G11 HL Chemistry Mad!Lab!

Chemical Kinetics

3/2/2018

Mission 1: Rate Expression & Reaction Mechanisms
Mission Objectives. You should be able to...
1. Deduce the rate equation from experimental data and solving problems involving the rate equation.
2. Sketch, identify and analyze graphical representations for zero, first and second order reactions.
3. Evaluate proposed reaction mechanisms to be consistent with kinetic & stoichiometric data.

Looking at the image below, a, b, c and d are stoichiometric coefficients. A, B, C & D are substance concentrations. The rate of a reaction depends on the concentrations of the reactants.

Image courtesy of mikeblaber.org.

The overall order of the reaction is defined as the sum of the m and n exponents. Rate equations can only be determined experimentally because orders can only be deduced empirically.

The word "order" conveys how sensitive the rate of reaction is to changes in the concentrations of A and B.

Here is a worked example from the Pearson text. Attempt to do the problems in the blue box.

I know there are a lot of videos, but they are all useful and each video explains the content in slightly different ways. It's worth it to watch them all (I do).

The sequence of reaction steps outlining the reaction pathway from reactants to the formation of products is called the reaction mechanism. Individual steps in reaction mechanisms are called elementary step, reaction or process. Elementary steps are classified by its molecularity, which represents the number of molecules or atoms involved as reactants in the elementary step.

Unimolecular: single molecule involved.
Bimolecular: two molecules or atoms involved in collision
Termolecular: three molecules or atoms involved in collision.

Each elementary step has its own rate constant, k, and its own activation energy.

In order to deduce the rate equation from a proposed reaction mechanism, (1) decide which step is the rate-determining step (the rate of the overall reaction is equal to the rate of the slow step), and (2), from #1, deduce the rate equation for the rate determining step.

The awesomely named Mike Sugiyama Jones breaks it down.

In this video, the instructor shows you how to suggest a mechanism using the rate law and how to determine the RDS.

Mission 2: Activation Energy
Mission Objectives. You should be able to...
1. Analyze graphical representations of the Arrhenius equation in its linear form.
2. Use the Arrhenius equation.
3. Describe the relationships between temperature and rate constant, frequency factor, and complexity of molecules colliding.
4. Determine and evaluate values of activation energy and frequency factors from data.

Image courtesy of IB Alchemy.

What is the effect of increased temperature on rate constant k? If you examine a Maxwell-Boltzmann distribution, particles at a higher temperature tend to have more collisions because the energy is greater than the activation energy. For most reactions, an increase in temperature of 10K doubles the rate of reaction. In the Arrhenius equation, e to the power of (-Ea/RT) is the fraction of molecules which have an energy equal to or greater than Ea at a particular temperature. So an increase in temperature increases the value of the rate constant k, and therefore the rate of reaction.

Watch the video. At the end, he does some practice problems. Put the calculations in your notes so you have a reference.

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Energetics & Thermochemistry

1/11/2018

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Energy Cycles

Mission Objectives:
1. Construction of Born-Haber Cycle for group 1 & group 2 oxides and chlorides.
2. Construction of energy cycles from hydration, lattice and solution enthalpy.
3. Calculation of enthalpy changes from Born-Haber or dissolution of energy cycles.
4. Relate size and charge of ions to lattice and hydration enthalpies.

Links: TutorVista

The Born-Haber Cycle is an energy cycle for the formation of an ionic compound. It is an application of Hess' Law that combines the enthalpy change with several steps in the formation of an ionic compound.

Below is a simple example of the concept. You'll want to take notes as we work through the example in class.

There are some different enthalpies that you need to be familiar with.

1. Lattice Enthalpy is the standard enthalpy change (SEC) that occurs on the formation of one mole of gaseous ions from the solid matter. Look at the example on page 358 in your text. The process is endothermic and data values are in the Data Booklet.

2. Enthalpy of Atomization. This is the SEC that occurs on the formation of one mole of separate gaseous atoms of an element in its standard state. "M" sublimates and "X" has an atom removed. Look at the example on page 358 in your text.

3. Ionization Energy. This is the SEC that occurs on the removal of one mole of electrons from one mole of atoms or cations in the gaseous phase. Metals will have multiple IEs depending on number of electrons removed. Look at the example on page 359.

4. Electron Affinity. The SEC on the ADDITION of one mole of electrons to one mole of atoms in the gaseous phase. See page 359.

These are combined to construct a Born-Haber Cycle and find either heat of formation or lattice energy of an ionic compound. To do this, add up the correct values for enthalpy changes. Keep your Data Booklet handy.

