Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved. E = q + w.
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Transcript of Thermodynamics Chapter 18. 1 st Law of Thermodynamics Energy is conserved. E = q + w.
ThermodynamicsChapter 18
1st Law of Thermodynamics
Energy is conserved.
E = q + w
SPONTANEOUS: occur without any outside intervention
Example: drop an egg
The REVERSE is not spontaneous!!
REVERSIBLE PROCESS: change can be restored to its’ original state by exactly reversing the change.
Example: ice water at 0o C
IRREVERSIBLE PROCESS: cannot simply be reversed to original state.
Example: gas expanding
Processes in which the disorder of the system increases tend to occur
spontaneously.
Ex: gas expanding, ice melting, salt dissolving
ENTROPY: (S) the change in disorder. (Change in randomness)
The more disorder, the larger the entropy.
S = Sfinal - Sinitial
S = > 0 when the final state is in more disorder
S = < 0 when the final state is more ordered than original
H2O (l) H2O (s)
Ag+(aq) + Cl-
(aq) AgCl(s)
- a solid melts
- a liquid vaporizes
- a solid dissolves in water
- a gas liquefies
For a process at constant temperature, the entropy change is the value of qrev divided by the absolute
temperature.S = qrev/T
Example: Calculate the entropy change when 1 mol of water is
converted into 1 mol of steam at 1 atm pressure. (Hvap = 40.67
kJ/mol)
(1 mole)(40.67 kJ/ mol)(1000 J/1 kJ)373 K
S = 109 J/K
The normal freezing point of mercury is -38.9oC, an its
molar enthalpy of fusion is Hfus = 2.331 kJ/mol. What is the entropy change when 50.0
g of Hg(l) freezes at the normal freezing point?
-2.48 J/KThe answer is negative because the
process brings more order
The normal boiling point of ethanol, is 78.3oC and its molar
enthalpy of vaporization is 38.56 kJ/mol. What is the change in
entropy when 25.8 g of C2H5OH(g) condenses to liquid at
the normal boiling point?
-61.4 J/K
The element gallium, Ga, freezes at 29.8oC, and its enthalpy of fusion is 5.59
kJ/mol. Calculate the value of S for the
freezing of 90.0 g of Ga(l).
S = -23.8 J/K
2nd Law of Thermodynamics: In any reversible process, Suniverse = 0. In any
irreversible (spontaneous) process, Suniverse > 0.
Suniverse = Ssystem + Ssurroundings
On a Molecular Level
TRANSLATIONAL MOTION: movement of molecules
VIBRATIONAL MOTION: the movement of atoms within the molecule.
ROTATIONAL MOTION: the molecules spinning
Increasing Temperature Increases Entropy
3rd Law of Thermodynamics: the entropy of a pure crystalline
substance at absolute zero is zero. S(0K) = 0
In general, the entropy increases when:
Liquids or solutions are formed from solids
Gases are formed from either solids or liquids
The number of molecules of gas increases during a chemical reaction.
CaCO3(s) CaO(s) + CO2(g)
N2(g) + 3H2(g) 2NH3(g)
Standard molar entropies: (So) absolute entropies for substances in their standard
state. (J/mol-K)
1. Unlike enthalpies of formation, the standard molar entropies of elements are not zero.
2. The So of gases are greater than those of liquids and solids.
3. The So generally increases with increasing molar mass.
4. The So generally increase with the number of atoms in the formula.
So = nSo(products) - mSo(reactants)
Calculate So for the synthesis of ammonia:
N2(g) + 3H2(g) 2NH3(g)
So = (2 mol)(192.5 J/mol-K) - [(1 mol)(191.5 J/mol-K) + (3
mol)(130.6 J/mol-K)] =
-198.3 J/K
Using Appendix C, calculate the standard entropy change, for the
following reaction:
Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
180.4 J/K
C2H4(g) + H2(g) C2H6(g)
NH3(g) + HCl(g) NH4Cl(s)
-120.5 J/K
-284.6 J/K
GIBBS FREE ENERGY
The spontaneity of a reaction involves both enthalpy and entropy. The relationship is
known as free energy.
G = H - TS
1. If G is negative, the reaction is spontaneous in the forward direction.
2. If G is zero, the reaction is at equilibrium.
3. If G is positive, the reaction in the forward direction is nonspontaneous; work must be supplied from the surroundings to make it occur. However, the reverse reaction will be spontaneous.
G
Go = nGfo(products) - mGf
o(reactants)
N2(g) + 3H2(g) 2NH3(g)
-33.32 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
-800.7 kJ
Assuming no change for Ho and So, what happens to Go
with an increase in temperature?
N2(g) + 3H2(g) 2NH3(g)
Calculate G at 298K for a reaction mixture that
consists of 1.0 atm N2, 3.0 atm H2 and 0.50 atm NH3.
N2(g) + 3H2(g) 2NH3(g)
Use standard free energies of formation to calculate the equilibrium constant
K at 25oC for the reaction involved in the Haber
process.