Class 3: Free Energy Continued

Question 1

Define enthalpy, entropy, and free energy as well as describe the relationship between them.

Answer 1

Enthalpy (H):
- A state function representing the total heat content of a system
- H = U + PV (internal energy + pressure-volume work)
- Useful for measuring energy changes at constant pressure

Entropy (S):
- A state function representing the molecular disorder/randomness of a system
- Higher entropy means higher disorder and lower energy availability
- S increases spontaneously for irreversible processes

Gibbs Free Energy (G):
- A state function useful for determining spontaneity
- G = H - TS (accounts for enthalpy and entropy)
- Negative ΔG means a spontaneous process

Relationships:
- ΔG = ΔH - TΔS (all terms temperature dependent)
- ΔH and TΔS contribute in opposite ways to ΔG
- Spontaneity driven by negative ΔG (either negative ΔH or positive TΔS)

Increasing Disorder:
- For a spontaneous process, entropy increases (ΔS > 0)
- This increases the -TΔS term, favoring negative ΔG
- Higher disorder/randomness means lower energy availability

Competing Factors:
- Enthalpy tends to make ΔG more positive (non-spontaneous)
- Entropy tends to make ΔG more negative (spontaneous)
- Their balance and temperature determine spontaneity

In summary, free energy (G) combines both the enthalpy and entropy driving forces to determine if a process will be spontaneous (ΔG < 0) or non-spontaneous (ΔG > 0).

Question 2

Explain the difference between spontaneous and non-spontaneous processes and how this connects to ΔG.

Answer 2

Spontaneous Processes:
- Occur naturally on their own without external driving forces
- Have a negative Gibbs free energy change (ΔG < 0)
- Are thermodynamically favored
- Examples: Ice melting, gas expansion, chemical reactions

Non-Spontaneous Processes:
- Require continual input of energy to occur
- Have a positive Gibbs free energy change (ΔG > 0)
- Are thermodynamically not favored
- Examples: Freezing water, compressing a gas, photosynthesis

Connection to ΔG:
- Sign and magnitude of ΔG determines spontaneity
- ΔG < 0 means a spontaneous, energetically favored process
- ΔG > 0 means a non-spontaneous, energetically unfavored process
- More negative ΔG indicates a greater driving force for spontaneity

Contribution of ΔH and ΔS:
- ΔG = ΔH - TΔS
- Negative ΔH (exothermic) favors spontaneity (ΔG < 0)
- Positive ΔS (increase disorder) favors spontaneity
- Competing effects balance based on temperature

Key Points:
- Spontaneous processes decrease free energy (ΔG < 0)
- Non-spontaneous processes increase free energy (ΔG > 0)
- Sign of ΔG determined by enthalpic and entropic driving forces
- ΔG is the criteria to determine if a process is spontaneous or not

Question 3

Predict changes in enthalpy and entropy for a given reaction.

Answer 3

Predicting ΔH:
- Consider bond breaking/formation in reactants/products
- Bond breaking is endothermic (ΔH positive)
- Bond formation is exothermic (ΔH negative)
- Compare strengths of bonds broken vs formed
- Stronger bonds formed = more negative ΔH

Predicting ΔS:
- Consider changes in number of gaseous molecules/ions
- Increased gas/ion particles = positive ΔS (more disorder)
- Decreased gas/ion particles = negative ΔS (less disorder)
- Solid/liquid phases contribute less to entropy changes

General Guidelines:
- Exothermic reactions have negative ΔH
- Increased gas particles gives positive ΔS
- Phase changes (solid/liquid -> gas) increase disorder
- Decomposition reactions increase both ΔH and ΔS

Examples:
1) 2Na(s) + 2HCl(aq) -> 2NaCl(aq) + H2(g)
- Bond formation is exothermic, so ΔH < 0
- Gas particles increase, so ΔS > 0

2) CaCO3(s) -> CaO(s) + CO2(g)
- Bond breaking is endothermic, so ΔH > 0
- Gas particles increase, so ΔS > 0

Question 4

Develop criteria for determining whether a reaction will be spontaneous or not at high or low temperatures

Answer 4

ΔG = ΔH - TΔS

At High Temperatures:
- The -TΔS term becomes more significant
- Entropy (ΔS) has a greater influence on spontaneity

