Thermodynamics Entropy, Energy and equilibrium
description
Transcript of Thermodynamics Entropy, Energy and equilibrium
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ThermodynamicsEntropy, Energy and
equilibrium
Chapter 19
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19.1 Thermodynamics
Thermo: heat Dynamics: power Study of energy flow and its
transformations (heat and energy flow) Determines direction of reactions
(spontaneous or nonspontaneous under given conditions)
State Functions: considers only initial and final states
Does not consider pathways or rate Organized into three laws:
1st Law (ΔU = qp + w = qp – PΔV ) 2nd Law 3rd Law
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19.1 Basic Definitions
System Surrounding Open system Closed system Isolated system State of system: defined by values of
composition, pressure, T, V. State Function: defined only by initial and
final condition of the system (Enthalpy, Entropy, Gibbs Free Energy).
Energy change signaled by: accomplishment of work and/or appearance or disappearance of heat.
system surroundings
universe
system surroundings
universe
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19.1 1st Law of Thermodynamics
1. A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy?
(21 J)2. The internal energy of a fixed quantity of ideal
gas depends only on its temperature. If a sample of an ideal gas is allowed to expand against a constant pressure at a constant temperature,a) What is ΔU for the gas? b) Does the gas do work?c) If any heat exchanged with the surroundings?
a) 0 b) w = -P ΔV c)no, only work
done by gas is energy leaving the system. Internal energy should decrease, so the temp; but temp is const.; therefore the int energy does not change; gas has to absorb enough heat from surroundings to compensate for the work. Q = -w; ΔU = -w+w
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19.1 First Law of Thermodynamics 1st Law: Energy can be neither created nor
destroyed, but it can be converted from one form to another or transferred from a system to the surroundings or vice versa. Energy of the universe is constant
Important concepts from thermochemistry Enthalpy Hess’s law
Purpose of 1st Law Energy bookkeeping
How much energy? Exothermic or endothermic? What type of energy? Δu = q + w q = heat; w = work system does on the
surroundings (-PΔV)
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19.2 Spontaneous Processes: Expansion
Spontaneous processes are those that can proceed without any outside intervention. Product favored at equilibrium. Product-favored at
equilibrium May be fast or slow May be influenced by
temperature The gas in vessel B will
spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously separate
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19.1 Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.
Examples: rusting, neutralization reaction, dissolution of sugar in water, heat flow, expansion of gas, spontaneous combustion (CH4 + O2), reaction of sodium with water
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19.1 Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0C it is spontaneous for ice to melt. Below 0C the reverse process is
spontaneous.
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19.2 Spontaneous vs. Nonspontaneous
3. Determine if the following processes are spontaneous or not and if they are exothermic or endothermic.(a) Gases expand into larger volumes at constant temperature ________________(b) H2O(s) melts above 0C ____________
(c) H2O(l) freezes below 0C ____________
(d) NH4NO3 dissolves spontaneously in H2O ___________
(e) Steel (iron) rusts in presence of O2 and H2O ________
(f) Wood burns to form CO2 and H2O __________
(g) CH4 gas burns to form CO2 and H2O __________
Evolution of Heat (Exothermicity) : not enough to predict spontaneity
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19.2 Spontaneous vs Nonspontaneous
Nonspontaneous process Does not occur unless there is outside
assistance (energy?) Reactants-favored at equilibrium
All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction Spontaneous processes have a definite
direction Spontaneous processes are irreversible. Can
be reversed with considerable input of energy.
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19.2 Factors That Favor Spontaneity
Spontaneous Processes driven by Enthalpy, H (Joules)
Many, but not all, spontaneous processes tend to be exothermic.
Entropy, S (Joules/K) Measure of the disorder of a system Many, but not all, spontaneous
processes tend to increase disorder of the system
Exothermicity favors spontaneity, but does not guarantee it.
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19.2 Factors That Favor Spontaneity: Enthalpy
Example of spontaneous reaction that is not exothermic:
NH4NO3(s) → NH4+ (aq) + NO3
-(aq) ΔH = 25 kJ/mol
Expansion of gas: energy neutral Phase changes: endothermic processes
that occurs spontaneously (ice to water). Chemical system: H2(g) + I2(g) ↔ 2HI(g)
Equilibrium can be approached from both sides (spontaneous both ways) even though the forward reaction is endothermic and the reverse is exothermic.
