Chapter 20: Thermodynamics - Linn–Benton Community...
Transcript of Chapter 20: Thermodynamics - Linn–Benton Community...
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Chapter 20: Thermodynamics
Thermodynamics is the study of energy
(including heat) and chemical processes.
First Law of Thermodynamics: Energy cannot be
created nor destroyed.
Euniverse = Esystem + Esurroundings
∆Euniverse = ∆Esystem + ∆Esurroundings
If ∆Euniverse = 0, then –∆Esystem = ∆Esurroundings
The first law of thermodynamics explains the energy exchange between the system and surroundings.
Does it explain the direction a reaction proceeds toward equilibrium?
For systems under constant pressure and no work, ∆H = ∆E. Does the sign of ∆H tell us the direction a reaction proceeds toward equilibrium?
Let’s try the Second Law of Thermodynamics.
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Second Law of Thermodynamics: The total entropy (S) of a system and its surroundings increases for spontaneous reactions.
A spontaneous reaction is one that occurs naturally under certain conditions without a continuous input of outside energy.
Note: If a reaction is spontaneous in one direction, it will be nonspontaneous in the reverse direction.
Consider the following diagram:
When the valve is opened some of the gas particles move spontaneously into the other side as shown in B.
Will all of the molecules spontaneously move back as depicted in A?
In B of the diagram above, the gas particles occupy twice the volume thus the gas particles can exist in twice as many locations. Thus greater disorder.
Disorder
Sdisorder > Sorder , in other words, the entropy of
disordered matter is greater than the entropy
of ordered matter.
Examples to consider:
Ice melting � disorder ∆Ssystem increases
Water freezing � order ∆Ssystem decreases
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All processes occur spontaneously in the
direction that increases the total entropy of
the universe.
∆Suniverse > 0,
Therefore,
∆Ssystem + ∆Ssurroundings > 0
Let’s consider two rxns, one exothermic and one endothermic.
4 Fe(s) + 3 O2(g) � 2 Fe2O3(s) + Heat
Now, let’s consider another reaction.
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Ba(OH)2.8H2O(s) + 2 NH4NO3(s) + Heat �
Ba2+(aq) + 2NH3(g) + 2 NO3
-(aq) + 10H2O(l)
Therefore changes in entropy of the system and the surroundings must be important to the spontaneity of reactions.
Third Law of Themodynamics: A perfect crystalline substance at 0 K has 0 entropy.
Perfect means that all of the particles are flawlessly aligned in the crystal with no imperfections.
Standard Molar Entropies (So) at 298 K are on pg A-5 of text.
Units: J/mol-K
Note: Standard Molar Enthalpies (∆Ho) are also located on pg A-5.
Let’s Make Some Comparisons
Temperature versus Entropy
Copper metal T(K) So (J/mol-K)
273 31.0
295 32.9
298 33.1
What general statement can you make about this comparison?
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Physical State Compared with Entropy
Values
Substance So (J/mol-K)
Ba(s) 62.5
Ba(g) 170.3
Br2(l) 152.2
Br2(g) 245.4
C(s) 5.7
C(g) 158.0
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Comparison of the Solid State with
Aqueous Solution
Substance Entropy (J/mol-K)
NaCl(s) 72.1
NaCl(aq) 115.1
AlCl3(s) 167
AlCl3(aq) -148
CH3OH(l) 127
CH3OH(aq) 132
Note: When gases are dissolved in liquids the system decreases in entropy. Explain.
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Compare Molar Mass and Entropy
Substance MW (g/mol) Entropy (J/mol-K)
Cu 63.6 33.1
Au 197 51
Ag 107.9 43
What general statement can you make about
this comparison?
Compare Compound Complexity with
Entropy
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Substance Entropy (J/mol-K)
MgCO3 65.9
Na2CO3 139
C2H6 229
CH4 186
What general statement can you make about
this comparison?
Calculating the Change in Enthalpy of a
Reaction
Recall the equation:
∆Horxn = Σ n∆Ho
products – Σ n∆Horeactants
We can use a similar equation to calculate the
change in entropy of a reaction.
∆Sorxn = Σ nSo
products – Σ nSoreactants
Problem Solving
Calculate the change in entropy of the combustion
of methane.
H2O(g) CO2(g) CH4(g)O2(g)
So (J/mol-K) 188.72 213.7 186.1 205.0
Did the entropy of the system increase or
decrease?
What does this mean for the surroundings?
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Entropy of the Surroundings
In order for ∆Suniverse to be increasing if ∆Ssystem
decreases for a spontaneous reaction, then
∆Ssurroundings must increase.
Let’s consider the following diagram.
Exothermic Rxn Exothermic Rxn Endothermic Rxn
Let’s Talk Heat Exchange (∆Η)
If heat is lost to the surroundings, then the
entropy of the surroundings should increase.
If heat is gained by the surroundings, then the
KE goes up and so does the entropy.
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If heat is gained by the system (lost by the surroundings), the system becomes more disordered (entropy increases).
Also,
If the temperature of the surroundings is low, the particles are more ordered and a change in heat of the surroundings would have an even greater effect on the change in entropy.
Then ∆Ssurroundings is directly related to an opposite sign of ∆Hsystem, and inversely related to temperature at which the heat is transferred.
Mathematically Speaking
∆Ssurroundings is proportional to - qsystem
∆Ssurroundings is proportional to 1/T
∆Ssurroundings = -qsystem/T
∆Ssurroundings = -∆Hsystem/T
So…, calculate ∆Ssurroundings for the combustion of
methane at 25oC from ∆Hfo values. ∆Hf
o(CH4) = -
74.87kJ/mol, ∆Hfo(CO2) = -398.5 kJ/mol, and
∆Hfo(H2O) = -241.83kJ/mol
Using your information, calculate ∆Suniverse for
the combustion of methane. Did the entropy
of the universe increase?