Criteria for Spontaneity at High T:
1) ΔS > 0 (increase in disorder favors spontaneity)
2) -TΔS term is large and negative, outweighing ΔH
3) ΔH positive can still allow ΔG < 0 if TΔS is highly negative

At Low Temperatures:
- The -TΔS term becomes less significant
- Enthalpy (ΔH) is the main driver of spontaneity

Criteria for Spontaneity at Low T:
1) ΔH < 0 (exothermic reaction favors spontaneity)
2) ΔH negative must outweigh small TΔS value
3) Positive ΔS may not overcome endothermic ΔH

General Criteria:
- ΔG < 0 is required for a spontaneous reaction
- At high T, maximize ΔS (increase disorder)
- At low T, minimize ΔH (make reaction exothermic)
- Very low T makes entropy effects negligible

The balance between ΔH and ΔS terms, weighted by T, determines if ΔG is negative for spontaneity or positive for non-spontaneity.

Question 5

Apply the concept of temperature dependence of spontaneity and describe what is meant by entropy- or enthalpy-driven reactions.

Answer 5

Temperature Dependence of Spontaneity:
- Spontaneity determined by sign of Gibbs free energy change (ΔG)
- ΔG = ΔH - TΔS
- ΔH = enthalpy change, ΔS = entropy change, T = absolute temperature
- At low temperatures, -TΔS term is small, so ΔH dominates ΔG
- At high temperatures, -TΔS term is large, so ΔS dominates ΔG

Entropy-Driven Reactions:
- Have positive ΔS (increase in disorder/randomness)
- Driven by favorable -TΔS term outweighing unfavorable ΔH
- Examples: Melting of solids, dissolution, gas expansion

Enthalpy-Driven Reactions:
- Have negative ΔH (release heat/energy)
- Driven by favorable ΔH term outweighing unfavorable ΔS
- Examples: Combustion reactions, formation of ionic compounds

Question 6

Describe how enthalpy changes with protein folding (considering both the interactions along the protein chain and those with the solvent).

Answer 6

Intramolecular Interactions (Along Protein Chain):
- Formation of hydrogen bonds stabilizes folded state (negative ΔH)
- Formation of disulfide bridges stabilizes folded state (negative ΔH)
- Breaking of hydrogen bonds during unfolding is endothermic (positive ΔH)
- Van der Waals interactions between nonpolar groups favor folding (negative ΔH)

Protein-Solvent Interactions:
- Folding buries nonpolar groups in protein interior, reducing solvent exposure (negative ΔH)
- Folding releases ordered water molecules from protein surface into bulk solvent (positive ΔH)
- Formation of intramolecular hydrogen bonds replaces hydrogen bonds with solvent (small ΔH change)
- Ion pairs buried in folded state have less solvent exposure (negative ΔH)

Overall ΔH for Folding:
- Typically small and negative due to favorable van der Waals, hydrogen bonding, hydrophobic effects
- Can be positive or negative depending on relative contributions of various terms

Question 7

Explain the hydrophobic effect and draw simple diagrams to back up your explanation.

Answer 7

• The hydrophobic effect refers to the tendency of non-polar molecules (or non-polar portions of molecules) to minimize contact with water molecules
• It arises due to hydrogen bonding between water molecules, which creates an ordered structure around non-polar molecules
• This ordering restricts the freedom of movement for water molecules, reducing entropy
• To maximize entropy, the non-polar molecules cluster together, minimizing their exposure to water
• This clustering effect drives the self-assembly of non-polar molecules and the folding of proteins into compact structures
• It underlies many biological processes like lipid bilayer formation, protein folding, and the action of detergents
• The hydrophobic effect is an entropic effect, driven by maximizing the overall disorder (entropy) of the system

Question 8

Define entropy and explain how it relates to free energy

Answer 8

Entropy is the measure of the hugeness of the number of possible configurations (velocities, energies and precise positions) of every molecule.
Change in S will be positive for spontaneous reactions, as the world trends towards disorder—moles of g created minus those started with
WHEN H is POSITIVE—S being positive is ENTROPY DRIVEN
WHEN H is NEGATIVE—S being negative is ENTHALPY DRIVEN
SOME RXNS ARE DRIVEN TO BE SPONTANEOUS BY -∆H or + ∆S

Question 9

Describe the difference between spontaneous and non-spontaneous processes

Answer 9

Spontaneous: it will eventually happen by itself, without an external agent (exergonic)
Nonspontaneous: it requires an external agent to happen (endergonic)

Question 10

Define what a state function is and give examples.

