Monday, 31 October 2011


Topics Covered
  • Physics 'n' Energy
  • Chemistry 'n' Energy
  • Enthalpy
  • Heat and Temperature
  • Hess's Law
  • Standard Enthalpies of Formation

Physics 'n' Energy

Thermochemistry is really all about energy. But first we've got to know what the hell energy is. The way that is used to describe it is mostly helpful in Physics. Therefore, I'm gonna describe it in physics first, and then the next section will be energy as it relates to our subject, chemistry. Heh...
Energy is the ability to do work. We've all heard that before. In actuality, it's sort of hard to define energy exactly. Even my normally adequate book admits that. Even though that's sorta fuzzy, you've used the term 'energy' before in everyday speech. It's... well... energy!!!
There are two flavors of energy; kinetic and potential. Let's start with kinetic energy. All matter that is moving has a certain amount of kinetic energy. In fact, if you actually visit my pages, I gave the formula for kinetic energy before:

It is dependant on mass and velocity. Note that it is proportional to the mass and the square of velocity. So if Ball A of some mass and velocity has a certain energy, and you have Ball B, with twice the velocity but only half the mass of Ball A, Ball B will still have twice as much kinetic energy, because velocity is squared. Kinetic energy can be changed, usually by a change in velocity (objects don't just change their mass, they're more likely to drop or gain in speed. Always exceptions, like if a rocket is losing fuel, then it's dropping in mass.) This change in kinetic energy is what is work. Like if a rocket is speeding up, then work is being done on it, because it's gaining kinetic energy. Likewise, if a rocket is slowing down, then the rocket is doing work, because it's losing kinetic energy. Yep, kinetic energy is some good stuff to have.
And then there's potential energy. It's harder to see, because it's the energy you can't see. You can 'see' the energy of a ball being thrown, you feel the energy transfer as you catch a ball. Potential energy is the energy available to the ball to convert to kinetic. In physics, the potential energy you talk about the most in mechanics is gravitational potential energy. This is simply the potential energy caused by gravity. Like if we take that same ball, and hold it off the side of the Empire State Building, it has the "potential" of gaining a lot of kinetic energy (by dropping it). It has a lot of potential energy, but no kinetic (it's not moving...yet.) But once you do, all that potential energy will be converted to kinetic, not all at once, but gradually. The ball speeds up faster and faster as it falls. A rule; the higher an object is, the more gravitational potential energy it has.
Now you don't have to know anything I have just said to do good in chemistry, just take away with you the Law of Energy Conservation. When you add the kinetic and potential energies of that ball at any time during its fall, they will be the same. Energy can't be gained or lost; it is always converted. Even when that ball hits the ground, and has lost all its kinetic and potential energy, that energy didn't really die, it just got converted to something else. The pavement (or car, wherever it landed doesn't matter) will have gained a slight increase in temperature, and that's where the energy went. It was dissipated as heat. The energy in a closed system always remains constant. I say closed system because if you considered the Earth a closed system, and then an outside source of energy (such as the falling of a great meteorite) came in, then the closed system will have more energy.
Energy comes in all sorts of shapes and sizes. There's heat (probably most important energy in chemistry), light, sound, mechanical, electrical, nuclear, matter/anti-matter reactions, the list goes on...
Let's talk chemistry.

Chemistry 'n' Energy

Finally, back to chemistry. In chemistry, reactions can give off heat, or absorb heat. Those that give off heat are called exothermic, and those that take in heat are called endothermic. In exothermic reactions, heat is a product (it's being formed), so a reaction of this kind might look like this:
A + B ---> C + D + heat
And similarly, if a reaction is endo, then it acts like a reactant (goes on the left side):
A + B + heat ---> C + D
Where does this heat flow from and to? In chem, you consider the entire Universe to be divided into two parts: the system, and the surroundings. The reaction is the system, and the surroundings is everything else. For the sake of pictures, pretty pictures, let's say the system is pink and the surroundings are purple. Here are illustrations of an exothermic and endothermic reaction.

Where is this heat coming from? I said that the energy in a closed system is constant, so in an exothermic reaction, how come this heat is coming from nowhere? It's not really. Before the reaction, there was some potential energy stored in the bonds that made up the chemicals. When the bonds were broken and new bonds were formed to make new things, the energy of the new bonds was less than the energy they had previously. So that potential energy that was 'lost' was actually converted into heat, and that's where the heat came from. So the total energy level of the system and surroundings has in fact remained equal. This is in fact the first law of thermodynamics.

In chemistry, you always look at the system's point-of-view. The energy of the system can be changed in two ways: either change the heat of the system, or make it do work or work done on it. The energy change of the system is equal to the amount of heat added to it, plus the amount of work done on it. In other words (or symbols):

DE is the change of energy that happens to the system. q is the heat added, and w is the work done to it.
In thermodynamics, there is the concept of + and - signs. Since we are talking about point-of-view of the system, let's see how a + or - q/w means. If q is positive, that means the system will gain heat. If it's minus, then heat is being removed. That's simple enough. If w is positive, then work must be done on something as to make the energy of the system gain. If you do work on the system, then w will be positive. If you are expending energy, doing work, to make the system happy, then the system will gain energy. Similarly, if you're tired of doing work for this system that has done nothing for you, you can let the system do work for you. In which case w is negative, since it will be losing energy.

The unit of energy is the joule (Symbol: J). It is equal to the amount of kinetic energy a 2 kilogram ball has when traveling at 1 m/s. If bigger units are needed, the kilojoule (kJ) is used, and it's obviously 1000 joules.
I think we can do a problem or two here.

Calculate the change in energy of a system if it did 14.3 kJ of work, while giving off 34.5 kJ of heat. Answer
The most important part of this problem is to figure out the sign of w and q. It's giving off heat, therefore losing energy, so you can expect q to be negative. If it is doing work, then it is also losing energy in that way too, so w is negative as well. All you gotta do is add them up (-14.3 kJ + -34.5 kJ) and you will learn that the system has lost a total of 48.8 kJ.
So the change of energy of a system is only dependant on the work a system does and the heat flow. Note the difference between heat and temperature: THEY ARE NOT THE SAME!!! Heat is energy. Temperature, is... well... how hot something is. It is the heat flow that causes a change in the temperature. If you add heat (energy), then the object's temperature will go up; by how much, you cannot tell by just knowing the heat involved. You'll learn more of this later.
Let's talk about a way some system can do work. There's many ways, so we'll just cover one; gases. Think about what you do if you squeeze a balloon. Aren't you doing work to it? You are using your hard-earned energy to push this balloon in, and since energy is always transferred, it has to go somewhere, so it goes into the system of the gas in the balloon. So, if a gas is being crushed (volume is getting smaller), then work is being done to it, and w in that case is positive.
And if the volume is getting bigger, then w is negative. Why? Because the gas is pushing against the walls of the container, to give itself more room. So it's doing work, or losing energy.
The work being done is the pressure times the change of volume, like this:

w = PDV

This equation is not done yet. If volume is increasing (DV is positive) , then by equation it says that w is positive. But we just said that the system will be expending work in increasing its volume, so w should be negative when the change in volume is positive. A simple negative sign should fix that. And your final equation for the work done by a gas changing volume is:

You can see how I could incorporate this into a problem involving the first and second equations, but I won't. You get the idea.


Aaah, yes, enthalpy. If you've just heard of enthalpy, you will have no idea what it is. When you're done with enthalpy, you'll still not have a clear good idea of what it is, but you will know how to do problems with it, and that's what is important.
The enthalpy of a system, H, is simply defined as:

H = E + PV

Enthalpy is equal to the total energy of the system, plus the pressure of the system times the volume of the system. It's sort of hard to grasp what enthalpy is, from that definition. Instead of enthalpy (H) itself, you will usually deal with a change of enthalpy (DH). And if pressure is constant, and the only work allowed to work on the system is through volume, then:

DH = q

Yep, you should think of enthalpy as sort of like heat. It's not heat exactly, but if those two conditions are met, then it is heat.
So if the change of enthalpy is increasing, that means it is gaining an increase of energy, and therefore is endothermic. If the change of enthalpy is decreasing, that means it is losing heat to the surroundings, or exothermic. Final thought: +DH = Endo, -DH = Exo.
How do you find the change of H? It's the enthalpy of the final products, minus the enthalpy of the reactants. Or...
DH = Hproducts - Hreactants
So if the products have less energy than reactants, then DH will be negative, indicating energy was lost. And vice versa.

