Class 2: Introduction to Free Energy

Question 1

Connect your understanding of potential energy from Chem 150 to the new concept of bond dissociation enthalpy.

Answer 1

Potential Energy Review:
- Potential energy is the energy stored within a system due to position/configuration
- Molecules/bonds have potential energy based on relative atomic positions
- Breaking bonds increases potential energy (endothermic)

Bond Dissociation Enthalpy:
- The energy required to break a specific covalent bond
- Quantifies the bond strength and potential energy stored
- Higher dissociation enthalpy = stronger bond, more energy to break

Connections:
- Bond dissociation enthalpy is a specific type of potential energy
- It represents the potential energy stored within a covalent bond
- Breaking a bond releases this stored potential energy
- The stronger the bond, the higher its dissociation enthalpy/potential energy

Energy Changes:
- Energy must be supplied to break bonds (endothermic)
- This energy overcomes the potential energy barrier of the bond
- Once broken, the atoms have higher potential energy in separated state
- Bond formation is an exothermic release of this potential energy

Key Points:
- Bond dissociation enthalpy quantifies potential energy of a covalent bond
- It’s the energy input required to overcome the bond’s potential energy barrier
- Stronger bonds have higher dissociation enthalpies/potential energies
- Bond breaking increases potential energy; bond forming releases it

Let me know if any part needs further clarification!

Question 2

Calculate the enthalpy of a reaction (ΔH°) from bonds broken and bonds formed.

Answer 2

Given:
- Bond dissociation enthalpies for all bonds broken in reactants
- Bond dissociation enthalpies for all bonds formed in products

Step 1: Sum the bond dissociation enthalpies for bonds broken
- This gives the total energy input to break reactant bonds

Step 2: Sum the bond dissociation enthalpies for bonds formed
- This gives the total energy released from forming product bonds

Step 3: Calculate ΔH°
- ΔH° = ΣBond Dissociation Enthalpies (Bonds Broken)
- ΣBond Dissociation Enthalpies (Bonds Formed)

Notes:
- Energies for breaking bonds are positive (endothermic)
- Energies for forming bonds are negative (exothermic)
- ΔH° positive means overall endothermic
- ΔH° negative means overall exothermic
- Units of ΔH° are kJ/mol

Example:
CH4 + 2O2 → CO2 + 2H2O
Bonds Broken: 4(C-H), 2(O=O)
Bonds Formed: 2(C=O), 4(O-H)
ΔH° = [4(413) + 2(498)] - [2(799) + 4(463)]
= 1652 + 996 - 1598 - 1852
= -802 kJ/mol

Question 3

Describe the difference between endothermic and exothermic reactions.

Answer 3

Endothermic Reactions:
- Absorb heat energy from the surroundings
- Temperature of the surroundings decreases
- Require input of energy to proceed
- Have a positive enthalpy change (ΔH > 0)
- Examples: Photosynthesis, melting, cooking

Exothermic Reactions:
- Release heat energy to the surroundings
- Temperature of the surroundings increases
- Release energy as they proceed
- Have a negative enthalpy change (ΔH < 0)
- Examples: Combustion, neutralization, respiration

Key Differences:
Endothermic | Exothermic
Absorbs Heat | Releases Heat
ΔH is positive | ΔH is negative
Energy is absorbed | Energy is released
Surroundings cool | Surroundings warm
Requires energy input | Releases energy

Energy Flow:
- Endothermic: Energy flows from surroundings → system
- Exothermic: Energy flows from system → surroundings

Overall:
- Endothermic reactions take in energy to break bonds
- Exothermic reactions release energy from forming bonds
- Enthalpy change determines endo/exothermic nature

Question 4

Define enthalpy and explain how it relates to free energy.

