Enthalpy Thursday

Today was a pretty easy day. We started off class by going over some new notes.

We learned that because most chemistry reactions take place at constant pressure, we can say the a change in enthalpy is equal to the heat supplied. The symbol for enthalpy is "H" so we can write: (the change in)H=q. This is good to know becasue it is not possible to measure the the change of H in a lab but we can measure q which is the same as H. Yay!

We also leanred about how exothermic and endothermic reactions relate to enthalpy. For an exothermic reaction, the heat released corresponds to a decrease in enthalpy, sooo...(the change in)H<0.>

For an endothermic process, an input of heat corresponds to an increase in enthalpy, which means (the change in)H>0. An example of this is photosynthesis.

We also learned that all chemical reactions either release or absorb heat. For exothermic reactions, the heat is written on the right side of the arrow, like it is a product: reactants --> products + energy as heat (H).

For endothermic reactions, the heat is written like a reactant and written on the left side of the arrow: reactants + energy as heat (H) --> products.

Mr. Lieberman also taught us how to use stoich in an equation to find the change in H.

We solved the sample in our notes.

When 1 mole of methane (CH4) is burned,

890 kJ of energy is released. Calculate the

(change in)H for a process that uses 5.8

grams of methane.

First, we need to make sure that it s a therodynamic reaction. The problem tells us that energy is released which means it is an exothermic reaction.

Then we write the chemical equation with the energy written as a product in this case:

CH4 + O2 --> CO2 + 2H20 +890kJ

Finally, we use stoich to solve the equation:

5.8 grams CH4 x 1mole CH4/12 grams x 890kJ/1 mole= 322.63kJ

The last ratio is the ratio of the amout of kJ used per mole which they give us in the

problem. You can write the amount of energy as a negative because it is an

exothermic reaction, but you don't need the negative because the problem already

says that the energy is "released."

So that is pretty much the notes we went over in class today. Mr. Lieberman gave us the rest of the class period to work on the labs from Tuesday and Wednesday which were due at the end of the period. For homework we had a WebAssign, a pre-lab, and a worksheet.

The next scribe will be Mollie M.

Systems, Surroundings, and Ice

Today, as we walked through the door, we picked up two sheets from Mr. Lieberman's desk:
A brand new notes sheet and a worksheet for homework to go with it.

The lesson for the day was all about ENERGY and where it gets distributed during a reaction. The reaction itself is called a system, which is what we look at for data. The surroundings are made up of everything else in the environment outside the system.
-There are then three types of systems that we went over quickly
-Open system: can exchange matter AND energy with the surroundings (ex: open beaker)
-Closed system: can exchange ONLY energy with the surroundings while matter remains fixed (ex: sealed flask)
-Isolated system: can't exchange energy or matter with its surroundings (ex: thermos)

Then there's the exothermic and endothermic reactions. Exothermic is when energy flows OUT of a reaction due to temperature difference in the system and surrounding (negative q). Endothermic is when energy flows INTO a system from the surroundings (positive q).
*Qsurroundings = -Qsystem* They are equal and opposite for distribution of heat/energy

these are graphs like the ones we made in class:


(the activation energy is the energy to start the reaction)













After notes, we did a mini lab called Heat of Fusion of Ice with our partners.
Each group got a 100 ml beaker which we needed to mass before putting ice in it and massing it with the ice. My partner, Kathryn, then got warm water from Mr. Lieberman which measured to be about 71 ml with our data.
The initial water temperature was 50.6 degrees C and the final temperature with the water and melted ice was 22.6 degrees C.
After finishing the procedure, we were assigned to find the molar heat of fusion of ice using ratio. Before that, however, you need to determine the MOLES of ice that were melted.

(this will be due friday, while we have a webassign due tomorrow with a lab and another webassign due friday. whew..)
The next scribe will be...Rebecca N.

Mystery Metal

Happy Tuesday!

Today, we started off class my going over the last two questions on last night's worksheet, which you can find...here. The answers are here! I shall explain the last two for any of you who are still wondering...

8. The key to this problem is that the heat lost by the copper is equal to the heat lost to the water. In the problem, they give you all three variables you would need to find the heat lost (the mass, the specific heat capacity, and the temperature change). However, to find the heat lost by the water, you still need to figure out the mass of the water. So, you set the two equations equal to each other, remembering that the heat lost by both of them is the same.

(110 g)(.20 J/g°C)(57.5°)=(m)(4.18 J/g°C)(2.6°C)

m= 116.4 g H20

9. Finding the specific heat capacity is simple in this question. You just must use the knowledge they give you to solve for that variable.

585 J=(125.6 g)(23.5°C)(shc)

shc=.14 g/°C

Then, using that answer, you just convert it to moles using the molar mass!

