong, for must know also how much acid is added to water.
A strong acid HA results in mostly conjugate base A^{−} relative to unreacted acid HA.
A weak acid HA results in very little conjugate base A^{−} relative to unreacted acid HA.
Therefore, it is the proportion of A^{−} to HA that distinguishes strong from weak acids.
In water, before any acid reacts, [H_{3}O^{+}] is present from the water autoionization reaction.
How to construct a weak acid ICE table.
Since a weak acid has K_{a} << 1, at equilibrium, the amount of unreacted will differ only a little from the amount originally present.
Therefore, c_{a} − x ≈ c_{a}.
Since a weak acid has K_{a} >> K_{w}, at equilibrium, the weak acid will be a much greater source of [H_{3}O^{+}] than water autoionization.
Therefore, 10^{−7} + x ≈ x.

Lecture slides and
lecture recording.

**Lecture 17, Wednesday, February 28, 2018**

pK_{a} is defined as −log(K_{a}).

Since [H_{3}O^{+}] = K_{a} [HA] / [A^{−}], pH = pK_{a} + log([A^{−}]/[HA]).

Therefore, a solution in which [A^{−}] = [HA] (an equimolar solution) will have pH = pK_{a}.

PDF notes: Overview of acid-base calculations.

Adding strong base to a weak acid solution will lower the hydronium ion concentration. Use two steps to get the resulting concentration:
Step 1: Assume all acid is present as HA only (that is, ignore its equilibrium with water), and calculate its reaction of the added hydroxide as a **limiting reagent problem**.
Step 2: Use results of step 2 to calculate the equilibrium concentration of hydronium ion.
If hydroxide ion is the limiting reagent in step 1 (“not enough” hydroxide added), then solve whichever of the following two equilibria has the larger equilibrium constant.

HA(\(aq)\) + H_{2}O\((l)\) \(\leftrightarrows\) H_{3}O^{+}(\(aq)\) + A^{−}(\(aq)\), K = K_{a}

A^{−}(\(aq)\) + H_{2}O\((l)\) \(\leftrightarrows\) HA(\(aq)\) + OH^{−}(\(aq)\), K = K_{b} = K_{w}/K_{a}

If K_{a} > K_{b} the resulting solution will be acidic; if K_{a} < K_{b}
the resulting solution will be basic.

Lecture slides and
lecture recording.

**Lecture 18, Friday, March 2, 2018**

Worked example of “not enough” OH^{−}(aq).
Reviewed that the conjugate base of a weak acid is a weak base, since K_{b} << 1, but a strong base compared to water, since K_{b} >> K_{w}.
Work example of “just enough” OH^{−}(\(aq)\).""
In this case, we must use the equilibrium

A^{−}(\(aq)\) + H_{2}O\((l)\) \(\leftrightarrows\) HA(\(aq)\) + OH^{−}(\(aq)\), K = K_{b} = K_{w}/K_{a}.

Work example of “too much” OH^{−}(\(aq)\).
In this case, we must use the equilibrium

A^{−}(\(aq)\) + H_{2}O\((l)\) \(\leftrightarrows\) HA(\(aq)\) + OH^{−}(\(aq)\), K = K_{b} = K_{w}/K_{a}.

Since the equilibrium begins with the left over OH^{−}(\(aq)\) from step 1, and since K_{b} << 1,
the equilibrium concentration of OH^{−}(\(aq)\) is just that from step 1.

Lecture slides and
lecture recording.

**Lecture 19, Monday, March 12, 2018**

Complete: Worked example of “too much” OH^{−}(\(aq)\).
Since the equilibrium begins with the left over OH^{−}(\(aq)\) from step 1,
and since K_{b} << 1,
the equilibrium concentration of OH^{−}(\(aq)\) is just that from step 1.
Practice determining whether a mixture is acidic, neutral, or basic.

Lecture slides and
lecture recording.