Answer 4

Enthalpy (H):
- A state function that combines internal energy (U) and pressure-volume work (PV)
- H = U + PV
- Useful for measuring energy changes in reactions at constant pressure

Free Energy (G):
- The energy available to do non-expansion work
- Accounts for enthalpy (H) and entropy (S)
- G = H - TS (T is absolute temperature)

Relationship between H and G:
- Enthalpy term (H) represents total heat energy of the system
- Entropy term (-TS) accounts for molecular disorder/randomness
- Free energy (G) is the usable portion of the enthalpy after entropy effects

Free Energy Changes:
- ΔG < 0 means a spontaneous, energy-releasing process
- ΔG > 0 means a non-spontaneous, energy-absorbing process
- ΔG is determined by both enthalpy (ΔH) and entropy (ΔS) changes

Significance:
- Reactions are driven by decreasing free energy (ΔG < 0)
- More negative ΔG means greater driving force
- Enthalpy and entropy changes contribute in opposite ways to ΔG

Key Point: Free energy relates the total enthalpy of a system to its randomness/disorder, determining if energy is available for useful work.

Question 5

Describe the difference between spontaneous and non-spontaneous processes.

Answer 5

Spontaneous Processes:
- Occur naturally on their own without outside assistance
- Have a negative Gibbs free energy change (ΔG < 0)
- Are thermodynamically favored
- Examples: Bomb explosions, combustion reactions, ice melting

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

Key Differences:
Spontaneous | Non-Spontaneous
ΔG < 0 | ΔG > 0
Favored by nature | Not favored
Releases free energy | Absorbs free energy
Occurs naturally | Requires energy input
Increases disorder | Decreases disorder

Driving Forces:
- Spontaneous: Driven by tendency to minimize free energy
- Non-Spontaneous: Driven by increasing free energy

Effects of ΔH and ΔS:
- For spontaneity, ΔG must be < 0
- ΔG = ΔH - TΔS
- Spontaneous if ΔH is negative OR if TΔS is large positive

The sign of ΔG determines if a process is spontaneous or non-spontaneous at that condition.

Question 6

Define what a state function is and give examples.

Answer 6

State Function:
- A property that depends only on the present state of the system
- Not dependent on how that state was reached or the path taken
- Has a precise value that can be calculated from system variables

Examples of State Functions:
- Internal Energy (U)
- Enthalpy (H)
- Entropy (S)
- Gibbs Free Energy (G)
- Volume (V)
- Pressure (P)
- Temperature (T)

Properties of State Functions:
- Their values are independent of the path/process
- Only depend on the initial and final states of the system
- Any exact differential is a state function (dU, dH, dS, dG, etc.)

Not State Functions:
- Work (W) - depends on the path
- Heat (q) - depends on the path
- Other process quantities that depend on path taken

Significance:
- State functions are convenient thermodynamic bookkeeping tools
- Allow calculating energy changes between two states directly
- No need to consider the complicated path details

Key Point:
State functions have values determined entirely by the system’s current state, regardless of how that state was reached. This simplifies thermodynamic calculations.

Question 7

Connect your understanding of potential energy to the new concept of bond dissociation enthalpy

Answer 7

Raising the PE to break a bond requires an input of KINETIC ENERGY—otherwise the bond will not break because it is at an energetically favorable position
U=internal energy of the system → changes in KE and PE add up to a change in U
Increases in KE of the system
Converts to PE, breaking bonds and/or deforming them away from minimum P.E/ lengths and angles
Some energy is lost as work, against external pressure, making the system expand (NOT IMPORTANT)
Heat entering the system (enthalpy) = internal energy + pressure x volume
H = U + pV
pV usually so small it is negligible
∆H > 0 for breaking bonds — energy added to the system — endothermic
∆H < 0 for forming bonds — energy leaves the system — exothermic
ALL OF THIS MUST BE TO A CONSTANT T [Wait for exothermic to cool down, or endothermic to get warm again]
Bond dissociation enthalpy is the heat input required to break one mole of bonds and maintain the same T at a constant pressure

Question 8

Calculate the enthalpy of a reaction from bonds broken and formed

Answer 8

BOND DISSOCIATION ENTHALPY = total bonds broken (reactants) - total bonds formed (products)
Will determine if H is positive or negative
Enthalpy change does not depend on path—only initial and final states—IT IS A STATE FUNCTION
Regardless of forward or back it does not matter