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Entropy and Equilibrium
What is the change in entropy of the universe
for rxns at equilibrium?
∆Suniverse = 0 @ Equilibrium
∆Ssurroundings + ∆Ssystem = ∆Suniverse
Thus, ∆Ssurroundings = -∆Ssystem @ Equilibrium
Problem
Calculate the change in entropy for the following
reaction:
If the change in enthalpy is 40,700 J/mol at 373 K, what
is the change in entropy of the universe?
H2O(l)H2O(g)
∆S(J/mol-K) 86.8 195.9 @ 373 K
Entropy, Free Energy, and Work
One criterion for spontaneity is Gibbs Free Energy (∆G)
Proof:
∆Ssurroundings + ∆Ssystem = ∆Suniverse
∆Ssurroundings = -∆Hsystem/T
-∆Hsystem/T + ∆Ssystem = ∆Suniverse
∆Hsystem - T∆Ssystem = -T∆Suniverse
Given that: -T∆Suniverse = ∆Guniverse
∆Hsystem - T∆Ssystem = ∆Guniverse
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Gibbs Free Energy and Spontaneity
If ∆G < 0, The rxn is spontaneous
If ∆G > 0, the rxn is nonspontaneous
If ∆G = 0, the rxn is at equilibrium
Note: ∆Gof values are on page A-5.
These values for an element in its most stable state are 0. Also, if a rxn has a particular value for ∆G in one direction, it will be the same value with reversed sign in the opposite direction. Just like ∆Η. ;)
Problem
Determine whether or not the following reaction is spontaneous. Use TWO methods and compare: (a) use only delta G values (kJ/mol), (b) use delta Ho
f
(kJ/mol) and So (J/mol-K) values
4 KClO3(s) � 3 KClO4(s) + KCl(s)
∆Hof -397.7 -432.75 -436.7
So 143.1 151.0 82.59
∆Gof -296.3 -303.2 -409.2
Most exothermic reactions are spontaneous. However, temperature will affect the T ∆S factor of the equation:
∆Hsystem - T∆Ssystem = ∆Guniverse
Therefore, reaction spontaneity can change with temperature.
Example: At 298 K, ∆Go = -141.6kJ, ∆Ho = -198.4 kJ, and ∆So = -187.9 J/K for a reaction.
(1) Is the rxn spontaneous at 298K?
(2) If temperature is increased what happens to ∆Go ?
(3) Assuming ∆Ηo and ∆So are constant with temperature, is the rxn spontaneous at 900.oC?
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Crossover Temperature
At what temperature does a reaction become
spontaneous?
+ ∆Go == � - ∆Go
At some point ∆Go equals 0 ( @ equil)
Thus, ∆Hsystem - T∆Ssystem = ∆Guniverse
∆Hsystem - T∆Ssystem = 0
∆Hsystem = T∆Ssystem or
∆Hsystem/∆Ssystem = Tcrossover
What is the crossover temperature for the
previous problem?
Problem: Determine if the following reaction is
spontaneous at 298K. Also determine the
crossover temperature.
Cu2O(s) + C(s) � 2 Cu(s) + CO(g)
∆So = 165 J/K and ∆Ho = 58.1 kJ
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Free Energy, Equilibrium, and Rxn Direction
Recall,
If ∆G < 0, rxn is spontaneous, product favored,
and Q is < K.
If ∆G > 0, rxn is nonspontaneous, reactant
favored, and Q > K.
If ∆G = 0, rxn is at equilibrium and Q = K.
Given that ∆Gorxn is proportional to ln (Q/K) and their
signs are identical for a given reaction, we have:∆Grxn = RT ln (Q/K)
∆Grxn = RTln Q - RTln K
If we set [ ]’s = 1.00 M for sol’ns and 1.00 atm for gases (std conditions),
then, ∆Go
rxn = RTln 1 - RTln K
Or ∆Gorxn = - RTln K
Note: K can be determined for any rxn at any temp if we know ∆Go
rxn
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Problem Solving
Determine the thermodynamic equilibrium constant at
25oC for the following rxn:
N2O4(g) 2 NO2(g)
∆Gof 97.82 51.30
(kJ/mol)
However, most reactions are not with 1.0 M solutions
and/or 1.00 atm for gases, thus we also need to
remember this form of the equation:
∆Grxn = RT lnQ - RT lnK
Since ∆Gorxn = - RT lnK
∆Grxn = ∆Gorxn + RTlnQ
The above relationship is very important to our next
chapter, you will definitely see it again.
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Problem SolvingConsider the following reaction:
2 SO2(g) + O2(g) 2 SO3(g)
∆Gorxn -141.6 kJ at 298K and -12.12 kJ at 973K.
(a) Determine Keq at 298K.
(b) Determine Keq at 973K.
(c) If a vessel contains 0.500M SO2 , 0.0100M O2, and 0.100M SO3 at 298K, what is ∆Grxn ?
(d) Under the conditions in (c) which way would the reaction proceed?
More Problem SolvingConsider the reaction:
N2(g) + 3H2 (g) 2 NH3(g)
So (J/mol-K) 191.5 131 193.0
∆Ho (kJ/mol) 0 0 -46.3
(a) Determine the equilibrium constants at 298 K and at 984 K.
(b) What is the cross-over temperature for this reaction in Kelvin.
(c) If, at 298K, a vessel contains 0.250M H2, 0.870 M N2, and 12.9M NH3, predict the direction the reaction will proceed at this temperature.