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Chemistry
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CH 301
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Dave A.Laude
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Midterm

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Exam #3 Question Type Notes
Color coordination:
Laude’s Notes: in blue Travis’s Notes: in green Course Packet Notes: in black
1. First law of thermodynamics
You need to know the first law in its various forms and apply it.
∆Euniverse0 = q + w
Energy is conserved; there can be no perpetual motion machine, etc.
∆E = ∆U = 0 [universe is the isolated system]
isolated system
Internal energy deals with the first law of thermodynamics
The sum of heat (q) and work (w)
A measure of the change in heat of a system at constant volume
2. Definition: enthalpy
You will be given a variety of statements about enthalpy (∆H). Some are true, some are not. Be able to list
4 or 5 ways you know ∆H.
Ex: ∆H = q at constant pressure
∆H = ∆E + P∆V, which is a correction for PV work at atmospheric pressure
*hint: if you see ∆H in an equation on the help sheet, it is fair game that you remember it here
∆G = ∆H + T∆S
Enthalpy is the measure of a system’s ability to change entropy of surroundings
∆E = q in a constant volume bomb calorimeter
∆H = q in a constant pressure Styrofoam cup container
When no gas is present, PV is negligible
o so ∆H ~ ∆E (values of these two state functions get close)
When gas is present at constant pressure: ∆H = ∆E + P∆V = ∆E + ∆nRT
3. Signs for thermodynamics quantities
I’ll give you a chemical or physical process and you tell me ∆S, ∆H, ∆G, w and q. You know the drill, “Be the
system”.
Sign Negative (-) Positive (+)
∆H Exothermic Endothermic
q Heat leaves Heat enters
w Work on surroundings (a bomb) Work on system (“arming a bomb”
∆S Increasing order Decreasing order
∆G Reaction happens (spontaneous) Reaction doesn’t happen (nonspontaneous)
4. Definitions: State functions
A gift. You have only been told of two path dependent values, q and w. All the capital letters are state
functions and independent.
Independent of path: internal energy, temperature, pressure, volume, entropy, enthalpy, G, n
Path dependent: q (heat) and w (work)
Definition of state function: its value depended solely on its final quantity minus the initial quantity.
5. Definition: Heats of formation
I’ll give you a bunch of chemical reactions. You have to pick which ones are ∆H
f
Remember:
Must be written as a formation
Must involve elements in standard state on the left
o H2(g), 2 (g),graphite, Hg(l), etc.
Must make a mole of product
Left side of chemical reaction right side of chemical reaction
Elements in standard state 1 mol of 1 product Exam #3 Question Type Notes
6. Definition: Heat capacity
Be able to apply the definition of heat capacity C = q/∆T to a variety of systems.
Note that C is per gram or mole, so the more stuff, the more heat
Note the inverse relationship of C and ∆T; as C gets large, ∆T changes little
o Ex: water
Heat capacity of water is 4.184 J/g*K
C and ∆T are inversely related (at constant q) *q = mC∆T+
7. Calculation: Bomb calorimeter
Classic bomb calorimetric problem
∆H system -∆H surrounding-mC∆T – C ∆cal
-mC∆T water
Ccal bomb calorimeter
Make sure to remember signs. The system is a combustion reaction and is negative.
Remember units. C is given in Joules usually but amounts of energy produced are kJ.
This is plug and chug, don’t make it hard.
Remember to have right amount for answer.
∆U = mC∆T + C ∆ca;
m = 1g/mL
∆T = T finalTinital
C will be provided
cal
Make sure units are correct (asking per gram!)
8. Calculation: Hess’ law and heats of formation
Classic plug and chug.
∆H °= ∑∆H °- ∑∆Hf °
rxn fproduct reaction
This can be especially easy. Remember to multiply by coefficients and double check signs.
Methane is CH 4
°
In this method, the molecules become elements in standard states (∆H ). f
9. Calculation: Hess’ law and combined reaction enthalpies
°
One of the harder questions on the test. Do at the end. I’ll give you a collection of ∆H rxnand reactions.
Combine them by flipping and multiplying through to cancel and yield ∆H rxn°of interest.
*hint: tragically, fractions are found in the coefficients you need to use
To make reaction backwards, multiply ∆H by -1.
10. Calculation: statistical mechanics determination of internal energy
I’ll give you a number of molecules. You tell me the kinetic energy on number of modes of motion for
translational, rotational, vibrational.
*hint: apply to a single molecule first
Translation = 3 3/2kT
Rotation = 2 kT (linear)
= 3 3/2kT (nonlinear)
Vibration = 3N – 5 (linear
= 3N – 6 (nonlinear)
After you find the energy for one molecule, multiply through by number of molecules in problem.
E = 1/2kT
Number of modes per molecule = 3N
K is for individual particles
R is for moles
Ex: SF6
3 modes * 7 molecules = 21modes
Translational = 3 (nonlinear)
Rotational = 3 (nonlinear)
Vibrational = 3(7) – 6 = 15 Exam #3 Question Type Notes
The average energy per motion is E = 1/3kT
Temperature is in Kelvin
K is Boltzmann constant
11. Calculation: Bond enthalpies
Classic calculation following Hess’ law.
∆H rxn°= ∑∆BE reaction∑∆BEfproduct
Make sure you draw out the structures so you don’t make a mistake on the number of bonds in a
molecule
Make sure you do stuff on left – stuff on right
As always, easy problems with lots of math to mess it up. Walk it carefully and neatly.
In this method, the molecules become gas atoms [only for gases].
12. Calculation: Work calculation
Classic work calculation
W = -P∆V = -∆nRT
I’ll give you a chemical reaction. You determine ∆n = nqright – q left, then multiply by RT.
Remember
RT = 2.5 kJ at room temperature
If ∆n grows, work done on surrounding is negative
If ∆n shrinks, work done on system is positive
Work done by system is negative, work done on system positive.
If Pext 0, no work is done [can happen in an open container with 1 atm of external pressure].
13. Definition: Interna

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