1/3/14 - 1/7/14

This past week in AP Chemistry we studied equilibrium. This included a range of topics including the equilibrium constant, Le Châtelier's Principle, the reaction quotient, RICE charts, as well as the relationship between thermodynamics and equilibrium.

Equilibrium is said to occur when a reaction and its reverse reaction proceed at the same rate. In a reaction that is at equilibrium, the amount of reactants and products remains constant. The rate of reaction is equivalent to the rate constant (K) multiplied by the concentration of the species. 

Equilibrium is reached when the amount of product and reactant becomes constant.

The equilibrium constant (Keq) is central to the concept of equilibrium. We simulated Keq) in the Phet simulation by taking the number of B molecules and dividing it by the number of A molecules. In reality, K is calculated in a similar manner. Consider the following hypothetical reaction:

Aa + bB <–> cC + dD

(where lowercase letters are stoichiometric coefficients and uppercase letters are hypothetical elements or compounds)

In this reaction, the equilibrium constant (in this case the concentration constant, Kc) is equivalent to the concentrations of the products raised to the power of their respective stoichiometric coefficients divided by the concentrations of the reactants raised to the power of their respective stoichiometric coefficients. This can be modeled by the following equation:

Kc = [C]^c[D]^d / [A]^a[B]^b

For gases, pressure is proportional to concentration in a closed system. Thus, there is also a pressure constant that is calculated in a similar manner:

Kp = (PC^c)(PD^d) / (PA^a)(PB^b)

If the K value for a reaction is much greater than one, the reaction is product favored. There is more product than reactant at equilibrium. If the K value is much less than one, the reaction is reactant favored. There is more reactant than product at equilibrium. The equilibrium constant is not affected by changes in number of moles, volume, or pressure. Keq is only affected by temperature. I found this concept to be fairly straightforward. The constant equals products over reactants. To cover this subject I completed the Phet simulation activity, the Equilibrium I lecture, as well as the Equilibrium I worksheet.

The equilibrium constant is similar to another aspect of equilibrium, the reaction quotient (Q). The reaction quotient gives the same ratio that the equilibrium constant gives (products over reactants) but for a system that is not at equilibrium. Q is calculated by substituting the initial concentrations for reactions and products into the equilibrium expression:

Qc = [C]^c[D]^d / [A]^a[B]^b

If Q > K, there is less reactant and more product in the initial conditions than at equilibrium. If Q < K, there is more reactant and less product in the initial conditions than at equilibrium. When Q = K, the reaction is at equilibrium. As it is very similar to calculating the normal equilibrium constant, I also understood the reaction quotient well. The reaction quotient was covered in the Equilibrium II worksheet as well as the Equilibrium Calculations I lecture.

Another central part of the concept of equilibrium is Le Châtelier's principle. Le Châtalier's principle states that if a system at at equilibrium is disturbed by a change in temperature, pressure, or the concentrations of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance. Once a reaction is at equilibrium is at equilibrium, it is possible to change the concentrations of the products and reactants by changing the external conditions in three ways: adding/removing reactants or products, expanding/contracting a reaction system, and changing the temperature. When equilibrium is disturbed, it is reestablished when the reaction proceeds in the direction where the number of moles or pressure has dropped. Le Châtelier's principle was covered in the Equilibrium Part II lecture as well as in the Equilibrium I and II worksheet. I found Le Châtelier's principle to be quite intuitive, especially after completing the Phet simulation. The Phet simulation really helped me to visualize the effects of a change in temperature or number of moles.

Additionally, this week I was introduced to the concept of RICE charts. RICE charts (standing for Reaction, Initial, Change, Equilibrium). It is a simple way of figuring out the number of moles of reactants and/or products at equilibrium, as well as the equilibrium constant. RICE charts were a central part of the latter half of the Equilibrium I worksheet.