**Lecture 20, Wednesday, March 14, 2018**

Strong acid equilibrium: The concentration of unreacted HCl can be calculated
by first assuming 100% of the HCl has reacted (revised initial conditions),
and then solving for the tiny amount of HCl that reforms to establish equilibrium.
When both strong base and strong acid are added to weak acid, carry out the limiting reagent reactions (step 1) successivley, until all acids and bases have reacted.
Buffer action: Solutions of a weak acid and its conjugate base (or a weak base and its conjugate acid)
strongly mute the change in pH that would otherwise result if strong acid or strong base were added to water.

Lecture slides and
lecture recording.

**Lecture 21, Friday, March 16, 2018**

Added NaOH solution to pure water changes hydroxide concentration
to the number of moles of NaOH added divided by the volume of the the water.
Added HCl solution to pure water changes hydronium concentration
to the number of moles of HCl added divided by the volume of the the water.

**Begin Mahaffy et al., chapter 15: Solubility, precipitation, and complexation.**

There are five kinds of Ksp problems:
(1) Given solubility in mass/volume, calculate Ksp;
(2) Given Ksp, calculate the ion concentration in a saturated solution in pure water;
(3) Calculate ion concentrations in a saturated solution that initially contains an amount of one (or both) of the ions (common ion);
(4) When solutions are combines, determine whether a precipitate will form; and
(5) If a precipitate will form, determine how much ionic solid will precipitate, how much of the ion initially present in excess will remain in solution after precipitation, and how much of the ion not initially in excess will reaming in solution after precipitation.
Examples of problem type (1) and type (2).
Sketch how to approach problem type (3)

Lecture slides and
lecture recording.

**Lecture 22, Wednesday, March 21, 2018**

Examples of problem type (3), type(4), and type (5).
In calculating precipitation, problem type (5), the key is to revise the initial conditions by
**assuming 100% precipitation**,
resulting in none of the limiting reagent ion remaining in solution.

Lecture slides and
lecture recording.

**Lecture 23, Friday, March 23, 2018**

**Begin Mahaffy et al., chapter 16: Electron transfer reactions and electrochemistry.**

PDF notes: Assigning oxidation numbers; balancing oxidation-reduction (redox) equations.

Electrochemistry harnesses the spontaneous transfer of electrons in redox (oxidation-reduction) processes as electrical work.
Electrochemical cells are summarized using cell line notation.

Lecture slides and
lecture recording.

**Lecture 24, Monday, March 26, 2018**

Electrical work is (charge transfered) × (cell voltage).
Charge transfered is (moles of electrons transfered) × (charge/mol).
The charge on 1 mol of electrons is the Faraday constant, \(F\) = 96485 C/mol.
Cell voltage is determined by the enthalpy change of the redox process and its spontaneity (entropy change).
Free energy change \(\Delta G\) is defined as the negative of the electrical work done in a redox process.

PDF notes: Standard reduction potentials.

Standard cell potentials are values for all cell components in their standard states, ad so \(Q\) = 1.
Standard reduction potentials are tabulated on a scale for which the standard hydrogen electrode (SHE) is 0 V.
Whichever half-reaction has the more positive reduction potential proceeds as a reduction, and so takes place at the cathode.
Whichever half-reaction has the less positive reduction potential proceeds as an oxidation, and so takes place at the anode.
The standard cell potential is
\(E^{\circ}\)_{cell} = \(E^{\circ}\)_{red}(more positive) \(-\) \(E^{\circ}\)_{red}(less positive).

Lecture slides and
lecture recording.

