Catalyst - MAEDA AP Chemistry
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Transcript Catalyst - MAEDA AP Chemistry
Catalyst
Egg Throwing
Justify – TPS
Which one of these represents a spontaneous process and which of
these represents a process that does not occur spontaneously?
Lecture 5.6 – 2nd and 3rd Law of
Thermodynamics
Today’s Learning Targets
LT 5.12 – I can analyze the Second Law of Thermodynamics and
how this relates to the idea of spontaneous process, entropy, and
whether a process is reversible or irreversible.
LT 5.13 – I can analyze the Third Law of Thermodynamics and
how it relates to entropy of a system.
LT 5.14 – I can interpret the entropy of a chemical reaction and
how standard entropies can be use to calculate the overall entropy
of the system.
Constraints of the 1st Law
1st Law and ΔH allows us to clearly study energy, but it has
limitations.
ΔH does not provide a good description of whether a reaction is
likely to occur
There are examples of both positive and negative ΔH values leading
to reactions that are favorable.
Spontaneous Process
A spontaneous process is one that proceeds without any outside
assistance.
A non-spontaneous process is one that requires assistance in
order to occur
Spontaneity does not mean a reaction occurs “fast”. Only tells us
the energy of a reaction
Spontaneous
Non – Spontaneous
Endothermic and Spontaneous
Why are reactions/processes that have a +ΔH able to occur (e.g.
reactions that feel cold)?
Only reactions that have –ΔH should be favored by the universe
The answer is due to:
Reversible vs. Irreversible Processes
Reversible Process is any process that can be restored to its
original state without no overall net change.
Irreversible Process is a process that cannot be simply reversed
to return system to its original state
Irreversible Process
Hot
Cold
Not Possible!
Isothermal Process
Major Idea
All real processes are irreversible
Therefore, all spontaneous processes are irreversible
ΔS and Spontaneity
Need to be able to characterize a reaction that we are unfamiliar
with as spontaneous or non-spontaneous
Entropy is a state function so:
ΔS = Sfinal – Sinitial
For a process under constant pressure, ΔS can be calculated by:
qrev
S
T
qrev represents the heat required to make the process reversible
ΔS and Phase Changes
Phase changes are isothermal processes.
Imagine examining the melting of a
substance. Therefore, we can say:
qrev = ΔHfusion
This means that for phase changes:
Sphase change
H phase change
T
Class Example
Elemental mercury is a silver liquid at room temperature. Its
normal freezing point is -38.9 oC, and its molar enthalpy of fusion
is ΔHfusion = 2.29 kJ/mol. What is the entropy change of the
system when 50.0 g of Hg (l) freezes at the normal freezing point?
Table Talk
The normal boiling point of ethanol, C2H5OH, is 78.3 oC,
and its molar enthalpy of vaporization is 38.56 kJ/mol. What
is the change in entropy in the system when 68.3 g of
C2H5OH (g) at 1 atm condenses to liquid at the normal
boiling point?
ΔS and the 2nd Law of Thermodynamics
Entropy is not conserved (unlike energy in the 1st law)
Example: 1 mole of water freezes at T < 0 oC
H2O (l) H2O (s)
What is ΔSsys? Assume the surroundings are around 310 K.
Ssystem Ssurrounding Suniverse 0
For all spontaneous processes
Did Lisa Simpson Violate the 2nd Law?
Homer claims that Lisa has violated the laws of thermodynamics
With your group, read the short overview of perpetual motion.
Applying your knowledge of thermodynamics, draw a conclusion
about whether the laws have been violated or not.
ΔSsurroundings and ΔHsystem
As we lose energy, entropy increases. So, we can say that:
Ssurr - Hsys
Furthermore, we know that entropy increases as T increases, so:
1
Ssurr
T
Therefore, it is possible to conclude that:
Ssurr
H sys
T
Two Important Equations
Ssurr
H sys
T
When does S equal 0?
When does ΔSsurr equal 0?
S klnW
3rd Law of Thermodynamics
The 3rd Law of Thermodynamics claims that the entropy of a pure
crystalline substance at absolute zero is 0.
There is only one possible microstate because all motion has
ceased, which means there is 0 entropy in the system.
This is one of the proofs that absolute zero is a quantity that must
exist.
