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!
Monday, January 31, 2011
Friday, January 27
Today, Mr. Henderson began class by announcing that it would be dedicated to preparing for Thursday's chapter test. He then highlighted the main ideas/sections this test will cover:
1. pp. 1-2
- Heat vs. Temperature (heat is energy)
- 0th Law of Thermodynamics (where the temperature of the system and the surroundings approach each other)
- Thermal Equilibrium
- Energy (differentiating between exothermic and endothermic reactions)
2. pp. 3-4
- Calorimetry (candle labs, etc.)
- Q=mC∆T
- Concept: Energy lost by one thing must be gained by another
- Lab
3. pp. 7-8
- Enthalpy change (∆H)
- Thermal Stoichiometry (how much heat gained/lost)
- 3 rules of Thermal Chemistry (I didn't catch all of them; one was that you have to multiply the moles given by the coefficients to get the proportional amount of heat gained/lost)
4. pp. 9-10
- Heat of Formation (∆Hf)
- Heat of reaction = sum of the heat of the products - sum of the head of the reactants
5. pp. 11-12
- Hess's Law
- The thing where we add, multiple, and cancel multiple chemical equations in order to figure out the amount of heat gained/lost for a given reaction
6. pp. 13-15
- Solids, liquids, gases (differentiating between them)
- Phase changes (melting, boiling, sublimation, etc.)
- Phase diagrams
7. pp. 17-18
- Spontaneity
- ∆H
8. -Entropy.
-Entropy ∆ (∆S)
These are the concepts that will be covered on the test Thursday. Mr. H warned us that this test tends to be challenging for the reason that it covers material from both semesters; and we'll have to remember things we learned before finals. Then, he reminded us that we have a Delicious assignment (a particularly IMPORTANT one, according to Mr. H) due Wednesday morning. For those of you that have forgotten, the tags are: hcp3y1011 unit7 nameL (separated by spaces not commas). We next reviewed Kon's blog and the demo. We spent good part of the class on pages 18 and 19 of our packets.
Page 18 discusses enthalpy, the amount of energy stored in the bonds of chemicals. We talked about how to increase a chemical's stability. Some ways the teacher listed are: a drop in energy, a negative enthalpy change, and as a result the opposite wasn't true, reactions which have an increase in energy generally do NOT increase the chemical's stability or occur naturally/spontaneously. We then reviewed the difference between exothermic and endothermic reactions. For the EN or EX? section, the answers are as follows: ex, en, ex, ex, en, ex, ex, ex. Letter "g" of that section involves calculating the heat of formation, and using the fact that it is negative to deduce that it is an exothermic reaction. Question 7 of the packet deals with diagrams which depict reactions (they mainly just tell us if the reaction is endothermic or exothermic), and Mr. H said that it "looks good for a question on the test".
We then moved on to packet page 19 and began to discuss entropy, which is a measure of the amount of energy dispersion. High entropy is a condition in which energy is NOT concentrated. We then tried to answer the question "What types of processes have an increase in entropy?" The first type is gas diffusion like this:
When the gas is diffused, or spread out, the molecules become less orderly, and less concentrated, which is an increase in enthalpy. The second type is dissolving. Again, this is the energy dispersing and becoming less orderly, so the entropy increases. Also, a note on spontaneity: this type of reaction is spontaneous, things dissolve naturally but this would never happen in reverse. The third type is cooling. The heat energy from a system disperses into the surroundings and spreads out, therefore increasing the entropy.
Mr. Henderson ended the class by reminding us that we have a Webassign reading sheet on chapters 17.1 and 17.2 due Monday.
Thursday, January 27, 2011
Thursday, January 27th, 2011
Today, Mr. H started out class by having us turn our packets to page 13. He told us that its important that we look over page 13 tonight because we have a QUIZ tomorrow that is has one math problem and is similar to problems #29 and #30 on page 14.
Next, Mr. H projected Emma's blog on the screen and went over the specific heat for each phase of thermochemistry and other main ideas. He also talked about how Emma's blog was very useful, because it explained how to do phase change calculations very well. If you are someone who is confused about phase change calculations and heating curves, i would recommend that you refer to Emma's blog.
After going over Emma's blog, Mr. H had the everyone open their packets to page 15. He reiterated the idea that if you lower the pressure around an element, then you can boil it; just like yesterday's demo. Mr. H then focused the rest of his time on page 15 to go over some of the answers and explain how to use the graph. He explained to the class, while using the graph, that at any given temperature, if you were to increase the pressure, then, an element would change its phase to either a solid or liquid (depending on the temperature of course). He also explained how if you were to decrease the pressure while at any temperature, then that element would turn to gas. For those of you who are still confused on how to use the graph on page 15, i would recommend trying to do problem #2 on page 15. For instance, if you want to find the phase type of an element when its pressure is 1atm and its temperature is 300°C, the answer would be a liquid. You would find this answer by using your finger and going along the x-axis and finding 300°C, then you would use your finger and go up the y-axis until you found 1atm. Then, you would slide your finger from the y-axis, to the right until you got to the point where 1atm and 300°C met. From there, you would look at which part of the graph you were in (solid,liquid,gas) and then record that answer. Its pretty simple once you understand how to do it.
After Mr. H explained how to use the graph, he then worked with the class on problems #3-5 and then he had the class do the rest on our own. the answers are as followed:
#3) Solid
#4)100°C
#5) 450°C
#6)175°C
#7) T=100°C P=0.70atm
#8) No, because although it is in the liquid state, its minimum temperature would have to be 100°C
#9) Melting
#10) 1st: Condensation 2nd: Freezing
After going over the answers in class, Mr. H then prepared a demo for the class. Prior to the demo occurring, he got dry ice (CO2 as a solid) from a container and then went around the class and had some students hold it. He warned that you can't hold the dry ice for too long because then it would burn your hand, so he advocated that students should hold it as if they were holding a hot potato. He then proceeded to crush it up, put the crushed up pieces in what seemed to be a pill capsule and then closed the lid of the capsule while using pliers. This demo was significant because it showed the transfer of heat from the water to the dry ice in the capsule (endothermic). Mr. H was also trying to demonstrate how an element could reach the triple point in a reaction. Mr. H then proceeded to put the capsule of dry ice in a beaker of water while holding the capsule with a pair of pliers. At first, bubbles began to escape from the capsule, after a couple seconds, there was a mini-explosion. That occurred because enough pressure built up inside the capsule and when it couldn't hold any more, it caused this mini explosion to occur.
After doing this lab, Mr. H had everyone turn their packets to page 17. Once everyone did, he began to talk about Spontaneity, what it means, and what it would look like with real world examples. It is important to note that Spontaneity= reactions that occur naturally (once activated). Also, there are 2 factors to be considered, and they are:
Next, Mr. H had us answer questions #1-2. The answers are as followed:
a) NS
b) S
c) NS
d)NS
e)S
f) NS
g)S
h)NS
i)S
#2) Tosh, because not all exothermic reactions are spontaneous
Once we finished talking about page 17, Mr. H did a special demo for the class. Mr. H prepared the demo by using Jovan's water bottle, then putting a lot of crushed up dry ice inside it and then closing the water bottle with the cap. Enough pressure built up inside the water bottle, and the outcome was similar to this:
Next, Mr. H projected Emma's blog on the screen and went over the specific heat for each phase of thermochemistry and other main ideas. He also talked about how Emma's blog was very useful, because it explained how to do phase change calculations very well. If you are someone who is confused about phase change calculations and heating curves, i would recommend that you refer to Emma's blog.
After going over Emma's blog, Mr. H had the everyone open their packets to page 15. He reiterated the idea that if you lower the pressure around an element, then you can boil it; just like yesterday's demo. Mr. H then focused the rest of his time on page 15 to go over some of the answers and explain how to use the graph. He explained to the class, while using the graph, that at any given temperature, if you were to increase the pressure, then, an element would change its phase to either a solid or liquid (depending on the temperature of course). He also explained how if you were to decrease the pressure while at any temperature, then that element would turn to gas. For those of you who are still confused on how to use the graph on page 15, i would recommend trying to do problem #2 on page 15. For instance, if you want to find the phase type of an element when its pressure is 1atm and its temperature is 300°C, the answer would be a liquid. You would find this answer by using your finger and going along the x-axis and finding 300°C, then you would use your finger and go up the y-axis until you found 1atm. Then, you would slide your finger from the y-axis, to the right until you got to the point where 1atm and 300°C met. From there, you would look at which part of the graph you were in (solid,liquid,gas) and then record that answer. Its pretty simple once you understand how to do it.
After Mr. H explained how to use the graph, he then worked with the class on problems #3-5 and then he had the class do the rest on our own. the answers are as followed:
#3) Solid
#4)100°C
#5) 450°C
#6)175°C
#7) T=100°C P=0.70atm
#8) No, because although it is in the liquid state, its minimum temperature would have to be 100°C
#9) Melting
#10) 1st: Condensation 2nd: Freezing
After going over the answers in class, Mr. H then prepared a demo for the class. Prior to the demo occurring, he got dry ice (CO2 as a solid) from a container and then went around the class and had some students hold it. He warned that you can't hold the dry ice for too long because then it would burn your hand, so he advocated that students should hold it as if they were holding a hot potato. He then proceeded to crush it up, put the crushed up pieces in what seemed to be a pill capsule and then closed the lid of the capsule while using pliers. This demo was significant because it showed the transfer of heat from the water to the dry ice in the capsule (endothermic). Mr. H was also trying to demonstrate how an element could reach the triple point in a reaction. Mr. H then proceeded to put the capsule of dry ice in a beaker of water while holding the capsule with a pair of pliers. At first, bubbles began to escape from the capsule, after a couple seconds, there was a mini-explosion. That occurred because enough pressure built up inside the capsule and when it couldn't hold any more, it caused this mini explosion to occur.
After doing this lab, Mr. H had everyone turn their packets to page 17. Once everyone did, he began to talk about Spontaneity, what it means, and what it would look like with real world examples. It is important to note that Spontaneity= reactions that occur naturally (once activated). Also, there are 2 factors to be considered, and they are:
- Enthalpy Δ (ΔH)
- Entrapy Δ (ΔS)
Next, Mr. H had us answer questions #1-2. The answers are as followed:
a) NS
b) S
c) NS
d)NS
e)S
f) NS
g)S
h)NS
i)S
#2) Tosh, because not all exothermic reactions are spontaneous
Once we finished talking about page 17, Mr. H did a special demo for the class. Mr. H prepared the demo by using Jovan's water bottle, then putting a lot of crushed up dry ice inside it and then closing the water bottle with the cap. Enough pressure built up inside the water bottle, and the outcome was similar to this:
^^this is similar to what Jovan's water bottle looked like after the reaction.
Mr. H advocated to the entire class not to do this type of experiment at home since it can have negative results. To conclude, today was mostly a review of phase calculations and phase diagrams and an introduction to Spontaneity. Tonight's homework: Webassign and study page 13 for the pop quiz.
Wednesday, January 26, 2011
Wednesday, January 26, 2011
Chemistry class today was extremely jam-packed and educational today. Mr. H began class by requesting us to take out our calculadoras and paquetos which is all we would need for the day. He told us we would begin the packet with page 13 and explained that the 2 problems we would do on page 14 directly pertained to the Webassign due Friday morning. If you did that Webassign by Thursday morning you would receive extra credit points. Next we went over Alex's blog and Mr. H briefly covered the main ideas of the past few days. We then went straight to the bottom of page 13 in the packet. Mr. H explained the calculations you would need to solve a quantity of heat problem when a substance changes temperature (from one state of being to another). For Q1, Q3, and Q5 which show positive change in both time and temperature, you use the basic q=m*c*ΔT equation. The mass is the mass of the substance that is changing, the c is the specific heat of the substance at that state, and the delta T is the change in temperature that the substance is experiencing as it changes state. Mr. H then clued us in the fact that the specific heat of water at the solid state is 2.05 J/g°C, at the liquid state is 4.18 J/g°C and at the gaseous state is 1.88 J/g°C. For Q2 and Q4 you use an enthalpy equation of either fusion or vaporization. The equation for fusion is q=nΔHfusion and the equation for vaporization is nΔHvaporization. To find the q values of Q2 and Q4 you convert the grams of the given substance to moles using the molar mass and then to kJ's using the heat of fusion or heat of vaporization values that are given to you. MAKE SURE YOU THEN CONVERT ALL THE JOULES VALUES TO EITHER kJ's OR J's. After finding each value you add up each quantity of heat value per step to find the quantity of heat for the entire change. Next we moved on to page 14 to do problems 29 and 30. Given the info, we set up three equations to find Q1, Q2, and Q3. The value of Q1 was 1045 J's which we converted to 1.04 kJ's. The value of Q2 was 10.9 kJ's and the value of Q3 was 4524 J's which we converted to 4.5 kJ's. Then adding each quantity together we got 16.5 kJ's of heat. Number 30 was directly related to the previous equation. We need to find the quantity of heat for a Q1, Q2, Q3, Q4, and Q5 change. However we already knew the values for Q1(1.04 kJ's) and Q2 (10.9 kJ's). We then solved for the new Q3 as well as Q4 and Q5. The new value of Q3 was 13.8 kJ's, the value of Q4 was 74.39 kJ's and the value of Q5 was 1.5 kJ's. Adding all 5 values together we got 102 kJ's of heat. While we were calculating number 30 Mr. H did a demo type experiment where he would rub water on the table and it would evaporate. He also did this with Acetone which is similar to nail polish remover and it evaporated in seconds. Mr. H explained that liquids turn to gases naturally even when they aren't heated and are able to "jump off" the table although it was obviously not 100 degrees Celsius in the room. Next Mr. H went over the vapor pressure of water curve. He told us that page 233 in the textbook has the best picture of it. Basically if you want to know the pressure at a specific degree you just follow the graph up and across. The graph stresses the point that depending what the pressure is, the point or temperature at which a substance changes to another state is different. Next we went to page 32 in the packet and Mr. H listed the things we should be able to note on a phase diagram. He also wrote that normal bp temperature at which the vapor pressure of a liquid=atmospheric pressure of 1 atm. Mr. H finished class with a demo. He had previously been boiling water in a beaker and then removed it from the heat and stoppered. He flipped it upside down on a rack and after adding ice cubes to the top of beaker (in contact with the gas not the liquid) the water began to boil again. It was apparent that the water was not at 100 degrees Celsius so the reason it boiled was because although the temperature went down the pressure did as well so we were able to change the boiling point. (see figure 9.5 in textbook). Mr. H then briefly mentioned that the reason people could walk on coals or other hot things was because they put water on their feet beforehand so the water absorbs the heat and their skin doesn't. Right before class ended we went over the top of page 15 about phase diagrams. First we labeled the different sections as either solid, liquid, or gas and then using the graph we found the answers to 2 which were vapor, solid, solid, and liquid. For homework we had a reading Webassign and a Webassign due Friday worth extra credit. There is a pop quiz on Friday.
