Ch.17 Flashcards
The Second Law of Thermodynamics - Entropy
The second law of thermodynamics tells us that spontaneous processes are those that raise the “entropy” of the universe.
• for any spontaneous process, the entropy of the universe increases (∆Suniv >0)
• Entropy is a state function so we normally concern ourselves with the change in entropy, ∆S = Sfinal- Sinitial
∆Suniv = ∆Ssys + ∆Ssurr > 0
• Processes which have ∆Ssys < 0 are not impossible, but must be coupled to some other process for which ∆S > 0 so that ∆Suniverse > 0
• Processes which have ∆Ssys > 0 may be non-spontaneous if they are coupled to other processes for which ∆S < 0 so that ∆Suniverse < 0
Entropy
• Entropy is often (over)simplified as a quantitative measure of “disorder”
• More correctly, entropy measures how dispersed the energy of the universe or a part of the universe is
• In other words, a state that requires a very specific arrangement of atoms (concentrated energy) is low entropy, while a state that allows for considerable variation (dispersed energy) is high entropy
• Entropy depends on how many different microstates can produce identical macrostates
Macro vs. Microstates
• Macrostate - describes the properties of the bulk system
– measurement parameters include things like pressure and temperature
— The container of gas has a specific volume, temperature and pressure
• Microstate - describes the properties of the individual molecules that make up the bulk system
– For example, the identity, orientation and velocity of every molecule in a sample of matter.
—At any given moment the gas molecules each have a velocity; va,vb,vc…
— Many combinations of velocities can satisfy the observed macrostate
Macro vs. Microstates – Phase Changes
- A sample of gas can have many different combinations of molecule positions and velocities that correspond to a given temperature, pressure and volume
- Condensing that gas into a liquid requires (nearly) all of the gas molecules to occupy a small portion of the container, giving fewer arrangements that correspond to the observed state functions
- Crystallizing the liquid into a solid further restricts the number of positions and orientations of molecules (microstates) that are consistent with the macrostate.
- The solid thus has a lower entropy than the liquid, and the liquid has a lower entropy than the gas.
Microstates & Entropy
• The more microstates that can describe a system, the higher the entropy of the system
• Opening the valve between the flasks of gas equalizes the pressure, because there are more ways of arranging the gas molecules across both flasks
Quantifying Entropy
• The Boltzmann equation describes the entropy of a system based on the number of microstates
S = k ln W
k = Boltzmann constant, R/NA
units are JK-1
W = # microstates
• The entropy of a state increases with the number of energetically equivalent ways to arrange the components of the system
• The state with the highest entropy also has the greatest dispersal of energy
• A state in which a given amount of energy is more highly dispersed has more entropy that a state in which the same energy is more highly concentrated.
• This allows us to define a state of zero entropy, when there is only one possible microstate
• A perfect crystal allows no deviations in the positions of the atoms or molecules, and zero temperature means none of the particles are moving.
Processes with ∆S > 0
• Melting, vapourization, sublimation
– Or reactions that turn solid starting materials into gaseous products
• Increase in temperature at a constant V
• Increase in the volume of a gas at constant P
• Increase in the number of moles of gas in a reaction
– Or any other process that increases the number of particles
Heat Transfer and Changes in Entropy
• Changes in enthalpy provide a means by which chemical reactions can be coupled to the rest of the universe
• Consider a process where entropy decreases, such as freezing of water
• We know that ∆Suniv must not decrease, so we must consider the entropy change of the system and
surroundings, ∆Ssys and ∆Ssurr
• ∆Suniv = ∆Ssys + ∆Ssurr
• Transfer of heat from a warmer place (the water) to a cooler place (the portion of the universe surrounding the water) results in an increase in entropy, which offsets the decrease in entropy resulting from the phase change
• Generally, exothermic processes increase Ssurroundings, while endothermic processes decrease Ssurroundings
– Hotter conditions = more possible combinations of individual molecular velocities (more microstates)
Temperature Dependence of ∆Ssurr
• The dispersal of energy to the surroundings and ∆Ssurr is temperature dependent
– The greater the temperature, the smaller the increase in entropy for a given amount of energy dispersed to the surroundings
– Adding heat to something very cold results in a larger increase in entropy than adding the same amount of heat to something warmer
Entropy is a measure of energy dispersal (joules) per unit temperature (kelvin)
Units: J/K
Quantifying ∆Ssurroundings
• A process that emits heat into the surroundings (qsys negative), increases the entropy of the surroundings (positive ∆Ssurr), exothermic reactions
• A process that absorbs heat from the surroundings (qsys postitive), decreases the entropy of the surroundings (negative ∆Ssurr), endothermic reactions
• The magnitude of the change in entropy of the surroundings is proportional to the magnitude of qsys and inversely proportional to temperature
∆Ssurr = qsurr/T = -qsys/T
• For a process occurring at constant temperature and pressure, the entropy change of the surroundings is equal to the energy dispersed into the surroundings (-∆Hsys) divided by the temperature of the surroundings in kelvin
∆Ssurr = -∆Hsys/T
• Note that this is not accounting for any work done on or by the system
Entropy of Phase Transitions
Consider a system that consists of a mixture of solid and liquid, or liquid and gas, that is at the phase change temperature and pressure
– The system is at equilibrium under these conditions
– At equilibrium, ∆Suniverse = 0, as the process is freely reversible
• A change to the lower entropy phase is exothermic
• A change to the higher entropy phase is endothermic
- For ∆Suniverse = 0, ∆Ssystem = - ∆Ssurroundings
• ∆Ssurr= -∆Hsys/T, and ∆Ssys = - ∆Ssurr,
• So ∆Sfusion = ∆Hfusion/Tm.p. and ∆Svaporization = ∆Hvaporization/Tb.p.