Entropy Change for Chemical Reactions
We use calorimetry to measure ΔH
We do not have a nice way to measure ΔS in a reaction
We do know that every substance must have a point where S = 0
due to the 3rd Law
As temperature increases from 0 K, the value of S also increases.
We therefore can determine standard entropies (So)
Standard Entropies is the entropy gained by taking 1 mol of
perfect crystalline substance at 0 K and bringing it to standard
conditions
Measured in J/(mol x K)
Important Points on So
Unlike ΔHf, So is not zero for elements
2. So is greatest in gas, than liquids, and least in solids
3. So increases as substances have larger molar masses
4. So increases as the number of atoms in a chemical formula
increases
1.
Increased number of possible microstates
Calculating ΔSo
We can use So values to calculate ΔSo through the equation:
S nS (products) mS (reactants)
o
o
o
Class Example
For the reaction:
N2 (g) + 3 H2 (g) 2 NH3 (g)
Calculate ΔSo for the reaction.You know that So (N2) = 191.5
J/(mol x K), that So (H2) = 130.6 J/(mol x K), and that So (NH3)
= 192.5 J/(mol x K).
Table Talk
Calculate ΔSo for the following reaction:
Al2O3 (s) + 3 H2 (g) 2 Al (s) + 3 H2O (g)
You know that So (Al2O3) = 51.00 J/ (mol x K), So (H2) = 130.6
J/(mol x K), So (Al) = 28.32 J/(mol x K), and So (H2O ) =
188.8 J/(mol x K).
Entropy Changes and Spontaneity
We now have a tool to calculate ΔSsystem, but how do we combine
this with ΔSsurroundings and ΔSuniverse?
We know that:
H sys H rxn
T
T
o
Ssurr
We then can add surroundings and system to determine if the
reaction is spontaneous.
Class Example
For the reaction:
N2 (g) + 3 H2 (g) 2 NH3 (g)
Determine whether or not the reaction is spontaneous under
standard conditions.You know that ΔHf for NH3 is -46.19 kJ.
Additionally you know that
So (N2) = 191.5 J/(mol x K)
So (H2) = 130.6 J/(mol x K)
So (NH3) = 192.5 J/(mol x K).
Table Talk
For the following reaction:
Al2O3 (s) + 3 H2 (g) 2 Al (s) + 3 H2O (g)
Determine whether or not the reaction is spontaneous under
standard conditions. You know that
ΔHf (Al2O3) = -1669.8 kJ/mol and ΔHf (H2O) = -241.82
kJ/mol
You know that:
So (Al2O3) = 51.00 J/ (mol x K)
So (H2) = 130.6 J/(mol x K)
So (Al) = 28.32 J/(mol x K)
So (H2O ) = 188.8 J/(mol x K).
White Board Questions
Question 1
Which molecule should have a higher So? Why?
C2H6 or C2H2
Question 2
Calculate ΔSo for the following chemical reaction:
C2H4 (g) + H2 (g) C2H6 (g)
So (C2H4) = 219 J/ (mol x K)
So (H2) = 114.60 J/ (mol x K)
So (C2H6) = 229.5 J/ (mol x K)
Question 3
Calculate ΔSo for the following chemical reaction:
N2O4 (g) 2NO2
So (N2O4) = 304.3 J/ (mol x K)
So (NO2) = 240.45 J/ (mol x K)
Question 4
Determine if the following reaction is spontaneous at 298 K:
Be(OH)2 (s) BeO (s) + H2O (g)
So (Be(OH)2) = 50.21 J/ (mol x K)
So (BeO) = 13.77 J/ (mol x K)
So (H2O) = 188.83 J/ (mol x K)
ΔHf (Be(OH)2) = -905.8 kJ/mol
ΔHf (BeO) = -608.5 kJ/mol
ΔHf (H2O) = -241.82 kJ/mol
Question 5
Determine if the following reaction is spontaneous at 298 K:
3 CH3OH (g) + O2 (g) 2 CO2 (g) + 4 H2O (g)
So (CH3OH) = 237.6 J/ (mol x K)
So (O2) = 205.0 J/ (mol x K)
So (H2O) =188.83 J/ (mol x K)
So (CO2) = 213.6 J/ (mol x K)
ΔHf (CH3OH) = -201.2 kJ/mol
ΔHf (CO2) = -393.5 kJ/mol
ΔHf (H2O) = -241.82 kJ/mol
Closing Time
Read 19.1, 19.2, and 19.4
Do book problems: