Transcript Chemistry 100 Chapter 19

Chemistry 100 Chapter
19
Spontaneity of Chemical and Physical
Processes: Thermodynamics
What Is Thermodynamics?
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Study of the energy changes that
accompany chemical and physical
processes.
Based on a set of laws.
In chemistry, a primary application
of thermodynamics is as a tool to
predict the spontaneous directions
of a chemical reaction.
What Is Spontaneity?
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Spontaneity refers to the ability of a
process to occur on its own!
Can the Niagara Falls suddenly
reverse?
“Ice will melt, water will boil,” Neil
Finn, Tim Finn of Crowded
House/Plant ‘It’s Only Natural’.
Water spontaneously freezes on a
cold winter day!
The First Law of Thermodynamics
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The First Law deals with the
conservation of energy changes.
E = q + w
The First Law tells us nothing
about the spontaneous direction of
a process.
Entropy and Spontaneity
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Need to examine
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the entropy change of the process as well
as its enthalpy change (heat flow).
Entropy – the degree of randomness
of a system.
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Solids – highly ordered  low entropy.
Gases – very disordered  high entropy.
Liquids – entropy is variable between that
of a solid and a gas.
Entropy Is a State Variable
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Changes in entropy are state
functions
S = Sf – Si
Sf = the entropy of the final state
Si = the entropy of the initial state
Entropy Changes for Different
Processes
S > 0 entropy increases (melting ice
or making steam)
S < 0 entropy decreases (examples
freezing water or condensing
steam)
The Solution Process
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For the dissolution of NaCl (s) in
water
NaCl (s)  Na+(aq) + Cl-(aq)
Highly ordered –
low entropy
Disordered or random
state – high entropy
The formation of a solution is always
accompanied by an increase in the
entropy of the system!
The Entropy Change in a Chemical
Reaction
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Burning ethane!
C2H6 (g) + 7/2O2 (g)  2CO2 (g) + 3H2O (l)
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The entropy change
rS   np S (products) -  nr S (reactants)
 np and nr represent the number of moles of
products and reactants, respectively.
Finding S Values
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Appendix C in your textbook has
entropy values for a wide variety of
species.
Units for entropy values  J / (K
mole)
Temperature and pressure for the
tabulated values are 298.2 K and
1.00 atm.
Finding S Values
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Note – entropy values are
absolute!
Note – the elements have NONZERO entropy values!
e.g., for H2 (g)
fH = 0 kJ/mole (by def’n)
S = 130.58 J/(K mole)
Some Generalizations
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For any gaseous reaction (or a
reaction involving gases).
ng > 0, rS > 0 J/(K mole).
ng < 0, rS < 0 J/(K mole).
ng = 0, rS  0 J/(K mole).
For reactions involving only solids
and liquids – depends on the
entropy values of the substances.
The Second Law of
Thermodynamics
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The entropy of the universe (univS)
increases in a spontaneous process.
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univS unchanged in an equilibrium
process
What is univS?
univS = sysS + surrS
sysS = the entropy change of the
system.
surrS = the entropy change of the
surroundings.
How Do We Obtain univS?
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We need to obtain estimates for both
the sysS and the surrS.
Look at the following chemical
reaction.
C(s) + 2H2 (g)  CH4(g)
The entropy change for the systems is
the reaction entropy change, rS.
How do we calculate surrS?
Calculating surrS
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Note that for an exothermic process,
an amount of thermal energy is
released to the surroundings!
Heat
System
surroundings
Insulation
Calculating surrS
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Note that for an endothermic process,
thermal energy is absorbed from the
surroundings!
Heat
System
surroundings
Connecting surrS to sysH
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For a constant pressure process
qp = H
surrS  surrH = -sysH
surrS = -sysH / T
For a chemical reaction
sysH = rH
surrS = -rH/ T
The Use of univS to Determine
Spontaneity
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Calculation of TunivS  two system
parameters
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rS
rH
Define a system parameter that
determines if a given process will be
spontaneous?
The Definition of the Gibbs Energy
The Gibbs energy of the system
G = H – TS
 For a spontaneous process
sysG = Gf – G i
Gf = the Gibbs energy of the final state
Gi = the Gibbs energy of the initial
state
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Gibbs Energy and Spontaneity
sysG < 0 - spontaneous process
sysG > 0 - non-spontaneous
process (note that this process
would be spontaneous in the
reverse direction)
sysG = 0 - system is in equilibrium
Note that these are the Gibbs
energies of the system under nonstandard conditions
Standard Gibbs Energy Changes
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The Gibbs energy change for a
chemical reaction?
Combustion of methane.
CH4 (g) + 2 O2 (g) 
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CO2 (g) + 2 H2O (l)
Define
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rG =  np fG (products) -  nr fG
(reactants)
fG = the formation Gibbs energy of the
substance
Gibbs Energy Changes
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fG (elements) = 0 kJ / mole.
Use tabulated values of the Gibbs
formation energies to calculate the
Gibbs energy changes for chemical
reactions.
The Third Law of Thermodynamics
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Entropy is related to the degree of
randomness of a substance.
Entropy is directly proportional to
the absolute temperature.
Cooling the system decreases the
disorder.
The Third Law of Thermodynamics
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The Third Law - the entropy of any
perfect crystal is 0 J /(K mole) at 0
K (absolute 0!)
Due to the Third Law, we are able
to calculate absolute entropy
values.
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At a very low temperature, the
disorder decreases to 0 (i.e., 0 J/(K
mole) value for S).
The most ordered arrangement of
any substance is a perfect crystal!
Applications of the Gibbs Energy
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The Gibbs energy is used to
determine the spontaneous
direction of a process.
Two contributions to the Gibbs
energy change (G)
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Entropy (S)
Enthalpy (H)
G = H - TS
Spontaneity and Temperature
H
S
G
+
+
< 0 at high
temperatures
+
-
> 0 at all temperatures
-
+
< 0 at all temperatures
-
-
< 0 at low temperatures
Gibbs Energies and Equilibrium
Constants
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rG < 0 - spontaneous under
standard conditions
rG > 0 - non-spontaneous under
standard conditions
The Reaction Quotient
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Relationship between QJ and Keq
Q < Keq
- reaction moves in the forward
direction
Q > Keq
- reaction moves in the reverse
direction
Q = Keq
- reaction is at equilibrium
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rG° refers to standard conditions
only!
For non-standard conditions - rG
rG < 0 - reaction moves in the
forward direction
rG > 0 - reaction moves in the
reverse direction
rG = 0 - reaction is at equilibrium
Relating Keq to rG
rG = rG +RT ln Q
rG = 0  system is at equilibrium
rG = -RT ln Qeq
rG = -RT ln Keq
Phase Equilibria
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At the transition (phase-change)
temperature only - trG = 0 kJ
tr = transition type (melting,
vapourization, etc.)
trS = trH / Ttr