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19.2 Spontaneity: Examples
4. Based on your experience, predict whether the following processes are spontaneous, are spontaneous in reverse direction, or are in equilibrium:
(a) When a piece of metal heated to 150 ºC is added to water at 40 ºC, the water gets hotter.
(b) Water at room temperature decomposes into hydrogen and oxygen gases
(c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 ºC.
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19.2 Reversible Processes
In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.Example: melting ice at its melting point
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19.2 Irreversible Processes
Irreversible processes cannot be undone by exactly reversing the change to the system. Different path has to be used.
Spontaneous processes are irreversible. Example: expansion a gas into vacuum.
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19.2 Entropy
Entropy (S) is a term coined by Rudolph Clausius in the 19th century.
Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q
T
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19.2 Entropy Direct measure of the randomness or
disorder of the system. Related to probability
describes # of ways the particles in a system can be arranged in a given state (position and/or energy levels)
The most likely state – the most random More possible arrangements, the higher
disorder, higher entropy Ordered state – low probability of occurring Disordered state: high probability of
occurring
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19.2 Entropy on the Molecular Scale
Ludwig Boltzmann described the concept of entropy on the molecular level.
Temperature is a measure of the average kinetic energy of the molecules in a sample.
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19.2 Entropy on the Molecular Scale
Molecules exhibit several types of motion: Translational: Movement of the entire molecule
from one place to another. Vibrational: Periodic motion of atoms within a
molecule. Rotational: Rotation of the molecule on about an
axis or rotation about bonds.
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19.2 Entropy on the Molecular Scale
Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the
molecules. He referred to this sampling as a microstate of
the thermodynamic system.
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19.2 Entropy on the Molecular Scale
Each thermodynamic state has a specific number of microstates, W, associated with it.
Entropy isS = k lnW
where k is the Boltzmann constant, 1.38 1023 J/K;W: number of microstates
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19.2 Entropy on the Molecular Scale
The change in entropy for a process, then, is
S = k lnWfinal k lnWinitial
lnWfinal
lnWinitial
S = k ln
• Entropy increases with the number of microstates in the system.
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19.2 Spontaneous Processes: Dispersal of Matter
Isothermal (constant temperature) expansion of gas
After opening stopcock the molecules could be in any arrangement shown (4 arrangements)Probability for each arrangement = (1/2)2
S = k (ln(4)
25% probability
Two molecules present:
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Gas Container = two bulbed flask
Gas Molecules
Ordered State
19.2 Spontaneous Process: Isothermal Gas Expansion
Consider why gases tend to isothermally (constant temp.) expand into larger volumes.
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Gas Container
S = k ln (W) = k (ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K
For 3 particles, probability = (1/2)3
For N particles, probability = (1/2)N
Ordered State
19.2 Spontaneous Process: Isothermal Gas Expansion
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More probable that the gas molecules will disperse between two halves than remain on one side
Disordered States
27Driving force for expansion is entropy (probability); gas
molecules have a tendency to spread out
Disordered States
28S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K
Disordered States
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Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K
Total Arrangements
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19.2 Entropy on the Molecular Scale
The number of microstates and, therefore, the entropy tends to increase with increases in Temperature. Volume. The number of independently moving
molecules.
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19.2 Entropy and Temperature
(a) A substance at a higher temperature has greater molecular motion, more disorder, and greater entropy than (b) the same substance at a lower temperature.
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19.2 Entropy and Physical States
Entropy increases with the freedom of motion of molecules.
Therefore,S(g) > S(l) > S(s)
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19.2 Entropy Changes
In general, entropy increases when Gases are formed from
liquids and solids. Liquids or solutions are
formed from solids. The number of gas
molecules increases. The number of moles
increases.
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19.2 Entropy in Temperature
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Sta
ndard
entr
op
y,
S(
J/K
)
Temperature (K)0
20
10
30
40
50
10050 250 300150 200
What kind of changes are represented here?
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19.2 Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
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19.2 Patterns of Entropy Change
6. Describe in words the entropy of the system
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19.2 Entropy
Like total energy, E, and enthalpy, H, entropy is a state function.
Therefore, S = Sfinal Sinitial
S > 0 represents increased randomness or disorder
Note: The magnitude of change in entropy depends on temperature.
