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REVIEW OF THE FIRST LAW OF THERMODYNAMICS AND ENTHALPY
More detail can be found in Topic 5.
A. Energy can be changed from one form to another, but cannot be created
or destroyed.
B. One way to track and measure these energy changes is the amount of
heat released or absorbed.
C. Enthalpy is a state function that is used to describe heat changes taking
place at constant pressure.
D. For constant pressure conditions the heat transferred equals the change
in enthalpy ((H).
THE SECOND LAW OF THERMODYNAMICS AND ENTROPY
A. Spontaneous processes
1. Definition
A spontaneous process is a physical or chemical process that occurs by itself without any outside assistance.
2. Description
A spontaneous process is capable of proceeding in a given direction, as written or described, without needing to be driven by an outside source of energy.
It occurs spontaneously, or naturally, until it reaches equilibrium.
Chemists need to know whether, under a given set of conditions (pressure, temperature, concentration), a reaction will or will not occur.
A reaction that does occur under a given set of conditions is said to be spontaneous.
A reaction that does not occur under a given set of conditions is said to be nonspontaneous.
3. Examples of spontaneous versus nonspontaneous processes
SpontaneousNonspontaneous2 H2 + O2 ( 2 H2O2 H2O ( 2 H2 + O2
requires the addition of electrical energy 4 Fe + 3 O2 ( 2 Fe2O3 2 Fe2O3 ( 4 Fe + 3 O2 requires the addition of
heat energy
B. Entropy
1. Definition
a. Obsolete definition
(1) The superseded definition
Entropy is a measure of the randomness or disorder of a system.
(2) Reasons it has been superseded
(a) Simplistic, intuitive applications in the
macrouniverse
The state of your room tends to disorder unless you work to keep it neat.
(b) The need for a new approach to
nonintuitive examples, such as the hot
watercold water problem.
In a thermally isolated system exactly one gram of water at exactly 20 (C is mixed with exactly one gram of water at exactly 30 (C.
The resulting mixture has a temperature of exactly 25 (C.
The average disorder is the same yet the process is spontaneous.
b. Current definition
Entropy is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy.
Note: The concept of entropy as disorder may still be useful at times, but needs to be applied with care.
2. Description
When spontaneous changes occur there is an
increase in entropy.
Spontaneous changes tend to smooth out differences that exist in a system:
Temperature
Pressure
Density
Chemical potential energy
etc.
Entropy can be described as a measure of how far along this smoothingout process has progressed.
C. Second law of thermodynamics
1. Verbal definition stated in terms of the universe
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
2. Spontaneous processes
For a spontaneous process at a given temperature, the change in entropy of the system is greater than the heat divided by the absolute temperature.
(S >qT
When a process is spontaneous in one direction it is nonspontaneous in the reverse direction.
3. Equilibrium processes
The entropy of the universe remains unchanged in an equilibrium process.
(S =qT
D. A molecular interpretation of entropy
1. Molecular motions and energy
a. Heating a substance increases its temperature, which
increases the kinetic energy of its particles.
b. Particles have three types of motion
(1) Translational
From place to place
(2) Vibrational
Oscillating back and forth along the axis of the bond
(3) Rotational
Spinning around their common center of gravity only applies to molecules
2. Microstate
A microstate is the state of a system at a particular instant, i.e., one of many possible states of the system.
A microstate describes one specific, very detailed microscopic configuration of a system.
The system comes and goes among the microstates due to thermal fluctuations.
Note: The macrostate of a system refers to its macroscopic properties such as its temperature and pressure.
3. Boltzmanns Equation
Ludwig Boltzmann showed that the entropy of a system is related to the natural log of the number of microstates.
