Transcript Document

THERMODYNAMICS
Moses Kool and Miguel Grizzly Bear
CONSERVATION LAWS
Energy and mass are interchangable
 Slopes: q=ms∆T or q=mC∆T
 q=quantity of energy
 m=mass of material
 C=s=constant=specific heat capacity
 ∆T=change in temperature
 Slopes are kinetic energy and plateaus are
potential energy

THERMO LAWS


1st law= energy cannot be created nor destroyed
2nd law= Universal randomness is always on the
rise. Irreversible processes result in increase in
randomness.
BRIEF INTRO TO HEATING CURVES

Plateaus: q=m∆H
q=quantity of potential energy
 ∆H= enthalpy of fusion or vaporization


Enthalpies of Water Szzzzzzzzz
∆Hf= +6.02 kJ/mol
 ∆Hv= +40.7 kJ/mol

INTERMOLECULAR BONDS
-When energy is applied and physical states are
altered, it is the intermolecular bonds that are
being broken
-Breaking requires input. Bonds that form release
energy
http://employees.csbsju.edu/hjakubowski/classes/ch111/olsg-ch111/statesmat/hbonds.gif
∆E WITH WORK AND HEAT
∆E = q + w
 q is energy liberated or added to the system while
w is the work done on or by the system
 The change in energy depends on the amount of
heat added or released from the system and the
amount of work done on or from the system
 When heat is added or work is done to the system
the energy increases
 Heat is hot!!!!!

A -25oC 5g block of ice is heated to the point where it is 150oC. What is the
quantity of the energy involved in this process?
H2O: Csolid=2.06 J/goC
Cliquid= 4.18 J/goC
ΔHfusion= 6.02 kJ/mol
Melting Point=0oC
ΔHvap= 40.7 kJ/mol
Boiling Point= 100oC
(5g)(2.06 J/goC )(25oC)=257.5 J= .2575 kJ
(.277 mols)(6.02 kJ/mol)=1.667 kJ
(5g)(4.18 J/goC)(100oC)= 2090 J= 2.09 kJ
(.227 mols)(40.7 kJ/mol)= 9.2389 kJ
(5g)(2.03 J/goC)(50oC)= 507.5 J= .5075kJ
ΔH=13.76 kJ
Cgas= 2.03 J/goC
CALCULATING ∆H AND ∆S IN REACTIONS
For a particular equation:
 CH4 + O2  CO2 + 2H2O
the ∆Hreaction is
 ∆Hreaction=∑ ∆Hproduct - ∑ ∆Hreactant


∆Hreaction= [1 mol(-395.5 kJ/mol) + 2mol (-285.83)]
–[1mol(-74.8) + 2 mol (0)]= -890.36 kJ
∆S IS ENTROPY
∆Sreaction=∑ ∆Sproduct - ∑ ∆Sreactant
 ∆S = [1mol(213.6 J/K) + 2mol(69.91)] –
[1mol(186.3) + 2 mol(205)] = -242.85 J/K
 Any change in energy increases the amount of
energy in the universe

SPONTANEOUS

Spontaneous = natural, happens normally in
nature
+ ∆H
-∆H
+ ∆S
Entropy driven.
Can be
spontaneous at
high temp.
Always
spontaneous
- ∆S
Never
spontaneous
Enthalpy Driven.
Can be
spontaneous at a
low temp.
GIBBS FREE ENERGY

∆G= ∆H - T∆S
∆H = enthalpy
 ∆S = entropy
 ∆G = free energy

∆G= ∆Gº + RTlnQ
 Q = rxn quotient
 From that we derive ∆Gº = -RT ln K

Equilibrium
Productfavored
Reactantfavored
∆G=0
Large - ∆G
Large +∆G
k= 1
k>> 1
k<< 1
MORE GIBBS
• 0 = ∆Gº + RT lnKa
• Q = e- ∆G/RT
• R is a constant
•T is the absolute temp
•Ka is the equilibrium constant
•Which is the ratio of the concentration of the products raised to
their respective powers all divided by the concentrations of the
reactants raised to their respective powers
We’re All Free!!!!
**Calculate ΔH, ΔS, and ΔG at 298K for the reaction: 2PCl3 + O2 
2POCl3
ΔH= 2(-542.2) - 2(-288.07)= -508.26 kJ/mol
ΔG= 2(-502.5) – 2(-269.6)= -465.8
ΔS= 2(325) – [2(311.7) + 205]= -178.4 J/mol K
**At equilibrium, what will be the temperature of the system?
ΔG= ΔH - T ΔS
0= ΔH - T ΔS
T ΔS= ΔH
T= (-508.26 kJ/mol)/(-.1784 J/ mol K)
T= 2848.99 K
**What is the K for this reaction occurring at 2009 K?
ΔG2009= -508.26 kJ/mol – 2009(-.1784 J/mol K)
-149.85 = (-8.314 J/ mol K)(2009) lnK
.00897 = lnK
e.00897= K
K=1.00
ΔG2009= -149.85
HESS’ LAW