Heat and Temperature

I hope by now you realize the difference between heat and temperature. If you add heat (energy) to something, it will eventually start getting hotter and hotter. But the difference is, some substances will get hotter than others, if given the same energy. For example, on a hot summer day, everything around you might seem very hot (like the pavement, or your car), but if you go into a pool of water, it will be much cooler. Both the pavement and the water have received about the same amount of energy from the sun, but the pavement has gone through a much bigger temperature change than the water. Water is resistant to a change of temperature, in comparison to the pavement. Let's assume the temperature change is proportional to the energy it has received, so we can come up with an equation relating temperature and energy.

C is called the heat capacity and is different for different substances. This number C will tell you how hard is it for this substance to get hot. As you can figure, the bigger C is for some substance, the more energy is needed to get it hot.
We are forgetting one important factor; how much of the substance there is. The more there is, the more energy is needed to change its temperature. Putting all three things into account, we can come up with a more useful equation.

This is the same equation as the one before, but notice there's an m in the bottom. This is the definition for specific heat capacity. It's equal to the heat added divided by the mass of it and the change in temperature. As with the normal heat capacity, the bigger this number for some substancee, the harder it is to change its temperature. Water's specific heat capacity (or simply specific heat) is around 4.18 J/oC g.
Important! The mass in the above equation must be in grams. Well, if you use kilograms consistently, I guess it doesn't matter, but since specific heats in tables in books are given with grams, you should use grams.
Rearranging the above equation, you can solve for energy, since you can figure out the other three by measuring but not always energy.
This starts a whole new area called constant-pressure calorimetry. You can figure out how much energy is given off by certain reactions, as long as the pressure is constant. Otherwise, some of the energy would be lost/gained by changing the pressure and your results would be off. You will do a lab with this, probably doing the reaction with styrofoam cups. Why? Because you don't want the energy escaping into the atmosphere or to the walls of the container; you want accurate results. Styrofoam is an insulator, doesn't absorb much energy. The basic setup is as follows:

You use two styrofoam cups to make sure all the energy involved in the reaction is directly related to the temperature change of the chemicals, not the container/air. The cover on top is to make sure energy isn't leaving from the top. The thermometer is to measure the temp change.
How exactly do you do this? Let's say you're going to measure the energy given off by adding 50.0 mL of Chemical A and 20.0 mL of Chemical B. You can start with either one, let's start with A. Measure the weight of the cups beforehand. Put the 50.0 mL of A into the cup. Measure the temperature before. And then add B to it, and cover it and mix periodically. Keep on look at the temp, it should be changing. When it's done changing and stays the same, record the new temp. That's all there is to it. (And then measure the mass after the two things were put together.)
Then in q = s x m x T, you can subtract the masses to find the mass of the chemicals (in grams!). You can subtract the final temp minus the initial temp to change delta-T. But what about s? You can use water's specific temp, 4.18 J / K m. Why? Because most chemicals you'll be using won't be pure chemicals, but rather solutions. In case u forgot, that means they're dissolved in water, and it's not enough to significantly change the 4.18. So now you can find out how much heat is gained or lost.
There is one more concept here. Since pressure is constant, we can figure that the heat lost/gained is directly related to the enthalpy change. You could say that q = DH, but you'd be WRONG! In actuality, q = -DH. Why? Remember that enthalpy looks at energy from the point of the system. If heat was released (indicated by positive q), that means the system itself has lost it. Or if heat was absorbed (negative q), that means the system has gained it. So heat evolved and enthalpy change are just opposite of each other.
We've talked about constant-pressure cal. in detail; I'll just touch on constant-volume calorimetry here. Instead of keeping constant pressure, it tries to keep volume constant. That means you need a container that won't change volume. They're usually big strong metal cubes called a bomb calorimeter. I won't go through the specifics, because you probably won't be doing a lab with it; just be aware.
One last thing; in addition to specific heat capacity, there's the molar heat capacity. Instead of using grams to measure amount of substance, it uses moles. I won't give you equation, or any examples, because using specific heat is by far what everyone uses. Of course it has different units; instead of J/ K g, it's J/K mol.

Hess's Law

This has to do with enthalpy again. Yay. Remember that reactions have a certain enthalpy change with it: if it's positive, then the reaction will absorb energy; if negative, the reaction will give off energy. We can write the equation, and then the delta-H associated with it to the right of that. Like for example, to show the enthalpy change for the boiling of one mole of water to water vapor, you can write like this:

H2O (l) ---> H2O (g)     DH = 44 kJ

So before one mole of water evaporates, 44 kJ of energy must be given to it.
I had to look up the 44 kJ in a table or something; there's no way anyone can just look at the equation and come up with it. But how about reactions that can't be found in a table? There are tons of reactions; you just can't carry a 30 pound book with you and look them all up. Hess's Law exists to make this easier for you. Technically, it means that the enthalpy change between two states is not dependant on the pathway it takes to get there. At first, this doesn't seem to help you one bit; but it means if you can add two or more equations to get the desired equation, then you can add their respective enthalpy changes to get the enthalpy change of this equation. There are two rules you will need to use in these problems. We'll use the equation above to illustrate these two rules:
  • If you have to multiply each side of a reaction by some number X, then multiply the respective enthalpy change of that reaction by X also. So, if we needed to use not
    H2O (l) ---> H2O (g), but twice that:
    2H2O (l) ---> 2H2O (g)
    Then we just multiply the delta-H of the original reaction by the same factor, two. Therefore the enthalpy change of 2 moles of water evaporating is 88 kJ.
  • If you have to flip the equation around (the right side on the left and the left side on the right; switching the products and reactants; you get the idea) then just take the negative of the delta-H to get the delta-H of your new reaction. So if we wanted to find the enthalpy change of one mole of water vapor condensing to liquid water, like this:
    H2O (g) ---> H2O (l)
    Then you just take the negative of the original delta-H, so the enthalpy change of the above equation is -44 kJ.
So now we can find enthalpies of certain reactions without a table. For example, if we needed to know the enthalpy change when 3 moles of water vapor condenses to water, all you do is flip the original equation and multiply by 3, or -132 kJ.
This is all you're going to learn in this section. But there are more involved problems. For the sake of simplicity, I'm not going to use real chemicals and enthalpies, because I don't know them. I'll just use chemicals "A", "B", and so forth. Have fun!
Please find the enthalpy change of
A + 2B ---> C
using the following information:
#1: E ---> D + 2A,        DH = 43.22 kJ
#2: B ---> F + [1/4]D,   DH = 342 kJ
#3: 4F + E ---> 2C,       DH = 2.2 kJ
This is about the hardest problem you might get. They can get harder, but I think they're past the scope of this course (I hope!) You have to add the equations below it to get the final equation. Where do we start? It's actually quite simple if you start with the beginning equation. We need a single A on the left. Well, look at the first given equation. It has a 2A in it, and no where else will you find an A. So you KNOW that the A on the left hand side MUST come from Equation #1.
You have to modify this equation so that the A's here look like what you want it to (the beginning equation). There's 2A's on the right. You want it to look like a single A on the left. So we can divide by two, and flip the equation.
A + [1/2]D ---> [1/2]E
Good! Now we have 1 A on the left. But remember that if we modify the equation, we have to modify the delta-H on this one too. You flipped it, so you can take the negative. So delta-H = -43.22 kJ. But we also divided by two, so divide this by two, and our new delta-H for #1 is -21.61 kJ. Keep this number in mind.
Let's move on to the next figure in the beginning equation: 2B on the left. Well, out of the three givens, only #2 has a B anywhere. So that's where we must get our B's from. Problem is, the B is on the left like we want it, but there's only one. So multiply by two:
2B ---> 2F + [1/2]D
Coolio. Now we got 2B on the left. That's it. But remember to multiply original delta-H by 2, so the new one is 684 kJ.
And finally, we need one C on the right. Since we used #1 and #2, we might suspect that C lies in #3. And of course, there it is, on the right of #3 like we want it. But there are 2 instead of one. What to do? Of course you divide by 2:
2F + [1/2]E ---> C
And then divide 2.2 kJ by 2, or 1.1 kJ for the new reaction.
Now what? Add the three new equations together (put all lefts on left, and all rights on right):
A + [1/2]D + 2B + 2F + [1/2]E ---> [1/2]E + 2F + [1/2]D + C
It looks like it's more fukt up than before, but look carefully and something magical happens; don't you have a [1/2]D on left and [1/2]D on right? You can subtract [1/2]D from both sides and cancel them. Same with 2F, and [1/2]E! So cancelling them, you have:
A + 2B ---> C
Wow, it's amazing, it's the beginning equation! Since all you did was add the modified equations to get the desired equation, all you have to do to find the desired enthalpy change is add the modified enthalpy changes:
DH = -21.61 kJ + 684 kJ + 1.1 kJ = 663 kJ.
Woohoo! That's the enthalpy change for A + 2B ---> C. It's strange how the given equations just coincidentally cancelled out to make exactly the equation we were looking for. It's not coincidence; these type of problems must be planned out. It's a LOT harder making one of these problems than it is to solve them!