Answer 10

A state function does not change regardless of the path taken to reach that specific function or value.
EX: the change in H or change is S or change in G

Question 11

Extra Notes:

Answer 11

Heat is a transfer of energy in the form of randomized kinetic energy

Question 12

Define enthalpy, entropy, and free energy as well as describe the relationship between them

Answer 12

Enthalpy: the change in potential energy of the system
Entropy: the change in disorder in the system (microstates)
Free energy: the capacity of the system to do work on the surroundings
Determines spontaneity
AT CONSTANT P, and TO A CONSTANT T

Question 13

Explain the difference between spontaneous and non-spontaneous processes and how this connects to a change in delta G

Answer 13

The difference between spontaneous and nonspontaneous reactions is the ability to do the reaction without an external help. Delta G being negative means that the reaction will be spontaneous because it creates more disorder and releases energy into the surroundings—increasing the work done on the environment, to a constant temp
If delta G is positive, then it is not spontaneous

Question 14

Predict changes in enthalpy and entropy for a given reaction

Answer 14

If the number of moles of gas of the product is greater than that of the reactants: ∆S is positive (vice-versa)
If there is a phase change in one substance from a solid to a liquid: ∆S is positive (vice-versa)

Question 15

Develop criteria for determining whether a reaction will be spontaneous or not at high or low temperatures

Answer 15

If they have the same sign, then it will matter what temp
If they’re both positive, then it will be entropy driven, and will be spontaneous at high temperatures
If they’re both negative, then it will be enthalpy driven, and will be spontaneous at low temperatures

Question 16

Apply the concept of temperature dependence of spontaneity and describe what is meant by entropy or entropy-driven reactions

Answer 16

Entropy is the reason that it is able to spontaneous/pushing the reaction along
If it is entropy driven, then it will be spontaneous under high temperatures

Question 17

Describe how enthalpy changes with protein folding (considering both the interactions along the protein chain and those with the solvent)

Answer 17

Some IMF enthalpy driven: H bonding, ion-ion
Hydrophobic: entropy driven

Question 18

Explain the hydrophobic effect and draw simple diagrams to back up the explanation

Answer 18

Entropy increase because the highly ordered water molecules are surrounding the nonpolar substance; when they form micelles, then water is freed up and is increasing entropy
ENTROPY DRIVEN (so must be endothermic)
Increases enthalpy as well
NORMALLY: when solutes disperse throughout a solvent, ∆S is positive. When solutes group away from solvent ∆S is negative

Question 19

Spontaneous change

Answer 19

WILL NOT reverse itself without external intervention. IT NEVER happens spontaneously in the reverse direction.

Question 20

2NH3(g) + C2O(g) → NH2CONH2(s) + H2O(l)

Answer 20

Evidence that this is an endothermic reaction? T decreases, so heat will have to be added to return to original temp after the reaction got colder

Question 21

Heat of reaction

Answer 21

The heat needed to bring the system back to its original temperature after a reaction occurs

Question 22

Initial and final states

Answer 22

At the start and end of a spontaneous reaction, the system is at the same temperature.

Question 23

Entropy-driven reaction

Answer 23

A spontaneous reaction where the entropy increase is the driving force, not the enthalpy change. For example, just having to mix solid powders increases entropy.

Question 24

How to predict entropy change

Answer 24

Gaseous > Liquid > Solid in entropy. More moles of gaseous products than reactants means entropy increases.

Question 25

Activation energy

Answer 25

Affects the rate/speed of a reaction, not whether it is spontaneous or not.

Question 26

RATE ≠ SPEED

Answer 26

Rate is the speed of a chemical process. Speed is the absolute velocity of an object.

Question 27

Stretched rubber band

Answer 27

When stretched, you put energy into it (endothermic). Relaxing is the spontaneous process (exothermic).

Question 28

HCl (g) → HCl (aq)

Answer 28

Endothermic (ΔH > 0) - lots of new ionic bonds forming between H+ and Cl- with water, requiring energy input.