In Hess’ law reactions that are done in one or
more steps can have the enthalpies changes
added up from each individual step to calculate
the total ∆H
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
2H2O(g)  2 H2O(l)
ΔH = −802 kJ
ΔH = − 88 kJ
--------------------------------------------------------------------CH4(g) + 2 O2(g) + 2 H2O(g)  CO2(g) + 2 H2O(l) + 2 H2O(g)
ΔH = − 890 kJ
Example From Chemistry: The Central Science, Ch.5 Sc. 6

Sometimes you may have to flip the equation, then you change the sign of
the enthalpy. Later on in this marvolous presentation you will be honored
with a example
Calculate the change in enthalpy for the reaction:
P4O6 + 2O2  P4O10
Using these enthalpies for reactions:
P4 + 3O2  P4O6
ΔH = -1640.1 kJ
P4 + 5O2  P4O10
ΔH = -2940.1 kJ
**Flip the first equation, as the product of that reaction is the reactant in our
problem reaction. At the same time, the sign in front of the ΔH needs to be
reversed as well.
P4O6  P4 + 3O2
ΔH = 1640.1 kJ
** The P4’s cancel on each side and 3O2’s also cancel. Thus the net reaction will
look like the problem reaction. Now that the reactions are equal, add the
enthalpies together to get the total change in enthalpy.
1640.1 kJ + -2940.1 kJ = -1300 kJ
LAWS OF THE THERMO-MAN
First law of thermodynamics states that any
energy lost by a system must be absorbed by its
surroundings and vice versa
 Internal energy is the sum of the kinetic and
potential energy in a system
 The change in energy …

∆E = Efinal - Einitial
 If the ∆E is positive then energy is gained from the
surroundings.
 If the ∆E is negative then energy is released into the
surroundings

**Calculate the ΔH for this reaction in kJ/mol.
HHH H
2 H-C-C-C-C-H + 13 O=O  8 O=C=O + 10 H-O-H
HH HH
20(413 kJ/mol) + 3(348) + 13(498) = 15778 kJ/ mol
16(799) + 20(463)= 22044 kJ/ mol
15778-22044= -6266 kJ/mol
http://www.justosphere.com/files/image/007logo-1.jpg
ENTHALPY
The word enthalpy come from the Greek word
enthalpein, which means hot in this famous
romance language
 It is the amount of heat that is absorbed or
released by a reaction
 Enthalpy is defined by H = E + PV
 Enthalpy is a state function means that it can
change. Thus the change in enthalpy equation is
∆H = ∆E + P ∆V
 Basically it is the thermal internal (it rhymes)
energy of a system (no it doesn’t)

ENTHALPIES OF FORMATION
The enthalpy of the reaction is the heat of a
reaction
 A negative ∆H would cause a exothermic reaction
 A positive ∆H would cause a endothermic
reaction
 The change in enthalpy in a reaction is equal to
the magnitude, but for a opposite sign of the
enthalpy the reaction is reversed.

CALORIMETRY
Calorimetry is the measure of heat flow
 C is heat capacity
 The heat capacity for an object is the amount of
heat energy required to raise the objects
temperature by 1 ºC

20g of water is heated up from 20oC to 85oC. Calculate the ΔH in kJ/mol for this
process.
q=mc(ΔT)
(20g)(65oC)(4.18 J/goC)= 5434 J/g
(5434 J/g)(18 g/mol)(.001 kJ/J)= 97.812 kJ/mol
---------------------------------------------------------------------------------------------------------------
“Random Mike Thing”
-qfuel=qnet · qwater
(-773261J)(12.25g/mol)=(1250J/K)(23.1-14.7) + m(23.1-14.7)(4.18J/g°C)
m= 270677.2 g that is a lot of agua!
The point of that was, -qrxn= qcal
THERMODYNAMICS DONE RIGHT WITH
MOSES AND MIKE