Standard Enthalpies of Formation

This has to do with enthalpy once again, as you might have brilliantly deduced from the title. Basically, the standard enthalpy change of formation of something is the enthalpy change of the reaction of the elements in their natural state coming together to form it under standard conditions. For example, the reaction of the formation of water is:
H2 (g) + [1/2]O2 (g) ---> H2O (l)
Note that hydrogen and oxygen must be in diatomic form and gaseous, because that is how they exist in a standard state. What is this standard state? In thermodynamics, the standard state is at 1 atmosphere and 25 oC (about room temperature).
The standard enthalpies of formation for many substances can be found in a table. What is so useful about this? Let's say you have a reaction that you have to find the enthalpy change in the standard state. But instead of before where we had all those given equations to mess around with, you are just given a table of standard enthalpies of formation. How will you do this? Here's the equation:
DHo = SnpDHof p - SnrDHof r
You might be staring at this mess and wondering what the hell it means, but basically it says that you add up the standard enthalpies of formation on the right side (multiplying with respective coefficients) and subtracting the standard enthalpies of formation on the left side (ditto.) You know, it just might be easier if you saw this one in an example. Once again, I am using A's, B's, C's, and D's, but in a real problem you'll get real things to use.
Find the change of enthalpy of the following equation (under standard conditions):
A + 2B ---> 2C + 3D
Using the following standard enthalpies of each substance:
A: 34 kJ/mol
B: 45 kJ/mol
C: 22.3 kJ/mol
D: 56 kJ/mol
Firstly, I should explain the kJ/mol. Enthalpies are supposed to be in joules, or kilo joules, right? But for formation, the enthalpy depends on how much you are forming. So, for A, it takes 34 kJ to form one mole of A. So if you were forming 2 moles of A, the enthalpy would be twice as much, or 68 kJ. If you were making 3.45 moles of B, the enthalpy of formation would be 155.25 kJ (3.45 mol x 45 kJ/mol).
Ok, let's start. The formula says to take the stuff on the right, and subtract the stuff on the left. So start with the first thing on the right, 2C. The standard enthalpy for C is 22.3 kJ/mol, but we have 2 of them. So the enthalpy for 2C is 44.6 kJ.
Next up on right; 3D. D's enthalpy of formation is 56 kJ/mol, and 3 of them makes 168 kJ.
So you add up to get total enthalpy on right, which is 44.6 kJ + 168 kJ = 212.6 kJ. Keep this in mind.
Now we go to the left. First thing on left: A. There's only one of A, so total enthalpy is 34 kJ.
Lastly, there are 2 B's. Enthalpy of 1 B: 45 kJ, but there are two, so total is 90 kJ.
Now you add up all the enthalpies on the left. So that is 34 kJ + 90 kJ = 124 kJ.
Last step: Subtract total of left from total on right. 212.6 kJ - 124 kJ = 88.6 kJ.
And there's your answer!
One note: You can only do this if the reaction is to take place in standard conditions. I don't know if I mentioned this before, but enthalpy is dependant on temperature and pressure. So if the reaction above took place at 10 degrees Celsius and at 30 atm's, you couldn't do it this way. Of course, if they gave you enthalpies of formation at those conditions, then go right ahead. (They have to match.)

Types of Chemical Reactions

Most reactions in chemistry take place with water as the solvent. Solutions where water is the solvent are called aqueous solutions. Water is known to be a polar molecule because the unequal distribution of charge caused by the electronegative oxygen. This polarity gives the water the ability to dissolve ions and other polar substance.
A useful method for characterizing an aqueous solution is its electrical conductivity. If a solution conducts electricity well, it is considered a strong electrolyte. If it only conducts slightly, it is considered a weak electrolyte; if it doesn’t conduct, it’s a nonelectrolyte.
Strong electrolytes are substances that are completely ionized in water. Such examples are strong acids, strong bases, and soluble salts. Acids ionize into H+ and A-; bases ionize into OH- and X+.
A weak electrolyte is a substance that only slightly ionizes when added to water. These substances include weak acids, weak bases, and slight soluble salts. Weak acids are those that only dissociate slightly into H+ and A-. Similarly, weak bases are those that dissociate only slightly into OH- and X+.
Nonelectrolytes are substances that dissolve in water, but that don’t break up into ions. These substances are mostly polar molecules; the reason they don’t conduct electricity is that no ions are formed
Solution Reactions
One type of reaction is called a precipitation reaction. This occurs when two solutions are mixed resulting, and a solid or precipitate forms. The precipitate contains ions that when combined are insoluble with water. However, these individual ions do dissolve in water. So, the result of mixing these ions is an insoluble solid. One example of a precipitation reaction is when a solution containing Ca2+ and a solution containing SO42-; the result of mixing these the formation of solid CaSO4.
Another type of reaction is an acid-base reaction. An acid-base reaction is one when the net reaction is the combination of a proton and hydroxide to form water. The acid base reaction is also called a neutralization reaction. When performing calculations fo an acid-base reaction follow these simple steps:
  1. List all the species before any reaction occurs and decide what reaction will occur.
  2. Write the balance net-ionic equation
  3. Calculate the moles of reactants
  4. Determine the limiting reactant where appropriate
  5. Calculate the moles of the required reactant or product.
  6. Convert to grams or volume, as required.
The final type of aqueous reaction is an oxidation-reduction reaction. An oxidation-reaction is characterized by the transfer of one or more electrons. These reactions are often used for energy production; in fact, these reactions are often used in the human body to provide energy.
The concept of oxidation states provides a means for keeping track of the movement of electrons in a redox reaction. The oxidation states can be assigned according to these rules:
  1. The oxidation state of an atom in an element is 0.
  2. The oxidation state for a single atom ion is the charge of the ion.
  3. Oxygen is assigned the oxidation state of –2 in covalent compounds, except in peroxides where oxygen is assigned a –1 state.
  4. In covalent compound hydrogen is assigned a +1 state.
  5. In a covalent compound, fluorine is always –1 state.
  6. The sum of the oxidation states must equal to overall charge of the molecule.
When balancing oxidation it is convenient to divide the reaction into two half-reactions; one reaction involves the oxidation, the other the reduction. Then, follow these steps if it’s in acid:
  1. write the half reactions
  2. for each half reaction balance the element except H and O, balance the O with water, balance the H with H+, and balance the charge with electrons.
  3. If necessary, multiple the reaction by an integer to equalize the number of electrons.
  4. Add the half reaction
If the reaction occurs in base repeat the above steps, but after balancing the hydrogen with H+, add hydroxides to cancel out the H+’s. Then, continue the above steps as before.

Periodic Trends

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  • Ionization energy is defined as the energy required to remove an electron from a neutral atom.
  • Electronegativity is a measure of the tendency of an atom to attract or gain an electron. Nonmetals are usually said to be electronegative; metals, electropositive.
  • Atomic size for the reactive elements is based on atomic radii that are based on bonding distances in compounds.
The inert gas sizes are determined in a different manner, so are not included. They do show the same vertical trend in size and ionization energy.Elements in the last horizontal row-elements 87 - 118 - are not included, although those for which data is available follow the same general trends.
Based on the observed trends, francium and cesium (on the bottom left), are the most metallic elements (i.e. the most likely to lose an electron) while fluorine (on the upper right) is the most nonmetallic (i.e. the most likely to gain an electron).