Question 9

Describe the difference between endothermic and exothermic reactions

Answer 9

Endothermic: takes energy in (immediately cold) to break bonds/create less strong bonds/configurations. KE converted to potential energy. (∆H > 0)
Exothermic: (immediately hot) releases potential energy as heat (kinetic energy) from the system when returned to a constant temperature—the energy of the products is less than that of the reactants (∆H < 0)
The new bonds formed/stronger bonds formed
Tend to be more likely spontaneous versus the other

Question 10

double concentration -> collisions double -> successful reactions prolly double

Question 11

collisions are independent , 2x collision with fixed fraction of energetic molecules means 2x reactions

Question 12

T affects Reactivity through 1) fraction of molecules with sufficient energy 2) rate of collisions

Question 13

Concentration affects reactivity through 1) rate of collisions

Question 14

Temperature -
measure of Kinetic energy at molecular level
Heat-transfer of energy in the form of molecular (randomized ) KE

Answer 14

If heat energy is added to system where does it go? INCREASES Kinetic energy of system (raising temp)

change in KE/PE - change in internal energy “V”

CONVERTS to PE , breaking bonds and deformation of bonds
CONVERTS to work against external pressure -> system expands

Question 15

Heat entering a system (at constant pressures) equals change in Enthalpy delta H

Question 16

H = U + pV
enthalpy = internal energy + pressure/volume

Question 17

breaking bonds INCREASES enthalpy

Answer 17

↳ requires input of heat to Maintain constant T

Question 18

breaking bonds LOWERS enthalpy

Answer 18

↳ requires release of heat to maintain T

Question 19

Bond dissociation enthalpy/bond enthalpy BE) -

Answer 19

heat input to break I mole of bonds or heat released to form ( mol of bonds

Question 20

Delta H = sum of BE of reactant bonds -
- sum of BE of all product bonds

Answer 20

Delta H= bonds broken-bonds formed

Delta H does not depend on intermediate states only initial and final state

Question 21

Enthalpy is a STATE FUNCTION-defined by current state and initial state

Answer 21

Heat is not state function -> depends on process and relies on pressure/temp

Question 22

Delta H = Delta PE

Answer 22

Equation:
- ΔH = ΔPE
- Where ΔH is the change in enthalpy
- ΔPE is the change in potential energy

Meaning:
- This equation relates the enthalpy change to the change in potential energy
- It states that the heat absorbed/released (ΔH) equals the change in potential energy

Potential Energy:
- Potential energy is the energy stored within a system due to its configuration
- For chemical systems, this includes bond energies and intermolecular forces

Bond Energy:
- Breaking bonds requires input of energy (endothermic, ΔH positive)
- This increases the potential energy of the separated atoms
- Forming bonds releases energy (exothermic, ΔH negative)
- This reduces the potential energy of the bonded atoms

Application:
- Used to calculate ΔH from bond dissociation enthalpies
- ΔH = ΣBond Energies (Bonds Broken) - ΣBond Energies (Bonds Formed)
- Applies to any process involving potential energy changes

Limitations:
- Only accounts for potential energy changes, not other energies like kinetic
- Assumes no other energy inputs/losses during the process

Question 23

Delta H of reaction - change in enthalpy reactants to products -> at same temp

Answer 23

REAL REACTION - temp Often changes

Question 24

Delta H Negative -> new bonds have lower potential energy
-increase in Ke goes up t raised goes up

to retain original original T -. transferred out
Delta H<0 exothermic

Answer 24

ΔH Negative:
- ΔH < 0 means an exothermic process
- Energy is released, usually from forming new bonds
- New bonds have lower potential energy than original bonds

Kinetic Energy Increase:
- Released energy goes into increasing kinetic energies of particles
- More kinetic energy = faster particle motions and higher temperatures

Temperature Increases:
- For an exothermic process in a closed system, temperature rises
- The kinetic energy increase causes an increase in average particle velocities
- This results in a measurable increase in temperature