**Lecture 25, Wednesday, March 28, 2018**

Cell potentials are positive and larger the farther to the reactants side of equilibrium the cell is,
that is, the smaller than 1 the ratio \(Q/K\) is. If there are only reactants, the cell potential is +∞.
Cell potentials are negative and larger the farther to the products side of equilibrium the cell is,
that is, the larger than 1 the ratio Q/K is. If there are only products, the cell potential is −∞.
The cell potential is 0 when the cell is at equilibrium, that is, when \(Q = K\) and so \(Q/K = 1\).
Cell voltage is proportional to −log(\(Q/K\)).
At 25 ^{o}C the constant of proportionality is approximately 0.06 V/n_{e},
where n_{e} is the number of moles of electrons transferred per reaction unit of the redox process.
(In a later lecture we will derive the expression for the constant of proportionality for any temperature.
The standard reduction potential is proportional to +log(K), with the same constant of proportionality.)
Therefore, at 25 ^{o}C the cell voltage is

\(E\) = \(E^{\circ}\) − (0.06 V/\(n\)_{e}) × log(\(Q\)).

The relation between \(E\), \(E^{\circ}\), and \(Q\) is known as the Nernst equation.

Lecture slides and
lecture recording.

**Lecture 26, Friday, March 30, 2018**

Qualitative calculations with the Nernst equation.
Changing how an equation is balanced (say, by multiplying all coefficients by 2)
change the values of the equilibrium constant, \(), and of the number of moles of electrons transferred per reaction unit, n_{e},
but these changes offset one another in the Nernst equation so that the standard cell voltage \(E\)^{o} is unaffected.
Changes in concentrations never change the the standard cell voltage \(E\)^{o} but they may change the cell voltage, \(E\).
Concentration cells use spontaneous mixing of unequal concentrations to generate electricity.
Equilibrium in a concentration cell has concentrations equal, and therefore the equilibrium constant of a concentration cell is always
\(K\) = 1 and so the standard cell voltage of a concentration cell is \(E\)^{o} = 0.

Lecture slides and
lecture recording.

**Lecture 27, Monday, April 2, 2018**

Chloride ion concentration cell: Half reactions require chlorine gas and platinum electrodes; cell line notation; calculation of cell voltage E.
If gas pressures are not equal, then the ratio of gas pressures, P(anode)/P(cathode) will affect the cell voltage.
Calculation of voltage of concentration cell made with fresh water and sea water.

**Begin Mahaffy et al., chapter 17: Spontaneous change: How far?**

PDF notes: Spontaneity: Second law of thermodynamics.

System spontaneously evolve to equilibrium.
The reason is that this evolution is statistically more likely to occur than any other change in the system.

Lecture slides and
lecture recording.

**Lecture 28, Wednesday, April 4, 2018**

To quantify the idea that equilibrium arrangements are more likely to occur than any other
we need formulas to count how many unique arrangements there are of different kinds of particles,
and of energy among different particles.
Counting particle dispersal:
The number of unique arrangements of \(j\) particles of one kind and \(k\) particles of another kind is

\(W(j, k) = (j + k)! / (j!k!)\).

Evaluating this formula is simplified by canceling the larger of the two factorials in the denominator with the same factor in the numerator.
For example,

\(W(2,120)\) = (122 × 121 × 120!) /( 2! × 120!) = (122 × 121) / 2 = 61 × 121.

Practice with particle dispersal.
Illustrate that maximum particle dispersal corresponds to uniform pressure in a gas.
Define entropy as proportional to the logarithm of the number of arrangements.
The logarithm makes entropy an extensive quantity, that is, a quantity that scales with the size of a system.
Counting energy dispersal:
The number of unique arrangements of \(q\) identical quanta among m identical molecules is

\(W(q, m) = (q + m - 1)! / (q!(m - 1)!)\).

The reason the number of molecules appears in the formula reduced by 1 is that to partition a collection things in m groups,
only \(m-1\) "slices" are needed.

Lecture slides and
lecture recording.