The final aspect of the equilibrium topics covered this week was the relationship between thermodynamics and equilibrium. I studied how Gibbs Free Energy (∆G) relates to equilibrium. At standard conditions, Qp = 1 and ∆G˚ = ∆G. However, conditions are not always standard for reactions. When conditions are not standard, change in Gibbs Free Energy can be calculated using this equation:

∆G = ∆G˚ + RTlnQ

(where R = 8.314 J/mol-K, T = temperature in kelvin, and Q = the reaction quotient at the moment)

At equilibrium, ∆G = 0. The standard state free energy of a reaction (∆G˚) is a measure of how far a reaction is from equilibrium. The smaller the ∆G˚ value, the closer the standard state is to equilibrium. The larger the ∆G˚ value, the closer the standard state is to equilibrium. When ∆G > 0 and Keq < 1, there are mostly reactants at equilibrium. When ∆G is much greater than 1 and Keq is much less than 1, there are almost solely reactants at equilibrium. When ∆G < 0 and Keq > 1, there are mostly products at equilibrium. When ∆G is much less than 1 and Keq is much greater than 1, there are almost all products at equilibrium. The relationship between thermodynamics and equilibrium was covered in the lecture of the same name as well as in the Equilibrium III worksheet. Of all of the concepts covered, this is the concept that I understood the least. I am able to solve the problems by blindly plugging in values to the given equations, but I do not really understand the concept. This website helped a bit, but I still do not have a very developed understanding of the topic.

 

1/13/14 - 1/17/14

This week in AP Chemistry we started the unit on gas laws. This covered a variety of issues including relations among variables such as pressure, temperature, moles of gas, and volume, as well as the ideal gas law, kinetic molecular theory (KMT), partial pressures, mole fractions, and real gases.

To begin, gases differ greatly from liquids and gases in that they expand to fill containers, are highly compressible, and have extremely low density. Temperature, pressure, volume, and number of moles are four of the most important aspects of gases that I had to learn. Temperature is the measure of the average kinetic energy of a sample of molecules. When dealing with gas laws, all calculations must be done in Kelvin (K). Pressure is the amount of force applied to an area by a sample. The common units of pressure are atmospheres (atm) and mmHg (a.k.a. torr). The number of moles is the number of molecules present (obviously) and the volume is the amount of space taken up. There are several basic gas laws that establish relationships between these variables, which are as follows:

Boyle's Law
Boyle's Law states that at fixed temperature, volume is inversely proportional to temperature. It is modeled by this equation:

P1V1 = P2V2

Charles' Law

Charles' Law states that at a fixed pressure, volume is directly proportional to temperature. It is modeled by the following equation:

V1/T1 = V2/T2

Gay-Lussac's Law

Gay-Lussac's Law states that at a fixed volume, pressure is directly proportional to volume. It is modeled by the following equation:

P1/T1 = P2/P2

Combined Gas Law

Boyle's Law, Charles' Law, and Gay-Lussac's Law can be combined to form the combined gas law:

P1V1/T1 = P2V2/T2

Avagadro's Law

Avagadro's Law states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas. It is modeled by the following equation:

V1/n1 = V2/n2

I first modeled these gas laws using the Java applet and the spreadsheet activity. Using the applet, I was able to produce data that proved the direct relationship between volume and temperature, pressure and temperature, the number of moles and volume, as well as the inverse relationship between pressure and volume. Additionally, the Ideal Gasses Parts I & 2 lectures covered this material. This subject matter was not hard conceptually. However, I sometimes have trouble matching the correct name with the correct law.

A slight mishap with the Java applet.

Combining the laws listed above produces perhaps the most important equation of the unit: the ideal gas law. The ideal gas law is as follows:

PV = nRT

This law states that pressure multiplied by volume is equivalent to the number of moles of gas multiplied by the gas constant multiplied by temperature. R, also known as the constant of proportionality, is equivalent to 0.08206 L-atm/mol-K (or 8.134 J/mol-K or 62.36 L-torr/mol-K). There are several variations on the ideal gas law which which allow for the calculation of density and molar mass. They are as follows:

PV = (m/M)RT

Pressure multiplied by volume is equivalent to the mass of the sample divided by the molar mass of the substance of the gas multiplied by the gas constant multiplied by temperature.

d = PM/RT

Density is equivalent to pressure multiplied by molar mass divided by the gas constant multiplied by temperature.

To help learn this concept, I completed several worksheets in class, including the Gases I and Gases II worksheets. The ideal has law was also featured in the Ideal Gasses lectures. Although this equation seems fairly straightforward, when I was asked to find pressure or temperature when given the mass of a sample in grams or given the density, I often had no idea what to do. To prepare myself for the upcoming test, I will find it necessary to familiarize myself with the variations on the ideal gas law.  

The ideal gas law ties in closely with Kinetic Molecular Theory (KMT). When a variable like pressure or temperature is being calculated using the ideal gas law, several assumptions are being made. These assumptions are the main ideas behind KMT. In KMT, it is assumed that gases consist of large numbers of molecules that are in continuous, random motion. It is assumed that the the volume of the molecules of gas is negligible. It is also assumed that attractive and repulsive forces between gas molecules is negligible. Kinetic energy is assumed to be conserved when gas molecules collide. It is also assumed that as long as the temperature of the gas remains constant, the average kinetic energy of the molecules remains constant. When the tenants of KMT are fully considered, a gas is considered to be "ideal." Under standard temperature and pressure (1.000 atm and 0˚C) one mole of an "ideal" gas has a volume of 22.4 L.

However, in real life, many aspects ignored in KMT such as intermolecular forces have an effect on the pressure and volume of a gas. Although the ideal gas equation is good for predicting pressure, temperature, volume, etc. at high temperatures and low pressures, these predictions break down at low temperatures and high pressures. This is because the intermolecular forces and volume have a substantial effect on the variables in the ideal gas equation. The ideal has equation fails to account for intermolecular forces. Pressure is calculated by measuring how hard gas particles hit the side of the wall in which they are being housed. Intermolecular forces between gas molecules pull gas molecules towards each other and away from the wall of the chamber. Thus, the pressure is lowered. Since the ideal gas equation fails to account for these intermolecular forces, the pressure values calculated using PV = nRT are often higher than the actual pressure under non ideal conditions. The ideal gas equation works poorly at low temperatures because at low temperatures molecules move slowly and thus have ample time to interact with one another. This exacerbates the effect of the intermolecular forces as the intermolecular forces now have more time to effect molecules, thus pulling gas molecules towards each other and away from the container wall. Furthermore, the ideal gas equation fails to account for the volume of the molecules. When the pressure is low and there are few molecules and/or there is a large volume, the volume of the gas particles is relatively negligible when compared to the volume of the container. However, when the pressure is high and there is a large number of molecules, the volume of the particles is significant when compared to the total volume of the container. Thus, in situations when pressure is high and the ideal gas law fails to take account for the volume of the molecules, the calculated volume is usually lower than the actual volume. In non ideal conditions, one can use a modified version of the ideal gas equation, the van der Waals equation. The van der Waals equation is as follows:

The van der Waals equation.

The (n^2 x a)/(V^2) portion of the equation adjusts the pressure upwards to what it should be if it acted ideally. The V – nb portion of the equation adjusts the volume downwards to what it would be if it was ideal. The concept of real gases and KMT was covered in the lecture on these subjects as well as in the concept tests that we covered in class. I found it a little hard to comprehend how the volume affects the ideal gas equation under non ideal conditions. This website helped bring some clarity to the topic.

Additionally, this week I learned the concept of partial pressures and mole fractions. They are fairly straightforward concepts. If there is mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. Partial pressures are governed by Dalton's Law of partial pressures, which is as follows:

Ptotal = P1 + P2 + P3 + … 

A mole fraction is the relative or percent composition by moles of a single component in a mixture, represented in its decimal. The mole fraction (Xa) is equivalent to the moles of one component in a mixture divided by the sum of all components in the mixture. When one has a mixture of gases and known mole fractions, one can multiply the mole fraction by the total pressure to calculate the partial pressure of one gas. This concept was covered in the partial pressures and mole fractions lecture as well as in the Gases II worksheet. I found this subject easy to understand.

Finally, I also learned the concept of effusion and diffusion this week. Effusion is the escape of gas molecules through a tiny hole into an evacuated space. The lighter the molecule, the faster it effuses. Effusion rates are governed by Grahm's Law of Effusion, which is as follows:

V1/V2 = (M2/M1)^0.5 

Diffusion is the spread of one substance throughout a space or throughout a second substance. It follows the same relationship as Grahm's Law.

