#### Transcript THERMODYNAMICS

```Chemical Thermodynamics
Chapter 19
(except 19.7!)
THERMODYNAMICS
The BIG question!!!
How do we know whether a
reaction will happen or not?
THERMODYNAMICS
Sample Problem 1
Which of the following processes are spontaneous
(happen naturally)?
(a) a ball rolling up a hill
(b) the freezing of water at -3oC and 1 atm pressure
(c) the freezing of water at 3oC and 1 atm pessure
(d) the decomposition of carbon dioxide into
carbon and oxygen at 25oC and 1 atm pressure
(e) the dissolving of NaCl in water
(a) not spontaneous
(b) spontaneous
(c) not spontaneous
(d) not spontaneous
(e) spontaneous
Processes that occur
spontaneously in one
direction are NOT
spontaneous in the
opposite direction.
THERMODYNAMICS
• The two driving forces for reactions are
(1) the release of energy (gain of
stability)
(2) increasing entropy
• In nature, reactions move toward
lowest energy and highest entropy
THERMODYNAMICS
• Sample Problem 2
At 25oC, the ΔH value for the reaction
2 H2(g) + O2(g)  2 H2O(l)
is - 572 kJ
Can you conclude from the sign of ΔH that
the reaction is spontaneous?
• NO! Many exothermic reactions are
spontaneous, but not all!
THERMODYNAMICS
• Entropy is a thermodynamic state function
represented by S.
• Entropy is a measure of the number of ways
energy is distributed in a system. The more
ways energy is distributed in a system, the
more random or the more disordered the
system is said to be. The greater the entropy
of a system, the greater is its randomness or
disorder.
• Entropies are usually positive for both
elements and compounds.
• The units of entropy are J/mole•K
THERMODYNAMICS
We usually look at the change in entropy
during a process:
• ΔS = Sfinal - Sinitial (state function)
• ΔS > 0 (is positive) if the system
changes to a state of greater disorder.
• ΔS < 0 (is negative) if the system
changes to a state of greater order
THERMODYNAMICS
2nd Law of Thermodynamics:
• In any spontaneous process, the entropy of
the universe always increases.
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
• A process that involves an increase in order
(ΔSsystem < 0) CAN be spontaneous provided
it causes an even greater positive entropy
changes in the surroundings.
• An isolated system (no exchange of energy
between the system and surroundings) always
increases its entropy when it undergoes a
spontaneous change. For such a change,
ΔSsurroundings = 0
THERMODYNAMICS
• Standard entropies (S°) have been
determined for many substances, and
these values are based on the 3rd law
of thermodynamics.
• 3rd Law of Thermodynamics
The entropy of a perfect, pure
crystalline substance at 0 K is zero.
THERMODYNAMICS
1) Entropy increases as one goes from a solid to a liquid,
or even more dramatically, a liquid to a gas
2) Entropy increases as the number of moles of gas
increases
3) The Entropy of any material increases with increasing
temperature
4) Entropy increases if a solid or liquid is dissolved in a
solvent.
5) Entropy is higher for weakly bonded compounds than
for compounds with very strong covalent bonds
6) Entropy increases as the number of particles in a
system increases
7) Entropy increases as the number of electrons increases:
Entropy increases as the mass, # of atoms, # of heavier
atoms, etc. of a molecule increases
THERMODYNAMICS
Entropy
changes
dramatically
at a phase
change.
