Wednesday, March 21, 2007

Figuring out E* at different pH's of water

There was one exam problem in which it asked you to calculate the E* of 2H+ + 2e- --> H20 at pH of 5. Will this type of problem be on the test, and if so what is the basic way to solve this problem.

Wednesday, March 7, 2007

Workshop post

sorry for the late post but here it is:

Oxidizing agent is in Cathode. Reducing agent is in Anode.

Negative ions (such as NO3-) leave salt bridge to the ANODE to prevent build up of positive charge (which would stop the system)

Positive ions (such as K+) leave salt bridge to the Cathode to prevent build up of negative charge (which would stop the system)

Electrons flow from anode to cathode.

finally (i know its been saiid but its important): REDUCTION POTENTIAL IS THE ABILITY OF SOMETHING TO BE REDUCED!!!!!!!!! HIGHER REDUCTION POTENTIAL WILL OXIDIZE ANYTHING WITH A LOWER REDUCTION POTENTIAL. IT IS A REALLY GOOD OXIDIZING AGENT. (and vice versa for a high oxidation potential).

Tuesday, February 27, 2007

Workshop 5 notes!

oxidizing agent- is being reduced, looses electrons + charge
reducing agent- gets oxidized, gains electrons - charge

If reduction potential is high, there is a higher potential to get reduced. They are also good oxidizing agents

If reduction potential is low, it will reduce anything and it will get oxidized

Ecell= Ecathode - Eanode

Wednesday, February 21, 2007

Chemistry Chapter 10 exam question

For ammonia (NH3) the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 j/kmol.
a. will the NH3 spontaneously melt at 200 K?
b. What is the approximate melting point of ammonia?


Question 47 from chapter 10.

Monday, February 12, 2007

Workshop #3 Fun

Ok, so here’s some fun information we learned today:

*Equations
∆Suniverse = ∆Ssystem + ∆Ssurroundings

a) Reversible:
∆S = qrev/tf = (nC∆T)/tf

b) Irreversible:
∆S = n C ln(Tf/Ti) = n R ln(Vf/Vi)

*For entropy changes: (like in the exploration in the workshop)
If the surroundings don't really change temperature, the process will be reversible. (exp. a(I))
If the surroundings do change temperature noticeably, the process will be irreversible. (a(II))
The same goes for the system. (the iron piece)
(The wonderful example of the millionaire vs. the college student with $50)

*Entropy is independent of path of process
There will be the same amount of entropy transferred if it goes back and forth randomly before it reaches the final, or if it goes directly to the final entropy.
(It doesn't matter how you get to the final point, just as long as you get there.)

Thursday, February 8, 2007

Week #2 Notes

Hey everyone,

I hope these notes help you out:

* A carnot cycle is the most efficient way heat can be converted to work. It is ideal and never actually occurs in real life because of friction and the loss of heat to the environment. The cycle is composed of two isothermal (no temp. change) and two adiabatic (no heat flow) reversible processes. These processes are either expansions (Won is negative) or compressions (Won is positive).

* A reversible process is one that can be reversed because it is ideal and none of the energy or heat involved is mistakenly lost forever due to friction and loss of heat to the environment. Reversible process never actually occur in real life.

* The efficiency of a carnot cycle is measured with the equation: efficiency = 1 - (Tlow/Thigh).

* Remember that all of the equations D-Rock wrote on the board should be used for engines. When you have a refrigerator problem, you should use these (I hope these are right):
- AB... Won = -qAB = nRTln(VA/VB)
- BC... Won = -qBC = Cv (Thigh - Tlow)
- CD... -Won = qDC = nRTln(VC/VD)
-DA... -Won = qDA = Cv (Tlow - Thigh)

* Work is always measured in terms of the work done on the system by the surroundings while heat is always measured in terms of heat gained by the system from the surroundings.

* Heat flows spontaneously from high to low temperature areas. If you want to make heat flow from low to high (like with a refrigerator), work needs to be done on the system.

Tuesday, February 6, 2007

Week #1 Highlights

Basically, to recap what we talked about in workshop last week:

-Laws of thermodynamics are broken down into the "system" and its "surroundings"

-The First Law of Thermodynamics states that internal energy for a system is the sum of heat absorbed by the system, and work done on the system (ΔE=q+w)

-Changes in a thermodynamic state happen when processes occur between the system and the surroundings.