Tuesday, January 25, 2011
Tuesday, January 25, 2011
This morning, we walked into class and Mr. Henderson passed out our semester grade reports. Once he was sure that everyone had received their grades, he asked us to turn to page 31 in our Thermochemistry unit packets. He gave us the answers for last night's WebAssign:
1. False
2. B
3 D
4. C
5. B
6. D
7. A
8. B
9. D
10. True
11. A, B, C
12. D, A, C, B
13. A
14. A
Subsequently, Mr. H told us to get out our lab notebooks while we looked at Daria's blog. As he read through it, he explained to us what was happening at the molecular level during state changes of water. First, the particles in the solid crystal lattice begin to vibrate when thermal energy (heat) is added (temperature increase). They wiggle more and more violently until they turn into a puddle (liquid). The fluid particles have small intermolecular forces and are relatively close together. When heat is added, they begin to vibrate and move about the puddle until so much kinetic energy is gained that the molecules "pop" out of the puddle and "fly" (as Matt said) because they are now a gas. Particles in a gas are far apart and if one were to walk through gas, there would be no resistance. Particles in a liquid are semi-close to each other and if one were to walk through liquid, there would be slight resistance. However, particles in a solid are very close together and if one were to walk through a solid...they wouldn't! Mr. H reminded us: "You can never walk through a solid."
1. False
2. B
3 D
4. C
5. B
6. D
7. A
8. B
9. D
10. True
11. A, B, C
12. D, A, C, B
13. A
14. A
Subsequently, Mr. H told us to get out our lab notebooks while we looked at Daria's blog. As he read through it, he explained to us what was happening at the molecular level during state changes of water. First, the particles in the solid crystal lattice begin to vibrate when thermal energy (heat) is added (temperature increase). They wiggle more and more violently until they turn into a puddle (liquid). The fluid particles have small intermolecular forces and are relatively close together. When heat is added, they begin to vibrate and move about the puddle until so much kinetic energy is gained that the molecules "pop" out of the puddle and "fly" (as Matt said) because they are now a gas. Particles in a gas are far apart and if one were to walk through gas, there would be no resistance. Particles in a liquid are semi-close to each other and if one were to walk through liquid, there would be slight resistance. However, particles in a solid are very close together and if one were to walk through a solid...they wouldn't! Mr. H reminded us: "You can never walk through a solid."
We then turned to our TC9 lab in our notebooks. Mr. Henderson told us that we must tape our data table and graphs into the lab. We also must write a conclusion/discussion, answering all 4 questions required in a fluid manner
1. Observe the plateaus on the two graphs. What could be happening during these plateaus?
2. Determine the melting point and freezing point of lauric acid. How do they compare?
3. Describe what happens at the particle level as heat is added to the substance and its temperature increases.
4. Describe what happens at the particle level as heat is added to a substance and its phase changes.
Henderson instructed us to take some notes:
- A plateau does not equal [Delta] T
- Melting point and freezing point are the same temperature! For lauric acid, it's 44 degrees Celsius
- Plateau = mix of solid/liquid or liquid/gas
After discussing this, Mr. H gave us 5 minutes to work on the conclusion. Then, he said that there would be a demo in the back of the class later on involving boiling water and capturing the emitted gas.
Even more after, we turned to page 13 in our Thermochemistry packets. First, we did numbers 1-13:
1. D E F
2. A
3. F
4. NONE
5. A
6. D
7. D E F
8. E
[oops, where's the 9?]
10. B
11. E
12. C
13. D
The bottom of page 13 is as Mr. Henderson said "basically lab TC9" except with water. The labels for the diagram are: 1 - solid, 2 - solid/liquid (0 degrees), 3 - liquid, 4 - liquid/gas (100 degrees), 5- gas. Then, we turned to the next page (14) where we completed questions 15-28:
15. 2, 4
16. 4
17. 1
18. 2
19. 2
20. 2
21. 4
22. 5
23. 2
24. 3
25. 4
26. 1, 3, 5
27. 2
28. 4
We migrated to the back of the room to finish the demo. Mr. H connected a thin copper pipe to the flask with the boiling water. We observed as steam spewed from the end. We thought this was gas. NO! It was water!!!!!!! As the gas traveled through the pipe, it cooled back to liquid state. Mr. H then put the burner under the pipe. He showed us that this was real gas. It was SO hot that he could light a match! Then he placed a paper at the end, which he burned "I'M HOT" into. Then the bell rang.
Homework: WebAssign Due Thursday the 27th, page 15 in packet
Monday, January 24, 2011
Monday January 24, 2011
Todays' class was relively easy. We started out class by quickly changing seats and getting new lab groups. Then people signed up to do the blog for the next two weeks. Also Mr. Henderson mentioned we will be starting a new unit thursday. Then Mr. Henderson explained the homelogic issue where the final grades were very low and wrongly posted; they are now fixed. There is also a 5% curve on the final grade. 
We then talked about signing up for courses next year as well as science electives that we are elegible to take. Mr. H explained the 5 electives; Medical Technologies(which is for people who are looking to go into the medical field, a little like house), Forensic Science(which is for people that are interested in CSI), Horticulture ( which is about plants, and anyone with a pulse can pass with an a), Brain Studies(which i for people who would like to be in the counciling and psychiatry fields of medicine) and last but not least astronomy (which Mr. H reccomends we take senior year after physics).