THERMODYNAMICS
Sample Problem 3
• For each of the following pairs, which
has the greater entropy:
(a) CO2(s) or CO2(g)
(b) NH3(l) or NH3(g)
(c) a crystal of pure Mg at 0 K or a crystal at
200 K
(a) CO2(g)
THERMODYNAMICS
• Sample Problem 3
• For each of the following pairs, which
has the greater entropy:
(a) CO2(s) or CO2(g)
(b) NH3(l) or NH3(g)
(c) a crystal of pure Mg at 0 K or a crystal at
200 K
(a) CO2(g)
(b) NH3(g)
THERMODYNAMICS
• Sample Problem 3
• For each of the following pairs, which
has the greater entropy:
(a) CO2(s) or CO2(g)
(b) NH3(l) or NH3(g)
(c) a crystal of pure Mg at 0 K or a crystal at
200 K
(a) CO2(g)
(b) NH3(g)
(c) crystal at 200 K
THERMODYNAMICS
• Sample Problem 4
• Predict whether ΔS is > 0 or < 0 for
each of the following:
(a) sugar + water  sugar dissolved in water
(b) Na2SO4(s)  2 Na+(aq) + SO4-2(aq)
(c) 2 H2(g) + O2(g)  2 H2O(g)
THERMODYNAMICS
• Sample Problem 4
• Predict whether ΔS is > 0 or < 0 for
each of the following:
(a) sugar + water  sugar dissolved in water
(b) Na2SO4(s)  2 Na+(aq) + SO4-2(aq)
(c) 2 H2(g) + O2(g)  2 H2O(g)
(a) > 0 (positive)
THERMODYNAMICS
• Sample Problem 4
• Predict whether ΔS is > 0 or < 0 for
each of the following:
(a) sugar + water  sugar dissolved in water
(b) Na2SO4(s)  2 Na+(aq) + SO4-2(aq)
(c) 2 H2(g) + O2(g)  2 H2O(g)
(a) > 0 (positive)
(b) > 0 (positive)
THERMODYNAMICS
• Sample Problem 4
• Predict whether ΔS is > 0 or < 0 for
each of the following:
(a) sugar + water  sugar dissolved in water
(b) Na2SO4(s)  2 Na+(aq) + SO4-2(aq)
(c) 2 H2(g) + O2(g)  2 H2O(g)
(a) > 0 (positive)
(b) > 0 (positive)
(c) < 0 (negative)
THERMODYNAMICS
• Entropy is a state function and can be
calculated for a reaction:
• ΔS°rxn = Σ n S°products - Σ m S°reactants
On AP Equation Sheet!
• values are tabulated in appendix C in
THERMODYNAMICS
• Sample Problem 5
Calculate ΔS°rxn for the following at 25oC
2 SO2(g) + O2(g)  2 SO3(g)
standard entropy values at 25oC (from book)
SO2(g)
248.5 J/K mole
O2(g)
205.0 J/K mole
SO3(g)
256.2 J/K mole
• ΔS°rxn = Σ n S°products - Σ m S°reactants
•
= 2 S° SO3 – (2 S° SO2 + S° O2)
•
= 2(256.2) – [2(248.5) + 205.0]
•
= - 189.6 J/K
• (decrease in # of gas molecules also indicates a
decrease in entropy)
THERMODYNAMICS
• Exothermic processes (-ΔH) are more
likely to be spontaneous than
endothermic processes (+ΔH).
• Processes that involve an increase in
entropy (ΔS > 0) are more probable
than those showing a decrease in
entropy (ΔS < 0).
THERMODYNAMICS
• Enthalpy and entropy are connected together
by the thermodynamic state function called
free energy (G). The change in free energy
for any process or reaction at constant
temperature and pressure is
On AP Equation Sheet!
•
ΔG = ΔH - T ΔS
• if ΔG < 0 (is negative), a reaction
spontaneously proceeds in the forward
direction
• if ΔG = 0, a system is at equilibrium
• if ΔG > 0 (is positive), a reaction is not
spontaneous as written, but the reverse
direction is spontaneous
THERMODYNAMICS
• ΔG is a state function and tabulated
values exist for standard free energy of
formation for substances - in
appendix C
• ΔG°f for any element in its most stable
state is zero.
• ΔG°rxn = Σ n ΔG°f (products) - Σ m ΔG°f (reactants)
On AP Equation Sheet!
