Spontaneous reactions
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Transcript Spontaneous reactions
AP Chemistry Chapter 17
Spontaneity of
Reaction
Spontaneous reactions
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
∆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!!