The above periodic trends are caused by the interactions of three factors: nuclear charge (the number of protons in the nucleus), the electron shell(s), and the shielding (the effect of the electrons between the outer electrons and the nucleus). These effects are summarized in Table 1 below. Similar arguments can be used to explain the trends in ionization energy and electronegativity.

Table 1. Factors affecting horizontal and vertical trends in atomic size:

Horizontal Behavior Vertical Behavior
Factors: Change Effect on Size Change Effect on Size
Atomic number: number of protons increases decreases increases decreases
Electron shell(s) stays same no effect shell added increases
Shielding (intervening electrons) stays same no effect increases increases
Net Effect:

* the increased shielding and the added shell(s) override the effect of the increasing nuclear charge.

Another helpful bit of information is a list of the common nonmetals in decreasing order of electronegativity - F, O, Cl, N, Br, I, S, C……H - (pronounced fossil n brisk to help memorize).
This, along with geometric considerations, will help to determine the polarity (another lesson) of small covalent compounds.

Saturday, 29 October 2011


A solution is defined as a homogeneous mixture or a mixture where the components are uniformly mingled.

Normally we thing of a solution is the liquid state, but this is a common misconception; a solution can be in any state. For example, air is a solution of oxygen, nitrogen, and a variety of other gases all in the gas state. Steel is also solution of carbon and iron, but this solution exists in the solid state. However, the most important solutions in chemistry involve water, so this section will focus of aqueous solution.
Before we begin discussing the properties of solutions, we must define ways to describe the concentrations of various mixtures:

Molarity (M): the number of moles per liter (moles/Liter)

Mole fraction (c ): the ratio of the number of moles of a compound to the number of moles of solution (c A = moles A / (Moles A + Moles B))

 Molality (m): the number of moles solute1 per kilogram solvent (m = moles solute/ kilograms solvent)

 Mass percent or weight percent: the percent by mass of a solute1 in a solution (Mass percent = mass solute / mass solution * 100)

1 A solute is the compound being dissolved in medium. The solvent is the medium in which the compound in being dissolved.

Forming a solution

Dissolving compound in liquids, such as water, is very common; so it would be valuable to understand how and why compound form solutions. To address these questions we must examine what happens when you add a compound to a liquid:
Step 1: the solute particles separate (overcoming the intermolecular attractive forces).
Step 2: the solvent particles separate (overcoming the intermolecular attractive forces).
Step 3: the solute and solvent interact to form the solution.
Each of these steps results in a change of energy. We can sum these energy changes to find the energy change of the solution, called enthalpy (heat) of solution (D Hsoln). Therefore, the D Hsoln is:
D Hsoln = D H1 + D H2 + D H3
The first two steps usually result in a large positive D H, but the third step is usually negative. If the first two enthalpies are greater that the third, the overall D Hsoln will be positive and vice versa. It is important to note, that process with a large positive D Hsoln tend not to occur; however, process with small positive D Hsoln may occur because nature strives for disorder (entropy), and a solution is more disordered than its individual components.

The above concepts can be applied to understand structure can effect solubility; let’s use water mixed with oil for an example. The first two steps from above result in a positive D H, because it requires energy to break water’s hydrogen bonding, and oil’s dispersion forces. The third step does not counter this with a large negative D H, because there is little interaction between oil and water. Thus the D Hsoln is large and positive, so the reaction is not favored. Salts do dissolve in water, because they have a significant interaction with water resulting in a low D Hsoln. Salts are describes as hydrophilic (water loving) where oils are called hydrophobic (water fearing).
Pressure can also effect the solubility of gasses; the relationship between pressure and the concentration of dissolved gas is described in Henry’s law; the amount of gas dissolved is directly proportional to the pressure of the gas above the solution:
Temperature effects the solubility of solutions. One normally thinks that by heating the solution, more solute can be dissolved: this is a misconception. There is come correlation between D Hosoln of a solution and the solubility (see Le Chatelier’s Principle), but the temperature dependence can only truly be seen by experiments.

Gases dissolved in a solution follow a simple rule; the higher the temperature, the less gas dissolved.
Effects of solutions
When a nonvolatile solute is added to a solvent, the vapor pressure of the solvent is lowered. Fancois M. Raoult studied phenomenon, and his results are desribes by Raoult’s Law: 

Psoln = Csolvent Posolvent

One is able to use a simple model to understand the phenomenon; think of the dissolved solute molecules evenly distributed in the solution, even at the surface. Then understand that this caused less molecules of volatile solvent to be at the surface and less molecules evaporating.

This decrease in vapor pressure as a result of the solution can help to explain the following colligative properties, properties that depend only on the number of molecules not he molecules themselves.
When a nonvolatile solute is added to a solvent, the boiling point of the solution is higher than that of the pure solvent. This can be explained by the decreased vapor pressure of the solution; the solution boils when the vapor pressure matches the outer pressure (usually 1 atm), but since the vapor pressure is lowered, a higher temperature is needed to match that outer pressure. This change is given by the equation:
D T = Kbmsolute
Similarly, when a nonvolatile solute is added to a solvent, the freezing point of the solution is lower than that of the pure solvent. This is also explained by the decreased vapor pressure; the solution freezes when the vapor pressure for the liquid and solid phases is the same. This occurs at a lower temperature because of the decreased vapor pressure. This change is given by the equation;
D T = Kfmsolute

The osmotic pressure of a solution is also a colligative property. Osmotic pressure (p ) is the pressure that must be applied to stop osmosis, or the flow of solvent down a concentration gradient. The solvent attempts to move to an area where there is a lesser concentration; the driving force of this movement is an increase in disorder. Experimentally, the osmotic pressure has been show to follow the equation:
p = MRT
where p is the osmotic pressure in atmospheres, M is the molarity of the solute, R is the universal gas law constant, and T is the temperature.
All these colligative properties must be slightly changed for electrolytes, which breakup into multiple pieces. The van’t Hoff factor i must be inserted into the equations. The van’t Hoff factor I is:
i = moles of particle in solution / moles of solute dissolved
Therefore, the new modified freezing point and boiling point equation is:
D T = iKm
The modifies osmotic pressure equation is:
p = iMRT
One should note that ion pairing may result is slightly lowered results than calculated. For example I for NaCl is 2, but the observed i for a .10 m solution is 1.87. This is caused by a few ions pairing together making a fewer number of molecules.


A colloid is similar to a solution, except that a colloid is a suspension of single large molecule or clumps of small molecules. Mud is an example of a colloid; the mud particles stick together and remain suspended in water for an extremely long period of time. The reason these particles remain in solution is that they have layers of oppositely charged articles. When the charges on the outside of these particles are the same, they experience electrostatic repulsion and remain suspended.

Colloids can be destroyed, called coagulation, by heating or adding an electrolyte.


