Hess's Law Lab

Friday was a pretty simple day in Chemistry. We started off class by going over the procedures for Hess's Law Lab. Here are some tips that might help you with the lab: When you are trying to find the heat of reaction for Reactions 1 and 2 you have to use q=mc∆t. Once you find J, all you have to do is divide that by 1000 to put it into kJ and then divide the kJ by the moles of magnesium/ magnesium oxide used in the experiment. Once you find the heat of reaction for both experiments, use Hess's Law to find the heat of reaction for the combustion of magnesium.

And the next Scribe is Rachel K.

Hess' Law

The class started off today with Liebs showing us the experiment that failed yesterday. He was showing us how the process of sublimation could change to fusion. In order to do so he put frozen carbon dioxide, or dry ice, under "intense" amounts of pressure.

Once the dry ice melted under the water, it would pop under the pressure and send water everywhere, which was terrifying to some of the people in our class... so this is is the same thing, but without the "pop".



Moving onto the lecture today, we spent the rest of class discussing something called Hess' Law. Hess' Law basically says that you can use multiple ways to find Hess' Law, and it will still work, as long as you come to same, correct conclusion.

"If you ski down a hill, it doesn't matter the path you take, you're still going to get to the bottom. As long as you don't hit a tree."

Basically you are given a formula, and you must add and subtract a series of other equations to match it, and then do the same process to series' delta H's in order to find the original missing delta H. To change the equations you can do one of two things: you can flip the formulas around or multiply their coefficients by a number. When adding the equations, if they are on the opposite side of the arrow and the same element/compound they cancel. If they are on the same side then they combine.

An example:

If at Once you don't Succeed Try, Try Again.

That saying above doesn't apply for our class today. We were learning about Phase Diagrams:

The Phase Diagram above is on Water!!! So the shaded blue part is where at a certain temperature and pressure the water is a Solid (such as Ice). The shaded green part is where the water is a liquid (Aka water). The shaded pink part (sure i guess you can call it peach) is the water as a gas. (The actual H2O molecule, not O2 or H2 molecules) Anyway...the places where the lines separate the three areas are where the change between the phases happen. The temperature is always on the X-axis and the pressure is always on the Y-axis. The lines tell the points at which the change takes place, but there is almost always a 'normal' melting or 'normal' boiling point. Which for water is 0 degrees Celsius and 100 degrees Celsius. The point on the graph that says "Triple Point" is where at this certain pressure and temperature the substance will be in all three states. We actually saw this happen in one of the demo's Mr. Liebs showed us... I couldn't find a picture of the triple point of nitrogen so I'll just use the picture of water...

As you can see, the Liquid is on the bottom, the Solid is in the center and the Gas is on the top. (trust me the gas is there). The critical point is where the liquid is not a solid, liquid, or gas actually. I know it's hard to believe...but it's true. Liebs tried to show us that it was true with dry ice, but we couldn't get a good enough seal on the container so it didn't work. We tried to seal it many, many times but in the end all we had was a broken ego and a wish to see the critical point. So as you can see, 'if at once you don't succeed try, try again.' doesn't always work. But being as optimistic as we are, we will try it again tomorrow. So either it will work, or we will fail...again.

http://http//www.youtube.com/watch?v=2Z6UJbwxBZI&feature=related This video posted here (look up) is about the critical point of helium. (sorry i still can't get the video into the actual blog)

We were trying to do the same thing with carbon (aka dry ice) but as I said earlier we couldn't get a good enough seal. Also in the video you will hear the word Viscosity. That means the thickness or thinness of a liquid (also used for oils, if anyone needs to put oil in their cars they look at the viscosity to make sure it's right).

Boiling occurs when the vapor pressure and the pressure from outside are equal, or when the vapor presser exceeds that of the outsides. During the Critical point the meniscus:

disappears and the point where liquid and gas are separated is nonexistent:


As you can see, the meniscus has disappeared.
Another thing we learned is the term Sublimation, this is when the solid changes to a vapor totally skipping the liquidation phase.
Please don't forget about the two web assigns. One has been extended from Monday, and is now due tomorrow, then also the web assign that is due on the 31st is up. We also have work sheets that are due on the day of the test.
The Scribe for tomorrow is...Erika G.!!!!!!!!! YAY!!!!! Good Luck Erika! And again sorry about the video, I'll figure it out by next time!!!!!

Heat

Well, today was the second day of second semester. We started off class by having Mr.Lieberman show us how to do the graph for the lab we did yesterday. Basically when doing your graph you will beableto notice three distinct regions on the graph. These regions can be identified on the heat curve shown below.

As you can see the first region is where the snow(solid) turned in to water. The second region is just the liquid stage. After this stage there is the region where the water turned into steam. The first and third region are both endothermic. When we finished discussing this Mr. Lieberman told us of the possible error that we might find with our lab results. He said that for some of the data we might have gotten the temperature of the hot plate rather than the water. This would be if we let the thermometer touch the bottom of the beaker when recording the temperature.

