AP Chemistry Chapter 17
Spontaneity of
Reaction
Spontaneous reactions
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What does that mean?
Some occur without any “help”
Others require some “help”
No help – ice cube melting
Help – wood burning
• If a reaction is spontaneous under a
certain set of conditions, the reverse
reaction must be nonspontaneous
• In any spontaneous change, the amount
of free energy available decreases toward
zero as the process proceeds towards
equilibrium.
Reactions tend to be spontaneous
when:
• it leads to lower energy = -∆H
• But not always!!!!
• Also tend to be spontaneous if the reaction
results in an increase in randomness
• Entropy S
• Greater entropy – more random the system is.
+∆S increase in entropy ∆S>0
•
- ∆S decrease in entropy ∆S<0
• Page 448 Example 17.1 Predict sign of ∆S
• In general, nature tends to move
spontaneously from more ordered to more
random state (less ordered)
• Entropy increases in the order:
• s <l < g
• g>l>s
• Increasing temperature of a substance
increases its entropy
Third Law of Thermodynamics
• A completely ordered pure crystalline
solid has an entropy of zero at 0K
∆S for reactions
• Pg. 450 Table of Standard Entropies
• Used to calculate the standard entropy
change, ∆So, for reactions.
• ∆So = ∑ So products – ∑So reactants
• Must remember to multiply by the
number of moles from balanced equation
• Note that So is a positive quantity for both
compounds and elements; can be negative
for ions in solutions
• Reactions which So is positive tend to be
spontaneous, at least at high temperatures.
• H2O(s)  H2O(l)
( ∆S > 0)
• H2O(l)  H2O(g)
(∆S > 0)
• Fe2O3(s) + 3H2(g)  2Fe(s) + 3H2O(g) (∆S > 0)
• All of these reactions are endothermic (∆H>0)
• They become spontaneous at high temperatures
A
reaction that results in an increase in
the number of moles of gas is
accompanied by an increase in entropy.
 If the number of moles of gas
decreases, ∆S is a negative quantity
 Elements
have nonzero standard
entropies
 Standard molar entropies of pure
substances are always positive
quantities
 Aqueous ions may have negative So
values
 Among
substances of similar
structure and physical state, entropy
usually increases with molar mass
 Molecule becomes more complex,
more ways for the atoms to move
about with respect to one another
(higher entropy)
• Pg. 449-451 samples
• Example 17.2
Second Law of Thermodynamics
• In a spontaneous process, there is a
net increase in entropy, taking into
account both system and
surroundings.
• ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0
• spontaneous
Gibbs Free Energy G
• Two quantities affect reaction
spontaneity;
• enthalpy, H and entropy, S
• Put them together in a way that the
signs will give us a clue
• G = H – TS
• T = kelvin temp
• ∆G – for a reaction at constant temp and
pressure, represents that portion of the total
energy change that is available to do useful
work – is a state property
• Depends only on the nature of products and
reactants and the conditions
(temp/pressure/concentration), not on the
path by which the reaction is carried out
 - ∆G = spontaneous
 + ∆G = not spontaneous (reverse is
spontaneous
∆G = 0 system is at equilibrium (no tendency
for reaction to occur in either direction)
∆G measure of the driving force
of a reaction
• Reaction, at constant pressure and
temperature, go in such a direction as to
decrease the free energy of the system
• Products have lower free energy, reaction
will go in that direction
• Reactants have lower free energy, reaction
will go in that direction (means the reverse
rxn spontaneous)
Gibbs-Helmholtz
equation
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∆G = ∆H - T∆S
To make ∆G negative;
Negative value for ∆H (exothermic)
Positive value for ∆S (less
ordered)
Gibbs-Helmholtz equation
• Valid under all conditions but we
will apply it only under “standard
conditions”
• Meaning: gases are at one
atmosphere partial pressure
• Ions or molecules in solution are at
one molar concentration
• ∆G = standard free energy change
• ∆Go = ∆Ho - T∆So
• now we can use the tables in the
book
• If ∆Go is negative = spontaneous at
standard conditions
• If ∆Go is positive = nonspontaneous
at standard conditions
• ∆G = 0 system is at equilibrium at
standard conditions
o
25 C
Calculation of ∆G at
Free Energies of Formation!!
 Make sure units are correct
 Use ∆H is kJ, convert ∆S for J/K to kJ/K
 Pg. 455 Example 17.3
 Pg. 456 Example 17.4, 17.5
 Pg. 458 IMPORTANT TABLE!!