All in all, I feel relatively good about these topics and the upcoming test. As always, I could just use a bit more practice on the topics. To study for the test on Wednesday, I will do over the Gases I and II worksheets. I felt as though I was comprehending the Taskchains well, with the exception of the forth one. If given the opportunity, I will cover that quiz again.

   

12/9/13-12/13/13

Previously on this blog, I have stated that there have been some challenging topics. However, those topics pale in their abstrusity when compared to the current subject matter. The unit at hand for the last several weeks in AP Chemistry is Thermodynamics, which covers a wide variety topics including energy, heat, work, the system and surroundings, state functions, enthalpy, entropy, and Gibbs free energy.

To start, energy is the ability to do work or transfer heat. Its SI unit is the joule (J). Heat is energy in transit. It is represented by the variable q. Work is equal to force multiplied by distance and is represented by the variable w. Additionally, work is equivalent to the negative pressure multiplied by change in volume (this will come into play later). Heat energy can be used to do work. Likewise, work can be transformed into heat. When referring to the transfer of energy, one must consider the system and the surroundings. The system is the actual chemical reaction that is taking place. The surroundings are everything outside of the reaction. The energy inside a system is referred to as a system's internal energy. Changes in a system's internal energy can be modeled by the following equation:

∆E = q + w

In other words, the change in internal energy is equivalent to heat plus work.

Enthalpy is related to internal energy. If a reaction takes place at constant pressure and the only work done is pressure-volume work, the heat flow of the process can be accounted for by measuring the enthalpy of the system. Simply put, enthalpy is the internal energy plus the product of pressure and volume. This can be modeled by the following equation:

H = E + PV

When the system changes at constant pressure, the change in enthalpy (∆H) is equivalent to:

∆H = ∆E + P∆V

Since it is known that ∆E = q + w, "q + w" (heat plus work) can be substituted into the equation for the change in enthalpy. It is also known that w = -P∆V, so this can also be substituted into the equation for change in enthalpy. Thus:

∆H = (w + q) - w
∆H = w - w + q 

∆H = q

At constant pressure, the change in enthalpy is equivalent to the heat gained or lost. When a reaction is endothermic (takes in heat) ∆H is positive. When ∆H is negative, the reaction is exothermic (heat leaves the system). There are four methods for determining the change in enthalpy, the first being using average bond energies. To do this, one must calculate the energy needed to break bonds in the reactants and then form bonds in the products. This process is modeled by the equation:

∆H = ∑BE(bonds broken) + ∑BE(bonds formed) 

Where the energy of the bonds broken is always negative

∑BE(bonds broken) is always positive as breaking bonds is an endothermic process, while ∑BE(bonds formed) is always negative as the formation of bonds in exothermic.

∆H can also be found using a calorimetry. Calorimetry is the measurement of heat transfer. Heat flow is measured using a calorimeter, which measures heat flow by measuring the before and after a reaction. Calculating change in enthalpy using this method requires values of heat capacity, the amount of energy required to raise the temperature of a substance by 1 K, as well as specific heat, the heat capacity of a substance per unit mass (side note: this website helped me to differentiate between specific heat and heat capacity). The heat capacity of a substance as well as the mass of the sample are two influential factors in temperature change. Thus, the change in enthalpy can be determined using this equation:

∆H = m x Cp x ∆T

The change in enthalpy is equivalent to the mass of the sample multiplied by the specific heat capacity of the substance multiplied by the change in temperature. However, calorimetry uses the energy gained or lost in the surroundings to calculate the energy gained or lost in the reaction. Due to the first law of thermodynamics (also known as the law of conservation of energy) the energy gained or lost by the system is is equal and opposite to the energy gained or lost by the surroundings.

∆Hreaction = -∆Hsurroundings  

The third method of obtaining the change in enthalpy value is through Hess' Law which states that if a reaction is carried out in a series of steps, the overall change in enthalpy will be equal to the sum of enthalpy for each of the individual steps. Hess' Law is closely related to the concept of state functions. A state function is a property that depends on the present state of the system, not the path taken to arrive at that state. 

An example of a state function. 

