Transcript AP Thermodynamics ppt.

CHEMISTRY
The Central Science
9th Edition
Chapter 19
Chemical Thermodynamics
David P. White
Prentice Hall © 2003
Chapter 19
Spontaneous Processes
• Thermodynamics is concerned with the question: can a
reaction occur?
• First Law of Thermodynamics: energy is conserved.
• Any process that occurs without outside intervention is
spontaneous.
• When two eggs are dropped they spontaneously break.
• The reverse reaction is not spontaneous.
• A process that is spontaneous in one direction is not
spontaneous in the opposite direction.
Prentice Hall © 2003
Chapter 19
Spontaneity & Temperature
The direction of a spontaneous process can
depend on temperature: ice turning to water is
spontaneous at T > 0C, water turning to ice is
spontaneous at T < 0C.
Packet Example Page 3
• Predict whether the following processes are spontaneous
as described, spontaneous in the reverse direction, or are
in equilibrium.
• a) When a piece of metal heated to 150C is added to
water at 40C, the water gets hotter.
• b) Water at room temperature decomposed into
hydrogen and oxygen gases.
• c) Benzene vapor at a pressure of 1 atm condenses to
liquid benzene at the normal boiling point of benzene.
Prentice Hall © 2003
Chapter 19
Reversible Processes
• Reversible processes can go back and forth
between states along the same path.
• Example: changes of state
• Path taken back to original state is exactly the
reverse of the forward process.
• No net change in system or surroundings when
cycle is completed.
Prentice Hall © 2003
Chapter 19
Reversible Processes
– When 1 mol of water is frozen at 1 atm at 0C
to form 1 mol of ice, q = Hvap of heat is
removed.
– To reverse the process, q = Hvap must be
added to the 1 mol of ice at 0C and 1 atm to
form 1 mol of water at 0C.
– Converting between 1 mol of ice and 1 mol of
water at 0C is a reversible process.
Irreversible Processes
An irreversible process cannot be reversed to restore the
system and surroundings back to their original state.
A different path with different values of q and w are needed.
Surroundings are not returned to original conditions.
Reversible & Irreversible
Processes
• Chemical systems in equilibrium are reversible.
• In any spontaneous process, the path between
reactants and products is irreversible.
• Thermodynamics gives us the direction of a
process but it cannot predict the speed at which
the process will occur.
Prentice Hall © 2003
Chapter 19
Spontaneous
Expansion of a
Gas
• Why does the gas
expand?
• Why is the reverse
process
nonspontaneous?
Spontaneous Expansion of
a Gas
• Consider 2 gas molecules in one flask (a).
• Once the stopcock is open, there is a higher
probability that one molecule will be in each flask
than both molecules being in the same flask (b).
Entropy
• Entropy, S, is a measure of the disorder of a
system.
• The more disordered or random the system, the
large value of S.
• Spontaneous reactions proceed to lower energy or
higher entropy.
Prentice Hall © 2003
Chapter 19
•In ice, molecules are
very well ordered
because of the H-bonds.
•Ice has a low entropy.
•As ice melts,
intermolecular forces are
broken , order is
interrupted.
•Water is more random
than ice, has higher
entropy.
• Ice spontaneously melts
at room temperature.
•A balance between energy and entropy considerations.
•When an ionic solid is placed in water two things happen:
–Hydrates form (entropy decreases)
–Ions dissociate (entropy increases)
Entropy, Cont.
• Generally, when an increase in entropy in one
process is associated with a decrease in entropy in
another, the increase in entropy dominates.
• Entropy is a state function.
• For a system, S = Sfinal - Sinitial.
• If S > 0 the randomness increases, if S < 0 the
order increases.
Prentice Hall © 2003
Chapter 19
Packet Example Page 6
• By considering the disorder in the reactants and
products, predict whether ∆S is positive or
negative for the following:
• a) H2O (l) → H2O (g)
• b) Ag+(aq) + Cl-(aq) → AgCl (s)
• c) 4Fe(s) + 3O2 (g) → 2Fe2O3 (s)
Prentice Hall © 2003
Chapter 19
Second Law of
Thermodynamics
• Second Law of Thermodynamics: In any
spontaneous process, the entropy of the universe
increases.
• Suniv = Ssys + Ssurr
• Entropy is not conserved: Suniv is increasing.
• To predict spontaneity, we must know the sign of
∆Suniv.
Prentice Hall © 2003
Chapter 19
Entropy and the Second
Law of Thermodynamics
• For a reversible process: Suniv = 0.
• For a spontaneous process (i.e. irreversible): Suniv > 0.
