Defining Entropy

After reviewing last night's blog, Mr. Henderson quickly went into giving examples and rules for when Entropy (S) increases and decreases. I found it very helpful to make a T-Chart to help organize these examples. Situations that had increased entropy included diffusion of gases, dissolving a solid in a liquid, phase changes in the order of solid --> liquid --> gas, and cooling. One common factor that all of these have in common is that each of them causes something to change from fairly compact to a more dispersed state. This is a very helpful tip to keep in mind when thinking about entropy; if it becomes more spread out, it has an increase in entropy. Then we considered this idea of increasing and decreasing entropies with chemical reactions. The reaction of N2O4--> 2NO2 caused an increase in entropy. This is because an increase of moles as well as the creation of simpler molecules causes an increase in entropy. A similar reaction was given but in reverse on the side of decreasing entropy (N2+3H2-->2NH3). We saw that not only do the molecules become more complex, but they also decrease in number of moles from 4 on the reactants side to 3 on the product side.

Next we went over pages 33-34 in our packets. This was once a reading sheet for webassign so the answers are as follows:
1) Independent of external agencies
2)C
3)B
4)B
5)A,D,E
6)B
7)A
8)A
9)C
10)B
11)B
12)C
13)A
14)B
15)D
16)C
17)A
18)iNCORRECT: The student did not take into account the number of moles that pertained to each molecule
19)E

While going over the reading sheet, Mr.H told us a few concepts that we should keep in mind for the test. One of which was that when there is a negative enthalpy change (-delta H) the reaction is most likely spontaneous. We also went over how to calculate change in entropy or delta S. This took no more than 2 minutes because as you will notice in the reading, the equation for delta H and delta S are identical! This means that find the sum of all products minus the sum of all reactants (in terms of entropy) will provide you with your delta S. Another concept that is important to take note of is that if the system does not lower its energy in a reaction and the universe does not higher its entropy, that specific reaction will never take place spontaneously.

Next, we did pages 19 and 20 as a class. The questions on this sheet were very straight forward with one key concept on page 19. This is that if a reaction has a positive delta S (or increase in entropy) then that reaction is favorable to occur spontaneously under the given conditions, and vice versa. Although this was true for problems 4 a b c d and e, a positive entropy change does not automatically mean a spontaneous reaction. The two driving factors that determine whether or not a reaction is spontaneous are 1) increase in entropy and 2) decrease in enthalpy. This was a main idea of today's lesson. We continued on to finish problems 6a and 7 on page 20. The process of finding the answer of 6a (-21.2 J/K) is the exact same as finding the change in enthalpy (sum of products entropy - sum of reactants entropy). Finally, number 7 was focused on determining whether or not a reaction was spontaneous. Those that had 1 of the 2 driving factors of determining spontaneity had insufficient information to determine whether or not the reaction would be spontaneous. Those that had none of the 2 factors were not spontaneous. Those that had both, of course, were spontaneous.

Lastly, we began our TC10) Hess's Law Lab that will utilize both entropy and enthalpy change. Today we obtained only general information regarding the reaction of Hydrochloric acid and magnesium, but tomorrow we will utilize the entropy and enthalpy equations. Good Luck!

Tuesday, October 19

Converting to Atoms, Moles, and other measurements
After going over last nights blog and talking to us about national mole week, Mr. Henderson began to focus in on the big idea for the day. Just as we have been doing for the past week or so, we started off by converting one unit to another. However, today we started with slightly more difficult conversion problems. Of course, I'm talking about converting moles to atoms, atoms to moles, atoms to grams, grams to moles and so on. Before I begin to go over pages 6 and 7 (the pages we worked on today), it is important to have a basic understand for some of the quantities we will be addressing.

First off, the one you will be hearing most about in this blog is the mole. 1 mole is 6.022x10^23 of something. This number was created to help us convert one atom to a useful unit, like the gram. This leads into my next point, the mass of atoms. On our periodic table, the mass of an element is known as XX amu's. This number can also be used to obtain the elements grams per mole. How you do this is actually very very very very simple. Take the atomic mass of the element and drop it into the g/mol unit! So, for example, oxygen's atomic mass is about 16 amu's. Therefore, oxygen is also 16g/mol. This is used so that we can speak of the units of each element in more relative terms, such as grams or grams per mole.