The fourth and final method of calculating ∆H is through enthalpies of formation. Enthalpies of formation are hypothetical values that indicate how much heat would be lost or gained during the formation of one mole of a compound from its elements in their standard states (standard states meaning 25˚C and 1 atm). Heat of formation reactions are always written so that all reactants exist as they would under standard conditions and if there was one mole of product. Using the method of enthalpy of formation, ∆H is calculate using the equation:

∆H = ∑n∆H˚f(products) - ∑n∆H˚f(reactants)

The heat gained or lost in the reaction is equal to the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the products minus the sum of the stoichiometric coefficients multiplied by the enthalpies of formation of the reactants. Methods of finding ∆H were covered in several thermodynamics lectures as well as in the Thermodynamics I and II worksheets.

Entropy is a statistical measure of the number of most probable distinguishable microstates or degrees of freedom available in a system. When there are a greater number of distinguishable microstates (i.e. greater entropy) the reaction is referred to as favorable or spontaneous. Thus, when the change in entropy, ∆S, is positive, the value is favorable. The change in entropy is equal to the sum of the change in entropy of the products minus the sum of the change in entropy of the reactants.

∆S = ∑S(products) - ∑S(reactants) 

The second law of thermodynamics states that the entropy of the universe is constantly increasing. Reactions may have a reduction in entropy as long as the entropy of the surroundings increase. Entropy is a state function. ∆S is greater than 0 when heat is added (such as in processes like melting and vaporization). ∆S is less than zero in certain cases, such as when substances are dissolved and when solutions of gases and liquids are made. The concept of Entropy was covered in a lecture as well as in the Thermodynamics III worksheet.

Entropy plays a part in determining thermodynamic favorability. A thermodynamically favored process (or a spontaneous process) is a reaction that proceeds without any assistance from the surroundings. In reversible reactions, the thermodynamic favorability determines the direction of the reaction. There are two methods of determining thermodynamic favorability. The first, ∆Suniverse states that ∆Suniverse = ∆Ssystem + ∆Ssurroundings. If ∆Suniverse is positive, then the reaction is favorable. If a reaction increases the entropy of the system (∆Ssystem > 0) and is exothermic, then the process must be thermodynamically favored. The second method of determining favorability is through Gibbs free energy. Gibbs free energy equals the change in enthalpy minus the system temperature multiplied by the entropy change.

∆G = ∆H - T∆S

If ∆G is negative, then the reaction is thermodynamically favored. Gibbs free energy is a state function. ∆G˚f is the energy change that occurs when one mole of a compound is made from its elements in their standard states. ∆G˚f values can be used to determine the ∆G value for an equation using the equation above. 

Combing over these concepts and writing this blog post, none of these concepts seem incredibly difficult. However, when it comes to applying these concepts and choosing which equation or formula to use, I find myself at a complete loss. I struggled to complete the review task chains. In the Thermodynamics II worksheet as well as in the task chains, I found myself struggling the most on problems that involved stoichiometry. I feel I have adequate understanding of Hess' law and finding ∆H based on enthalpies of formation. I have had trouble determining whether or not reactions are endothermic or exothermic. The concept of specific heat and calorimetry also baffled me a bit. As I do not feel prepared in the slightest for the upcoming test this week, I will be going over the task chains multiple times and revisiting worksheets that were completed in class. I'm sure the test will be a sobering experience that will test the self inflated image of my self worth and knowledge.  

11/4/13-11/8/13

This week in AP Chemistry I continued to study intermolecular forces. The topics covered this week included viscosity, surface tension, cohesive and adhesive forces, different forms of solids, vapor pressure, and lattice energy.

Viscosity is a liquid's resistance to flow. It is the ease by which the molecules in a substance move past each other. The higher the viscosity, the less a liquid moves. Viscosity increases with stronger intermolecular and decreases with higher temperature. This topic was covered in the Liquids lecture.

Surface tension results from an imbalance of intermolecular forces at the surface of a liquid. Molecules on the surface are subject to only inward forces, whereas molecules in the center of a substance are surrounded on all sides by molecules and thus are subject to a full cadre of intermolecular forces. Of all the topics covered this week, I had the the weakest understanding of surface tension. I did not comprehend the portion in the Liquids lecture about surface tension well. I also feel as though I am unable to complete the questions in the worksheets I was given in class that regard surface tension. This website gave me a little clarity on the subject, although I was unable to find a good source that could give me a more in depth explanation that I could understand well.