The IB requires that you know the B-H Cycle for sodium chloride, magnesium fluoride, and magnesium oxide. Examine the worked examples on page 360.

Enthalpy changes in solution look at the relationship between enthalpy change of solution, hydration enthalpy, and lattice enthalpy.

The SEC of solution is a change in enthalpy when one mole of a substance is dissolved in an excess of a pure solvent. The enthalpy change of hydration is the enthalpy change when one mole of gaseous ions are added to water to form a dilute solution.

Terms to know: solvation (solvents used other than water), hydration (water as the solvent), and dissolution (homogeneous phase after solute has been added).

Entropy & Spontaneity

Mission Objectives:
1. Predict whether a change will result in an increase or decrease in entropy by considering the states of the reactants and the products.
2. Calculate entropy changes from standard entropy values.
3. Application of Gibbs' Free Energy equation in predicting spontaneity and calculations of various conditions of enthalpy and temperature that will affect this.
4. Relate G to position of equilibrium.

A reaction is said to be spontaneous when it moves towards completion or equilibrium under a set of conditions without external intervention. Spontaneous reactions can be endothermic or exothermic. Reactions that do not take place under a given set of conditions are said to be non spontaneous.

Recall the laws of thermodynamics. The second law focuses on entropy, which is a measure of the distribution of total available energy between particles. The greater the shift from localized to widespread, the lower the chance of the particles returning to their original state and the higher the entropy of the system.

Watch the below videos. Make sure you take notes, especially when he shows the different flowcharts showing how these terms relate to one another.

Entropy is represented by delta S. It can be calculated from thermodynamic data from the Data Booklet. The standard values relate to standard conditions of temperature and pressure.

Delta S (rxn) = Delta S (products) - Delta S (reactants) (page 367)

When completing entropy calculations, remember that values are specific for different states of matter. Also recall that the coefficients used to balance the equation must be applied to molar entropy values when calculating overall entropy change.
Be sure to examine the chemical reaction and predict whether you expect the reaction to have a positive or negative entropy change based on the degree of disorder in the products and reactants.

Gibbs' Free Energy is a state function, along with entropy, temperature and enthalpy. The below video goes into detail in the relationship between entropy, enthalpy and Gibbs' Free Energy.

The Gibbs' Free Energy provides an effective way of focusing on a reaction system at constant temperature and pressure to determine its spontaneity. For a reaction to be spontaneous, the GFE must have a negative value. Examine Table 3 on page 369.

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14: Covalent Bonding & Hybridization

8/28/2017

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Mission 1: Charge it Up!
Mission Objectives. You should be able to...
1. Predict whether sigma or pi bonds are formed from the linear combination of atomic orbitals.
2. Deduce the Lewis structures of molecules and ions showing all valence electrons for up to six electron pairs on each atom.
3. Apply Formal Charge to ascertain which Lewis structure is preferred from different Lewis structures.
4. Deduce using VSEPR Theory of the electron domain geometry and molecular geometry with five and six electron domains and associated bond angles.

Formal charge is a way to keep up with the available electrons in a structural formula. The way to calculate formal charge is as follows:

# of valence electrons - 1/2 # of bonding electrons - # of free electrons.

Expanded Octets. When the central atom is an element from the 3rd period or below, compounds can form with more than eight electrons around the central atom. This is called an expanded octet. This arrangement is possible because the d orbitals available in the valence shell of these atoms have energy values relatively close to those of the p orbitals. Promoting electrons from 3p to empty 3d orbitals allows additional electron pairs to form. This happens with phosphorous and sulfur.

For example: Phosphorous pentachloride has five electron domains and forms a shape called trigonal bipyramidal. It contains three different bond angles: 90, 120, and 180 degrees.

There are tables on pages 333 and 334 that illustrate the shape and bond angles for molecules with expanded octets. Lone pairs of electrons occupy equatorial positions. Basically, just count the number of electron pairs on the central atom, both shared and lone. If there are five pairs of electrons total, then that is 5 electron domains that form a trigonal bipyramidal shape. Six pairs of electrons around the central atom leads to an octahedral shape. There are other shapes to get familiar with: see-saw, linear, T-shape, square planar and square pyramidal.

Below is a short video that talks about delocalization. Basically, if a molecule has resonance, delocalization occurs and stabilizes the molecule.

Mission 2: Ozone Depletion
Mission Objectives. You should be able to...
1. Explain the wavelength of light required to dissociate oxygen and ozone.
2. Describe the mechanism of the catalysis of ozone depletion when catalyzed by CFCs and NOx.