Suniv< 0 means spontaneous is opposite direction.
• Second law states that the entropy of the universe must
increase in a spontaneous process. Entropy of a system
can decrease as long as the entropy of the surroundings
increases.
• For an isolated system, Ssys = 0 for a reversible process
and Ssys > 0 for a spontaneous process.
Entropy and the Second
Law of Thermodynamics
Entropy
• Suppose a system changes reversibly between
state 1 and state 2. At constant T where qrev is the
amount of heat added reversibly to the system, the
change in entropy is given by
 S sys 
q rev
(constant T )
T
• Example: a phase change occurs at constant T
with the reversible addition of heat.
Prentice Hall © 2003
Chapter 19
Example Packet p 7
• The element mercury, Hg, is a silvery liquid at
room temperature. The normal freezing point of
mercury is -38.9C and its molar enthalpy of
fusion is ∆H = +2.29 kJ/mol. Calculate the
entropy change of the system when 50.0 g liquid
mercury freezes at the normal freezing point.
• Remember: fusion = melting
• Use:  S  q rev (constant T )
sys
T
Practice Example
• The normal boiling point of ethanol (C2H5OH) is
78.3C and its ∆H vaporization= +38.56 kJ/mol.
Calculate ∆S when 68.3 g of ethanol at 1 atm
condense to liquid at the normal boiling point.
Prentice Hall © 2003
Chapter 19
Example Packet p 9
• Consider the reversible melting of 1 mol of ice in
a large, isothermal water bath at 0C. The
enthalpy of fusion is 6.01 kJ/mol. Calculate the
entropy change in the system and in the
surroundings and the overall change in entropy of
the universe for this process.
Prentice Hall © 2003
Chapter 19
Practice Example
• ∆Hvap = 30.71 kJ/mol for liquid bromine.
Calculate the change in entropy for the system,
the surroundings, and the universe for the
reversible vaporization of liquid bromine at its
normal boiling point of 59C.
Prentice Hall © 2003
Chapter 19
Effect of Temperature on
Spontaneity
• ∆Ssurr are determined primarily by flow of energy into
and out of system as heat.
• An exothermic process in system increases entropy in
surroundings.
• Exothermicity is a driving force for spontaneity BUT it
depends on the temperature at which the process occurs!
• Impact of transfer of a given quantity of energy as heat
to or from surroundings will be greater at lower
temperatures.
Prentice Hall © 2003
Chapter 19
Entropy Changes in the
Surroundings
• Sign of ∆Ssurr depends on direction of heat flow.
• Magnitude of ∆Ssurr depends on the temperature.
• ∆Ssurr depends directly on quantity of heat
transferred and inversely on temperature.
• ∆Ssurr = -∆H / T
• (We need a minus sign here to change the point
of view from the system to the surroundings.)
Prentice Hall © 2003
Chapter 19
Examples from Zumdahl
pp 782 & 783
• Calculate ∆Ssurr for the following reactions at 25C and
1 atm.
• Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ
• Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(s) ∆H = +778 kJ
• Use ∆Ssurr = -∆H / T
• +419 J/K
• -2.61 x 103 J/K
Prentice Hall © 2003
Chapter 19
Determining the Sign of
∆Suniv
Signs of Entropy Changes
• ∆Ssys
• +
• • +
• -
∆Ssurr
+
+
Prentice Hall © 2003
∆Suniv
+
?
?
Chapter 19
Process Spontaneous?
Yes
No, rxn goes in reverse
Yes, if ∆Ssys  ∆Ssurr
Yes, if ∆Ssurr  ∆Ssys
Atomic Modes of Motion
• Movement of molecules is related to energy and entropy.
• Three atomic modes of motion:
– translation (moving from one point in space to
another),
– vibration (shortening and lengthening of bonds,
including the change in bond angles),
– rotation (spinning around an axis).
The Molecular
Interpretation of Entropy
• Energy is required to get a molecule to translate,
vibrate or rotate.
• The more energy stored in translation, vibration
and rotation, the greater the degrees of freedom
and the higher the entropy.
• In a perfect crystal at 0 K there is no translation,
rotation or vibration of molecules. Therefore, this
is a state of perfect order.
Prentice Hall © 2003
Chapter 19
Third Law of
Thermodynamics
• Third Law of Thermodynamics: the entropy of a perfect
crystal at 0 K is zero.
• As we heat a substance from absolute zero, the entropy
must increase.
• Entropy changes dramatically at a phase change.
• Entropy increases when
--Liquids or solutions are formed from solids
--Gases are formed from solids or liquids
--The number of gas molecules increases.