Cohesive forces are intermolecular forces between molecules of the same kind. Adhesive forces are forces that are intermolecular forces between two molecules of a different kind. The strength of cohesive and adhesive forces can determine how a substance acts on a certain surface. For example, if you put a drop of water on a waxed surface and a drop of water on glass, the water on the waxed surface beads up (the cohesive forces between water molecules are greater than the adhesive forces between water and the waxed surface), while the water on the glass spreads out (the cohesive forces between water molecules are weaker than the adhesive forces between the water and the glass). Cohesive and adhesive forces were covered in the Liquids lecture as well as in the Intermolecular Forces II worksheet. I felt I had a strong understanding of this topic.

Another example of cohesive vs. adhesive forces. In the tube with water, the adhesive forces between water and glass are greater than the cohesive forces between the water molecules. Thus, water has a concave meniscus. In the tube with mercury, the cohesive forces between mercury molecules are greater than the adhesive forces between mercury and glass. Therefore, mercury has a convex meniscus.

Additionally, I studied various properties of solids this week. Solids generally fall into two groups: crystalline and amorphous. Crystalline solids are solids in which the molecules have a highly ordered arrangement. Amorphous solids are solids in which there is no order in the arrangement of particles. Furthermore, there are other classifications of solids including molecular solids, ionic solids, and covalent network solids. Molecular solids are solids where molecules are held together by van der Waals forces. They tend to be soft and usually have low melting and boiling points. Ionic solids are the solid forms of ionic compounds. In ionic solids, ions pack themselves in a position that maximizes attraction between ions. Ionic solids typically have very high melting and boiling points. Covalent network solids are solids in which atoms are covalently bonded to each other in a large network. Substances like diamond, graphite, quartz, and glass are all network solids. These solids tend to be strong and have high melting and boiling points.

A vapor is the gas form of a substance that is normally liquid at room temperature. Pressure is the force by which gas particles hit a surface. Vapor pressure is simply the pressure of a vapor at a given temperature. The greater the number of gas particles, the greater the pressure of a gas. As temperature increases, the kinetic energy of molecules in a liquid increases and more particles "escape" the liquid and evaporate. Thus, the pressure of the surrounding atmosphere is increased. When the vapor pressure equals the atmospheric pressure, a substance boils (at sea level, this pressure is 1 atm). However, intermolecular forces influence boiling point. The greater the intermolecular forces, the greater the boiling point. Boiling point and vapor pressure are linked; the greater the boiling point, the lower the vapor pressure. The lower the boiling point, the higher the vapor pressure. All substances have the same vapor pressure at their boiling point. Initially, this concept threw me off. When we studied properties like viscosity and boiling point, as the intermolecular forces grew stronger, the viscosity and boiling point of that given substance increased as well. But with vapor pressure the relationship is different. Other than that small caveat, I understood this concept well. I completed several assignments that related to vapor pressure, including a lecture and the Intermolecular Forces II worksheet.

The final subject I studied this week was lattice energy. Lattice energy is the energy required to completely separate a mole of a solid ionic compound into its gaseous ions. Lattice energy is associated with coulomb's law; it increases as charge magnitude increases and increases as the size of ions decreases (distance between oppositely charged ions decreases, so the charge between them is stronger). This process is very exothermic; it takes a considerable amount of energy to break ions from the solid.

I felt I had a strong grasp on most of the topics covered this week with the exception of surface tension. I will continue to look for a source that could give me some clarity in that subject. I also feel relatively prepared for the upcoming test on Tuesday, but none the less I will be furiously studying and stressing on Monday night.

10/28/13-11/1/13

This week in AP Chemistry I learned about intramolecular forces and intermolecular forces. Intramolecular forces are forces between atoms in a molecule such as covalent and ionic bonds. However, intramolecular forces were not the main focus this week. Intermolecular forces were.

Intermolecular forces are attractive and repulsive forces between molecules. Intermolecular forces are weaker than intramolecular forces but are strong enough to influence physical properties of substances. There are several different types of intermolecular forces. Most fall under the category of "van der Waals forces." In class, I learned of four important van der Waals forces: dipole-dipole interactions, dipole induced dipole interactions, induced dipole-induced dipole interactions (a.k.a. London dispersion forces), and Hydrogen bonding.