Calculating ∆S°rxn
• We’ve already seen the standard states for ∆H°
• Gas : 1 bar
• Liquid or Solid : pure substance at 1 bar and 25°C
• Substance in solution : concentration of 1 mol/L
• Standard entropy change for a reaction (∆S°rxn) is the change in entropy for a process in which all of the reactants and products are in their standard states
• We can use standard molar entropies (S°) to calculate ∆S°rxn
• ∆S°rxn = ∑nS°(products) - ∑nS°(reactants)
Standard Molar Entropies, S°
• There is a relative zero enthalpy, which was the standard enthalpy of formation for an element in its standard state
• There is no absolute zero for enthalpy
• Entropy has an absolute zero, so standard entropies for elements in their standard state is not zero
– Entropy can only be zero at 0K
• Standard molar entropies have the units JK-1mol-1
• Note that entropy (like enthalpy) is an extensive property and depends on the amount of the substance
Third Law of Thermodynamics
— The entropy of a perfect crystal at absolute zero (0K) is zero
• A perfect crystal at absolute zero has only 1 microstate (W=1), so the ln(W) term in the Boltzmann equation is zero and the entropy (S) is zero
S=k ln(W)
Entropy – Structure Relationships
Gaseous states of compounds have higher entropies than liquid or solid
• Entropy increases with increasing molar mass of elements in the same state
• Allotropes - materials with two or more forms (ie. graphite and diamond) can have different entropy values
• In a given state entropy usually increases with increasing molecular complexity
– N2(g) has more entropy than Ar(g) even though Ar(g) has a higher molecular mass
Gibbs Free Energy
• We have seen that ∆Ssurr = -∆Hsys/T
• We have also seen that ∆Suniv = ∆Ssys + ∆Ssurr
• Thus, ∆Suniv is related to both ∆Ssys and ∆Hsys
∆Suniv = ∆Ssys + ∆Ssurr
∆Suniv = ∆Ssys - ∆Hsys/T
-T∆Suniv = -T∆Ssys + T∆Hsys/T (multiply by -T)
-T∆Suniv = ∆H - T∆S (rearrange)
∆G = ∆H - T∆S (Gibbs energy)
∆G = -T∆Suniv
• If ∆G is negative then ∆Suniv is positive and the reaction is spontaneous
What does Gibbs Free Energy Tell Us?
• DG is negative for spontaneous reactions, positive for non-spontaneous reactions, and 0 for reactions at equilibrium
• When DG of a reaction is negative, it represents is the maximum amount of energy available to do work
• When DG of a reaction is positive, it is the minimum amount of energy required to make the reaction occur
• In real life, the efficiency with which free energy can be used to do work is usually less than 100%
– energy is lost to other processes such as friction, or heat escapes to the surroundings before being harnessed
• Regardless the efficiency, the ΔG value tells us the maximum amount of energy available or minimum required
∆H, ∆S, and T
Based on the equation for ∆G there can be 4 different combinations that effect spontaneity
- DH Negative, DS Positive
- DH Positive, DS Negative
- DH Negative, DS Negative
- DH Positive, DS Positive
Depending on the temperature we can see different effects on spontaneity
∆G°rxn and Temperature
∆G°rxn=∆H°rxn - T∆S°rxn
∆G is temp dependent
ENTHALPY CONTRIBUTION (∆H): exo, ∆H < 0
ENTROPY CONTRIBUTION (- T∆S): ∆S > 0
SPONTANEOUS? (∆G): yes at all temps, ∆G always -
ENTHALPY CONTRIBUTION (∆H): exo, ∆H < 0
ENTROPY CONTRIBUTION (- T∆S): ∆S < 0
SPONTANEOUS? (∆G): YES AT LOW TEMPS WHEN T∆S < ∆H
ENTHALPY CONTRIBUTION (∆H): endo, ∆H > 0
ENTROPY CONTRIBUTION (- T∆S): ∆S > 0
SPONTANEOUS? (∆G): YES AT HIGH TEMPS WHEN T∆S > ∆H
ENTHALPY CONTRIBUTION (∆H): endo, ∆H > 0
ENTROPY CONTRIBUTION (- T∆S): ∆S < 0
SPONTANEOUS? (∆G): no at all temps, ∆G always +
DH, DS and DG vs. DH°, DS° and DG°
• The ° indicates standard conditions:
• Temperature = 25°C or 298.15 K
• Gas pressure of 1 bar
• Pure liquids or solids
• Solution concentrations of 1 mol/L
• Sometimes the ° will be used when a temperature other than 25°C is specified
• This indicates that the pressures and concentrations are still standard even though the temperature is not.