THERMODYNAMICS
• Sample Problem 6
• Given the ΔG value for the following phase
changes at 1 atm, predict whether each
change is spontaneous:
• (a) at 283 K, ΔG = -250 kJ/mole for H2O(s)  H2O(l)
• (b) at 273 K, ΔG = 0 kJ/mole for H2O(s)  H2O(l)
• (c) at 263 K, ΔG = 210 kJ/mole for H2O(s)  H2O(l)
THERMODYNAMICS
• Sample Problem 6
• Given the ΔG value for the following phase
changes at 1 atm, predict whether each
change is spontaneous:
• (a) at 283 K, ΔG = -250 kJ/mole for H2O(s)  H2O(l)
• (b) at 273 K, ΔG = 0 kJ/mole for H2O(s)  H2O(l)
• (c) at 263 K, ΔG = 210 kJ/mole for H2O(s)  H2O(l)
(a) spontaneous since ΔG < 0
THERMODYNAMICS
• Sample Problem 6
• Given the ΔG value for the following phase
changes at 1 atm, predict whether each
change is spontaneous:
• (a) at 283 K, ΔG = -250 kJ/mole for H2O(s)  H2O(l)
• (b) at 273 K, ΔG = 0 kJ/mole for H2O(s)  H2O(l)
• (c) at 263 K, ΔG = 210 kJ/mole for H2O(s)  H2O(l)
(a) spontaneous since ΔG < 0
(b) at equilibrium – no net reaction
THERMODYNAMICS
• Sample Problem 6
• Given the ΔG value for the following phase
changes at 1 atm, predict whether each
change is spontaneous:
• (a) at 283 K, ΔG = -250 kJ/mole for H2O(s)  H2O(l)
• (b) at 273 K, ΔG = 0 kJ/mole for H2O(s)  H2O(l)
• (c) at 263 K, ΔG = 210 kJ/mole for H2O(s)  H2O(l)
(a) spontaneous since ΔG < 0
(b) at equilibrium – no net reaction
(c) non spontaneous since ΔG > 0
THERMODYNAMICS
• Sample Problem 7
Given:
ΔG°f for C6H12O6(s) = -907.9 kJ/mole,
ΔG°f for CO2 (g) = -394.6 kJ/mole
ΔG°f for H2O(l) = -237.2 kJ/mole
calculate ΔG° for the oxidation of glucose
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
ΔG°rxn = Σ n ΔG°f (products) - Σ m ΔG°f (reactants)
= [6(ΔG°f CO2) + 6(ΔG°f H2O)] – [ΔG°f C6H12O6 + 6(ΔG°f O2)]
= [6(-394.6) + 6(-237.2)] – [1(-907.9) + 6(0)]
= - 2882.9 kJ (spontaneous)
THERMODYNAMICS
Effect of temperature on Reaction Spontaneity
ΔG = ΔH - T ΔS
ΔH
ΔS
ΔG
Rxn Characteristics
-
+
always -
always spontaneous as written
+
-
always +
always nonspontaneous as written
reverse rxn is spontaneous
-
-
- at low T
spontaneous at low T
+ at high T nonspontaneous at high T
+
+
+ at low T
- at high T
nonspontaneous at low T
spontaneous at high T
THERMODYNAMICS
• Sample Problem 8
What is the temperature at which sodium chloride
reversibly melts (solid and liquid states are in
equilibrium)? The enthalpy of melting is 30.3 kJ/mole
and the entropy change upon melting is 28.2 J/mole K.
ΔG = ΔH - T ΔS
at equilibrium ΔG = 0 = ΔH - T ΔS
so ΔH = T ΔS
and T = ΔH / ΔS
30300 J/mole
T = -------------------- = 1070 K (797oC)
28.2 J/mole K
above this temp (1070 K), melting is spontaneous
below this temp, melting would be nonspontaneous
THERMODYNAMICS
• Sample Problem 9
At 298 K, ΔG° = -190.5 kJ and ΔH° = -184.6 kJ
for the reaction
H2(g) + Cl2(g)  2 HCl(g)
calculate the standard entropy ΔS° for the
reaction
ΔG° = ΔH° - T ΔS°
Δ S° = (ΔH° - ΔG° ) / T
- 184.6 kJ – (-190.5 kJ)
ΔS° = ------------------------------- = .020 kJ/K = 20 J/K
298 K
Typically, S values
or S values are in
J/K not kJ/K
THERMODYNAMICS
• Sample problem 10
Given the following data:
Substance
Hof (kJ/mol)
So (J/K mol)
Fe2O3(s)
-826
90.0
Fe(s)
0
27.0
O2(g)
0
205.0
Calculate ΔGo for the reaction: 4 Fe(s) + 3O2(g)  2 Fe2O3(s)
Go = Ho - TSo
Ho = prod - react = [2(-826)] – [0 + 0] = - 1650 kJ
So = prod - react = [2(90.0)] – [3(205.0) + 4(27.0)] = -543 J/K
Notice that H is kJ and S is J/K
-543 J/K = -0.543 kJ/K
Go = Ho - TSo
= - 1650 kJ - (298 K)(-0.543 kJ/K)
= - 1490 kJ
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