**Lecture 29, Friday, April 6, 2018**

Illustrate that entropy change for a given amount of heat flow is greater
the smaller the amount of energy present (the smaller the absolute temperature \(T\)) and
the larger the amount of heat transferred (Δ\(H\)): Δ\(S\) = Δ\(H\) / \(T\).
Spontaneity requires Δ\(S\)_{tot} = Δ\(S\)_{sys} + Δ\(S\)_{sur} > 0.
The spontaneity of phase transitions is a neat application of
Δ\(S\)_{tot} = Δ\(S\)_{sys} + Δ\(S\)_{sur}.
The key ideas are (1) that Δ\(S\)_{sys} does not change with temperature
and (2) that
Δ\(S\)_{sur} = −Δ\(H\)_{transition}/\(T\).
At the transition temperature, \(T\)_{transition}, the two phases are in equilibrium, and so
Δ\(S\)_{tot} = Δ\(S\)_{sys} − Δ\(H\)_{transition}/\(T\)_{transition} = 0
and so that
Δ\(S\)_{sys} = Δ\(H\)_{transition}/\(T\)_{transition}.

Lecture slides and
lecture recording.

**Lecture 30, Monday, April 9, 2018**

At other temperatures other than \(T\)_{transition}
Δ\(S\)_{tot}
= Δ\(H\)_{transition}/\(T\)_{transition}
− Δ\(H\)_{transition}/\(T\).
From this result we can understand why, for example, steam condenses to water when \(T\)
is below \(T\)_{transition} = \(T\)_{bp}
but does not condense when T is above \(T\)_{transition} = \(T\)_{bp}.

PDF notes:
Entropy change perspective of colligative properties.

In freezing of pure water from an aqueous solution of non-volatile solute,
the change in entropy required to go from the solution to pure ice is larger than that
to go from pure water to pure ice. This requires that \(T\)_{freezing} is smaller (lowered) for the solution,
so that Δ\(S\)_{freezing} = Δ\(H\)_{fus}/\(T\)_{freezing} is larger.
For the example of freezing point depression, the freezing temperature must be lower in order for the
entropy change to reach to that of the solid phase.

Lecture slides and
lecture recording.

**Lecture 31, Wednesday, April 11, 2018**

Absolute entropy: At \(T\) = 0 K, the number of arrangements is \(W=1\), and so the entropy of each substance at 0 K is \(S=0\).
Adding small amounts, \(dq\), of heat will add a small amount entropy \(dS = dq/T\) and increase the temperature a little bit, \(dT\).
Adding up all of these small amounts allow us to determine the total entropy at a final temperature.
If there is a phase transition, say melting, it will add entropy
Δ\(S\)_{fus} = Δ\(H\)_{fus}/\(T\)_{fus}.
In this way, so-called absolute entropies of substances have been found at room temperature.
Using these, entropy change for chemical reactions can be found as
\(\Delta_{\rm r} S\) = \(S\)_{products} - \(S\)_{reactants}.
For chemical reactions not involving gases, entropy change is usually small and can be either positive or negative.
For chemical reactions in which the amount of gas moles increases, entropy change is large and positive.
For chemical reactions in which the amount of gas moles decreases, entropy change is large and negative.
For chemical reactions in which the amount of gas moles does not change or in which there are no gases, entropy change is usually small and can be either positive or negative.
Free energy change is defined as
\(\Delta G\) = \(-T \Delta S_{\rm tot}\) = \(-T \Delta S_{\rm sur} - T \Delta S_{\rm sys}\) = \(\Delta H_{\rm sys} - T \Delta S_{\rm sys}\).
Usually the subscripts sys are omitted, and so free energy change is written as
\(\Delta G = \Delta H - T\Delta S\).
The negative sign in the definition of the free energy change means that free energy change is negative for spontaneous processes.

Lecture slides and
lecture recording.

**Lecture 32, Friday, April 13, 2018**

Effect of temperature on equilibrium:
\(
{\rm ln}(K) = -\Delta H^{\circ}/R(1/T) + \Delta S^{\circ}/R.
\)
This relationship can be used is two ways, which make it an exceptionally important tool.