Thermal Energy Transfer:
- If the system is open, the thermal energy can transfer out
- This prevents the full temperature rise that would occur in a closed system
- Some of the exothermic energy leaves as heat instead of raising particle KEs

Key Points:
- ΔH < 0 means exothermic, energy released in bond formation
- Released energy increases particle kinetic energies
- This kinetic energy increase raises the temperature
- In an open system, some energy transfers out as heat flow

Question 25

when ΔH is positive:

ΔH Positive:
- ΔH > 0 means an endothermic process
- Energy must be absorbed, usually to break existing bonds
- Bonds being broken have lower potential energy than new bonds formed

Energy Absorption:
- The required energy is absorbed from the surroundings
- This energy goes into increasing the potential energies of the separated atoms/molecules
- It overcomes the potential energy “well” of the original bonds

Temperature Decreases:
- For an endothermic process in a closed system, temperature drops
- Absorbing energy removes thermal kinetic energy from particles
- This results in a measurable decrease in temperature

Supplying Energy:
- The absorbed energy can come from various sources
- Common sources are heat transfer from the surroundings or supplying light/radiation
- This energy input drives the endothermic bond-breaking process forward

Key Points:
- ΔH > 0 means endothermic, energy absorption required
- Energy absorbed increases potential energies by breaking bonds
- In a closed system, this draws energy away, lowering particle KEs
- Thus endothermic processes lower temperatures unless energy is continually supplied

Question 26

Spontaneity-reaction will proceed (maybe slowly) without active intervention

Question 27

microstate-defined by precise positions, velocities , energy of every molecule

Question 28

ENTROPY (S) -
amount of disorder in a system (ex - microstates , moles , etc)

Question 29

ΔG = ΔH - TΔS

Answer 29

Equation:

ΔG = ΔH - TΔS
Where ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the absolute temperature
ΔS is the change in entropy

Gibbs Free Energy:

ΔG indicates the spontaneity of a process
ΔG < 0 means a spontaneous process
ΔG > 0 means a non-spontaneous process

Enthalpy (ΔH) Term:

Accounts for energy changes due to breaking/forming bonds
Negative ΔH favors negative ΔG (spontaneous)

Entropy (ΔS) Term:

Accounts for changes in molecular disorder/randomness
Positive ΔS favors negative ΔG (spontaneous)

Temperature (T) Dependence:

At higher T, the -TΔS term has a larger value
Entropy becomes more important driver of spontaneity

Question 30

∆H = sum of all bonds broken – sum of all bonds formed

Answer 30

• make sure you consider the number of moles combusted (ex. if it says 1 mole of methanol and the chemical equation shows 2 mol, so divide every coefficient by 2)
• The stoichiometric coefficients in a balanced chemical equation directly determine the number of moles of bonds that are broken and formed during the reaction. To calculate the enthalpy change correctly using bond energies, you must multiply the number of bonds of each type by the corresponding coefficient from the balanced equation. This gives you the precise number of moles of bonds broken and formed, which you then substitute into the enthalpy change equation.

Question 31

negative enthalpy

Answer 31

exothermic
PE reactants > products

Here’s an explanation in bullet points about what the sign of ∆H tells you regarding the potential energy of the reactants compared to the products:

• The sign of the enthalpy change (∆H) indicates whether a chemical reaction is endothermic or exothermic.
• A positive value of ∆H (∆H > 0) means that the reaction is endothermic:
  
  • It requires an input of energy from the surroundings to proceed.
  • The potential energy of the products is higher than the potential energy of the reactants.
  • The reactants have a lower potential energy than the products.
• A negative value of ∆H (∆H < 0) means that the reaction is exothermic:
  
  • It releases energy to the surroundings as it proceeds.
  • The potential energy of the products is lower than the potential energy of the reactants.
  • The reactants have a higher potential energy than the products.
• If ∆H = 0, the reaction is considered to be at equilibrium:
  
  • The potential energy of the reactants is equal to the potential energy of the products.
  • There is no net energy exchange with the surroundings.
• Therefore, when ∆H < 0 (an exothermic reaction), it implies that the potential energy of the reactants is greater than the potential energy of the products (PE reactants > PE products).
• Conversely, when ∆H > 0 (an endothermic reaction), it implies that the potential energy of the reactants is lower than the potential energy of the products (PE reactants < PE products).