Question 29

Breaking bonds

Answer 29

Breaking covalent bonds requires energy input (endothermic). Breaking intermolecular forces like hydrogen bonds and London forces requires lower energy input.

Question 30

Entropy of rubber band

Answer 30

When stretched, the rubber band has more surface area and more available conformations, so higher entropy than in the relaxed state.

Question 31

Phase transitions

Answer 31

For freezing (liquid to solid), ΔH < 0 and ΔS < 0. ΔG depends on temperature - spontaneous if ΔG < 0.

Question 32

Worksheet

Answer 32

Rubber bands are polymers made of long molecular chains that are coiled and tangled together. When stretched, these chains become more ordered and aligned. When released, the chains can return to their more disordered, tangled state.
The contraction of the rubber band involves breaking some intermolecular interactions (like van der Waals) between the polymer chains as they uncoil and re-tangle (ΔH > 0). It also involves forming new intermolecular interactions as the chains re-arrange (ΔH < 0). There may also be changes in bond angles, lengths and chain conformations (ΔH ≠ 0).

The overall ΔH could be exothermic (ΔH < 0) if the new intermolecular interactions formed outweigh the ones broken. Or it could be endothermic (ΔH > 0) if breaking interactions dominates. An athermic process (ΔH = 0) is less likely.

Touching the stretched band to your lip allows you to sense if the stretching was an endothermic (lip feels cool) or exothermic process (lip feels warm). Releasing and re-touching tests if the contraction was endo or exothermic.
If the contracted band feels cooler than the stretched band, it suggests an overall exothermic contraction process, releasing heat.
A cooler contracted band contradicts a prediction of ΔH < 0 (exothermic). An exothermic process should release heat and warm its surroundings slightly before that heat dissipates.
When the rubber band contracts, the disordered molecular chains have more possible arrangements (microstates), so entropy increases (ΔS > 0).
Since ΔH is not large and negative for the spontaneous contraction, it must be entropy-driven (ΔS > 0) rather than enthalpy-driven.
The spontaneous entropic driving force (TΔS term) increases with higher temperature. So the force of contraction will be greater at higher temperatures.

In summary, the contraction involves increasing entropy by forming more disordered chains, overcoming a smaller unfavorable enthalpy change from breaking some intermolecular interactions. Higher temperatures enhance this entropy-driven process.

Question 33

different types of bonds and molecular chains in polymers like rubber

Answer 33

Covalent Bonds:
- These are strong bonds formed by sharing electron pairs between atoms
- In polymers, covalent bonds form the backbone of each polymer chain, linking the monomers together
- Breaking/forming covalent bonds requires a lot of energy (ΔH&raquo_space; 0)
- Covalent bonds in the polymer backbone are not significantly broken/formed during stretching/contracting of rubber

Ionic Bonds:
- Attractions between positive and negative ions
- Not present in rubber or other non-ionic polymers
- Their breaking/forming can impact enthalpy, but is not relevant here

Intermolecular Interactions:
- Weaker attractive forces between molecules/chains
- Include van der Waals forces, hydrogen bonding, etc.
- In rubber, these occur between the coiled polymer chains
- Stretching breaks some of these intermolecular interactions (ΔH > 0)
- Contracting allows new ones to form between re-tangled chains (ΔH < 0)
- Net ΔH depends on balance of breaking vs forming interactions

Polymer Chains:
- Each chain is a long, flexible covalent backbone
- In unstretched rubber, chains are coiled and tangled
- Stretching causes chains to uncoil and align (lower entropy)
- Releasing allows chains to re-tangle in a disordered state (higher entropy)
- More disordered chains = more possible arrangements = higher entropy

So in summary, stretching/contracting rubber does not break the strong covalent bonds holding the chains together. Instead, it disrupts/re-establishes the weaker intermolecular interactions between coiled chains, changing their entropy.