Following concepts will be discussed in this post.
  • Pressure
  • Temperature
  • R, the Gas Constant
  • A Few Gas Laws
  • Back to Stoich
  • Kinetic Molecular Theory of Gases
  • Effusion and Diffusion
  • An Alternative to the Ideal Gas Law
This is a different section. The sections before this just built upon each other, all having to do with chemicals and reactions and formulas. The first part of this chapter will have almost nothing to do with that. Then of course, we must tie it back in. But it's a good break from stoich, you must agree.
Before anything, I hope I don't have to define "gas" for you. Well, just in case, a gas is a substance that doesn't have a definite shape nor volume. Obviously gases will form the shape of their container. And their volume is simply the volume of the container they're in.
Gases push and expand. It's in their nature. That's why they will evetually take the shape of their container: they push and expand, hitting the walls and being stopped. That's because gases have particles that have lots of motion, flying around, in comparison to solids, whose molecules are almost fixed. Gases push against the walls of their container, with pressure. Pressure is defined as a force per unit area. It is simply not enough to say how much it pushes, but also how much surface area it is pushing on.
For example, if you press a book against your hand with a relatively strong force, it doesn't hurt. That force is acting on a large area, so you don't feel much pressure. But what would happen if you push a needle pointwise first into your hand, with the same force? You will definitely feel a pressure. Because that force is acting on a very small area, the pressure will be greater.
You can easily see the pressure by a gas in a balloon. The reason it stays round is because the gas inside is under pressure, pushing at the walls in every direction. If you try to squeeze it, you can feel the gases' pressure increasing, pushing against you. (That's one of the gas laws.)
So what are units of pressure? Pressure is force divided by area, or newtons per square meter. The SI system doesn't like to have different units' names in other units, so they call a newton per square meter a pascal (Symbol: Pa). So 1 pascal of pressure is equal to a force of 1 newton pressing down on a square meter. If you don't know force, a newton is pretty small. Imagine a small force delivered to a relatively big area of 1 m2. A pascal is very small. A more common unit is the kilopascal (Symbol: kPa). Obviously 1 kPa = 1000 Pa.
That's the SI version. Obviously, here in the great United States, we like to use different units that run different to the rest of the world, and pressure is no exception. We use pounds per square inch (Symbol: psi), which is the same concept as the pascal, just with different units. More common units to use in chemistry are the torr and the atmosphere. To discuss any of these requires a discussion of atmospheric pressure:
Air doesn't weigh much. Yet it does have a little weight. Look at all the air above us. There's a ton of it in the atmosphere. It's hard to imagine all that air, and that we don't feel it crushing us down. Air does weigh a lot. But since it's force is spread out over such a large surface as the earth, it's not as much. Still, the pressure is equal to about 101325 pascals, or almost 15 pounds per square inch. That means that a force on a 8.5 by 11 inch piece of paper should be almost 1400 pounds! How come we don't feel all this atmosphere pressure? Because it acts in all directions. That piece of paper feels 15 psi pushing down, pushing up, pushing left, pushing right, pushing in all directions. So the net force, and pressure, is zero. The point is, wherever there is air, it will be pushing at 15 psi in all directions. So, let's take a can. There is air inside and outside a can, and the air inside is pushing outward at 15 psi, and the air outside is pushing inward at 15 psi, and the can doesn't implode nor explode.
Would the atmospheric pressure be greater or less at higher altitudes than at lower altitudes? Well, as you go up, there is less and less air pushing down on you, since a lot of it is now below you. So air pressure will decrease as you go up. That's why they pressurize air inside an airplane, to keep the pressure at normal 15 psi inside, while it's much lower outside. If you thought someone could open the door of the plane, it would take a couple of tons of force to get that door open. Plane doors swing inward, and the much greater pressure on the inside would oppose the door moving in.
One more thing about atmospheric pressure: why vacuums suck. If you have a hollow cube, and the air is taken out of the middle, you have an enormous force crushing all sides of the cube. If you make a tiny hole in this cube, then all this air will try to rush in, because it pushes, and you get a sucking effect.
Anyway, I hope you get what atmospheric pressure is. They named a unit of pressure after it, appropriately called the atmosphere (Symbol: atm). 1 atm is equal to the pressure of the atmosphere, at sea level. So, 1 atm = 101325 Pa. To explain torrs (or mmHg, same thing), we must set up a lab.
Fill a dish with some mercury. Fill a tube completely with mercury. Turn this tube really quickly upside down into this dish. Let's see what happens:
This is how it looks like. Immediately after you flip it, the mercury will want to fall out of the tube, due to gravity. But, there is one force acting on the liquid that will oppose that: atmospheric pressure. It is pushing on the surface of the liquid, and pushing the column back up, like so:
Those arrows are pushing on the liquid, and forcing the column to go back inside the tube. These two forces on the column will come to an equilibrium (they will balance). The result: the distance between the top to the tube and the top of the mercury is always 76.0 cm (or more commonly in mm, 760 mm), at least at sea level. That is the concept of the mm Hg (pronounced millimeters of mercury): 1 mm Hg is the amount of pressure that would make that column of vacuum above the mercury be 1 mm tall. Therefore, the atmospheric pressure is 760 mm Hg. Also, another name for mm Hg is the torr. Just another way of saying it. So 1 mmHg = 1 torr.
There are other units of pressure, like bars or mm H2O, but they aren't used commonly, at least not in chemistry. So that's about it for pressure. It will come back to haunt you later, in fact, probably in the next section. So beware!
You might feel silly to be reading about temperature. You know what it is. It's how hot or cold something is. But in terms of particles and chemistry and such, what exactly is temperature? It's a measure of the average kinetic energy of the particles in a gas. What is kinetic energy? The definition is as follows:
m stands for mass, and v stands for velocity. So if something is moving faster than before, it will have more kinetic energy than before. And so temperature is really the measure of how fast the particles are moving around. (You generally ignore the mass part of it in temperature, since particles are known to change speed but usually not mass... :)
So now you can see why if you heat a liquid enough, it will turn into a gas (evaporate). As it gets an increase in temperature, its getting an increase in kinetic energy (and speed of particles). If you get high enough, the particles will be moving fast enough and be going crazy, and becomes a gas.
Note that temperature is a measure of the average kinetic energy of all the particles. Not all particles will be traveling at the same speed. But the average of the energies is related to the temperature.
There is no upper limit on temperature. You can always give something more and more energy, and the particles will just get faster and faster. There is a lower bound tho. Since the slowest a particle can move is 0 mph, temperature will be at its lowest when the particles are absolutely still. The temperature at which this theoretically happens is called absolute zero. This is very very very cold, much more cold than the coldest day in the North Pole. It's -459.67 oF, or -273.15 oC. You know, that's cold. This is the temperature at which all motion of particles will stop. This temperature has never been reached by anyone anywhere, but people have come close to it (within .00000001 oC or something). Wouldn't it make sense if we started a temperature scale that had this temperature as 0? So that we wouldn't need to deal with negative numbers? The Kelvin scale is the scale we need. The symbol is K (without the stupid circle; not oK.) 0 K is absolute zero. Therefore there are no negative Kelvins. The good thing is, 1 K = 1 oC. The Kelvin and the degree Celsius are the same unit, equal. The only thing different is where they start. So, converting is easy. Let's assume that we can round -273.15 to -273. This is the number commonly used to convert. All you do to the Celsius temperature is add 273 to it, and that's how many Kelvins you have. A quick few examples:
  • 0 oC = 273 K.
  • 10 oC = 283 K.
  • -30 oC = 243 K.
  • 100 oC = 373 K.
You get the idea. And to go the other way; if you wanted Celsius from Kelvin, just subtract 273. You generally don't do that, unless a question asked for units of Celsius or something.
Why go through all this? What was wrong with the Celsius system, it never failed us! But it did, because the numbers weren't proportional. Something at 20 degrees Celsius did not have twice as much kinetic energy as something at 10 degrees Celsius. Back when they invented the Celsius and Fahrenheit scales, they had no idea that there was a bottom of the scale. They thought it went infinite in both directions. So they just picked a temperature and called it zero. (For celsius, it was the freezing point of water.) But with the Kelvin scale, you can truly say that something at 20 K has twice as much kinetic energy as something at 10 K. All of the equations that involve temperature in this section will need to be in Kelvins. So remember to convert!
And one last note: I always said that temperature was related to the average kinetic energy, but not exactly how. Here is the equation (you'll probably never use it, but if you're interested... knock yourself out.)
KE is the average kinetic energy, T is temperature (in Kelvins of course!) and R is...well...let's talk about that next.
R, the Gas Constant
R is a gas constant. It has a constant value. It's used in many equations. There are two forms of this constant (meaning different units). There's a version that involes energy, and the version that involves volume and pressure. Learn to use which one in each equation! (For example, since the equation above in the last section had energy, but not pressure nor volume, you would use the energy form.)
Pressure-Volume form:
Energy form:
In case you don't have a graphics browser, the pressure-volume form is .08206 L atm / (K mol), and the energy form is 8.3145 J / K mol. These two values are exactly the same, they are just used in different equations.
A Few Gas Laws
There are laws all over in science. There just so happens that gases must follow some laws of their own. Let us go over some relationships between some properties of gases.
Boyle's Law
This law says: The pressure and the volume of a gas are inversly proportional to each other, temperature kept constant. What this means, is that if you keep the temperature constant, then if you increase the pressure on the gas, the volume will decrease, and if you lessen the pressure on a gas, the volume will increase.
If you crush a gas, it's gonna shrink.
Real world example: that balloon thing again. If you squeeze a balloon enough, it pushes back, until the gas is too pressurized and the balloon pops. Bigger balloons are easier to squeeze. (Note: I don't know anything about squeezing balloons. I have never done it. I'm just assuming this stuff, because it's the only example I could think of. If I am misinformed about balloon squeezing, please tell me and I will fix it, hehe.)
Boyle's Law can be expressed like this: for every gas, assuming temperature is constant, there exists a constant, say, b, and this equation will hold true for all values of P and V obtained simultaneously: b = P / V.
Believe it or not, you can do some example problems with this!
Please solve the following problems.
  1. The value of P / V for a certain gas is 3.42 atm / L, at a temperature of 300 K. How much space is needed to contain the gas at 5 atmospheres at 300 K?
  2. At 273 K, 10 liters of a gas was pressurized at 3.00 atmospheres. How pressurized would that gas be if it only took up 5 liters of space at 273 K?
  1. This is middle-school stuff. Sorry! So P / V = 3.42 atm/L. You know V = 5 L, so just solve for P. P = (3.42 atm/L) x V
    P = (3.42 atm/L)(5 L)
    P = 17.1 atm.
  2. Here is a way that my Chem I teacher taught how to do this stuff. Boyle's Law can also be written as:
    VoPo = VNPN
    This is pronounced "Vopo equals Vinpin." V stands for volume, P for pressure, and subscripts 'o' and 'N' denote whether it's the volume/pressure of the old state or the new one. In this equation, the temperature must remain constant. Since the temperature stays at 273 K in this equation, let's try it.
    What are we looking for? We have two states: one at 10L/3 atm, and a new one, at which V = 5L, but pressure is unknown. So we are looking for pressure of new state, "Pin", or PN.
    Solving for PN in that equation, we have:
    PN = VoPo / VN
    PN = (10 L)(3.00 atm) / (5 L) = 6.00 atm.
    The new pressure will be 6.00 atm. That makes sense; since we reduced volume (crushed it), it should be more pressurized. Furthermore, since we halved the volume, we should expect the pressure to double.