After that talk we went over a new topic that we have a webassign as homework tonight on, Chapter 9 states of matter. Mr. H explaines mostly the names of the changes in matter, liquid to gas, solid to liquid ect. but he also mentions that gas has no orderly arrangement, has translational and vibrational motion and no intermolecular forces. Which basicly means we can walk through gas without even feeling it and one gas particle can go from one end of the room to another without a problem. Liquid also has no orderly arrangement, it has as well translational motion and vibrational motions but, differently from gas, it has a weak intermolecular force. That means that even though it shapes to its container it almost always comes together, so if we so a karate chop through it, it will come back together. Solids are a whole different factor, they have a vary orderly arrangement of particles, NO translational motion(only sometimes vibrational) and they have strong molecular forces. That means that solids are one shape and if we do a karate chop through it, its going to hurt.
After our quick lesson Mr. H gave us a data sheet and we were sent off to the computer lab to do graphs based on the data he gave us. That is lab TC9 and i will provide a picture of all the nessesary objectives we need to do and answer.
At the end of class Mr. H gave us a print out of our semester one grades. That was the end to a very lovely chemistry class.
Wednesday, January 12, 2011
Today in chemistry we started by looking at the answers for the 6.5 reading sheet. Some tips that Mr. Henderson left for us were that you should remember to pay attention so see whether H2O is in a liquid state or a gas because the energy for each one is different. He also reminded us to use coefficients of balanced chemical equations as multipliers and also that for elements in standard states ΔHf=0
On the board were notes on using ΔHf values to calculate heats of reaction. They were the following:ΔH RXN = Sum of ΔH°f products – Sum of ΔH°f reactants
What the above basically means is that the heat of reaction from any reaction is equal to sum of the heats formation values for reactants subtracted from those of the product.
Next we moved to page 9 in our packets and we started on number 2. The problem was the following:
The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide can be used as an oxidizing agent. The balanced equation for the reaction is:
5N2H4(l) + 4N2H3CH3(l) -----------------> 3N2(g) +4H2O(g),
Determine the heat of the reaction.
What we did first was move the product side in the reaction over. Next we looked at the chart and we found the ΔH°f values of the compounds and in this N2 was 0 because it was an element in its natural state. For H2O in a GAS state the chart said it was -242 kJ/mol. Then we moved on to the reactant part of the equation. We found the ΔH°f values for the reactants and we multiplied them by the coefficients attached to each one of the compounds. So for this problem we had to multiply -20 kJ/mole by 5 for the 5N2H4(l) and that is -100 kJ/mole. Next we found that the heat formation values of N2H3CH3(l) were 54 kJ/mole, and because the compound was accompanied with a molar number of four, you have to multiply 54 by 4 to get 216 kJ/mole. When you add the two heat formation numbers you get -46 kJ/mole. Finally to finish the problem you subtract the first sum from the second one. So it would be -242 - -46 and you get -196 kJ/mole and that would be your answer.
Then we also did #3 following and if you weren’t here that one would be good practice. Next we looked at Wills blog and Mr. Henderson briefly went over its content.
We then went back to our packets and looked at page 11 and started with #2. The problem was the following:
Calculate the ΔH for: C(s) + 2 H2(g) → CH4 (g)
Given: C(s) + O2(g) → CO2 (g) ΔH = -394 kJ
2 H2(g) +O2(g) → 2 H2O (l) ΔH = -572 kJ
CH4(g) + 2O2 (g) → CO2 (g) + 2 H2O ΔH = -890 kJ
To solve the problem you basically have match up variables that aren’t in the model equation (the one on top) and cancel out the ones that aren’t by either switching around the product and reactants or by multiplying the whole equation by a number to put coefficients to the numbers could match up and cancel out. Then by adding up the ΔH of the equations to finally end up with the model one you wanted and that will be the answer to the problem which in this case is ΔH = -76 kJ. We finished the class by working on our labs TC4.
On the board were notes on using ΔHf values to calculate heats of reaction. They were the following:ΔH RXN = Sum of ΔH°f products – Sum of ΔH°f reactants
What the above basically means is that the heat of reaction from any reaction is equal to sum of the heats formation values for reactants subtracted from those of the product.
Next we moved to page 9 in our packets and we started on number 2. The problem was the following:
The space shuttle orbiter utilizes the oxidation of methyl hydrazine by dinitrogen tetroxide can be used as an oxidizing agent. The balanced equation for the reaction is:
5N2H4(l) + 4N2H3CH3(l) -----------------> 3N2(g) +4H2O(g),
Determine the heat of the reaction.
What we did first was move the product side in the reaction over. Next we looked at the chart and we found the ΔH°f values of the compounds and in this N2 was 0 because it was an element in its natural state. For H2O in a GAS state the chart said it was -242 kJ/mol. Then we moved on to the reactant part of the equation. We found the ΔH°f values for the reactants and we multiplied them by the coefficients attached to each one of the compounds. So for this problem we had to multiply -20 kJ/mole by 5 for the 5N2H4(l) and that is -100 kJ/mole. Next we found that the heat formation values of N2H3CH3(l) were 54 kJ/mole, and because the compound was accompanied with a molar number of four, you have to multiply 54 by 4 to get 216 kJ/mole. When you add the two heat formation numbers you get -46 kJ/mole. Finally to finish the problem you subtract the first sum from the second one. So it would be -242 - -46 and you get -196 kJ/mole and that would be your answer.
Then we also did #3 following and if you weren’t here that one would be good practice. Next we looked at Wills blog and Mr. Henderson briefly went over its content.
We then went back to our packets and looked at page 11 and started with #2. The problem was the following:
Calculate the ΔH for: C(s) + 2 H2(g) → CH4 (g)
Given: C(s) + O2(g) → CO2 (g) ΔH = -394 kJ
2 H2(g) +O2(g) → 2 H2O (l) ΔH = -572 kJ
CH4(g) + 2O2 (g) → CO2 (g) + 2 H2O ΔH = -890 kJ
To solve the problem you basically have match up variables that aren’t in the model equation (the one on top) and cancel out the ones that aren’t by either switching around the product and reactants or by multiplying the whole equation by a number to put coefficients to the numbers could match up and cancel out. Then by adding up the ΔH of the equations to finally end up with the model one you wanted and that will be the answer to the problem which in this case is ΔH = -76 kJ. We finished the class by working on our labs TC4.
Monday, January 10, 2011
Monday, January 10
Today we started class by grabbing our lab notebooks. We would need them for the lab that we would soon be doing. Next on our agenda, going over the webassign due today. It was a reading sheet, so we turned to page 27 (32) in our packets, and filled in the answers.