Dipole-dipole interactions are attractive forces between the partially negative end of a polar molecule and the partially positive end of another molecule. Although the strength of dipole-dipole interactions are dependent on the magnitude of a substance's dipole moment and the proximity of its molecules, in general dipole-dipole attractions are relatively weak. Dipole-dipole interactions are only present in polar substances.

Hydrogen bonds are closely related to dipole-dipole interactions. Hydrogen bonds occur when partially positive hydrogen atoms are attracted to very electronegative partially negative atoms like oxygen, fluorine, and nitrogen. Although in essence they are the same as dipole-dipole interactions, they have their own classification because they are much stronger than typical dipole-dipole forces. Hydrogen bonds are the strongest of all of the van der Waals forces.

When a polar molecule approaches a nonpolar molecule, the electrons in the nonpolar molecule, usually shared evenly, are attracted to the polar molecule. This results in an induced dipole in the nonpolar molecule. This interaction is aptly named a dipole-induced dipole interaction. Much like the dipole-dipole interaction, it too is relatively weak.

Induced dipole-induced dipole interactions are the most common. Electrons are constantly in motion. Even in nonpolar molecules, there is a probability that at some point there will be an imbalance of electrons (this distortion in electron clouds is called polarizability). When such an imbalance occurs, a temporary dipole is produced. This temporary dipole induces a temporary dipole in other molecules. This intermolecular force, also called the London dispersion force, is present in all substances.

Additionally, I learned this week about ion-dipole interactions. Ion-dipole interactions are forces between ions and molecules. They are not van der Waals forces. Ion-dipole interactions make it possible for some ionic substances to dissolve in water. If the forces of attraction between atoms in an ionic compound are greater than the ion-dipole force between the ionic compound and water, the substance will not dissolve in water. If the forces of attraction between atoms in an ionic compound are lesser than the ion-dipole force between the ionic compound and water, the substance will dissolve in water.

Intermolecular forces have a great effect on the physical properties of substances. For example, the stronger the intermolecular forces, the higher the boiling and melting point of a substance. Intermolecular forces also affect the vapor pressures and viscosity. To help learn the concepts associated with intermolecular forces, I completed three lectures (one on dipoles and induced dipoles, one on hydrogen bonds, and one on ion-dipole interactions) as well as a the Intermolecular Forces POGIL and the Water POGIL. In addition, we went over the Intermolecular Forces I worksheet. Initially I found the concept of intermolecular forces to be confusing. I found the scanned textbook chapter you gave us in class on Tuesday to be challenging reading. I did not comprehend the text well (the fact that I was running on a minuscule amount of sleep certainly did not help in this regard). However, I found that the lectures brought me to understand these topics well.

A demonstration from class this past week. Butane is lit on fire. In the can, butane is in liquid form because it is under very high pressure. However, when it was sprayed into a test tube, it immediately started to boil, as butane normally has a very low boiling point. When lit on fire, the butane gas burned.  

10/21/13-10/25/13

This week in AP Chemistry we had the third test of the year. This exam covered the topics of covalent bonding, electron domain geometry, as well as and molecular geometry. To study for the test, I had to complete a task chain. We also went over a packet of review worksheets in class on these subjects. At home, I did the task chain multiple times and completed extra problems on the review worksheets. On Tuesday, I took the test and got an 80 percent on the multiple choice and a 98 percent on the free response. I was very satisfied with my performance on the free response but was a bit disappointed with my multiple choice test. I felt I had a fairly good grip on this subject matter (better than the material on the stoichiometry test in fact), and yet I got a score that I was not very satisfied with. Considering the fact that I have gotten relatively iffy scores on the last two multiple choice exams, I've surmised that I need to pay more attention to the multiple choice portions on tests and make a better effort to practice the types of the questions that may be found on them. All in all, I was a bit irked with my performance on the multiple choice. However, because of my results on the short answer and the fact that the multiple choice did not significantly affect my grade, I found some satisfaction in my performance on this test.