When we finished talking about the lab, we worked on the worksheet that we picked up in class today. If you did not get it, you can get it on moodle. Also, if you finished the worksheet and want to review your answers you can get that on moodle as well.On this worksheet two new concepts were introduced, Heat of fusion and Heat of vaporization. The heat of fusion is used for the melting point, and the heat of vaporization is used for the boiling point. These can be found in our textbook.

Mr. Lieberman also did a demo today. Basically it proved if steam or liquid water could get hotter. To do this Mr. Lieberman burned a piece of paper only using steam. This helped to prove that vapor burns were more sever than liquid ones.

Remember that we have a WebAssign due tomorrow as well as the worksheet and lab from Monday.

The next scribe is Danielle D. Have fun!

Heating Curve Lab

    Welcome back!  Today was the first day of the second semester.  Besides getting new seats and some old work back, we also began a new lab.  If you don't have it you can find it here (however this format is slightly different).  We will no longer be using the pre-lab, data, post-lab format.  The new outline for our labs is:


Title:
Lab Goal:
Data/Collections:  (includes data, labeled graph, and percent error)
Conclusion:  (includes claim (according to your data...), evidence (what data did you use 
                   that helped you make this claim?  How did you identify what ever you were 
                   looking for), and reasoning (does your data make sense?  Why might it not be
                   what you expected it to be?)


    In the lab, we filled a 400 mL beaker with about 150 mL of snow/crushed ice.  The goal of the lab was to find out the temperature at which ice melts and boils (using the heating curve).  The next step was to take the initial temperature of the ice.  My group got about -.5º C (however you should turn this number to Kelvin [add 273] because when calculating the percent error it will be difficult to work with Celsius whose freezing point is 0º).  
    
    Next, the beaker was placed on a hot plate and the heat was turned on halfway.  The crushed ice was stirred and every minute it's temperature was taken.  After all the ice melted the hot plate was turned on to full value.  We continued to stir the liquid and take its temperature every minute.  The hot plate was turned off after 3 minutes of boiling (100º C).  


    The data you got (temperature and time of the the different phases of the ice) will now be used to make a graph (recall earlier this year when we used excel to make tables and graphs with curves/lines).  Instead of writing a post lab, you will write up a conclusion summarizing the lab and explaining at what temperature the ice melted and boiled at and why this is so.  


    That was about it for today.  This lab is due wednesday.  Our test will be next thursday (02/03/11).  There is a webassign due tomorrow (tuesday the 25), and don't forget to start the second set of book problems which will be due on the day of the test.  We will also be turning in the first set of book problems that we did before finals.


The next scribe will be Ann Marie C.


Good luck!

Heat of Formation Lab

Hey you guys! Today was basically a lab/work day. In the beginning of class we started by having Mr. Lieberman explain to us how to to the questions after the lab. If you don't remeber, or you didn't write them here it is:

Question 2. Use q=m x c x \Delta T

3. The equal opposite of #2 and convert into kilojoules

4. Do this in the same format as #2 from the pre-lab

5. Use 986.1 kJ as your theoretical amount

The lab itself was pretty easy to follow. After massing the the calcium metal, you had to place it in the calorimeter with aproximately 100 mL of water. After doing that a temperature had to be taken find the amount of heat created by the reaction. When everyone was done we did any work that we had not finished. All of this work will be turned in tomorrow. This includes 2 labs, 4 worksheets, book problems, and a webassign. For review here is a link to all of the notes we have done this chapter:

Thermo#1

View more presentations from gbsliebs2002.

Thermo#2

View more presentations from gbsliebs2002.

As well here is a link to all of the answers for the worksheets in case you are unsure about how to do something. http://gbs-moodle.glenbrook225.org/moodle/mod/resource/view.php?id=1054

Well this is it for today! Good luck to everyone for tomorrow. The next scribe is Stephanie K.

Heat of Formation

On Friday, we first started off with a Heat of Combustion lab. So basically, we added 100 mL of water into the calorimeter which is the can. We also had to mass the candle. We then quickly lit then candle and placed the calorimeter over the fire. Then we waited for the water to reach 35 celsius. Then we blew out the candle and recorded the highest temperature. Then we massed the candle once more to find out how much wax was used in the reaction. The lab is due on Tuesday with the rest of the worksheets and book problems. Ok, this is not cool. It's not letting me upload the pictures, so just look at Flickr to see the pictures of the lab.

http://www.flickr.com/photos/hc1011/5334103360/
http://www.flickr.com/photos/hc1011/5333487119/
http://www.flickr.com/photos/hc1011/5334103324/
http://www.flickr.com/photos/hc1011/5333486983/

After doing the lab, we started learning about Heat of Formation! How exciting. So basically heat of formation is the change in enthalpy that accompanies the formation of one mole of a compound from its elements. Here's a link if you want to learn more about enthalpy! http://en.wikipedia.org/wiki/Enthalpy
By the way, there was no notes sheet about this, so I'll try my best at explaining this. So there's an equation for finding the heat of formation.