Answers:
1. C
2. B
3. C
4. C
5. B & C
6. False - The enthalpy value is relative to those values of other substances
7. B
8. A
9. To give off 185 kJ of heat
10. B
- B
- 370 kJ
- 1998 kJ
11. a. 6232.1 kJ
b. 91.0 kJ
12. D
13. B
14. D
15. 790 kJ
16. A
17. 40.64 kJ
18. (Delta)H4 = 40 kJ
Then it was time to learn about something new, Hess's law. We flipped to page 11 of our packet. Mr. H told us that this section of the unit would be one of the harder sections. But he also mentioned that it was a bit like algebra. 2 of my favorite things! Chemistry and algebra! But the way you would solve the problems went something like this.
1. Calculate (Delta)H for: 2C(s) + O2(g) --> 2CO2(g) (Delta)H = -393.5 kJ
Now to do this, we had to find a way to combine the formulas, like in algebra. In this situation, we multiplied all of the coefficients in the first formula by 2, and we flipped the second formula around making the (delta)H positive 566.
2C(s) + 2O2(g) --> 2CO2(g) (delta)H = -787 kJ
2CO2(g) --> 2CO(g) + O2(g) (delta)H = +566 kJ
and the result.
2C(s) + O2(g) --> 2CO2(g) (delta)H = -221kJ
2. Calculate (Delta)H for: 4Al(s) + 3MnO2(s) --> 2Al2O3(s) + 3Mn(s)
Now to solve this one we flipped the second formula and multiplied it by 3.
4Al(s) + 3O2(g) --> 2Al2O3(s) (delta)H = 3352 kJ
3MnO2(s) --> 3Mn(s) + 3O2(g) (delta)H = 1563 kJ
and the result.
4Al(s) + 3MnO2(s) --> 2Al2O3(s) + 3Mn(s) (delta)H = 4915 kJ
We finished the day with a lab. The lab was the "Heat of Formation Lab"
Purpose: To use calorimetry to determine the heat of formation of calcium hydroxide (knowing that the heat of formation of H2O(l) is -286 kJ)
Tomorrow in class we will discuss the calculations, but these are the results my group got.
Volume of H2O: 100 mL
Mass of Ca: 2.0g
Initial temp. of H2O: 21.6 degrees Celsius
Final temp of H2O: 53.3 degrees Celsius
And here's a picture of what the reaction looked like. It is a bit unclear, but the reaction was a bit hard to see in real life due to all the steam.
Today would not have been a very good day to miss, because I think it's we all had fun doing this lab today.
Answers:
1. C
2. B
3. C
4. C
5. B & C
6. False - The enthalpy value is relative to those values of other substances
7. B
8. A
9. To give off 185 kJ of heat
10. B
- B
- 370 kJ
- 1998 kJ
11. a. 6232.1 kJ
b. 91.0 kJ
12. D
13. B
14. D
15. 790 kJ
16. A
17. 40.64 kJ
18. (Delta)H4 = 40 kJ
Then it was time to learn about something new, Hess's law. We flipped to page 11 of our packet. Mr. H told us that this section of the unit would be one of the harder sections. But he also mentioned that it was a bit like algebra. 2 of my favorite things! Chemistry and algebra! But the way you would solve the problems went something like this.
1. Calculate (Delta)H for: 2C(s) + O2(g) --> 2CO2(g) (Delta)H = -393.5 kJ
Now to do this, we had to find a way to combine the formulas, like in algebra. In this situation, we multiplied all of the coefficients in the first formula by 2, and we flipped the second formula around making the (delta)H positive 566.
2C(s) + 2O2(g) --> 2CO2(g) (delta)H = -787 kJ
2CO2(g) --> 2CO(g) + O2(g) (delta)H = +566 kJ
and the result.
2C(s) + O2(g) --> 2CO2(g) (delta)H = -221kJ
2. Calculate (Delta)H for: 4Al(s) + 3MnO2(s) --> 2Al2O3(s) + 3Mn(s)
Now to solve this one we flipped the second formula and multiplied it by 3.
4Al(s) + 3O2(g) --> 2Al2O3(s) (delta)H = 3352 kJ
3MnO2(s) --> 3Mn(s) + 3O2(g) (delta)H = 1563 kJ
and the result.
4Al(s) + 3MnO2(s) --> 2Al2O3(s) + 3Mn(s) (delta)H = 4915 kJ
We finished the day with a lab. The lab was the "Heat of Formation Lab"
Purpose: To use calorimetry to determine the heat of formation of calcium hydroxide (knowing that the heat of formation of H2O(l) is -286 kJ)
Tomorrow in class we will discuss the calculations, but these are the results my group got.
Volume of H2O: 100 mL
Mass of Ca: 2.0g
Initial temp. of H2O: 21.6 degrees Celsius
Final temp of H2O: 53.3 degrees Celsius
And here's a picture of what the reaction looked like. It is a bit unclear, but the reaction was a bit hard to see in real life due to all the steam.
Today would not have been a very good day to miss, because I think it's we all had fun doing this lab today.
Sunday, January 9, 2011
Friday, January 6th
Today started off as a normal day in Mr.H's 3rd period chemistry class. We went over Tim's blog that was on enthalpy that we had done the previous day. Next Mr.H directed us to page 4 in our packets where we found an ordinary problem about determining the heat released in the combustion of a walnut, determining the heat content of a walnut and determining the amount of calories in 50 walnuts.
This problem started with the heat released by the combustion of a walnut. We began with the equation q=mCΔT where q=the amount of heat released, m=the mass of the heated substance C=the specific heat(4.18 for water) and T for the temperature change. In this problem we needed to find the amount of heat that was released into the water causing the water to change temperature. We began plugging in the known values in the equation and ended up with a multiplication problem of:q=(200g)(4.18)(39.1-25). Then after the multiplying ended up with 11787.6J. This was the amount of heat gained by the water. Since the water gained energy and the walnut lost it this makes the heat lost by the walnut -11787.6J. this can be expressed in the equation: qwalnut=-qH2O, meaning as the walnut looses energy the walnut gains it.
The next part of the problem was to find the energy content of the walnut in kJ/gram. for this we set up a conversion factor that converted from J to kJ then we divided by the amount of grams in the problem to determine the kJ/gram. This conversion factor looked like:(11787.6J)(1kJ/1000J)= 11.7876kJ the diving by 2.6 making the answer 4.53 kJ/ gram.
The Final part of the problem was the amount of calories in 50 walnuts. This problem consisted of a few simple conversions from kJ to Calorie. First, we know that 11787.6 J are released in one walnut so then in 50 walnuts there will be 50 times as much heat:(50 walnuts)(11787.6/1 walnut)=58930J of heat. Then we must convert this back to kJ because it as given in the problem that 4.18 kJ=1 calorie:(58930J)(1kJ/1000J)= 58.930kJ of heat. Last we set up the conversion factor to finish the problem:(58.930kJ)(1 cal/4.18kJ)= ~141calories in 50 walnuts.