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19.2 Entropy
For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:
S =qrev
T
Units: Joule/K
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19.2 Entropy: Example
5. ΔS = q/TThe element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 ºC, and its molar enthalpy of fusion is ΔHfusion = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg(l) freezes at the normal freezing point?
(-2.48 J/K)
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19.2 Entropy: Examples
7. Predict if ΔS increases, decreases or does not change
(a) Freezing liquid mercury(b) Condensing H2O(vapor)
(c) Precipitating AgCl(d) Heating H2(g) from 60.0 ºC to 80 ºC
(e) Subliming iodine crystals(f) Rusting iron nail
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19.2 Entropy - Examples
8. Predict which substance has the higher entropy:a) NO2(g) or N2O4(g)b) I2(g) or l2(s)
9. Predict whether each of the following leads to increase or decrease in entropy of a system If in doubt, explain why.a) The synthesis of ammonia:
N2(g) + 3H2(g) ↔ 2NH3(g)
b) C12H22O11(s) → C12H22O11(aq)
c) Evaporation to dryness of a solution of urea, CO(NH2)2 in vapor.
CO(NH2)2(aq) → CO(NH2)2(s)
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19.3 Second Law of Thermodynamics
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19.3 Second Law of Thermodynamics: System
10. Predict the sign of ΔS0 for each of the following reactions:a) Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s)
b) MgCO3(s) → MgO(s) + CO2(g)
d) H2(g) + Br2(g) → 2HBr(g)
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Second Law of ThermodynamicsThe second law of thermodynamics states that theentropy of the universe increases for spontaneousprocesses, and the entropy of the universe doesnot change for reversible processes.
In other words:For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
For nonspontaneous Process:
ΔS univ= ΔS syst. + ΔS surr. <0
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Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.
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19.3 Second Law: Entropy Changes
EQUILIBRIUM PROCESSES (reversible)
♦ΔS universe= ΔS syst. + ΔS surr. =0
♦ΔS syst = ΔS surr
♦ΔS syst = ΔSº final - ΔS0 initial
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19.3 Entropy Changes in a System (Reactions)
Entropy changes in a system aA + bB → cC + dD Standard entropy change ΔSº (25 ºC, 1atm). Only changes in entropy can be measured. Each element has an entropy value (compare to
enthalpy). Absolute value for each substance can be
determined. For a chemical system:
S° = nS°(products) - mS°(reactants)
where n and m are the coefficients in the balanced chemical equation.
Standard Molar entropy, S0, is the entropy of one mole of a substance in its standard state (298 K)
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19.2 Standard Entropies
These are molar entropy values of substances in their standard states.
Standard entropies tend to increase with increasing molar mass.
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19.2 Standard Entropies
Larger and more complex molecules have greater entropies.
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11. Using standard molar entropies, calculate S°rxn for the following reaction at 25°C:
2SO2(g) + O2(g) → 2SO3(g)
S° = 248.1 205.1 256.6 (J · K-
1mol-1)
(Ans.: -187.9 J/K)
19.3 Second Law: Example
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19.3 Entropy of Reactions (System)
12. Using thermodynamic tables, calculate the standard entropy changes for the following reactions at 25 ºC
a) Evaporation of 1.00 mol of liquid ethanol to ethanol vapor.
b) The oxidation of one mole pf ethanol vapor (combustion reaction)c) Are the reactions spontaneous under the given conditions.
Answers:a) 122.0 J/K b) 96.09 J/K
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19.3 Entropy Changes in the System
In a reaction More gas molecules produced:
entropy increases Less gas molecules produced:
entropy decreases No net change of # of gas molecules
produced: entropy changes, but slightly
Liquid, solid products: hard to estimate-needs calculations
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19.3 Entropy of Surrounding
(a)When an exothermic reaction occurs in the system (ΔH < 0), the surroundings gain heat and their entropy increases (Δ Ssurr > 0).
(b) When an endothermic reaction occurs in the system (Δ H > 0), the surroundings lose heat and their entropy decreases (Δ Ssurr < 0).
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19.3 Entropy Changes in Surroundings
Heat that flows into or out of the system changes the entropy of the surroundings.