Mission 3: Hybridization
Mission Objectives. You should be able to...
1. Explain how sp3, sp2 and sp hybrid orbitals are formed in methane, ethene and ethyne.
2. Identify and explain the relationship between Lewis structure, electron domains, molecular geometry, and types of hybridization.

Hybridization. A hybrid orbital results from the mixing of different types of atomic orbitals on the same atom. There are three types of hybridization: sp3 (ex: methane), sp2 (ex: ethene), and sp (ex. ethyne). For these examples, the second electron in carbon's 2s orbital gets excited to fill in the 2pz orbital. As a result, the 2s and 2p combine to form a set of orbitals.

Molecules that have tetrahedral geometry are predicted to have sp3 hybridization. This means that there is a combination of 1 s orbital and 3 p orbitals. The molecule has 25% s character and 75% p character.

Molecules that have trigonal planar geometry are predicted to have sp2 hybridization. This means that there is a combination of 1 s orbital and 2 p orbitals. The molecule has 33% s character and 66.7% p character.

Molecules that have linear geometry are predicted to have sp hybridization. This is a combination of 1 s and 1 p orbital. The molecule has 50% s character and 50% p character. See the table on page 352.

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13: Transition Metals & Colored Complexes

8/28/2017

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Mission 1: First Row d-Block Elements
Mission Objectives. You should be able to...
1. Explain the ability of transition metals to form variable oxidation states from successive ionization energies.
2. Explain the nature of the coordinate bond within a complex ion.
3. Deduce the total charge given the formula of the ion and the ligands present.
4. Explain the magnetic properties in transition metals in terms of unpaired electrons.

The IUPAC defines a transition metal as an element that has an incomplete d-sublevel, or can give rise to cations with incomplete d-sublevels. Based on this definition, group 12 elements (Zn, Cd, Hg, Cn) are not considered to be transition metals because they have full d-sublevels. This means that groups 3-11 are transition metals.

Characteristics of transition metals:
--variable oxidation states
--compounds and ions are colored
--form complexes with ligands
--often used as catalysts
--magnetic properties depend on oxidation states and coordination numbers

Look at elements manganese and chromium. Manganese can have oxidation states from +1 to +7. Mn+2 is colorless while Mn+7 is purple. See the below image from thesavvychemist.com.

Your textbook lists some common colored compounds. Zinc compounds are usually colorless (due to the completely filled d-sublevel) unless the complex has a chromophore. The chromophore is a group of atoms responsible for the absorption of electromagnetic radiation, which can be absorbed in the visible region of the e-mag spectrum.

Transition metal complexes have a central metal atom bonded to a group of molecules or ions. These are called ligands. Ligands are molecules or ions with a lone pair of electrons that can attach themselves to ions with a positive charge density. Examples of ligands include water (H2O), ammonia (NH3), the cyanide ion (CN-), the hydroxide ion, (OH-), the oxalate ion (C2O42-), and EDTA. All of these molecules and ions have lone electron pairs.

Metal ions with a high charge density: Sn4+, Al3+, Cr3+, Cu2+, Zn2+, Be2+, Sn2+

Compounds are described as coordination compounds to signify the coordinate bonding present between ligands and the central metal ion. Monodentate ligands use one lone pair of electrons that form CCBs (coordinate covalent bond; also called dative covalent bond) with a central metal ion. Polydentate ligands have more than one lone pair of electrons to form CCBs to a central metal ion.

Coordination numbers are the total number of points at which a central atom or ion attaches ligands.

Read up on EDTA on p. 315. EDTA (EDTA4-) is a polydentate ligand that can form 6 CCBs to the central metal ion. It has a coordination number of 6. EDTA can remove metal ions from solution and inhibit enzyme-catalyzed reactions.

Transition metals are often used as catalysts in chemical reactions. Examples: the Haber Process, the decomposition of hydrogen peroxide, hydrogenation of alkenes, and hydrogenation of oils. They can be used in catalytic converters in cars and are biological catalysts. An example is heme, which is the iron center of hemoglobin. Hemoglobin is the protein that transports oxygen in the blood. Blood is red due to the presence of heme. Each subunit of hemoglobin has an iron atom at its center to which oxygen binds.

Magnetic properties of transition metals and their complexes depend on oxidation state of the metal, coordination number, and the geometry of the complex. Ferromagnetic metals are iron, cobalt and nickel. Paramagnetic materials contain unpaired electrons that behave as tiny magnets and are attracted by an external magnetic field. Diamagnetic materials do not contain unpaired electrons and therefore are repelled by external magnetic fields. See the below video.