Next we moved on to the next problem on page 7 of our packets problem number 1 which consisted of STOICHIOMETRY. YAYYYY. SWEEEETTTTT
This problem was the combustion of iron oxide.
4Fe(s)+3O2(g) -> 2 Fe2O3(s) ΔH=-1652kJ
a. For part a, we needed to find the energy released by the combustion of 1 mol Fe. This was a simple conversion factor that converted 1 mole Fe to the amount of energy that would be theoretically produced. this conversion factor looked like (1 mole Fe)(1652kJ/4 mole). this resulted in a answer of 413 for the first problem.
b. Next for part b, this was almost the same thing as part a. but with a different amount of Fe. the conversion looked like (12 mole Fe)(1652kJ/4 mole). this parts answer was 4956 kJ.
c. Last part c. this one was a little harder because of the conversion of grams to moles, but still was simple. this conversion looked like (110g Fe)(1 mole/55.845g Fe)(1652kJ/4 mole)= 813.5 kJ as the final answer.
This was the end of the lesson part of the class. Next, we were on the the lab test part of class. This lab test was to combust paraffin in the for of candle. Due to the Acedemic dishonesty policy at GBS. I may not divulge more information than that. (i dont want to cheat or help someone cheat). The lab took us to the end of the period.
Overall it was a good day. =P
Thursday, January 6, 2011
Thursday, January 6, 2011
Mr. H started the day by showing us the answers to Chapter 8.2 reading sheet. Next, we moved on to review the blog by Neil. He reviewed the exothermic and endothermic reactions, but added entholpy, which is a form of chemical energy. Entholpy is basically the total energy or heat and is symbolized by a H. In an exothermic reaction, the heat is always on the product side and ΔH is a negative value. In an endothermic reaction, the heat is always on the reactant side and ΔH is a positive value.
Next we moved onto a demo. Mr.H started it off by asking people to feel the beaker, the wood, and the water to prove that this wasn't a trick. He started it off by putting water on the wood. Next, he put the beaker on top of the wood. He then mixed two compounds together to create an endothermic reaction. The mixture inside the beaker pulled the heat from the water and caused it to freeze onto the wood.
After that, we moved on to pg 5 of our packets. We worked on #1 which required us to figure out what kind of reaction it was. The answers are..
a.EX b.EN c.EX d.EN e.EX f.EN g.EN h.EX i.EN j.EX k.EX l.EX
Next, Mr. H showed us this concept on the board.
If A → B + 100J
...then 100J + B → A
...then 2A → 2B + 200J
...then 300J + 3B → 3A
After learning this we moved on to work on problem 2 on pg 5. It required us to find the missing value. The answers are..
a. -2408kJ b. 1204kJ c. 1652kJ d.2.70kJ e. -13.5kJ
Next, we moved onto page 4 #4. This problem was to be a sample problem for our TC3 lab. The answers are
a.Q= -27692.5
b. (Determine the Moles of ice that melted) 3.9111
c. 7.08kJ/mol
Finally, we finished our day by finishing up our TC2 and TC3 labs.
Wednesday, January 5, 2011
Wednesday, Janurary 5, 2011
Today Mr. H started off class by complimenting Emma on the blog post that she had completed the previous night with a quick review of the previous day using her blog as a guide. To begin class Mr. H started off class by giving us the answers to the reading sheet on pg. 36, but really pages 23 and 24. The answers are:
1-A
2-B
3-A
4-A,E
5-B
6-A
7-False
8-D
9-D
10-A,C
11-B,C
12-B
13-B
14-A,B,C,D
15-False
16-False
17-B
18-B
19-B
As usual, Mr. H walked us through the reading sheet by answering any questions we had or highlighting the key points on the reading sheet. The main focus of the discussion was on endothermic and exothermic reactions. We also learned about the difference between the system and surroundings. Next, we began to work in the packet for the second day and it was pretty exciting. The answers are below in the picture.
The first few questions weren't difficult because they started off distinguishing between basic endothermic and exothermic reactions. The top half of this page discusses this topic. Endothermic reactions are where the system absorbs energy from the surroundings cooling down the surroundings and exothermic reactions release heat/energy from the system heating up the surroundings. The second half of the page was a little more complex because it involved distinguishing what graphs represent which type of reaction. Also, more real world examples were used to help us grasp the concept of endothermic and exothermic reactions. Next, we moved onto pg. 3 of the packet and we got to use our calculators for the first time this unit, SWEEEEET!!
The questions on the following page involved understanding a mathematical concept. The equation that was essential was:
q=m*c*delta-t
q=a measure of the heat flow
m=mass of the substance that you are trying to find the heat flow for
c=heat capacity of the substance
delta-t=change in temperature over the reaction of the substance
We began to work on the problem, which was going to help us work on the lab TC2 that we had begun the day before. We were trying to find the heat flow of water to later find the specific heat of iron. In the equation, m was substituted with the mass of the water which was 100g, c was substituted with the heat capacity of water which is 4.18 and chart with heat capacity numbers can be found in our book in ch. 8. The change in temperature was from 21.9 degrees Celsius to the highest value of 29.5. So the equation looked like this:
100*4.18*(29.5-21.9)=q
Once we found q which was 2758.8 joules were needed to calculate the kilojoules lost by iron in the reaction, so we were able to conclude that the amount of energy water gained was the amount of energy that iron and that was -2.7588 KILOJOULES. The following equation was derived to find out the specific heat or heat capacity of the iron.
C= q/m*delta-t
When the known variables were substituted the equation looked like this:
-2758.8 J/(6.52)*(28.5-86.6)=c
The answer was 7.41 J/g*c.
Next, Mr. H gave us the directions, title and purpose Lab TC3: Heat of Fusion Lab. The goal of the lab was to determine the molar heat of the fusion of ice. We came back to the front of the room for a little bit as Mr. H explained to us how to complete the calculations to the Lab TC2 and TC3. It made it a lot easier to do. Most people didn't finish Lab TC3 and Mr. H said we would finish up the next day from help from him. That was the end of the class and Mr. H reminded us of our Webassign that was due which was a reading sheet.
1-A
2-B
3-A
4-A,E
5-B
6-A
7-False
8-D
9-D
10-A,C
11-B,C
12-B
13-B
14-A,B,C,D
15-False
16-False
17-B
18-B
19-B
As usual, Mr. H walked us through the reading sheet by answering any questions we had or highlighting the key points on the reading sheet. The main focus of the discussion was on endothermic and exothermic reactions. We also learned about the difference between the system and surroundings. Next, we began to work in the packet for the second day and it was pretty exciting. The answers are below in the picture.