For an isothermal process:
Ssurr =qsys
T
• At constant pressure, qsys is simply H for the system. (negative sign needed to make entropy positive in an exothermic process)
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19.3 Second Law of Thermodynamics (System)
13. Examples: H2O(s) melts above 0C (endothermic). What
about entropy? Steel (iron) rusts in presence of O2 and H2O
(exothermic). What about entropy? 4Fe(s) + 3O2(g) 2Fe2O3(s)
CH4 gas burns to form CO2 and H2O (exothermic). What about entropy?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
Each process increases entropy of the universe.
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19.3 Second Law of Thermodynamics Consider transferring the same amount of heat to two
different systems, one at 298K and another at 500K.
More entropy (disorder) is created in the system at lower temp.
At high temperatures the system is already disordered
T
qS rev
298K 500KQ Q
K
q
K
q
500298
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19.3 Second Law of Thermodynamics To determine Suniv for a process, both Ssystem and
Ssurroundings need to be known:
Ssystem
related to matter dispersal in system Ssurroundings
determined by heat exchange between system and surroundings and T at which it occurs
Sign of Ssurr depends on whether process in system is endothermic (Ssurr< 0) or exothermic (Ssurr >0)
Magnitude of Ssurr depends on T
Ssurr = -Hsystem/T
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19.3 Second Law of Thermodynamics
14. Reaction: N2(g) + 3H2(g) → 2NH3(g)
ΔHº = -92.6kJ ΔSsys = -199 J/K at 25 ºC
ΔSsurr = -(-92.6 x1000)J/298K = 311 J/K
ΔSuniv = -199 J/K + 311 J/K = 112 J/K
Reaction spontaneous at 25 ºC
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ΔSsystem ΔS surr ΔSuniv Spontaneity
+ + + Yes
- - - No, reaction towards reactants
+ - ? Yes, if ΔSsys> ΔSsurroun
- + ? Yes, if ΔSsurr> ΔSsyst
19.3 Conditions for Spontaneity
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19.3 Entropy and Spontaneity
15. Consider the vaporization of liquid water to steam at a pressure of 1 atm. a) Is this process endothermic or exothermic, explain.b) In what temperature range is the process spontaneous?c) In what temperature range is it a nonspontaneous process?d) At what temperature are the two phases at equilibrium?
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19.3 Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero is 0.Perfect crystal: its internal arrangement is absolutely regular. Nothing is in motion (vibrations, rotations and translations)
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Third Law of Thermo
Gives us a starting point, S at 0K is equal to zero.
All others must be >0.
Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.
Products - reactants to find Sº (a state function)
More complex molecules higher Sº.
19.3 Entropy and Third Law
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19.4 Gibbs Free Energy
Heat that flows into or out of the system changes the entropy of the surroundings.
For an isothermal process:
Ssurr =qsys
T
• At constant pressure, qsys is simply H for the system.
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19.4 Gibbs Free Energy
The universe is composed of the system and the surroundings.
Therefore,Suniverse = Ssystem + Ssurroundings
For spontaneous processes
Suniverse > 0
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19.4 Gibb’s Free Energy
This becomes:
Suniverse = Ssystem +
Multiplying both sides by T,
TSuniverse = Hsystem TSsystem
Hsystem
T
TΔS universe is defined as the Gibbs free energy, G.When Suniverse is positive, G is negative.Therefore, when G is negative, a process is spontaneous.
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19.4 Gibb's Free Energy at any Conditions
G=H-TS Never used this way. At constant temperature
G=Hsys –TSsys
G function eliminates the need to deal with entropy of the surroundings
If G is negative at constant T and P, the process is spontaneous.
We deal only with the SYSTEM.
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19.4 Gibbs Free Energy
The Gibbs Free Energy is a measure of the maximum amount of work, at a given temperature and pressure, that can be done on the surroundings by a system.
Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.
Wmax = G For a spontaneous process:
Maximum amount of energy released by the system that can do useful work on the surroundings
Energy available from spontaneous process that can be used to drive non-spontaneous process.
For a nonspontaneous process: Minimum amount of work that must be done to
force the process to occur
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19.4 Gibbs Free Energy
Summary of Conditions for Spontaneity G < 0
reaction is spontaneous in the forward direction
(Suniv > 0) G > 0
reaction is nonspontaneous in the forward direction
(Suniv < 0) G = 0
SYSTEM IS AT EQUILIBRIUM (Suniv = 0)
Remember- Spontaneity tells us nothing about rate.