S = k ln W
S is the entropy of a system in J/K
k is the Boltzmann constant
1.38 x 10(23 J/K
W is the number of microstates
(S = Sfinal ( Sinitial
(S = k ln Wfinal ( k ln Winitial
(S = k lnWfinalWinitial
If the final condition has more microstates
than the initial condition, then Wfinal > Winitial
andlnWfinal> 0Winitial
Therefore (S > 0
4. Entropy and disorder
a. Example where increased disorder correlates to
increased entropy
Melting
In a solid, the particles are locked in place, and the number of microstates is small.
When the solid melts, the particles can now be arranged in many more ways, and the number of microstates increases.
The predicted order ( disorder increase in entropy applies.
b. Examples where increased disorder does not always
correlate to increased entropy
(1) The solution process usually, but not always,
leads to an increase in entropy.
(2) For molecular substances, the solution formation
process leads to an increase in entropy.
C6H12O6 (s) ( C6H12O6 (aq)
The liquid water has a limited number of microstates, and the solid glucose has an even more limited number of microstates.
The contribution to the increase in entropy comes from the solution process the mixing of the solute and the solvent.
When the sugar dissolves, the solution has a greater number of microstates than the sum of the microstates of the liquid water and the solid glucose.
(3) For salts whose cations come from Groups I A
and II A, the solution formation process leads
to an increase in entropy.
NaCl (s) ( Na+ (aq) + Cl( (aq)
There are two contributions to the increase in entropy:
The solution process the mixing of the solute and the solvent
The dissociation process the
separation of the cations from the anions
The dissociation process results in a larger number of particles and thus a greater number of microstates.
(4) For salts whose cations are small and highly
charged (such as Fe3+, Al3+, etc.), the solution
formation process does not always, lead to
an increase in entropy.
The two contributions to the increase in entropy (the solution process and the dissociation process) are still in play, but
the hydration process can offset the effect of those two.
For example, Fe3+ takes its hydrated form when dissolved in water:
Fe3+ + 6 H2O ( Fe(H2O)63+
The bonding of the water molecules to the small, highly charged cation reduces the
number of microstates available after mixing and dissociation increased them.
Depending on the actual numbers, the entropy change for the system can be negative and increased disorder may not correlate to increased entropy.
5. Entropy and temperature
Heating always increases the entropy.
As temperature increases, the kinetic energy increases (translational, vibrational, and rotational).
This increase in energy is distributed among many more possible energy levels.
Thus, at higher temperature, more microstates become available, and entropy increases.
6. Qualitative rules for comparing entropy
a. Four rules of thumb
(1) Entropy is larger from solid to liquid to gas.
Each phase has more opportunities for
motion than the one before it.
Therefore, each phase has more microstates than the one before it.
(2) Entropy is larger with a less ordered structure.
The less order there is to the structure, the more microstates that are available.
(3) Entropy is larger with greater complexity
of structure (more atoms in the molecule
or particle).
The greater complexity of the structure provides more opportunities for a greater number of molecular motions.
Therefore, there are more microstates and a higher entropy.
(4) Entropy is larger with larger molar mass
(or higher atomic number).
As the number of electrons increases the energy levels become more closely spaced and more levels become more available.
Therefore, there are more microstates and a higher entropy.
b. Examples
Which substance has the larger entropy?
Liquid water or solid water?
Liquid water liquid vs. solid
Methane (CH4) or ethane (C2H6)?
Ethane greater complexity of structure
Solid carbon (diamond) or solid carbon (graphite)?
Graphite diamond has the more highly ordered structure, so graphite would have the higher entropy.
Bromine vapor or liquid bromine?
Bromine vapor gas vs. liquid
Helium or neon?
Neon higher atomic number
THE THIRD LAW OF THERMODYNAMICS AND ENTROPY
A. The third law of thermodynamics
1. The third law of thermodynamics stated
The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero is zero.
2. The third law of thermodynamics described
At absolute zero all of the particles in the crystal lattice have no thermal motion at all.
Therefore, there is only one microstate.
Since S = k ln W, and since there is only one microstate
W = 1, and since ln(1) = 0, then at absolute zero S = 0.