Question 32

Breaking a bond

Answer 32

put energy to break
endothermic

Question 33

The sign of the enthalpy change (∆H) indicates whether a chemical reaction is endothermic or exothermic.
A positive value of ∆H (∆H > 0) means that the reaction is endothermic:

It requires an input of energy from the surroundings to proceed.
The potential energy of the products is higher than the potential energy of the reactants.
The reactants have a lower potential energy than the products.

A negative value of ∆H (∆H < 0) means that the reaction is exothermic:

It releases energy to the surroundings as it proceeds.
The potential energy of the products is lower than the potential energy of the reactants.
The reactants have a higher potential energy than the products.

If ∆H = 0, the reaction is considered to be at equilibrium:

The potential energy of the reactants is equal to the potential energy of the products.
There is no net energy exchange with the surroundings.

Therefore, when ∆H < 0 (an exothermic reaction), it implies that the potential energy of the reactants is greater than the potential energy of the products (PE reactants > PE products).
Conversely, when ∆H > 0 (an endothermic reaction), it implies that the potential energy of the reactants is lower than the potential energy of the products (PE reactants < PE products).

Question 34

Evidence of increasing entropy in the methanol cannon demo:

Answer 34

Gas expansion (molecules occupy greater volume)
Greater number of available microstates/arrangements

Here’s a deeper explanation on microstates and how they relate to entropy:

• Microstates refer to the different possible arrangements or configurations of particles (atoms, molecules, etc.) in a system.
• The entropy of a system is directly proportional to the number of accessible microstates.
• In the methanol cannon demo, when the methanol combusts to form gaseous products (CO2 and H2O), the gas molecules can occupy a much larger volume compared to the initial liquid methanol.
• This larger volume provides a significantly greater number of possible arrangements or microstates for the gas molecules to exist in.
• Each unique position and momentum that a gas molecule can have within the larger volume constitutes a different microstate.
• The more microstates available to a system, the higher its entropy.
• Therefore, the expansion of gas and the resulting increase in accessible microstates for the gas molecules lead to an increase in entropy for the overall system.
• This increase in entropy is evidence that the combustion reaction is spontaneous and favored by the second law of thermodynamics.
• The larger the volume available to the gaseous products, the greater the number of microstates, and consequently, the higher the entropy change associated with the reaction.

So, in summary, the expansion of gas molecules into a larger volume provides access to a vastly greater number of microstates or arrangements, directly translating to a significant increase in entropy for the system.

Question 35

Balanced equation: 2NH4Cl(s) + Ba(OH)2*8H2O(s) → 2NH3(g) + 10H2O(l) + BaCl2(aq)
a. Evidence of endothermic reaction:

Temperature of surroundings decreases (water freezes to the box)

b. “Entropy-driven” reaction:

Spontaneous reaction (∆G < 0)
Endothermic (∆H > 0)
Must have ∆S > 0 to satisfy ∆G = ∆H - T∆S < 0
High entropy increase (T∆S > ∆H) drives the spontaneous endothermic process

Key points:

Gas expansion increases entropy
Endothermic reactions can be spontaneous if entropy increase is significant
Entropy-driven reactions are spontaneous due to positive ∆S, despite endothermic ∆H

Question 36

What is the difference between an exothermic and exergonic reaction?

Answer 36

The main difference between an exothermic reaction and an exergonic reaction lies in the thermodynamic quantity they describe:

Exothermic Reaction:
- Describes the enthalpy change (ΔH) of a reaction.
- An exothermic reaction releases heat to the surroundings, resulting in a negative enthalpy change (ΔH < 0).
- It means that the products have lower potential energy than the reactants, and the excess energy is released as heat.
- Exothermic reactions can be either spontaneous or non-spontaneous, depending on the overall Gibbs free energy change (ΔG).

