Chemical Thermodynamics
Name: ____________________
AP Chemistry Lecture Outline
Recall that, at constant pressure, the enthalpy change equals the heat transferred between
the system and its surroundings.
Thermodynamics involves enthalpy changes and changes in order / disorder.
Spontaneous Processes
--
First Law of Thermodynamics:
E = q + w
changes in system’s
internal energy
If q is (–), system released heat.
If q is (+), system absorbed heat.
work done
If w is (–), system did work.
If w is (+), system had work done on it.
heat transferred
reversible process: “undo exactly what you did and you’ve got what you started with”
-- both system and surroundings go back to original states
-- e.g.,
irreversible process: getting back what you started with requires more than just an “undo”
-- we can restore the original system, but the surroundings will have changed
-- e.g., a gas expanding into an evacuated space
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Points of note:
1. Whenever a chemical system is in equilibrium, we can go reversibly between reactants
and products.
2. In any spontaneous process, the path between reactants and products is irreversible.
3. Thermodynamics refers to the direction of a reaction, not its speed.
4. In general, exothermic processes are more likely to be spontaneous.
Entropy and the Second Law of Thermodynamics
Processes in which the system’s disorder increases tend to occur spontaneously.
S = Sf – Si
+ S 
– S 
For a system in which heat is
transferred at constant temperature…
EX. The normal boiling point of ethanol (CH3CH2OH) is 78.3oC; its molar enthalpy of
vaporization is 38.56 kJ/mol. Find S when 25.8 g of ethanol condense at the NBP.
The entropy of the universe increases in any _____________________ process.
Entropy is NOT conserved;
For an isolated system…
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The Molecular Interpretation of Entropy
There are three types of motion, each having kinetic energy (KE).
--- The more KE we have, the more S we have.
In general…
and…
Process
freezing
melting
condensing
boiling
Sign of S
Third Law of Thermodynamics:
The entropy of a pure crystalline substance at absolute zero is…zero.
--
Entropy usually increases when:
1. number of gas particles increases
2. liquids or solutions are formed from solids
3. gases are formed from liquids or solids
EX. Which has the greater entropy?
1 mol NaCl(s) @ 200 K
or
1 mol HCl(g) @ 200 K
1 mol HCl(g) @ 200 K
or
2 mol HCl(g) @ 200 K
(same volume)
1 mol HCl(g) @ 200 K
or
1 mol Ar(g) @ 200 K
(same volume)
1 mol N2(g) @ 200 K
or
1 mol N2(g) @ 298 K
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Calculation of Entropy Changes
standard molar entropies, So: molar entropy values of substances in their standard states
(i.e., pure substances at ~1 atm)
actually, at 1 bar = 105 Pa = 0.987 atm
So values…
---In a chemical reaction…
EX. Calculate the standard molar entropy change for…
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
Gibbs Free Energy
Enthalpy changes (H) and entropy changes (S) both determine whether or not a reaction is
spontaneous. Spontaneity can be determined using the equation for Gibbs free energy…
If G < 0…
If G > 0…
If G = 0…
As an example, for the Haber process…
N2(g) + 3 H2(g)
spontaneous
spontaneous
2 NH3(g)
Q<K
G < 0
pure
N2 + H2
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Q>K
G > 0
equilibrium mixture
(Q = K, G = 0)
pure
NH3
standard free energies of formation, Gfo
-- are tabulated for pure solids, pure liquids, gases at ~1 atm pressure, and 1 M solutions
-- For elements in their standard states…
-- For a reaction, the standard free-energy change is found by…
EX. Calculate the standard free-energy change for…
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Free Energy and Temperature
From G = H – TS, we see that G varies with temperature.
-- H and S change little with temperature
EX.
(a) Calculate Ho and So at 298 K for…
(b) Estimate Go at 400 K.
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2 SO2(g) + O2(g)  2 SO3(g)
EX. Estimate the normal boiling point of CCl4.
Free Energy and the Equilibrium Constant
Go is the standard free-energy change (i.e., for a reaction at standard conditions). Under
any other conditions…
EX. Calculate G at 298 K, given the following:
N2(g) + 3 H2(g)
1.0 atm
3.0 atm
2 NH3(g)
0.50 atm
At equilibrium, G = 0 and Q = K, so we get…
EX. Calculate Go and Kp at 298 for…
H2(g)
+
Br2(g)
2 HBr(g)
What if we want G and K, but NOT at 298 K?
1. Use tabulated values to calculate Ho and So. These are ~ constant with
temperature, so we set them equal to H and S.
2. Use G = H – TS and your answers above to approximate G at given temperature.
3. Use K = e–G/RT to approximate K at given temperature.
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