DG°rxn and Temperature
If ∆H° and ∆S° for the reaction are known, we can estimate the transition temperature at which the reaction reverses spontaneously
• At the transition temperature, the point of transition from non-spontaneous to spontaneous, the equilibrium is standard state
– Thus DG°=0
– DG°=DH°-TDS° = 0 at Ttrans
- Rearranging gives: TTrans = DH° / DS°
Calculating DG°
There are three different ways to calculate DG°
– Using DG° = DH° - TDS°
– Using stepwise reactions
and combining their DG°
– Using DG°f, Gibbs energy
of formation
• The latter two are analogous to using Hess’s law with enthalpies
Estimating ∆G° at T other than 25 oC
• Assume that DH° and DS° do not change with temperature and continue to use the Gibbs- Helmoltz equation
– This is not strictly true, but is generally close if there are no phase changes associated with the temperature change
• Unlike DH and DS, DG is strongly dependent on temperature
• DG° calculated from tabulated data only applies at 298.15 K
Free Energy, Equilibrium, and the Reaction Direction
• If we begin with pure reactants, the reaction will spontaneously move towards equilibrium Q<K, ∆G<0
• When the reaction is at equilibrium Q=K, ∆G=0
• If we begin with pure products, the reaction will spontaneously move towards equilibrium Q>K, ∆G>0
• The relationship between DG and the direction of the reaction can be summarized in the following equation:
DG = RTIn(Q/K)
• If Q/K < 1 (Q<K), then ln(Q/K) < 0 and DG is negative:
—Reaction will proceed to the right (∆G<0)
• If Q/K > 1 (Q>K), then ln(Q/K) > 0 and DG is positive:
—Reaction will proceed to the left (∆G>0)
• If Q/K=1, then Q = K and DG = 0:
—The reaction is at equilibrium
• A large absolute value of DG indicates that a reaction is far from equilibrium, while a small DG indicates that a reaction is close to equilibrium
Standard Free Energy and K
• Under standard conditions, Q = 1
DG = RTIn(Q/K)
• Recall that ln(x) = -ln(1/x) and that ln(x/y) = ln(x) - ln(y)
• Thus: ∆G° = −RTln K
• And: DG = DG° + RTln(Q)
• We have just determined a relationship between the standard free energy and the equilibrium constant, as well as a relationship between the reaction quotient and the free energy under non-standard conditions
Significance of the Sign and Magnitude of DG°
DG° negative
• If DG°reaction is negative, G°products < G°reactants
– Reaction will proceed to the right (forward direction) to bump up the concentrations of products at the expense of reactants
– At equilibrium, K will be greater than one (product concentrations are larger than reactant concentrations
• When K>1, ln K is positive and ∆G°rxn is negative.
• Under standard conditions (when Q=1), the reaction is spontaneous in the forward direction
• Reaction will continue until total free energy is minimized
• The more negative the DG°, the larger the K
Significance of the Sign and Magnitude of DG°
DG° positive
• If DG°reaction is positive, G°products > G°reactants
– Reaction will proceed to the left (reverse direction) to bump up the concentrations of reactants at the expense of products
– At equilibrium, K will be less than one (reactant concentrations are larger than product concentrations)
• When K<1, ln K is negative and ∆G°rxn is positive.
• Under standard conditions (when Q=1), the reaction is spontaneous in the reverse direction
• Reaction will continue until total free energy is minimized
• The more positive the DG°, the smaller the K
Significance of the Sign and Magnitude of DG°
DG° = 0
• If ∆G° = 0, reaction is at equilibrium under standard conditions (K = 1)
• The smaller the absolute value of ∆°Grxn, the closer the standard state is to equilibrium
• The larger the absolute value of ∆°Grxn, the further the reaction has to go from standard.
• When K=1, ln K is zero and ∆G°rxn is zero.
• The reaction is at equilibrium under standard conditions.
Temperature Dependence of K
∆G°= -RT ln K & ∆G°= ∆rH° - T∆rS°
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| ∆Ho - T∆So = -RT ln K rearrange
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| V
| ln K = -∆H°/RT + ∆S°/R Hoff equation
V
ln K2/K1 = -∆H°/R(1/T2 - 1/T1) Two point curve