Exergonic Reaction:
- Describes the Gibbs free energy change (ΔG) of a reaction.
- An exergonic reaction is a spontaneous reaction, where the Gibbs free energy decreases (ΔG < 0).
- It means that the products have lower free energy than the reactants, and the reaction is thermodynamically favored to proceed spontaneously.
- Exergonic reactions can be either exothermic (releasing heat) or endothermic (absorbing heat), depending on the enthalpy change (ΔH) and entropy change (ΔS).

In summary:
- Exothermic reactions release heat (ΔH < 0), while exergonic reactions are spontaneous (ΔG < 0).
- An exothermic reaction may or may not be spontaneous, depending on the overall Gibbs free energy change.
- An exergonic reaction can be either exothermic or endothermic, as long as the decrease in free energy (ΔG < 0) is fulfilled, accounting for both enthalpy and entropy changes.

Therefore, exothermic reactions describe the heat exchange, while exergonic reactions describe the spontaneity and overall feasibility of the process based on the free energy change.

Question 37

Is the combustion of acetylene spontaneous or non-spontaneous? Make a
qualitative argument for your answer

Answer 37

Qualitative argument:

Doesn’t require supply of external energy to keep it going
The combustion of acetylene is an exothermic reaction, meaning it releases heat to the surroundings (ΔH < 0).
Exothermic reactions are often spontaneous because they proceed with a decrease in enthalpy, which is thermodynamically favored.
However, spontaneity also depends on the entropy change (ΔS) of the reaction.
In this case, the reactants (2 C2H2 + 5 O2) are in the gaseous state, while the products (4 CO2 + 2 H2O) include both gases and a liquid.
The formation of the liquid water product (H2O(l)) results in a decrease in entropy, counteracting the increase in entropy from the formation of gaseous products (CO2).
Nevertheless, the overall entropy change is likely to be positive (ΔS > 0) due to the significant increase in the number of gaseous product molecules.
Since the reaction is exothermic (ΔH < 0) and has a positive entropy change (ΔS > 0), the Gibbs free energy change (ΔG = ΔH - TΔS) is likely to be negative (ΔG < 0).

Question 38

When 500.0 g of acetylene burns, what amount of heat is given off?

Answer 38

500 g C2/H2 * (1 mol (C2H2/26.04g C2H2) * (-1259 KJ/1mol C2H2)

26.04g - molar mass
-1259 KJ - enthalpy

Question 39

The following reactions are not balanced.
a) Balance each one.
b) Predict the sign of ΔS° for these reactions, without detailed calculations.
c) Also, name all reactants and products in all of the reactions.
(i) CaCO3(s) + HCl (aq) ® CaCl2 (aq) + H2O (l) + CO2 (g)
(ii) NO (g) + O2 (g) ® NO2 (g)
(iii) KClO3 (g) ® KCl (s) + O2 (g)

Answer 39

(i) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
a) Balanced
b) ΔS° is positive (increase in gas molecules)
c) Reactants: Calcium carbonate, Hydrochloric acid
Products: Calcium chloride, Water, Carbon dioxide

(ii) 2NO(g) + O2(g) → 2NO2(g)
a) Balanced
b) ΔS° is negative (decrease in gas molecules)
c) Reactants: Nitric oxide, Oxygen
Products: Nitrogen dioxide

(iii) 2KClO3(s) → 2KCl(s) + 3O2(g)
a) Balanced
b) ΔS° is positive (increase in gas molecules)
c) Reactants: Potassium chlorate
Products: Potassium chloride, Oxygen

Question 40

Entropy Change

Answer 40

ΔS = Σ Smol,products - Σ Smol,reactants

2NO(g)+1O2->2NO2

ΔS = 2(Smol,NO2)- 2(Smol,NO)- 1*(Smol,O2)

negative ΔS mean entropy decreases