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19.4 Gibbs Standard Energy Free Energy
The standard free energies of formation, G°, are the free energy values for the formation of a substance under standard conditions.
The standard free energy of formation for any element in its standard state is zero. (Like enthalpy, but not entropy)
Compare: G = Hsys - TSsys (any conditions)
Gº = H0sys - TS0
sys (standard conditions)
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19.4 Convention for Standard States
State of Matter Standard State
Gas 1 atm pressure
Liquid Pure liquid
Solid Pure solid
Elements Gºf = 0
Solution 1 molar
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19.4 Free Energy in Reactions
Gº = standard free energy change. Free energy change that will occur if reactants in their
standard state turn to products in their standard state. Can’t be measured directly, can be calculated from
other measurements. The reaction:
aA + bB → cC + dD
Gºrxn = ΣnGº (products) - ΣmGº (reactants)
or Gº=Hº-TSº
Use adapted Hess’s Law with known reactions.
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19.4 Free Energy Changes
At temperatures other than 25°C,G ° = H TS How does G change with temperature? There are two parts to the free energy
equation: H — the enthalpy term TS — the entropy term
The temperature dependence of free energy, then comes from the entropy term.
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19.4 Gibbs free Energy: Example
16. For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0?
Look at the equation G=H-TS Spontaneity can be predicted from the
sign of H and S.
(ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C)
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ΔH ΔS -|TΔS| ΔG = ΔH-TΔS Reaction Characteristics
Example
- + -Always negative Spontaneous at
all temperatures2O3(g) →3O2(g)
+ - +Always positive
Nonspontaneous at all temperatures
3O2(g) →2O3(g)
- - +
Negative at low T; positive at high T
Spontaneous at low T; nonspontaneous at high T. Enthalpy driven
H2O(l) →H2O(s)+
+ + -
Positive at low T; negative at high T
Nonspontaneous at low T; becomes spontaneous at high T. Entropy driven.
H2O(s) → H2O(l)
19.4 Gibbs Free Energy and SpontaneityEffect of Temperature on the Spontaneity of
Reactions
1
2
3
4
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Free energy change as a function of temperature.
19.4 Gibbs Free Energy and Spontaneity
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19.4 Gibbs Free Energy and Temperature
Spontaneous Processes H2O(s) melts above 0C (endothermic)
H > 0 (endo process), S > 0 ENTROPY DRIVEN
Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) 4Fe(s) + 3O2(g) 2Fe2O3(s) H < 0 (Exo) , S < 0 (entropy decreases) ENTHALPY DRIVEN
G < 0 for each process T determines sign of G: G = H -TS
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19.4 Gibbs Free Energy: Example (p. 748)
17. Predict which of the four cases in Table (slide #78) you expect to apply to the following reactions:a) C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(g) ΔH = -2540 kJ
b) Cl2(g) → 2Cl(g)
18. Calculate ΔG0 at 298 K for the reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) ΔH0 = 114.4 kJ
a) using Gibbs free Energy equationb) from standard free energies of formation.
( -76.0 kJ)
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19.4 Gibbs free Energy: Example
18. For the reaction SO2(g) + 2H2S(g) →3S(s) + 2H2O(g)
Calculate the temperature at which ΔG0 = 0Values of ΔH0, kJ/mol ΔS0 (kJ/(K mol) SO2 -296.8 0.2481
H2S -20.6 0.2057S 0.0 0.0318H2O -241.8 +0.1887
Answer: 780K
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19.4 Temperature and Chemical Reactions – Problem Set
19. Calculate the temperature at which the reaction:
CaCO3(s) → CaO(s) + CO2(g) becomes spontaneous.
ΔS º = 160.5 J/K; ΔH º = 177.8 kJ
Answer: 835 ºC
20. At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should be its approximate normal boiling point temperature be? ΔS0 of vaporization is 87 J mol-1 K-1
(570 K)
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19.4 Gibbs Free Energy
21. Consider the reaction C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH0 = -2220
kJ(a) Without using data from the Thermodynamic
Tables, predict whether ΔG0 for this reaction is more negative, or less negative than ΔH0.