Heat and Entropy
As we heat a substance from
absolute zero, the entropy must
increase.
Boiling corresponds to a
much greater change in
entropy than melting.
Example Packet p 11
• Choose the sample of matter that has the greater entropy
in each pair and explain your choice:
• a) 1 mol of solid NaCl or 1 mol of gaseous HCl at 25C
• b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25C
• c) 1 mol of HCl(g) or 1 mol of Ar(g) at 25C
• d) 1 mol of N2(s) at 24 K or 1 mol of N2 at 25C
Prentice Hall © 2003
Chapter 19
Example Packet p 12
• Predict whether the entropy change of the system
in each of the following isothermal reactions is
positive or negative.
• a) CaCO3(s) → CaO(s) + CO2(g)
• b) N2(g) + 3H2(g) → 2NH3(g)
• c) N2(g) + O2(g) → 2NO(g)
Prentice Hall © 2003
Chapter 19
Entropy Changes in
Chemical Reactions
• Absolute entropy can be determined from complicated
measurements.
• Standard molar entropy, S: entropy of a substance in its
standard state. Similar in concept to H.
• Units: J/mol-K. Note units of H: kJ/mol.
• Standard molar entropies of elements are not zero.
• For a chemical reaction which produces n moles of
products from m moles of reactants:
 S    nS   products
   mS   reactants 
Gibbs Free Energy
• For a spontaneous reaction the entropy of the universe
must increase.
• Reactions with large negative H values are spontaneous.
• How to we balance S and H to predict whether a
reaction is spontaneous?
• Gibbs free energy, G, of a state is
G  H  TS
• For a process occurring at constant temperature
G  H  TS
Prentice Hall © 2003
Chapter 19
Gibbs Free Energy
• There are three important conditions:
– If G < 0 then the forward reaction is spontaneous.
– If G = 0 then reaction is at equilibrium and no net reaction
will occur.
– If G > 0 then the forward reaction is not spontaneous. If G >
0, work must be supplied from the surroundings to drive the
reaction.
• For a reaction the free energy of the reactants decreases
to a minimum (equilibrium) and then increases to the free
energy of the products.
Prentice Hall © 2003
Chapter 19
Gibbs Free Energy
Prentice Hall © 2003
Chapter 19
Gibbs Free Energy
• Consider the formation of ammonia from nitrogen and
hydrogen:
N 2 (g) + 3H 2 (g)
2N H 3 (g)
• Initially ammonia will be produced spontaneously (Q <
Keq).
• After some time, the ammonia will spontaneously react to
form N2 and H2 (Q > Keq).
• At equilibrium, ∆G = 0 and Q = Keq.
Prentice Hall © 2003
Chapter 19
Gibbs Free Energy
Standard Free-Energy Changes
• We can tabulate standard free-energies of formation, Gf
(c.f. standard enthalpies of formation).
• Standard states are: pure solid, pure liquid, 1 atm (gas), 1
M concentration (solution), and G = 0 for elements.
• G for a process is given by
 G    n  G  f  products    m  G  f  reactants 
• The quantity G for a reaction tells us whether a
mixture of substances will spontaneously react to produce
more reactants (G > 0) or products (G < 0).
Prentice Hall © 2003
Chapter 19
Free Energy and
Temperature
• Focus on G = H - TS:
– If H < 0 and S > 0, then G is always negative.
– If H > 0 and S < 0, then G is always positive. (That is, the
reverse of 1.)
– If H < 0 and S < 0, then G is negative at low temperatures.
– If H > 0 and S > 0, then G is negative at high
temperatures.
• Even though a reaction has a negative G it may occur
too slowly to be observed.
Prentice Hall © 2003
Chapter 19
Free Energy and
Temperature
Prentice Hall © 2003
Chapter 19
Free Energy and The
Equilibrium Constant
• Recall that G and K (equilibrium constant) apply to
standard conditions.
• Recall that G and Q (equilibrium quotient) apply to any
conditions.
• It is useful to determine whether substances under any
conditions will react:
 G   G   RT ln Q
Prentice Hall © 2003
Chapter 19
Free Energy and The
Equilibrium Constant
• At equilibrium, Q = K and G = 0, so
 G   G   RT ln Q
0   G   RT ln K eq
 G    RT ln K eq
• From the above we can conclude:
– If G < 0, then K > 1.
– If G = 0, then K = 1.
– If G > 0, then K < 1.
Prentice Hall © 2003
Chapter 19
End of Chapter 19
Chemical Thermodynamics
Prentice Hall © 2003
Chapter 19