Charles's Law
This law says: The temperature and the volume of a gas are directly proportional, assuming pressure is kept constant. That means as the temperature goes up, the volume will go up as well.
This is easy to picture. If the temperature goes up, the particles get faster, and they want to move outward more. The gas expands, and the space it takes up increases. The opposite also happens; if temperature is lowered, volume will be less.
This is easily observed in a hot air balloon. To make a balloon go up, they light a fire inside of it. This will heat up the gas inside the balloon. So the volume will go up. Since mass stays the same, the density of the air inside will become less. The reason a balloon floats is because the air inside is less dense than the air outside. Wow...
Just as we could write Boyle's Law with some sort of equation, you can write Charles's Law too.
For every gas there exists a constant, say, d, at which V = dT, pressure being held constant.
So here are some examples! (Actually, just one.)
Solve: A sample of gas is to be kept at a constant 3.00 atm. The gas is at 265 K, and takes up 2.45 L of space. If the temperature is to be increased 30 degrees, what should the new volume of the gas be, to keep the pressure equal?
Just like the above shortcut, there is a nice little equation that will compare two states of the gas. The way to say it is "Votin equals Vinto." Equationwise, it looks like this:
For an explanation of subscripts and such, see previous example.
Solving for the new volume, or Vn, you should have:
VN = VoTN / To
Plugging in, your answer should be:
VN = (2.45 L)(265 K + 30 K) / (265 K)
VN = 2.73 L

So Boyle's Law and Charles's Law have to do with relationships between different properties of gases, how they vary with each other. There is one more, and this one's easy: Avogadro's Law.
Avogadro's Law
This one is easy: the moles of gas is directly proportional to the volume. Easy; if you have twice as many moles of gas, it will take up twice as much space, assuming you keep that other crap constant (such as pressure and temperature).
This also leads to something interesting. Before we go into this "interesting" thing, I introduce to you STP. Standard Temperature and Pressure. When a gas is at STP, it is understood that it is at O oC, and at 1 atm.
Now for the interesting thing. At STP, one mole of gas, any gas, will occupy exactly 22.4 liters. Doesn't matter how big the particles are, and such.
If you must need an example on this suck!
Combining Boyle's and Charles's Laws
This one is combining volume, pressure and temperature into one easy equation to remember. It is pronounced, "Vopotin equals Vinpinto", and you can find one of them, given the other five. The equation is written as follows.
This is one step below the almighty Ideal Gas Law. Try a problem out with this:
A very cautious scientist (perhaps Mieze) has measured very precise values for a gas he has. The pressure was found to be 2.349202 atms, the volume to be 2423.53929 mL, and the temperature was 24.23411 oC. Just as he finished, a very clumsy student walked in (I won't say who he is, but his initials are J.A.P.), and screwed it up, breaking the scientist's thermometer in the process. The scientist frantically tries to measure the new information (finding the pressure has changed to 3.423103 atms, and the volume at 2.34131248 L) but can't use his broken thermometer. What did he calculate the temperature to be? Answer
Sorry for the extra-long question, but I have to keep you interested! We see five numbers, so obviously we can use the VoPoTn=VnPnTo equation to get the sixth. We are looking for the new temperature, so solving for Tn, you get:
TN = VNPNTo / ( VoPo )
Just plug-'n'-chug. Make sure you convert to proper units.
TN = (2.34131248 L)(3.423103 atm)(24.23411 + 273 K) / (2.42353929 L)(2.349202 atms)
TN = 418.415 K = 145.415 oC = 293.747 oF.
That J.A.P. really screwed up.

Dalton's Law of Partial Pressures
This is the easiest law. The pressures of gases are additive. If you put a gas at A atms into a container, and then put another gas of B atms into it, then the total pressure in that container is A + B. Now it's time for the big daddy of all Gas Laws...the Ideal Gas Law.
Ideal Gas Law
Well, here it comes, the law that will tie it all together: the Ideal Gas Law. It ties in all the quantities together we have discussed so far, and a simple and easy-to-memorize equation is what you get.
(Sorry for overdoing the pic!)
This equation is called "Piv-Nert". That is how you pronounce it. You will be using this for many many things, so get used to it! Basically, this equation will let you find the property that you don't know, given that you know the other three.
P - pressure (in atms).
V - volume (in liters).
n - quantity of gas (in moles).
R - the gas constant discussed earlier. Hopefully, you recognize this R as being the pressure-volume form, and is therefore .08206 L atm / K mol.
T - temperature, IN KELVINS!!! The biggest mistake people make early on is forgetting to convert to Kelvins. So do it!
And that is it. This equation is probably most useful for finding the moles of gas, since all the other quantities can be measured easily. Pressure is found by some kind of gauge, volume is simply the volume of the container, and temperature can be found with a thermometer (I hope you knew that.)
I won't give any examples, because you can do plug-'n'-chug, and because any other problem that might require some tinkering with the Ideal Gas Law can be easily solved with the VoPoTn = VnPnTo equation.
Note that this is called the "Ideal" Gas Law. Real gases don't exactly follow this law, mostly because the pressure and volume aren't as they seem to be. So what is an ideal gas? There isn't such a thing in the real world, it's simply a model that tries to mimic experiment data as much as possible, and still be simple. Let's go back and see what's wrong with this model.
Pressure. The Ideal Gas Law assumes that there are no interactions between molecules. They just bounce around, with some kinetic energy. But in reality, the molecules are attracted to each other. Therefore, they wouldn't be hitting the walls of the container as hard as are expected. Therefore the actual pressure will be greater than the observed pressure.
Volume. The Ideal Gas Law assumes that the molecules are points that don't take any actual space themselves. But in reality...molecules take up space!!! (Don't act surprised.) So the actual volume that is there is less than what is observed, since the molecules take some space up which the gas should be taking up according to the Ideal Gas Law.
These factors will be taken into consideration when coming up with a more realistic gas law.
Back to Stoich
I'm so sorry, but we do have to return back to the world of moles and LRs and formulas. It was a good break from it tho. This section will not teach you anything new, just help you connect stoich and this new junk you had to learn. The key is...moles!!! Stoich is always about the moles. So if you get something in grams... convert to moles! If you get a volume, pressure, and temperature... use Pivnert to get moles!
Here's a typical problem you might have to do.
Solve the following, oh pretty please? A 100 mL of CH3OH (Density = .850 g/mL) is mixed with some oxygen, 32.21 liters at 3.00 atmospheres and 28 oC. The products of the reaction are carbon dioxide and water. Find the number of moles of water formed.
Write the reaction first:
CH3OH + O2 ---> CO2 + H2O
Check to see if it is balanced. It isn't, so balance it!
2CH3OH + 3O2 ---> 2CO2 + 4H2O
Now you must use all those numbers above to get moles for some of these things. Let's begin with CH3OH. There is 100 mL, and density is .850 g/mL, so we can easily find the mass, 85.0 grams (no calculator there, folks!) Now to get moles, simply calculate the molar mass (12 + 3 + 16 + 1 = 32 g/mol), and divide.
85.0 / (32 g/mol) = 2.65625 moles CH3OH.
The second one is the part we learned in this section. You can just use the Ideal Gas Law to find the moles needed.
PV = nRT
n = PV / RT
n = (3.00 atm)(32.21 L) / [(.08206 L atm / K mol)(28 + 273 K)]
n = 3.91214 moles.
Now we must find the limiting reactant. Let's assume oxygen is the limiting reactant. Then let's see if we have enough of that other stuff:
3.91214 moles O2 x [2 moles CH3OH / 3 moles O2] = 2.60809 moles of other stuff.
2.60809 moles of stuff is what is required. We have more than that (2.65625), so oxygen is indeed what will run out first. From here on out you know what to do.
2.60809 moles CH3OH x [ 4 moles H2O / 2 moles CH3OH ] = 5.22 moles H2O.