The day after the test was mole day. We were assigned an essay on the subject of polarity and hydrogen bonding as it related to an article we were given about paintball. I found the assignment to be difficult at times, as I found the prompt to be a bit vague. As only one section of the paintball article really related to polarity and hydrogen bonding, it was hard to find excerpts in the writing that supported the points in my essay. With that being said, I did learn about hydrogen bonding and how polarity relates to solubility. Hydrogen bonding is when a partially positive hydrogen atom bonded to an electronegative atom is attracted to a partially negative electronegative atom in a separate molecule. Unlike covalent bonding, there is no physical bond present, but there is attraction between the two atoms.

A hydrogen bond, represented by the dashed line.

After these two assignments, I was reintroduced to the topic of ionic bonding via a POGIL that I completed on class. Ionic bonds are bonds between a metal and a nonmetal. I learned that the size and individual charge of each ion has a profound effect on the strength of an ionic bond. The stronger the bonds in an ionic compound, the higher the melting point. For the most part, I felt I had a decent understanding of this topic. I was a bit confused about the sizes of cations and anions.

The final topic we covered this week was metals. Metals are elements that conduct heat an electricity well. They are typically malleable and ductile. They exhibit several other unique physical properties, such as luster. Metals bond together through metallic bonds. In metals, electrons called conduction bond electrons form a "sea" of electrons around atoms. These conduction bond electrons are d-orbital electrons that are promoted into the outer p-orbitals of atoms. Conduction electrons are the reason that metals conduct heat and electricity well. The more conduction electrons, the higher the melting point of the metal and the harder the metal is. Most ordinary uses of metals involve alloys. Alloys are mixtures of elements that have the properties of metals. There are two main types of alloys: substitutional alloys (solute particles take the place of solvent metal atoms) and interstitial alloys (solute particles fit in holes in between solvent metal atoms). To help learn about metals, I completed a POGIL on metals on Friday and watched a lecture on metals this weekend. I found the subject straightforward and easy to understand.

    

10/14/13-10/18/13

This past week in AP Chemistry was the most confusing so far. I was introduced to the concept of hybridization and sigma and pi bonds.

In covalent bonds, electrons can only be shared if the electron orbitals of two atoms overlap. Hybridization occurs when these orbitals combine to form new, hybrid orbitals. For example, the s and p orbitals combine to form an sp orbital. There are three widely accepted forms of hybridization: sp, sp², and sp³. Hybridization is a theory; whether or not it works beyond the second period is questioned. Additionally, hybridization is central to valence bond theory, a.k.a. sigma and pi bonding, which is explained later. 

sp hybridization.

I did not understand the concept hybridization well, which would explain my rather lackluster explanation. I do know that you can determine the hybridization of a molecule by simply counting the number of electron domains around the central atom. If there are two electron domains, the hybridization is sp. If there are three electron domains around the central atom, the hybridization is sp². If there are four electron domains, the hybridization is sp³. Whether hybridization occurs beyond that is hotly contested within the scientific world and does not concern me. However, conceptually I do not understand hybridization beyond the notion that orbitals can combine to make "hybrid" orbitals. To cover these concepts, I watched a lecture on hybridization and completed the Hybrid Orbitals POGIL for homework. While what I know may be enough to get by on the test, I still would like to understand the concept of hybridization more.


I found the concept of sigma and pi bonds to be more straightforward. Sigma bonds are bonds with head to head overlap and cylindrical symmetry of electron density about the internuclear axis. Pi bonds are bonds with side to side overlap. Electron density is above or below the internuclear axis. In a single bond there is one sigma bond. In a double bond there is one sigma bond and one pi bond. In a triple bond there is one sigma bond and two pi bonds. I was confused about this concept until I completed the task chain quiz that involved sigma and pi bonds. When I got an answer wrong regarding a sigma and pi bond, the information window that popped up gave a good explanation of sigma and pi bonds. To cover sigma and pi bonds, I watched a lecture on the subject. This website also gave me some clarity on the subject.

There is a pi bond between the blue p orbitals, as there is side to side overlap between them. There is a sigma bond between the green orbitals, as there is head to head overlap between them. 

Throughout the week in class I worked on a lab/activity involving the program Web MO in which I modeled molecules from the VSEPR Theory lab with my table group. Using the program, I determined the bond angles, polarity, dipole moment, and more for several molecules. The activity really helped me better understand the concepts of bond and molecular polarity, as well as how unbonded electron pairs affect the shape of molecules.