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

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.

Workshop #3 Fun

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

*Equations
∆S_universe = ∆S_system + ∆S_surroundings

a) Reversible:
∆S = q_rev/t_f = (nC∆T)/t_f

b) Irreversible:
∆S = n C ln(T_f/T_i) = n R ln(V_f/V_i)

*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.)

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.

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.

Notes from workshop 1

First, our icebreaker.

We're all underclassmen
We're all sceince-y
We're all in freshmen housing
We're all on facebook
We're all jealous that it's 77 degrees in California.

Now the meat of the workshop

Work done on the system + heat absorbed on the system, is the 1st law of thermodynamics.

ΔE=q+w

In an exothermic reaction the system will lose energy but temperature will rise, because your thermometer measures the released heat.

If work is done by system, the heat is negative (there is a loss of heat) Imagine exercising, you lose heat through your skin.

ΔH=q under constant pressure

Hess' law
ΣΔH(total) = ΣΔH(products) - ΣΔH(reactants)

q=mCΔT

Exam Question 1

Sorry this is a little late, technology hates me!

(a). How much heat is evolved when 1 mole of propane is burned?
(b). What is the enthalpy of combustion per gram of propane?

Hf (CO2(g)) = -394kJ
Hf (H2O(l)) = -286 kJ
Hf (C3H8(g)) = -104kJ