Question 34

melting and freezing

Answer 34

Intermolecular Forces (IMFs):
- Attractive forces between molecules like hydrogen bonding, van der Waals, etc.
- In solid phase, molecules are held together by strong IMFs in an ordered arrangement
- In liquid phase, molecules have fewer/weaker IMFs and can move more freely

Enthalpy (ΔH) Changes:
- Breaking IMFs requires input of energy (ΔH > 0), e.g. melting ice
- Forming IMFs releases energy (ΔH < 0), e.g. freezing water
- The magnitudes depend on the strengths of the specific IMFs

Entropy (ΔS) Changes:
- Solids have fewer possible arrangements/microstates than liquids
- Going from solid -> liquid increases molecular disorder/entropy (ΔS > 0)
- Going from liquid -> solid decreases molecular disorder/entropy (ΔS < 0)

Melting (Solid -> Liquid):
- Requires breaking IMFs, so ΔH > 0 (endothermic)
- Increases molecular disorder, so ΔS > 0
- At melting point, ΔG = 0, so ΔS must outweigh ΔH (entropy-driven)

Freezing (Liquid -> Solid):
- Involves forming IMFs, so ΔH < 0 (exothermic)
- Decreases molecular disorder, so ΔS < 0
- Below melting point, ΔG < 0, so ΔH must outweigh TΔS (enthalpy-driven)

So in summary, the entropy increases for melting, favoring it above the melting point. While entropy decreases for freezing, the negative enthalpy makes it spontaneous below the melting point.

Question 35

Problem set

Answer 35

c) At 0°C, neither melting ice nor freezing water is spontaneous, because this is exactly at the melting/freezing point.

At the melting point:
ΔG = 0, since the solid and liquid phases are at equilibrium.
Melting is not spontaneous (ΔG ≠ < 0)
Freezing is not spontaneous (ΔG ≠ < 0)

that ΔG = 0, since G is a state function that only depends on the initial and final states, not the path taken between them.

d) We are given:
ΔH = +6.0 kJ/mol for melting ice
T = 273.15 K
ΔG = 0 at the melting point

We want to calculate ΔS for melting 1 mol of ice.

Using ΔG = ΔH - TΔS and setting ΔG = 0:
0 = (6000 J/mol) - (273.15 K * ΔS)
ΔS = 6000 J/mol / 273.15 K
ΔS = 21.97 J/mol*K

The entropy increase ΔS = 21.97 J/mol*K when 1 mol of ice melts at its melting point

Question 36

1) Will an ice cube melt spontaneously in a 1.0 M sucrose solution at 0°C?
2) At what temperature would the direction of spontaneous change reverse?

Answer 36

The question asks two things:
1) Will an ice cube melt spontaneously in a 1.0 M sucrose solution at 0°C?
2) At what temperature would the direction of spontaneous change reverse?

To answer:
1) At 0°C (273.15K):
- ΔH = 6000 J/mol (for melting ice)
- ΔS = 21.97 J/mol·K (pure water) + 0.15 J/mol·K (for sucrose solution)
= 22.12 J/mol·K
- Substitute into ΔG = ΔH - TΔS
ΔG = 6000 J/mol - (273.15K * 22.12 J/mol·K)
= -40.97 J/mol

Since ΔG is negative, melting is spontaneous in the sucrose solution at 0°C.

2) To find the temperature (T) where spontaneity reverses:
- Set ΔG = 0
- ΔG = ΔH - TΔS
- 0 = 6000 J/mol - T(22.12 J/mol·K)
- Solving for T:
T = 6000 J/mol / 22.12 J/mol·K
T = 271.3 K or -1.85°C

So spontaneity reverses below -1.85°C. Above this temperature ΔG is negative, so melting is spontaneous. Below -1.85°C, ΔG is positive, so melting is non-spontaneous.

Key Concepts:
- Entropy increase is higher for dissolving, so ΔS is larger for melting into solution
- Larger ΔS makes ΔG more negative, increasing spontaneity
- At some low temperature, TΔS becomes too small to overcome ΔH, so ΔG switches sign

Question 37

More PS

Answer 37

2a) ΔH > 0 for breaking bonds to dissociate Br2 into atoms (endothermic)

2b) ΔS > 0 as atoms have more microstate orientations than molecules

2c) At high temperatures, the TΔS term becomes larger, eventually outweighing the positive ΔH, making ΔG negative and the reaction spontaneous.

3) Key points:
- Hydrophobic effect refers to non-polar molecules/groups clustering in water to minimize unfavorable entropy
- This clustering is entropically driven (ΔS > 0) to increase solvent molecule orientations
- To break apart micelles, we want to decrease temperature
- Lower T decreases the TΔS term, eventually making ΔG positive and the clustering non-spontaneous