So for calculating the Hf (heat of formation), you either have to look it up online, or in the chemistry text book. It's on page 209 or pg 638. Here's a table of what it looks like.
I'm pretty sure Mr. Lieberman said we don't have to memorize it. You only need to use this if it's a compound of two elements. For elements, the heat of formation in its stardard state is 0. Here's an example problem.
Calculate Delta H for reaction Kj/mole. Here's something that you should also consider. The form of the compound (example: gas, liquid, solid) matters.

So first, we find each heat of formation. The heat of formation for NH3 is (-46 kj/mole) You multiply it by 4 because in the equation, there are 4 moles of NH3. Then the heat of formation for O2 is just 0, because oxygen is is in its standard state. Then NO2 is (34 kj/mole) then multiply that by 4. Then finally, H20 is (-286 kj/mole) then multiply that by 6. So then when you multiple it out, the 6 moles of H20=-1716, 4 moles of NO2=136, O2=0, and then 4 moles of NH3=-184. You then take the products-reactants=delta H equation, so (H20+NO2)-(0+ -184)= delta H. Then your answer should equal to -1396. Remember, you always have to do products- reactants, not reactants-products.

After our notes, we had a quick discussion on rare earth metal magnets. To answer Mr. Lieberman's question "do magnets release or absorb energy when pulling them apart, or connecting them together", you have to think about it. Well, you have to use up energy when pulling apart the magnets, and then release energy to combine the magnet. Now think about this. Breaking apart two magnets is like breaking bonds which is an endothermic reaction and combining the magnets is like forming bonds which is an exothermic reaction...
It seems like we'll be learning more about this some other time. Okay, that's it for me. The homework is the pre-lab, webassign, post lab of the heat of combustion, worksheets, AND book problems. Cool. The next scribe is AnnMarie C.

Endothermic Vs. Exothermic

Hey there!

Alright so today in class we picked up some papers, they are The Heat of Combustion Lab, the Energy Diagram Worksheet, then finally Energy and Reactions. We also picked up some notes but we didn't go over them in class, instead Mr. Lieberman talked to us about Endothermic and Exothermic graphs. The Exothermic graph looks like this:

The reason why the products are lower is beacuse Exo means out, so the reactants lose energy. The reaction Coordinate is just the run time.

The Endothermic Graph looks like this: (Ignore the Delta E it's supposed to be Delta H)

The reason why the products are higher is because Endo means in, so the reactants gain energy. The reaction is the run time. E is the amount of energy. So I bet you're wondering about the Ea and the Delta H right about now. So I'll explain them to you.

Ea stands for Activation enery. That means the energy used to create the product.

Delta H stands for the energy released. That is the energy that was released by the reactant to make the product. The H stands for Enthalpy. The Delta H and q have equal values. So Delta H=q. The q stands for heat gained (or lost). But we cannot just switch q out for Delta H in the formula (to the right)

Here's the reason why.

q can be determined by an experiment: get the mass the specific heat and the temperature change and you can find q. But Delta H is theoretical, and q is actual. So you cannot find Delta H through an experiment. But one way you can find Delta t is the products (on graph) minus the reactants (on graph) no matter which type of transfer there is, it is always products minus reactants.

Mr. Lieberman showed the class some magnets today and was asking about the energy transfer as they connect and as they detach. What do you think? When put together is the energy absorbed or released, and when detached is the energy absorbed or released? How can you tell? I won't answer, just think about it.

Well I lied earlier when I said we didn't go over the notes... We went over one slide but we didn't even look at it on the projector screen. It was the last slide, the sample problem. All we did was copy down two ways to write the change of energy using a methane reaction. Here are the two ways:

(there's supposed to be a space between the H2O and Delta H).

The first way has the equation that was the reaction and product and to the right of the equation was the release of energy. Since the release had a negative effect (the Methane lost 890KJ) the number is a negative. But if instead of losing 890 it gained it, then the number would be positive.

The second way is to add the energy release directly into the problem. The + 890 KJ is just saying the reaction included 890 KJ of energy released.

The pre-lab for Heat if Combustion is due tomorrow, also there is a web assign due tomorrow as well. The worksheets are, like always, due on the last day of the chapter, which this time is the day of the quiz (1/11/11).

Have you figured out the questions I asked earlier? No? Google it!

The next scribe is...Lauren C.!!!! YAY!

Good Luck.

P.S. This took me a lot of time to work out (I had to redo a ton of stuff) So please like it!