Now lets begin going over the packet pages starting with pg.6. The top of page 6 asks how many donuts or in 2 dozen donuts. This question is there to help us understand that a mole is nothing more than a way to help us speak more easily about atoms, like a baker uses the dozen to talk about the number of donuts he is baking. The next question asks how many atoms are in 2 moles of water. Since we know that 1 mole of atoms is 6.022x10^23, we automatically know that 2 moles of water would be 12.044^23 atoms, or 2 x 1 mole of atoms. Questions 4-6 are very similar and are used to help us understand just how large a mole is. The first question asks how much mass would 1 mole of .62kg basketballs have? So, as we would start of all conversions, we begin with the unit we are given which is 1 mole of basketballs, in this case. Now, our next step would be to put 6.022x10^23 basketballs / 1 mole of basketballs. Our final calculation would be to multiply the mass of a basketball / 1 basketball. We are now left with (6.022x10^23)x.62 kg. After doing these calculations, we get the answer......... 3.7kg x 10^23! If you still don't understand just how massive this is, the mass of the of the moon is also given. It is only 7.4x10^22. 1 mole of basketballs is 5 times as massive as the very moon that orbits our Earth!

(Yup, that moon)

Next, on Questions 7-9, we began to use the mole for what it is truly meant for; Converting atoms to more useful units of measurement. Our first question is How many atoms of lead(Pb) are in 5.0x10^-4 grams of lead? Now lets begin our conversion with 5.0x10^-4 g of Pb. Because we are looking to obtain a number of atoms as our answer, we will start with the next corresponding conversion factor, the mole. 1mole/207.2 g of Pb should be your next step because this is the number of grams that would result from 1 mole of lead atoms. Also, as we discussed earlier, it is the same amount as the atomic mass of lead on the periodic table. Finally, we will multiply 6.022x10^23 atoms/ 1 mole to cancel out our moles and get an answer. Although 0.0005 grams of lead may seem like a small amount relative to our everyday lives, It is still 1.45 x 10^18 atoms of lead! A massive number of atoms for a minuscule amount of grams.


The answer for pages 6 are as follows:

#1)24 donuts
#2)12.044x10^23 molecules
#3)10 (every power added to 10 multiplies the given amount by 10, in this case, moles)
#4)3.7 x 10^23kg
#5)3.8 x 10^ 14 years
#6)3.6 x 10^26 cubic centimeters
#7)1,45 x 10^18 atoms
#8)9.0 x 10^18 atoms
#9) 2.1 x 10^-22 grams (that's 0.000000000000000000000021 grams!)

As for pg 7, we just began to discuss the molar mass of compounds and the way to do this is probably the way you would have guessed. Just add together the mass of each elements in the compound. To obtain the molar mass of Barium Chloride (BaCl2), add the mass of 1 mole of barium atoms(137.33 g/mol) to the mass of 2 moles of chloride atoms (2 x 35.34 g/mol). Your answer should add up to 208.23 g/mol. We will continue on to finish this sheet tomorrow but all the answers we have gotten in class so far are as follows.

#1) 32 g/mol
#2) 58.45 g/mol
#2.5) H20= 18 g/mol (IMPORTANT FOR LAB MR1.)
#4) 208.23 g/mol

With the final 10 minutes of class, Mr. Henderson let us get back to our lab groups and finish up the lab with the knowledge that we had gained from question #2.5 on pg.7. Now we know that water is about 18 g/mol. Next, each group would have to find the amount of moles that was assigned to your group. This could be done by dividing the amount of molecules assigned to each group and dividing it by 6.022x 10^23 Luckily, all our group were assigned a number of molecules that could easily be divided by a mole. Finally, all that had to be done to solve this seemingly impossible conversion was to multiply the number of moles your group had by the grams per mole (g/mol) of water, which we know is 18 g/mol. Now you would be left with xx grams of water, and since we know water's density is 1 gram per milliliter, the last step was to change grams into mL and present your findings to Mr.Henderson.

I hope this blog helped clear up any confusion about molar mass and atomic conversions. Today was definitely a jam packed day. Good Luck!

Monday 27, September

To start off class, Mr. Henderson clarified any information that we may have missed concerning the TLC and the tutors there. He told us that if we were struggling in Chemistry, we could visit the TLC for help from actual science teachers or AP students as well as visit Mr. H early in the morning. We then moved on to the subject of the day; The history of the atomic model and the various scientists that helped create each one in their respective times.