Here's a special formula. Nothing out of the ordinary, just derived from the Ideal Gas Law. It's how to find the molar mass of the gas, given pressure, temperature, and density.
d is density, in grams per liter, instead of milliliter, because R has liters in it.
There's some things to notice in this. If the temperature goes up of a gas, then density goes down. That's how a hot air balloon works, by decreasing the density of the gas inside by increasing temperature.
Mole fractions. I thought I mentioned them before, looks like I didn't. This is how much a mixture is made up of a particular gas. The symbol for it is a greek 'x' thing, like X. It's very easy. Let's say you have x moles of gas A, and y moles of gas B put together. The mole fraction of gas A will be:
x moles
Total moles (x + y)
It's like a percentage, but you don't multiply by 100. Mole fractions have no units, since mol cancels out.
Quikie: You have 2 moles of helium mixed with 5 moles of xenon. The mole fraction of helium is 2 moles / (2 moles + 5 moles) = 2/7, or 0.286. The mole fraction of xenon is .714. Note that all the mole fractions of all the gases making up a mixture must add up to 1. (Just like all percentages add up to 100%). You can think of the mole fraction as the probability of picking up one random particle and getting that gas.
Now, the cool thing is that pressure and mole fraction are related. When two or more gases are put together in the same container, their mole fractions are proportional to the pressures, and vice-versa. Why? Because when gases are put together, their volumes are the same, and so is their temperatures. The only thing that is different is their pressures and mole fractions, and so they are proportional.
What does this mean for you? It means you can do stupid little problems like... (Two gases, with .432 and .568 for mole fractions, have a total combined pressure of 3.00 atms. What is the pressure of each gas?)
Note that the pressures of the individual gases added together will be the total pressure observed, because of Mr. Dalton's law.
Let's talk about vapor pressure, since it is needed in some problems.
Vapor Pressure
Let's say you want to collect the gas of a reaction. One way of doing this is like so:
You let the gas bubble out to the top, and there's a little space above the water where the gas will reside. One very important thing to realize is that the pressure of the 'stuff' above there isn't the pressure of the gas being collected. There is also water vapor in that space, which is contributing to the pressure.
You say, 'Where is this vapor coming from?' From the water, of course! This means the pressure observed is greater than the actual pressure of the gas being collected, since the pressure measured includes both the water vapor pressure and the gas pressure.
So what do you do? You will get the pressure of water vapor as a given, with the problem. Then just take the pressure you measured, and subtract the water vapor pressure. And there's your true pressure of the gas.
I'm very sorry for the lack of examples in this section, since that is what it was intended for; to help you solve problems. But the truth is, it's very hard to make up a problem without copying it from the book, and I don't like to do that. Plus, it's hard on the fingers. Maybe when this whole page is done, I will come back and do these. Of course, by then, you will have graduated college and wouldn't care, would you?

Kinetic Molecular Theory of Gases

The Kinetic Molecular Theory of Gases (henceforth the "KMT") is the model we've been using, in these Laws, including Ideal Gas Law. It's a model, which tries to mimic real-world observations as much as possible, but also keeping it simple. As a result, you get a really simple equation to work with, but it makes some assumptions that aren't really true. Here is what this model is assuming:
  • The molecules of gas don't take up any volume. They are simply points of mass.
  • The molecules are always moving, and the average of the energy is related to temperature.
  • The molecules don't have any forces exerted upon each other.
These are wrong! But... they can be ignored, resulting in a formula that actually does a good job of predicting gases in the real world, at least in certain situations.
Now my book goes through the trouble of deriving the Ideal Gas Law, but I don't want to bother you. You can thank me later.
Perhaps the only thing you might need to know is the root mean square velocity. This equation will give you the average of the squares of the speeds of the particles. This can be easily derived, since you know the mass of a particle, and the energy (you learned the equation way back up there). Here it is:
Capital M? Doesn't this mean molarity? Nope, not here. There are 26 letters, actually 52 both UPPER and lower, and almost as many Greek letters, and it's funny why they reuse the same letters... M means kilogram per mole. So just divide the molecular weight of a particle of gas by 1000.
Effusion and Diffusion
Tired of gases? We still have two quick topics to cover until you become a full-fledged PhD in AP Gases. Effusion and diffusion wasn't really covered in my class. They basically compare the rates at which gases travel from one place to another. This is only relative rates; you'll see more later.
Effusion is the rate at which particles will travel from one compartment to another totally empty compartment (vacuum), separated by only a small hole. To help ya visualize this, of course here's a visual (that's what they're for):
The particles go in the direction of that yellow arrow, towards the empty chamber. It does so at a constant rate, until they are balanced. Different gases will do this at different speeds. Generally, the lighter the particle is, the faster it will effuse. All right, all right, I'll stop yakking and just give you the formula, that's all you ingrates want anyway.
Note that this will not give you any rates. It will tell how much faster Gas A will effuse than Gas B. Note the capital 'M' again, meaning the kilograms per mole. But in this case, it doesn't really matter, since units cancel. Just keep molecular weight units consistent with both gases.
So let's find out how fast hydrogen gas will effuse compared to oxygen. Since hydrogen is lighter, we know that it will be faster, but by how much? Let's call hydrogen Gas B, since its rate will be on top on the left part. Therefore, helium's molecular weight will go on top on the right (note that it's flip-flopped?)
Rate of H2 / Rate of O2 = Ö(32 / 2) = 4.
This means hydrogen effuses 4 times as fast as oxygen.
Diffusion is when a gas travels across a distance. Like in a room, when someone farts at one end of the room, and it travels to the other end, it's diffusing. Some gases travel faster than others. Generally, lighter gases travel faster, and the whole concept is exactly the same as effusion. Same formula, that's good news 4 u.
An Alternative to the Ideal Gas Law
We already said that the Ideal Gas Law could be better, accounting for the fact that molecules take up space, and have attractions to each other. Well, here it comes. We know that we must subtract from volume, and add to pressure to account for those things. Therefore, here's the "real" (or at least better) Ideal Gas Law.
There are no new variables except for this 'a' and 'b'. They are constants that depend only on the identity of the gas. This formula will be more accurate than the Ideal Gas Law. The only thing is, you can't use them unless you are given the constants a and b. Once you do, tho, you can solve for anything.