The first few questions weren't difficult because they started off distinguishing between basic endothermic and exothermic reactions. The top half of this page discusses this topic. Endothermic reactions are where the system absorbs energy from the surroundings cooling down the surroundings and exothermic reactions release heat/energy from the system heating up the surroundings. The second half of the page was a little more complex because it involved distinguishing what graphs represent which type of reaction. Also, more real world examples were used to help us grasp the concept of endothermic and exothermic reactions. Next, we moved onto pg. 3 of the packet and we got to use our calculators for the first time this unit, SWEEEEET!!
The questions on the following page involved understanding a mathematical concept. The equation that was essential was:
q=m*c*delta-t
q=a measure of the heat flow
m=mass of the substance that you are trying to find the heat flow for
c=heat capacity of the substance
delta-t=change in temperature over the reaction of the substance
We began to work on the problem, which was going to help us work on the lab TC2 that we had begun the day before. We were trying to find the heat flow of water to later find the specific heat of iron. In the equation, m was substituted with the mass of the water which was 100g, c was substituted with the heat capacity of water which is 4.18 and chart with heat capacity numbers can be found in our book in ch. 8. The change in temperature was from 21.9 degrees Celsius to the highest value of 29.5. So the equation looked like this:
100*4.18*(29.5-21.9)=q
Once we found q which was 2758.8 joules were needed to calculate the kilojoules lost by iron in the reaction, so we were able to conclude that the amount of energy water gained was the amount of energy that iron and that was -2.7588 KILOJOULES. The following equation was derived to find out the specific heat or heat capacity of the iron.
C= q/m*delta-t
When the known variables were substituted the equation looked like this:
-2758.8 J/(6.52)*(28.5-86.6)=c
The answer was 7.41 J/g*c.
Next, Mr. H gave us the directions, title and purpose Lab TC3: Heat of Fusion Lab. The goal of the lab was to determine the molar heat of the fusion of ice. We came back to the front of the room for a little bit as Mr. H explained to us how to complete the calculations to the Lab TC2 and TC3. It made it a lot easier to do. Most people didn't finish Lab TC3 and Mr. H said we would finish up the next day from help from him. That was the end of the class and Mr. H reminded us of our Webassign that was due which was a reading sheet.
Tuesday, January 4, 2011
Tuesday, January 4, 2011
Mr. H began class by greeting us and explaining our agenda for the day. We immeditely were asked to get out our lab notebooks to write down the title and purpose of the TC2 Lab: Specific Heat of a Metal Lab. Purpose: To determine the specific heat of a sample of metal. After going over Kendall's blog we went to the back of the class room to begin the lab. Our lab groups split into two groups to complete to tasks. One group had to Prepare the Metal. Step 1: Measure the mass of a clean, dry, test tube. Record. Step 2: Empty metal (beads) into test tube. Step 3: Measure the mass of the test tube with the beads. Record. [Then subtract the mass of the test tube from the mass of the test tube with metal beads to find m the mass for your equation.] Step 4: Put the test tube into a hot water bath and heat water to about 80 degrees Celsius. The thermometer should be in the water (not the test tube.) Remember the 0th law. The other group had to Prepare the Water Calorimeter. Step 1: Obtain some cool/chilled water. Step 2: Place 50 mL into a dry styrofoam cup. After finishing our preparations we went back to the front of the classroom. Mr. H then reviewed with us our graphs from yesterday. He explained how after the system and surroundings met at thermal equilibrium their temperatures began to go down and would continue to go down indefinitely until the room temperature cooled and then the outdoor air would cool and then the universe would cool and we would all die. But that won't happen for billions of years so were alright. Next we went over Page 1 in our packets. First we were asked to state the zeroeth law of thermodynamics in our own words. I wrote two objects with a different temperature in contact with each other will transfer energy(heat) from high temp to low temp until they have the same temp. Then Alex explained what would happen if a hot piece of dishware was placed in a cold bowl of water. She said heat from the hot dishware would transfer to the cold water until they reached an equal temp
erature or equilibrium. Then we went over the chart at the bottom of page 1. We were given two situations ans asked to state the system and surroundings, the directional flow of heat, and what happened to the tempeatures (see picture left). Then we moved into the left side of our notebook. Mr. H went over an equation that pertains to our lab (see picture left). Then he went over exothermis process which pertains to our lab. Energy is lost from the system and transfered to the surroundings. The change of T is positive, the q of the surroundings is positive and the q of the system is negative. Next he went over endothermic process. The change of T is negative, the q of the surroundings is negative, and the q of the system is positive. Mr. H a
lso noted through graphing that to being a chemical process there has to be a spark or actication. Finally we went back to the back of the room to finish the lab. We measured the temperatures of the hot water and the cold water then carefully put the metal into the cold water and measured the temperature until we got the highest temperature. Mr. H dismissed us. Our homework was Reading Webassign on 8.2.
erature or equilibrium. Then we went over the chart at the bottom of page 1. We were given two situations ans asked to state the system and surroundings, the directional flow of heat, and what happened to the tempeatures (see picture left). Then we moved into the left side of our notebook. Mr. H went over an equation that pertains to our lab (see picture left). Then he went over exothermis process which pertains to our lab. Energy is lost from the system and transfered to the surroundings. The change of T is positive, the q of the surroundings is positive and the q of the system is negative. Next he went over endothermic process. The change of T is negative, the q of the surroundings is negative, and the q of the system is positive. Mr. H a
lso noted through graphing that to being a chemical process there has to be a spark or actication. Finally we went back to the back of the room to finish the lab. We measured the temperatures of the hot water and the cold water then carefully put the metal into the cold water and measured the temperature until we got the highest temperature. Mr. H dismissed us. Our homework was Reading Webassign on 8.2.
Monday, January 3, 2011
Monday, January 3, 2011
We began class today by passing out our lab notebooks, old papers, as well as a graph and the new unit packet. The next unit is Thermochemistry. We then discussed the last test. The results were unfortunatley not so great and Mr. Henderson gave us the option to come in to review the test. As an opportunity to help our grades, there are bonus points if the Webassign due the 13th is completed before the due date. For the Webassign, questions 1-6 need to be completed by Friday January 7th; 7-10 by Monday January 10th; 11-15 by Tuesday January 11th; and 16-19 by Wednesday January 12th. Tonights homework is a Webassign Reading Sheet on 8.1 due Tuesday.