First way: By measuring the equilibrium constant at two different temperatures,
\(T\)_{1} and \(T\)_{2},
and the extrapolating the straight line through the points
\(\{1/T_1, {\rm ln}(K_1)\}\) and \(\{1/T_2, {\rm ln}(K_2)\}\)
the standard enthalpy change can be determined from
slope \(m = -\Delta H^{\circ}/R \),
and the standard entropy change can be determined from
intercept \(b = \Delta S^{\circ}/R\).

Second way: By calculating the standard enthalpy change, using standard enthalpies of formation,
and the standard entropy change, using standard absolute entropies, ln(\(K\)),
and so the equilibrium constant, \(K\), can be evaluated at any specified temperature.

Practice predicting the effect of temperature on equilibrium

Lecture slides and
lecture recording.

**Lecture 33, Wednesday, April 18, 2018**

**Begin Mahaffy et al., chapter 18: Spontaneous change: How fast?**

Equilibrium means forward and reverse rates are equal.
The equilibrium constant is the ratio of the forward rate constant to the reverse constant,
\(K=k_{\rm for}/k_{\rm rev}\).
Definition of reaction rate.

Lecture slides and
lecture recording.

**Lecture 34, Friday, April 20, 2018**

Rate versus concentration from experiment.
Making sense of rate versus concentration: Reaction mechanism.
Making sense of rate constants: Arrhenius relation.

Lecture slides and
lecture recording.

**Lecture 35, Wednesday, April 25, 2018**

Making sense of rate constants: Arrhenius relation.

CDF animation:
Maxwell-Boltzmann energy distribution.

The rate constant can be expressed as \( k = A\ {\rm exp}[-E_a/(RT)\) where
\( A \) is proportional to the probability that collisions have correct geometry, and
\( {\rm exp}[-E_a/(RT)] \) the the fraction of collisions that have kinetic energy at least \( E_a \).
The fraction \( {\rm exp}[-E_a/(RT)] \) is 0 at \( T = 0 \) and 1 at \( T = \infty \).
Therefore at \( T = 0 \), \( k = 0 \), and at \( T = \infty \), \( k = A \).
Given the rate constant at two different temperatures, the slope of ln\(k\) vs 1/\(T\) is \( -E_a/R \).
Once \( E_a \) is determined, \( A \) can be calculated as \( A = k\ {\rm exp}[+E_a/(RT)] \).

Lecture slides and
lecture recording.

**Lecture 36, Friday, April 27, 2018**

The larger \( E_a \), the more sensitive reaction rate is to temperature.
Catalysis speeds a reaction by providing a path from reactants to products with a lower \( E_a \).
This means a catalyzed reaction is less affected by temperature than is the non-catalyzed reaction.
Half-life is the time required for the amount of a reactant to be reduced to half its original value.
For first order processes, half-life does not depend on how much is present.
Here are the essential equations of half-life calculations for first-order processes:
\( (1/2)^n = [{\rm A}]_n/[{\rm A}]_0 \),
\( {\rm ln}([{\rm A}]_t/[{\rm A}]_0) = -kt \), and
\( nt_{\rm half} = t \).

Lecture slides and
lecture recording.

**Lecture 37, Monday, April 30, 2018**

Example half-life calculations.
(1) Given amount consumed in time \( t \), how many half-lives have elapsed?
(2) Given how many half-lives have elapsed in time \( t \), what is \( t_{\rm half} \)?
(3) Given \( t_{\rm half} \), how much remains after time \( t \)?
(4) Given amount consumed in time \( t \), what is the rate constant \( k \)?

Lecture slides and
lecture recording.

**Lecture 38, Wednesday, May 2, 2018**

Putting it all together: First law, second law, equilibrium, and kinetics.
\( \Delta S^\circ = R\ {\rm ln}(A_{\rm for}/A_{\rm rev}) \).
\( \Delta H^\circ = E_{\rm a,for} - E_{\rm a,rev} \).
At very high temperature the equilibrium constant depends only on the entropy change of the system.

Lecture slides and
lecture recording.

Copyright © 2018 Dan Dill | Contact | Department of Chemistry | Boston University