B. Corollaries of the third law of thermodynamics
It is impossible to reach absolute zero in a finite number of operations.
The entropy of a pure substance approaches zero as the absolute temperature approaches zero.
A reference point, zero entropy at 0 K, exists that can be used to determine entropy relative to this point.
C. The third law of thermodynamics, absolute entropy, and standard
entropy
1. Absolute entropy
The entropy determined relative to this reference point of zero entropy at 0 K is the absolute entropy.
Absolute entropy represents the entropy change of a substance taken from absolute zero to a given temperature.
2. Standard entropy
a. Definition
Standard entropy is the entropy value for the standard state of a species.
Standard state is defined as 1 Molar for any aqueous ion; and 25 degrees Celsius (298.15 K) and a pressure of 1 atm for all species.
State of MatterStandard StateGas1 atmLiquidpure liquidSolidpure solidElementmost stable allotropic form at 25( C and 1 atmSolution1 M
b. Symbol
S(
c. Determination of standard entropy
In an equilibrium process
(S =qT
By making very, very small changes in temperature, and measuring the molar heat capacity (Cp) for that temperature change, (S for that temperature change can be approximated by
(S =CpT
where T is the average absolute temperature for that temperature range.
By starting as close to absolute zero as possible, and then making a series of molar heat capacity determinations with very, very small temperature changes one will eventually reach 298.15 K.
S( = 0 + (S1+(S2 +(S3
Since Cp/T may not be linear over large ranges for some substances, curve fitting is needed to produce a mathematical function.
By determining a mathematical function
we can say that S( is the integral from 0 to 298.15 K of (Cp/T)dT.
EMBED Excel.Chart.8 \s
If the substance is a solid at 298.15 K then:
S( = EMBED Equation.3
If the substance is a liquid at 298.15 K then:
S( = EMBED Equation.3 + EMBED Equation + EMBED Equation.3
If the substance is a gas at 298.15 K then:
S( = EMBED Equation.3 + EMBED Equation + EMBED Equation.3 + EMBED Equation
+ EMBED Equation.3
ENTROPY CHANGE FOR A REACTION
A. Standard entropy of reaction
1. Definition
The standard entropy of reaction is the difference in standard entropies between the products and the reactants.
2. Symbol
S(rxn
3. Formula
(S(rxn = S( (products) S( (reactants)
for the reaction:
aA + bB ( cC + dD
this would be:
(S(rxn = [c S( (C) + d S( (D)] ( [a S( (A) + b S( (B)]
B. Predicting the sign of the standard entropy of reaction
1. Rules of thumb for predicting entropy increases ((S > 0)
a. Entropy usually increases in a reaction in which a larger
molecule is broken up into two or more smaller
molecules.
b. Entropy usually increases in a reaction in which there
are more moles of gas on the product side than on the
reactant side.
c. Entropy usually increases in a process in which a solid
is changed to a liquid or a gas, or in which a liquid is
changed to a gas.
d. Entropy usually increases in a process in which the
temperature increases.
e. Entropy usually increases in a process in which the
volume increases.
f. Entropy usually increases in a process in which the
number of independently moving particles increases.
2. Examples
Would the entropy increase or decrease in the following
reactions or processes?
2 SO2 (g) + O2 (g) ( 2 SO3 (g)
Decrease(S negative
Fewer moles of gas on the product side
CaCO3 (s) ( CaO (s) + CO2 (g)
Increase(S positive
Larger molecule broken up
More moles of gas on the product side
Ag+ (aq) + Cl( (aq) ( AgCl (s)
Decrease(S negative
The number of independently moving particles decreases.
CS2 (l) ( CS2 (g) at constant temperature
Increase(S positive
Liquid changed to a gas
A 1liter container with 1 mole of N2 is connected to a 1liter container with 1 mole of O2 and the gases are allowed to diffuse.
Increase(S positive
The volume increases.
The number of microstates of each gas is increased as each has a greater volume in which it is found.
