Ch.17

Question 1

The Second Law of Thermodynamics - Entropy

Answer 1

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

Question 2

Entropy

Answer 2

• 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

Question 3

Macro vs. Microstates

Answer 3

• 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}

Question 4

Macro vs. Microstates – Phase Changes

Answer 4

• 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.

Question 5

Microstates & Entropy

Answer 5

• 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

Question 6

Quantifying Entropy

Answer 6

• 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.

Question 7

Processes with ∆S > 0

Answer 7

• 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

Question 8

Heat Transfer and Changes in Entropy

Answer 8

• 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)

Question 9

Temperature Dependence of ∆Ssurr

Answer 9

• 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

Question 10

Quantifying ∆Ssurroundings

Answer 10

• 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

Question 11

Entropy of Phase Transitions

Answer 11

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.

Question 12

Calculating ∆S°rxn

Answer 12

• 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)

Question 13

Standard Molar Entropies, S°

Answer 13

• 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)

Question 14

Entropy – Structure Relationships

Answer 14

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

Question 15

Gibbs Free Energy

Answer 15

_{• 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

Question 16

What does Gibbs Free Energy Tell Us?

Answer 16

• 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

Question 17

∆H, ∆S, and T

Answer 17

Based on the equation for ∆G there can be 4 different combinations that effect spontaneity

1. DH Negative, DS Positive
2. DH Positive, DS Negative
3. DH Negative, DS Negative
4. DH Positive, DS Positive

Depending on the temperature we can see different effects on spontaneity

Question 18

∆G°rxn and Temperature

Answer 18

∆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 +

Question 19

DH, DS and DG vs. DH°, DS° and DG°

Answer 19

• 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.

Question 20

DG°rxn and Temperature

Answer 20

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°

Question 21

Calculating DG°

Answer 21

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

Question 22

Estimating ∆G° at T other than 25 oC

Answer 22

• 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

Question 23

Free Energy, Equilibrium, and the Reaction Direction

Answer 23

• 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

Question 24

Standard Free Energy and K

Answer 24

• 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

Question 25

Significance of the Sign and Magnitude of DG°

DG° negative

Answer 25

• 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

Question 26

Significance of the Sign and Magnitude of DG°

DG° positive

Answer 26

• 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

Question 27

Significance of the Sign and Magnitude of DG°

DG° = 0

Answer 27

• 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.

Question 28

Temperature Dependence of K

Answer 28

∆G°= -RT ln K & ∆G°= ∆rH° - T∆rS°
|
| ∆Ho - T∆So = -RT ln K rearrange
| |
| V
| ln K = -∆H°/RT + ∆S°/R Hoff equation
V
ln K2/K1 = -∆H°/R(1/T2 - 1/T1) Two point curve