Heat of Fusion of Ice Lab

Hello. Today we started class by going over notes. First we learned about systems and surroundings. There are three systems: open, closed, and isolated. An open system can exchange both matter and energy with the surroundings. A closed system can exchange only energy with the surroundings. An isolated system can exchange neither energy nor matter with its surroundings.

Next we learned about exothermic and endothermic processes. They both can be chemical reactions or physical changes.

An exothermic process loses eneregy. The energy flows out of the system and to the surroundings. Exothermic processes have a negative q.

An endothermic process gains energy. The energy flows into the system from the surroundings. Endothermic processes have a positive q.

Remember: The q of a system is equal and opposite to the surroundings.



Then we did a short lab.

We had to fill a beaker with about 25g of ice and obtain a beaker with about 100mL of warm water. We recorded the initial temperature of the water. Then we mixed the ice with the water and waited for the ice to melt. Once it had melted we recorded the final temperature of the water. We then used the data we had to find the molar heat of fusion of ice.
In order to find the molar heat of fusion of ice, you have to find the q of the water. Next convert the grams of ice into moles. Then make a ratio (J/mol)

The next scribe is Danielle

Specific Heat of Metal Lab

Hey everyone~

So we started class by looking over two of the problems from the worksheet: 8 and 9.

Mr. Liebs explained that problem 8 can be done by setting the amount of energy for copper equal to the amount of energy for water. Knowing that, you can use the law of thermodynamics to solve for the mass of water:

q Copper = q Water

(110)(0.2)(57.5)=(4.18)(2.6)(m)

m= ______

In problem 9, you find the specific heat capacity using the law of thermodynamics again. Once you have calculated the specific heat capacity, you convert moles (because the question asks for you to solve for the molar heat capacity of mercury). You can easily convert it by knowing that the unit for specific heat capacity is J/g C and that the grams of mercury is over 1 mole:

585J=(125.6g)(53.5-20.0) c

c= 0.14 J/gC

0.14J/gC x 200.51g/1 mole = ______

Check the answers on moodle!

After the worksheet, we moved onto our lab: The Specific Heat of Metals Lab

We were assigned to work with our partners and follow the procedures from the lab, which was basically just to...

1. collect a dry test tube and measure the mass

2. mass the test tube with a given metal

http://www.flickr.com/photos/hc1011/5324135903/

3. heat up a beaker full of water until it is boiling

4. Put the test tube inside the boiling water for 10 minutes

http://www.flickr.com/photos/hc1011/5324740964/

5. Fill a calorimeter with about 50 mL of water and measure its volume

6. Then measure the initial temperature

http://www.flickr.com/photos/hc1011/5324740468/

7. After the 10 minutes are up, place the test tube inside the calorimeter and measure its final temperature.

http://www.flickr.com/photos/hc1011/5324136951/

After recording each of these measurements, complete the rest of the lab for homework.

The next scribe is... Paige H.

Thermodynamics

Hey everybody, and unfortunately welcome back from winter break. Today we started class by going over our unit 6 test. Then we took some notes about energy for the rest of the period. If you missed class today then you can get the notes off Moodle.

First we started talking about two forms of energy, kinetic energy, and potential energy. Kinetic energy is produced with motion. Potential energy is stored energy, and it is produced when not in motion. To relate these two, remember that when kinetic energy goes up, potential energy goes down, and vice versa.

Next we talked about temperature and heat. Temperature is the measurement of the kinetic energy of particles, or how fast the particles are moving. Temperature is measured in degrees of celcius, farenheight, or kelvin. Heat is the measure of energy that transfers from an object of high temperature to an object of low temperature. Heat is measured in units of joules, or the stupid american unit of calories, as Liebs tells us. Remember that heat always transfers from high temperature to low temperature. It never transfers the other way around.

To relate temperature and heat we use heat capacity. Heat capacity is different with every object. It is the amount of heat required to raise the temperature of an object by 1 degree celcius. The formula for heat capacity is:

heat supplied (J) / temperature (C) = heat capacity (J/C)

For the homework we need to figure out how much heat is required to raise an object or a substance by a certain temperature. So use these simple formulas to help you out:

heat required = mass x heat capacity x change in temperature
change in temperature= final temperature - initial temperature

Here is a sample problem:
How much energy is required to increase the temperature of 112 g of water from 35 degrees celcius to 80 degrees celcius, if the heat capacity of water is 4.18 J/gC?
Well this pretty easy, all you do is is find the change in temperature by doing 80-35=45. Then you just plug and chug from here.

energy required = (112 g) x (4.18 J/gC) x (45 C).
so the answer is 21067.2 joules

Well that is all for today. The homework tonight is to finish the Heat Calculations Practice worksheet and maybe review a little for the final if you can. The next scribe is Rachel K.