(b) Use data from the Tables to calculate ΔG0 at 298 K. Is your prediction correct?Values of ΔG0 (in kJ/mol): C3H8(g), -23.5; O2(g), 0; CO2(g) -394.4; H2O(l) -237.2
(a) less negative; (b) -2108 kJ
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19.5 Gibbs free Energy and Equilibrium
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19.5 Free Energy and Equilibrium
ΔG and ΔGº are not the same. ΔG = ΔH – T ΔS
ΔGº = ΔH º – T ΔS º ΔGº is only at standard conditions
(values obtained from Tables)ΔGº = Gº(products) - ΔGº
(reactants)
ΔG any conditions (no Tables of values available)
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19.5 Free Energy and Equilibrium
Predicament: we start a reaction with all reactants in standard state (1 atm, 25ºC, 1M solution). Is the standard state preserved as the reaction progresses?
It can be shown mathematically that at non-standard conditions:
]tan[
[products] quotient reaction ,
tsreacQ
ΔG = ΔGº +RTlnQ
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If ΔG is negative, the forward reaction is spontaneous.
If Δ G is 0, the system is at equilibrium.If G is positive, the reaction is spontaneous in the
reverse direction.
K>Q
K<Q
K=Q
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19.5 Free Energy and the Equilibrium Constant
At equilibrium:
eq
eq
KRTG
KRTG
ln
ln0
From the above we can conclude:If G < 0, then K > 1 Product favoredIf G = 0, then K = 1 EquilibriumIf G > 0, then K < 1 Reactants favored
K = e-(ΔGº /RT)
Kc used for solutions and molarities, Kp used for gases
Useful equation to determine small
Keq
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19.5 Temperature Dependence of K
Gº= -RTlnK = Hº - TSº
ln(K) = Hº/R(1/T)+ Sº/R
A straight line of lnK vs 1/T Slope?
Y-intercept?
19.5 Free Energy and Chemical Equilibrium
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19.5 Reaction Path and ΔGº
ΔGº = -RTln K
Value of ΔGº Sign of K Path of reaction
Negative, <0 K>0 Products are favored
Positive, >0 K<0 Reactants are favored
zero, =0 K = 1 Equilibrium
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19.5 Keq: Problem Set – 19.6 (p. 755)
22. Determine the value of Keq at 25 ºC for the reaction:
2NO2(g) ↔ N2O4(g).
Note: calculate ΔG0 from tables Use ΔG0 = -RTln Kp
( 6.9)
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19.5 Keq: Problem Set 19.7 (p. 755)
23. Using the solubility product of of silver iodide at 25 ºC (8.5 x 10-17), calculate ΔGº for the process:
AgI(s) ↔ Ag+(aq) + I-(aq)Answer:24. Estimate the value of ΔS0
298 for the dissociation of copper (II) oxide.
CuO(s) ↔ 2Cu2O(s) + O2(g) ΔH0298 = 283 kJ
( 0.203 kJ K-1
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19.5 Gibbs and Equilibrium: Example
25. Calculate Grxn for the reaction below:
2A(aq) + B(aq) C(aq) + D(g) if G°rxn = 9.9 x 103 J/mol and 150 °C(a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05
M, and PD = 0.05 atm, and (b) (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5
M, and PD = 0.5 atm. (c) Is the reaction spontaneous under
these conditions?
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26. Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard
state conditions? (ans.: 27 kJ)
27. Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions? For the reaction below, K = 1.0 x 1014.
H+ + OH- H2O
(ans.: -80 kJ)
19.5 Gibbs and Equilibrium: Example
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28. Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored?
(ans.: a) 0.15; b) 216; 250C)
19.5 Gibbs and Equilibrium: Example
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19.5 Summary
First Law of Thermodynamics: ΔU = Δq + Δw
Second Law: ΔSuniv = ΔS sys + ΔSsurr
ΔS sys= ΔH/TΔG = ΔH – T ΔSΔG = ΔGº +RTlnK
ΔGº = -RTlnK
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19.5 Gibbs Free Energy
Many biological reactions essential for life are nonspontaneous Spontaneous reactions used to “drive”
the nonspontaneous biological reactions Example: photosynthesis
6CO2 + 6H2O C6H12O6 + 6O2 G > 0
What spontaneous reactions drive photosynthesis?
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19.5 Entropy and Life Processes
If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible? growth of a complex life form
represents an increase in order (less randomness) lower entropy
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19.5 Entropy and Life Processes
CO2
H2
Oheat
Organisms “pay” for their increased order by increasing Ssurr.
Over lifetime, Suniv > 0.
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