CO (g) + H2O (g) ( CO2 (g) + H2 (g)
Unknown by our rules
Whichever it is, the magnitude of the entropy change has to be small.
C. Calculating the standard entropy of reaction
Calculate the standard entropy change for the following reactions at 25( C
Values are taken from the CRC Handbook 20052006 Edition
1. CaCO3 (s) ( CaO (s) + CO2 (g)
(S(rxn = [1(CaO (s)) + 1(CO2 (g)) ( [1(CaCO3 (s))]
= [1 mol(38.1 J/mol K) + 1 mol (213.8 J/mol K)]
( [1 mol (91.7 J/mol K)]
= [251.9 J/K] ( [91.7 J/K]
= +160.2 J/K
2. 2 Na2O2 (s) + 2 H2O (l) ( 4 NaOH (aq) + O2 (g)
(S(rxn = [4(Na+ (aq) + OH( (aq)) + 1(O2 (g))]
( [2(Na2O2 (s)) + 2(H2O (l))]
= {4 mol [(58.45 J/mol K) + (( 10.90 J/mol K)]
+ 1 mol (205.15 J/mol K)}
( {2 mol (95.0 J/mol K) + 2 mol (69.95 J/mol K)}
= [4 mol (47.55 J/mol K) + 1 mol (205.15 J/mol K)]
( [2 mol (95.0 J/mol K) + 2 mol (69.95 J/mol K)]
= [190.2 J/K) + 205.15 J/K] ( [190.0 J/K) + 139.90 J/K)]
= [395.35 J/K] ( [329.9 J/K]
= 65.45 J/K
= 65.4 J/K
GIBBS FREE ENERGY
A. Was a proposal of Willard Gibbs to predict spontaneity.
1. Enthalpy and entropy were both known and studied state
functions.
Some reactions are spontaneous because they give off energy in the form of heat ((H < 0).
Other reactions are spontaneous because they lead to an increase in entropy ((S > 0).
However, what if enthalpy IS favorable, but entropy is NOTor vice versa
2. Gibbs proposed a new state function now called the Gibbs
free energy.
Gibbs related enthalpy and entropy
G = H TS
At constant temperature and pressure:
(G = (H T(S
The Gibbs free energy reflected the balance between enthalpy and entropy.
B. Spontaneity and free energy
1. Rule of thumb for reacting systems
Every system seeks to achieve a minimum of free energy.
2. Moving toward this minimum
When a reaction moves towards this minimum, G is negative.
This reaction is favored and energy will be released.
The energy released equals the maximum amount of work that can be performed as a result of the chemical reaction.
3. Moving away from this minimum
When a reaction moves away from this minimum, G is positive.
Energy, in the form of work, must be added in order for the reaction to go in the forward direction.
4. Spontaneity and the sign of (G
a. (G < 0
Forward reaction is spontaneous
b. (G = 0
The system is at equilibrium.
c. (G > 0
Reverse reaction is spontaneous
Forward reaction is nonspontaneous
For the reaction to occur in the forward direction as written, work must be supplied from the surroundings.
C. Temperature and free energy
1. The temperature at which a reaction occurs can affect its
spontaneity.
CaCO3 (s) (( CaO (s) + CO2 (g)
At 25( C (298.15 K)
From table of enthalpies and calculations:
(H( = 177.8 kJ/mol
From table of entropies and calculations:
(S( = 0.1605 kJ/mol
(G( = (H( T(S(
= (+177.8 kJ/mol)
( [(298.15 K)(0.1605 kJ/mol)]
= + 129.9 kJ/mol
nonspontaneous
At 925( C (1198.15 K)
(Assuming (H( and (S( do not vary with temperature which they do somewhat!)