States of Matter

Matter: States of Matter

by Anthony Carpi, Ph.D.
water - boiling
As a young boy, I remember staring in wonder at a pot of boiling water. Searching for an explanation for the bubbles that formed, I believed for a time that the motion of the hot water drew air down into the pot, which then bubbled back to the surface. Little did I know that what was happening was even more magical than I imagined - the bubbles were not air, but actually water in the form of a gas.
The different states of matter have long confused people. The ancient Greeks were the first to identify three classes (what we now call states) of matter based on their observations of water. But these same Greeks, in particular the philosopher Thales (624 - 545 BCE), incorrectly suggested that since water could exist as a solid, liquid, or even a gas under natural conditions, it must be the single principal element in the universe from which all other substances are made. We now know that water is not the fundamental substance of the universe; in fact, it is not even an element.
To understand the different states in which matter can exist, we need to understand something called the Kinetic Molecular Theory of Matter. Kinetic Molecular Theory has many parts, but we will introduce just a few here. One of the basic concepts of the theory states that atoms and molecules possess an energy of motion that we perceive as temperature. In other words, atoms and molecules are constantly moving, and we measure the energy of these movements as the temperature of the substance. The more energy a substance has, the more molecular movement there will be, and the higher the perceived temperature will be. An important point that follows this is that the amount of energy that atoms and molecules have (and thus the amount of movement) influences their interaction with each other. Unlike simple billiard balls, many atoms and molecules are attracted to each other as a result of various intermolecular forces such as hydrogen bonds, van der Waals forces, and others. Atoms and molecules that have relatively small amounts of energy (and movement) will interact strongly with each other, while those that have relatively high energy will interact only slightly, if even at all, with others.
How does this produce different states of matter? Atoms that have low energy interact strongly and tend to “lock” in place with respect to other atoms. Thus, collectively, these atoms form a hard substance, what we call a solid. Atoms that possess high energy will move past each other freely, flying about a room, and forming what we call a gas. As it turns out, there are several known states of matter; a few of them are detailed below.
ice - cubes
Solids are formed when the attractive forces between individual molecules are greater than the energy causing them to move apart. Individual molecules are locked in position near each other, and cannot move past one another. The atoms or molecules of solids remain in motion. However, that motion is limited to vibrational energy; individual molecules stay fixed in place and vibrate next to each other. As the temperature of a solid is increased, the amount of vibration increases, but the solid retains its shape and volume because the molecules are locked in place relative to each other. To view an example of this, click on the animation below which shows the molecular structure of ice crystals.

water - liquid
Liquids are formed when the energy (usually in the form of heat) of a system is increased and the rigid structure of the solid state is broken down. In liquids, molecules can move past one another and bump into other molecules; however, they remain relatively close to each other like solids. Often in liquids, intermolecular forces (such as the hydrogen bonds shown in the animation below) pull molecules together and are quickly broken. As the temperature of a liquid is increased, the amount of movement of individual molecules increases. As a result, liquids can “flow” to take the shape of their container but they cannot be easily compressed because the molecules are already close together. Thus liquids have an undefined shape, but a defined volume. In the example animation below we see that liquid water is made up of molecules that can freely move past one another, yet remain relatively close in distance to each other.

Gases are formed when the energy in the system exceeds all of the attractive forces between molecules. Thus gas molecules have little interaction with each other beyond occasionally bumping into one another. In the gas state, molecules move quickly and are free to move in any direction, spreading out long distances. As the temperature of a gas increases, the amount of movement of individual molecules increases. Gases expand to fill their containers and have low density. Because individual molecules are widely separated and can move around easily in the gas state, gases can be compressed easily and they have an undefined shape.

Solids, liquids, and gases are the most common states of matter that exist on our planet. If you would like to compare the three states to one another, click on the comparison animation below. Note the differences in molecular motion of water molecules in these three states.

Plasmas are hot, ionized gases. Plasmas are formed under conditions of extremely high energy, so high, in fact, that molecules are ripped apart and only free atoms exist. More astounding, plasmas have so much energy that the outer electrons are actually ripped off of individual atoms, thus forming a gas of highly energetic, charged ions. Because the atoms in plasma exist as charged ions, plasmas behave differently than gases, thus representing a fourth state of matter. Plasmas can be commonly seen simply by looking upward; the high energy conditions that exist in stars such as our sun force individual atoms into the plasma state.
As we have seen, increasing energy leads to more molecular motion. Conversely, decreasing energy results in less molecular motion. As a result, one prediction of Kinetic Molecular Theory is that if we continue to decrease the energy (measured as temperature) of a substance, we will reach a point at which all molecular motion stops. The temperature at which molecular motion stops is called absolute zero and has been calculated to be -273.15 degrees Celsius. While scientists have cooled substances to temperatures close to absolute zero, they have never actually reached absolute zero. The difficulty with observing a substance at absolute zero is that to “see” the substance, light is needed, and light itself transfers energy to the substance, thus raising the temperature. Despite these challenges, scientists have recently observed a fifth state of matter that only exists at temperatures very close to absolute zero.
Bose-Einstein Condensates represent a fifth state of matter only seen for the first time in 1995. The state is named after Satyendra Nath Bose and Albert Einstein who predicted its existence in the 1920’s. B-E condensates are gaseous superfluids cooled to temperatures very near absolute zero. In this weird state, all the atoms of the condensate attain the same quantum-mechanical state and can flow past one another without friction. Even more strangely, B-E condensates can actually “trap” light, releasing it when the state breaks down.
Several other less common states of matter have also either been described or actually seen. Some of these states include liquid crystals, fermionic condensates, superfluids, supersolids and the aptly named strange matter. To read more about these phases, visit the Phase page of Wikipedia, linked to below in the Further Exploration section.

Phase transitions

water - boiling
The transformation of one state of matter into another state is called a phase transition. The more common phase transitions even have names; for example, the terms melting and freezing describe phase transitions between the solid and liquid state, and the terms evaporation and condensation describe transitions between the liquid and gas state. Phase transitions occur at very precise points, when the energy (measured as temperature) of a substance in a given state exceeds that allowed in the state. For example, liquid water can exist at a range of temperatures. Cold drinking water may be around 4ºC. Hot shower water has more energy and thus may be around 40ºC. However, at 100°C under normal conditions, water will begin to undergo a phase transition into the gas phase. At this point, energy introduced into the liquid will not go into increasing the temperature; it will be used to send molecules of water into the gas state. Thus, no matter how high the flame is on the stove, a pot of boiling water will remain at 100ºC until all of the water has undergone transition to the gas phase. The excess energy introduced by a high flame will accelerate the liquid-to-gas transition; it will not change the temperature. The heat curve below illustrates the corresponding changes in energy (shown in calories) and temperature of water as it undergoes a phase transition between the liquid and gas states.
graph2 - heat curve
As can be seen in the graph above, as we move from left to right, the temperature of liquid water increases as energy (heat) is introduced. At 100ºC, water begins to undergo a phase transition and the temperature remains constant even as energy is added (the flat part of the graph). The energy that is introduced during this period goes toward breaking intermolecular forces so that individual water molecules can “escape” into the gas state. Finally, once the transition is complete, if further energy is added to the system, the heat of the gaseous water, or steam, will increase.
This same process can be seen in reverse if we simply look at the graph above starting on the right side and moving left. As steam is cooled, the movement of gaseous water molecules and thus temperature will decrease. When the gas reaches 100ºC, more energy will be lost from the system as the attractive forces between molecules reform; however the temperature remains constant during the transition (the flat part of the graph). Finally, when condensation is complete, the temperature of the liquid will begin to fall as energy is withdrawn.
Phase transitions are an important part of the world around us. For example, the energy withdrawn when perspiration evaporates from the surface of your skin allows your body to correctly regulate its temperature during hot days. Phase transitions play an important part in geology, influencing mineral formation and possibly even earthquakes. And who can ignore the phase transition that occurs at about -3ºC, when cream, perhaps with a few strawberries or chocolate chunks, begins to form solid ice cream.
Now we understand what is happening in a pot of boiling water. The energy (heat) introduced at the bottom of the pot causes a localized phase transition of liquid water to the gaseous state. Because gases are less dense than liquids, these localized phase transitions form pockets (or bubbles) of gas, which rise to the surface of the pot and burst. But nature is often more magical than our imaginations. Despite all that we know about the states of matter and phase transitions, we still cannot predict where the individual bubbles will form in a pot of boiling water.

A video tutorial