We then began a new lab (TC1) called Zeroing in On Heat Lab. The purpose is to describe and explain the temperature changes experience by two objects of different temperatures when place din close contact with one another. In order to do this, we marked the temperatures of hot water and cold water every 10 seconds. We made a graph and the times started at 0 and went to 190. The temperatures for hot water (in celcius) were 84.6, 81.6, 75.2, 69.1, 64.9, 62.9, 59.3, 56.0, 51.8, 51.8, 51.3, 50.8, 50.2, 50.0, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. The temperature of the cold water was as follows: 7.3, 17.8, 24.7, 28.3, 32.7, 37.8, 39.3, 43.4, 46.1, 47.0, 47.9, 48.3, 49.5, 49.5, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. We then graphed our results, which should look something like this:
There is a line through 140 because that is when the hot and cold temperatures were equal. We then made our lab conclusion, which described how the temperature changes over time for both sets of data. My results were that the temperatures are inversely proportional until 140 seconds have passed. The difference between the seconds decreases over time. At the end, the released the same temperature. This constitutes a heat flow (exchange of energy). The hot warms up the cold and continues until they are at the same temperature. The graph should be taped into the lab notebook. Mr. Henderson also mentioned that there is a quest on Wednesday January 12th.
We then began a new lab (TC1) called Zeroing in On Heat Lab. The purpose is to describe and explain the temperature changes experience by two objects of different temperatures when place din close contact with one another. In order to do this, we marked the temperatures of hot water and cold water every 10 seconds. We made a graph and the times started at 0 and went to 190. The temperatures for hot water (in celcius) were 84.6, 81.6, 75.2, 69.1, 64.9, 62.9, 59.3, 56.0, 51.8, 51.8, 51.3, 50.8, 50.2, 50.0, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. The temperature of the cold water was as follows: 7.3, 17.8, 24.7, 28.3, 32.7, 37.8, 39.3, 43.4, 46.1, 47.0, 47.9, 48.3, 49.5, 49.5, 49.7, 49.7, 49.4, 49.2, 49.0, and 48.9. We then graphed our results, which should look something like this:
There is a line through 140 because that is when the hot and cold temperatures were equal. We then made our lab conclusion, which described how the temperature changes over time for both sets of data. My results were that the temperatures are inversely proportional until 140 seconds have passed. The difference between the seconds decreases over time. At the end, the released the same temperature. This constitutes a heat flow (exchange of energy). The hot warms up the cold and continues until they are at the same temperature. The graph should be taped into the lab notebook. Mr. Henderson also mentioned that there is a quest on Wednesday January 12th.
Sunday, January 2, 2011
Partial Pressures and Working through a Word Problem
To begin class, Mr. H asked about the webassign that would be due on Friday and whether or not we had any questions on it. The main question was on how to recognize what equations we should use on each problem because, unlike stoichiometry, all our gas equations are word problems that have about 5 key equations. These include:
P1V1=P2V2,
P1/n1=P2/n2,
P1/T1=P2/T2,
PV=nRT (R as a constant),
P1V1/T1=P2V2/T2.
All others can be derived from these core equations and concepts. It is also important to know that if P is divided by a variable, like temperature, then it varies directly with that variable (P/T). If P is multiplied by a variable, like volume, then it varies indirectly with that variable (PV). (look at pgs. 23-24)
After doing a few gas word problems on the white board, we continued on to pg.20. On pg. 20 was a question that seemed very difficult because it incorporated the mixing of two gases and their resulting pressure. We know from the start that the volume of H2 gas is 2L with a pressure of 1.2atm and the volume of the N2 gas is 1L with a pressure of 4.6atm. Temperature and moles stay constant. Although this equations looks very difficult, it is really just a P1V1=P2V2 equation. We begin by solving for 1 gas, H2. First, rearrange the equation using a bit of algebra and make it so that you are solving for P2, our unknown variable. Now, you should end up with P2=P1V1/V2. Next, simply plug in the information we are given to get P2= (1.2atm)(2L)/V2. Because we know that the new volume will be the old volumes added together, V2 is 3L. Now we solve and get that P2=.80atms. Repeat the same process for the N2 gas and you should end up with 1.53atms. Finally, to solve the last problem of the resulting pressure, just add together the two P2 values to get 2.33 atms. The equations should have a diagram that looked something like this:
This is the diagram from the webassign that was due on friday, so no you will not get a free answer!
Finally, we looked at pg. 19, which had similar partial pressure problems (although they are a bit easier). The key to these equations was to use your textbook to look up the vapor pressure of water at certain temperatures. This was given to us by Mr. H. He told us that the V.P. of H20 for problems 1 and 2 where 31.8 and 149 mmHg respectively. This left us with 868.3 mmHg and 151.02 mmHg for the O2 values of questions 1 and 2. Then, to solve parts 3+4 and Q 1-2, we put each pressure value over the total pressure. After converting each kPa pressure to mmHg, the correct answers for Parts 3-4 were 3.53% and 96.46% for Q1 and 49.66% and 50.34% for Q2. The rest of the problems could be solved the same way.
That was all for Thursday's class. Thank You and see you in class!
P1V1=P2V2,
P1/n1=P2/n2,
P1/T1=P2/T2,
PV=nRT (R as a constant),
P1V1/T1=P2V2/T2.
All others can be derived from these core equations and concepts. It is also important to know that if P is divided by a variable, like temperature, then it varies directly with that variable (P/T). If P is multiplied by a variable, like volume, then it varies indirectly with that variable (PV). (look at pgs. 23-24)
After doing a few gas word problems on the white board, we continued on to pg.20. On pg. 20 was a question that seemed very difficult because it incorporated the mixing of two gases and their resulting pressure. We know from the start that the volume of H2 gas is 2L with a pressure of 1.2atm and the volume of the N2 gas is 1L with a pressure of 4.6atm. Temperature and moles stay constant. Although this equations looks very difficult, it is really just a P1V1=P2V2 equation. We begin by solving for 1 gas, H2. First, rearrange the equation using a bit of algebra and make it so that you are solving for P2, our unknown variable. Now, you should end up with P2=P1V1/V2. Next, simply plug in the information we are given to get P2= (1.2atm)(2L)/V2. Because we know that the new volume will be the old volumes added together, V2 is 3L. Now we solve and get that P2=.80atms. Repeat the same process for the N2 gas and you should end up with 1.53atms. Finally, to solve the last problem of the resulting pressure, just add together the two P2 values to get 2.33 atms. The equations should have a diagram that looked something like this:
This is the diagram from the webassign that was due on friday, so no you will not get a free answer!
Finally, we looked at pg. 19, which had similar partial pressure problems (although they are a bit easier). The key to these equations was to use your textbook to look up the vapor pressure of water at certain temperatures. This was given to us by Mr. H. He told us that the V.P. of H20 for problems 1 and 2 where 31.8 and 149 mmHg respectively. This left us with 868.3 mmHg and 151.02 mmHg for the O2 values of questions 1 and 2. Then, to solve parts 3+4 and Q 1-2, we put each pressure value over the total pressure. After converting each kPa pressure to mmHg, the correct answers for Parts 3-4 were 3.53% and 96.46% for Q1 and 49.66% and 50.34% for Q2. The rest of the problems could be solved the same way.
That was all for Thursday's class. Thank You and see you in class!
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