(.14 g)(200.59 g)=30.67 J/mol

Next, we moved onto the lab called the Specific Heat of Metal Lab. Mr. Liebs (follow him on Twitter-...@davejlieberman)checked in all our pre-labs while we set up the lab that was to be worked on with our partners (mine being Michelle). Get the lab here if you missed it.

For the pre-lab, number one asked you to calculate the energy transfer.

1. (300 g)(4.18 J/g°C)(37°C)=46,398 J.

Number two asked why you shouldn't eat snow if you're trapped in Alaska in place of water. That is because your body uses up so much energy trying to melt the snow that it dehydrates you. All of your body's energy is being transferred to the snow! So just remember, melt the snow before you attempt to consume it.

First, we turned up the hot plate all the way so that we could go the water in the 600 mL beaker to a boil.

Then, we got a dry test tube and measured the weight, which was 12.01 g. We took a random container from the counter, which was labeled with a "B" and poured all the contents into the test tube.

We then measured the total amount, which was 60.0 g.



While we were waiting for the water on the hot plate to boil, we got a calorimeter, which strongly resembled a styrofoam cup, from the counter and poured 51 mL of water into it. The temperature of the water was 22.6°C. We placed the top over the calorimeter to insulate the water.

While we waited for the water to boil, I snapped a couple of pictures of some of my peers, which they promptly made me delete. I thought they were good! But, whatever...

When the water finally came to boil, Michelle and I, along with the other duo at our table, Ben W. and Aaron, placed our test tubes filled with mystery metal into the beaker.

After waiting, 10 minutes, we took the beakers out carefully, turned the hot plate off and spilled the metal into the calorimeter. While it was still hot, we recorded the new temperature of the water inside the calorimeter. It came out to be 30.6°C.

We then cleaned up our lab tables and head back to our desks to try out the post-lab.

I don't know why there's this weird space below, but I don't know how to fix it, so...this is my data table!

My Data Table:

Mass of test tube	12.01 g
Mass of test tube & metal	60.0 g
Label on container	B
Volume of water	51 mL
Initial temp of water	22.6°C
Initial temp of boiling water	100°C
Final temp in calorimeter	30.6°C

These answers are according to my data, but I hope it helps!

1. The heat gained by the calorimeter

30.6°C-22.6°= 8°C

2. The heat lost by the metal

100°C-30.6°C= 69.4°C

3. The specific heat of the metal and what it is- for this question, you must set the heat lost by the metal to the heat gained by the calorimeter because you do not have the specific heat for the metal.

(51 g)(8°C)(4.18 J/g°C)=(47.99 g)(69.4°C)(specific heat)

specific heat= .512 J/g°C

This number is not exactly close to any metal on the worksheet, but it is closest to zinc and copper, and since the metal looked more silvery and not reddish, I assume it was zinc.

4. Percent error. Which is the absolute value of actual-theoretical divided by actual times 100 if anyone forgot.

|.385-.512|/.385*100= 33%

Yikes! This is obviously not a very good percent error, but there were things in the lab that could've been done slightly wrong. We may have not kept the thermometer in the water for long enough, or maybe we did a bad massing job. Some of our measurements could've been off, but in the end, we were able to figure out our mystery metal.

That's about it for now! I hope you guys had a great break and happy 2011!

The next scribe is...my fabulous partner Michelle T.! Have fun :)

SO. whats up.

Today in chem we got our tests back. Not too many of us did very well on that. DLiebs himself said the average score was "not too good". But still, life goes on.

After taking our questions on the test and refusing to give any sort of curve on the test, Liebs began our new unit. Thermodynamics.

We began by discussing energy. We divided energy into kinetic energy (energy due to motion) and potential energy (energy due to position). To demonstrate this concept, Liebs used the example of a ball thrown in the air; as the ball slows on its ascent, it loses the initial kinetic energy it had in movement, and gains potential energy with its height. As it begins its decent it loses potential energy and gains kinetic energy as it falls faster and faster.

New Law: Conservation of Energy. Same deal as most other conservation laws. Energy cannot be created or destroyed; it can only be transformed. There is a limited amount of energy in the universe, but there's a lot of it.

We also learned that heat and temperature are two different things.

Temperature is the measure of kinetic energy of particles' random motion. (Measured in °C, °F or K)

Whereas heat is the total amount of energy transferred from an object of high temperature to one of low temperature. (Measured in J or Cal)

Another good thing to know is heat capacity. Heat capacity is the amount of heat required to raise a substance by 1° celsius. Heat capacity is different for each sample. For example a metal has a lower heat capacity than water, and a larger sample has a greater heat capacity than a smaller sample.

Specific heat capacity is a physical property measured in J/g°C. Basically, it is the amount of heat required to raise 1 gram of a substance by 1°C. (denoted by C_p

The Ultimate formula is as follows: q=m*C_p* ΔT.

Or heat = mass * specific heat * change in temperature

Thats it. I need to make cookies.

Good night y'all. Do the prelab and the worksheet he gave us.

Kathryn J can scribe next.