On Pg. 15 in your chemistry Unit2 packet, there is a very general history concerning the atomic model starting with Dalton. However, Mr. Henderson told us that the first ever recorded theory concerning the atom dates all the way back to Ancient Greece and a philosopher by the name of Democritus. Democritus argued that he could chop a piece of papyrus into halves and halves and halves until he would reach some sort of building block that could not be broken down any further into a smaller or simpler form. He called these building blocks atomos and he believed they made up all matter. Democritus had no idea how accurate he actually was, but for nothing more than an educated guess, this was a remarkable theory. It is important to note that Democritus never actually tested this theory using science and therefore, he is regarded as a philosopher rather than a scientist. It was not until the late 1700's and early 1800's that the scientific revolution ushered in a new wave of interest in the field of atomic structure.

The next man to take a stab at an accurate atomic model was John Dalton. This scientist used the scientific method of developing a hypothesis and testing it to form a theory. He called these theories postulates, or basic rules. The five postulates that Dalton formed are as follows.
#1) All matter consists of atoms

#2) Atoms are indivisible (cannot be broken down any further)

#3) In a chemical change or reaction, atoms simply rearrange themselves but never turn into any other atoms. This is demonstrated in the picture below in which Hydrogen and Oxygen combine to form H20.



#3 (continued) Also, atoms never gain or lose mass in any way during any chemical or physical change.

#4) Compounds have a determinable amount of atoms and always have the same amount of each atom.

#5) Law of Constant Composition. All Compounds contain the same amount of each atom no matter where the compound is found or under what circumstances.

These 5 postulates are still seen today as a basic level in understanding the structure of atoms and compounds. Dalton never did create a finite model that he believed was what an atom might look like. He did however represent an atom as a circle with a symbol on it, such as H for Hydrogen.

Proceeding John Dalton, J.J. Thomson was at the forefront of the atomic model issue. J. J. Thomson believed that atoms did not only consist of alpha particles, but also possessed some other feature. By separating the alpha particles from a beam of light and firing this beam, known as a cathode ray, through a glass tube with positive or negative plates on either side Thomson was able to test this. After seeing the cathode ray bend towards the positive plate, Thomson was able to determine that the are other particles that make up an atom and that they are negatively charged. Thomson called these negatively charged particles electrons. This lead Thomson to relate his atomic model to the favorite desert of England at the time, earning it the name The Plum Pudding Atom. Thomson believed that atoms consisted of moving negatively charged electrons that are immersed in a sea of positively charged stationary particles. The model looked something like this picture, with the dots being electrons and the red material being the positively charged "sea".





For many years, A scientist by the name of Ernest Rutherford studied this model and debated it's accuracy. To test the Thomson models validity, Rutherford set up an experiment in which he took a thin (only a few hundred atoms thick) piece of gold foil and placed it in a container which had sensors on either side while shooting a stream of atoms in its direction. If Thomson's model was correct, the atoms should have gone through the gold foil with no problem and triggered only the sensors on the other side. What Rutherford found was that although most of the atoms did behave this way, 1 out of every 10/20,000 deflected off of the foil and reflected back at an acute angle. Rutherford was quoted saying "Its as if you shot 20,000 bullets at a tissue with most of them going through the tissue with no problem, but 1 out of every 20,00 of those bullets would reflect back at you!" Rutherford was amazed at this and determined that atoms were not full of positive particles, but were rather made up mostly of empty space with 99% of its mass concentrated in its nucleus, earning his model the name The Empty Space model. Rutherford still did not know how the electrons were distributed across the atom, leaving his model still somewhat inaccurate

Finally, one of Rutherford's colleagues, Niels Bohr, measured light energy to more accurately determine where the electrons would be relative to the nucleus of the atom. By measuring each layer of electrons of an atom and observing the light they emitted, Bohr was able to establish the general amount of electrons and their area relative to the nucleus. The downfall to Bohr's method was that once he began getting into the atoms with more electrons, such as Gold, his estimates were off by more than 20-40% at times. However, Niels Bohr was able to develop an atomic model, known as The Planetary Model.

Neils Bohr called it this because he thought that the electrons orbited the nucleus much like moons orbit a planet. This is the most accurate of all the depictions discussed so far.

To wrap up the class, Mr. Henderson gave us the answers to page 15 which are as follows:

1: b
2: d
3: c
4: a
5: c, b, a

We also drew sketches of what each scientist, (Dalton, Thomson, Rutherford, Bohr) believed
that atom looked like and those can be seen by asking me or any other student before or after class. Tomorrow, we will incorporate verbal explanations with these visuals models to describe each scientist views on concerning the structure of an atom.