(G( = (H( T(S(
= (+177.8 kJ/mol)
( [(1198.15 K)(0.1605 kJ/mol)]
= ( 14.5 kJ/mol
spontaneous
2. Comparing the enthalpy term and the entropy term in the Gibbs
free energy equation at various temperatures
(G = (H T(S
(G =(H+( T(S)Enthalpy
TermEntropy
Term
The sign of (G will tell us whether the reaction as written is spontaneous or not.
Both the enthalpy term and the entropy term will contribute to (G.
If (S is positive (meaning more microstates or greater disorder) then ( T(S) will be negative.
If (S is negative (meaning fewer microstates or less disorder) then ( T(S) will be positive.
If both (H and ( T(S) are negative, then (G is always negative, and the reaction is spontaneous at all temperatures.
If both (H and ( T(S) are positive, then (G is always positive, and the reaction is nonspontaneous at all temperatures.
If (H and ( T(S) have opposite signs, then temperature becomes critical because the value of T affects the magnitude of ( T(S).
An unfavorable (H can be overcome by a favorable
( T(S), and vice versa.
3. Rules of thumb regarding the signs of (H, (S, (G
and spontaneity.
(H(S(GComments++dependsHigh temperatures
(G < 0
Forward reaction spontaneous
Low temperatures
(G >0
Reverse reaction spontaneous((dependsHigh temperatures
(G > 0
Reverse reaction spontaneous
Low temperatures
(G < 0
Forward reaction spontaneous+((G > 0Reverse reaction spontaneous at all temperatures(+(G < 0Forward reaction spontaneous at all temperatures
D. Standard free energy changes
1. Definition
The standard free energy change of a reaction ((G(rxn) is the free energy change for a reaction when it occurs under standard state conditions when reactants in their standard states are converted to products in their standard states.
Standard state is defined as 1 Molar for any aqueous ion; and 25 degrees Celsius (298.15 K) and a pressure of 1 atm for all species.
State of MatterStandard StateGas1 atmLiquidpure liquidSolidpure solidElementmost stable allotropic form at 25( C and 1 atmSolution1 M
2. Standard free energy equation
For a reaction:
aA + bB ( cC + dD
(G(rxn = ( EMBED Equation ( ( EMBED Equation
(G(rxn = [c EMBED Equation (C) + d EMBED Equation (D)]
( [a EMBED Equation (A) + b EMBED Equation (B)]
E. Standard free energies of formation
1. Definitions
a. Standard free energy of formation
The standard free energy of formation ((G(f) of
a substance is the free energy change that occurs when 1 mol of a substance is formed from its elements in their most stable states at 1 atm and
25( C.
b. Standard free energy of formation of an element
The standard free energy of formation of an element is its most stable form is defined to be zero.
F. Calculating free energy changes
1. Calculating standard free energy from standard enthalpy and
standard entropy
Given standard enthalpies and standard entropies, calculate the standard free energy for the following reaction at 25(C:
CH4 (g) + O2 (g) ( CO2 (g) + H2O (g)
Dont forget to balance the equation:
CH4 (g) + 2 O2 (g) ( CO2 (g) + 2 H2O (g)
(H(
kJ/mol(S(
J/mol(KCH4 (g)(74.6+186.3O2 (g)0+205.152CO2 (g)( 393.5+213.8H2O (g)( 241.8+188.8
Values are taken from the CRC Handbook 20052006 Edition
(H(rxn = [c EMBED Equation (C) + d EMBED Equation (D)] ( [a EMBED Equation (A) + b EMBED Equation (B)]