Mission 2: Colored Complexes
Mission Objectives. You should be able to...
1. Explain the effect of the identity of the metal ion, the oxidation state of the metal, and the identity of the ligand on the color of transition metal ion complexes.
2. Explain the effect of different ligands on the splitting of the d-orbitals in transition metal complexes and colour observed using the spectrochemical series.

Before we proceed, let's figure out how to determine the oxidation state of the metal ions in a complex. Mike S. Jones to the rescue.

In a complex ion, d orbitals split into two sublevels. The electronic transitions between these sublevels leads to absorption and emission of photons of light, which are responsible for the color of the complex. dxy, dxz, and dyz orbitals are stable because their orbital lobes lie at 45 degrees to the Cartesian axes. dx2-y2 and dz2 orbitals are destabilized because their lobes are directed along the Cartesian axes. This is called a 3:2 split, with the last two orbitals having more energy than the first three.

Light passes through the complex. Some of it is absorbed. The absorbed energy causes electrons to get excited and jump to the higher d orbitals. The change in energy is referred to as delta E. The color that you see is what is left behind, and is the opposite on the color wheel to what was absorbed. For instance, if you're seeing red, the absorbed energy falls in the green range on the color wheel.

The energy separation between the two split degenerate sets of orbitals is defined as the crystal field splitting energy. There are five theories, but your book focuses on Crystal Field Theory (CFT). Factors that affect CFT: (1) identity of the metal ion, (2) oxidation state of the metal ion, (3) nature of the ligands, and (4) geometry of the complex ion. You need to know the first three.

Identity of the metal ion. This influences the extent of the crystal field splitting. In general, the splitting field energy parameter increases as you go down a group with the metal in the same oxidation state. Oxidation state increases as the parameter increases. Distances between the metal and the ligands decrease, resulting in a better overlap between the metal orbitals and the ligand orbitals.

Nature of the ligands. They may have different charge densities. The greater the charge density, the greater the crystal field splitting. Ammonia's charge density is greater than that of water. Ligands can be arranged into a spectrochemical series, based on the order of increasing field energy parameters. Using the series, the further to the right, the bigger the split. A big split = a big jump = big amount of energy absorbed (and falls in the violet range). A small split = a small jump = a small amount of energy absorbed (and falls into the red range).

From the Pearson text:

Oxidation state of the metal ion. The lower the oxidation state, the smaller the split because there is less interaction with the ligands.

Learn more about colored complexes with the below videos.

Practice worksheet, courtesy of MSJChem. We will also work the problems in the text starting on page 325. Make sure you have your data booklets.

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12.1: Electrons in Atoms

8/17/2017

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Mission 1: Those Doggone Electrons
Mission Objectives. You should be able to...
1. Solve problems using E = hv
2. Calculate value of the first IE from spectral data which gives wavelength or frequency of the convergence limit.
3. Deduce the group of an element from its successive IE data.
4. Explain trends and discontinuities in data on first IE across a period.

Ionization energy is related to the process X(g) --> X+(g) + e-. Successive ionizations are possible: X+(g) --> X2+(g) + e-. This can go on, as represented by: nth = X(n-1)(g) --> Xn+ + e-.

For a given element, the ionization energy increases for successive ionizations in the order IE1 < IE2 < IE3 < IE4 < IE5... This is because it requires more and more energy to remove electrons as you get closer to the nucleus.

Emission lines converge at higher energies. Look at the figure below. At the limit of convergence, the lines merge and form a continuum. Beyond this limit, the electron can have any energy because it is no longer under the influence of the nucleus; the electron is outside of the atom, which means ionization has occurred.

There's some problem solving that goes along with this, so you'll need your Data Booklets and calculators.

Periodic Trends in IE. The image below shows successive ionization energies for calcium and titanium. In the case of calcium, there is a significant jump going from IE2 to IE3; the third IE corresponds to the removal of an electron from the fully occupied 3p sublevel. As a result, there aren't any Ca3+ ions. Titanium exhibits oxidation states of +2, +3 and +4, with the most stable being +4. The figure shows that there is a large jump in IE for titanium going from IE4 to IE5, which corresponds to the removal of a fifth electron, supporting the observation that titanium doesn't form ions with +5 oxidation states. The IEs increase more gradually than for calcium because electrons are being removed from the 3d and 4s orbitals which are much closer in energy compared to 3p and 4s.

Image courtesy of slideplayer.

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