= [1 mol(( 393.5 kJ/mol) + 2 mol(( 241.8 kJ/mol)]
( [1 mol((74.6 kJ/mol) + 2 mol(0)]
= [(877.1 kJ] ( [(74.6 kJ]
= ( 802.5 kJ
(S(rxn = [c S( (C) + d S( (D)] ( [a S( (A) + b S( (B)]
= [1 mol(+213.8 J/mol(K) + 2 mol(+188.8 J/mol(K)]
( [1 mol(+186.3 J/mol(K) + 2 mol(+205.152 J/mol(K)]
= [+591.4 J/K] ( [+596.6 J/K]
= ( 5.2 J/K
= ( 0.0052 kJ/K
(G( = (H( T(S(
= [( 802.5 kJ] [(298.15 K)((0.0052 kJ/K)]
= [( 802.5 kJ] + [1.55 kJ]
= ( 801.0 kJ
2. Calculating standard free energy from standard free energies
of formation
Given standard free energies of formation, calculate the standard free energy for the following reaction at 25(C:
C2H5 (l) + O2 (g) ( CO2 (g) + H2O (g)
Dont forget to balance the equation:
C2H5OH(l) + 3 O2 (g) ( 2 CO2 (g) + 3 H2O (g)
(G(rxn = [c EMBED Equation (C) + d EMBED Equation (D)] ( [a EMBED Equation (A) + b EMBED Equation (B)]
Values are taken from the CRC Handbook 20052006 Edition
= [2 mol((394.4 kJ/mol) + 3 mol((228.6 kJ/mol)]
( [3 mol(0)] + 1 mol((174.8 kJ/mol)]
= [((788.8 kJ) + ((685.8 kJ)] [(174.8 kJ]
= [((788.8 kJ) + ((685.8 kJ)] [(174.8 kJ]
= (1474.6 kJ + 174.8 kJ
= (1299.8 kJ
G. Equilibrium and free energy
1. The approach to equilibrium and the change in the free energy
In any spontaneous process at constant temperature and pressure, the free energy decreases.
This decrease in free energy continues until it reaches a minimum value.
When this minimum value is reached, the system is at equilibrium.
2. The thermodynamic equilibrium constant
a. Definition
The thermodynamic equilibrium constant (K) is the equilibrium constant in which the concentrations of gases are expressed in partial pressure in atmospheres, whereas the concentrations of solutes in liquid solutions are expressed in molarities.
b. The relationship among K, Kc, and Kp.
For a reaction involving only solutes in liquid solution K is identical to Kc.
For a reaction involving only gases K is identical
to Kp.
c. Free energy change involving nonstandard states
When reactants are in nonstandard states and the products are in nonstandard states the following equation can be used:
(G = (G( + RT ln Q
R = 8.3145 x 10(3 kJ/K(mol
T = temperature Kelvin
Q = reaction quotient
At equilibrium the free energy ceases to change, so (G = 0 and the reaction quotient becomes the equilibrium constant.
0 = (G( + RT ln K
(G( = ( RT ln K
Also useful is an equation allowing the calculation of K from (G(.
ln K =(G(( RT
K = e((G(/RT
3. Calculating K from (G(
a. Find the value of K at 25 (C for the following reaction:
N2O4 (g) (( 2 NO2 (g)
(G(rxn = [c EMBED Equation (C) + d EMBED Equation (D)] ( [a EMBED Equation (A) + b EMBED Equation (B)]
Values are taken from the CRC Handbook 20052006 Edition
(G(rxn = [2 mol(+51.3 kJ/mol)] ( [1 mol(+99.8 kJ/mol)]
= 2.8 kJ
K = e((G(/RT
= e((2.8)/RT
= EMBED Equation
K = 0.32
b. Find the Ksp for the following:
Mg(OH)2 (s) (( Mg2+ (aq) + 2 OH( (aq)
(G(rxn = [c EMBED Equation (C) + d EMBED Equation (D)] ( [a EMBED Equation (A) + b EMBED Equation (B)]
= [2 mol((157.3 kJ/mol) + 1 mol((454.8 kJ/mol)]
( [1 mol((833.7 kJ/mol)]
= [(769.4 kJ] ( [(833.7 kJ]
= + 64.3 kJ
Ksp = e((G(/RT
= e((64.3)/RT
= EMBED Equation
Ksp = 5.43 x 10(12
PAGE
PAGE 22
Topic 20 Thermodynamics
2008 Lloyd Crosby
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