Class 19: Introduction to Kinetics and the Chemistry of Atmospheric Pollutants

Question 1

Define what a greenhouse gas is, give examples, and explain their environmental impact.

Answer 1

Definition:
- Greenhouse gases are atmospheric gases that absorb and emit infrared radiation, trapping heat in the Earth’s atmosphere.
- They allow sunlight to pass through the atmosphere but prevent some of the infrared radiation from escaping back into space.

Examples:
- Carbon dioxide (CO₂) - Released by burning fossil fuels, deforestation, and industrial processes.
- Methane (CH₄) - Produced by livestock, landfills, natural gas extraction, and wetlands.
- Nitrous oxide (N₂O) - Emitted from agricultural activities, combustion of fossil fuels, and industrial processes.
- Fluorinated gases (HFCs, PFCs, SF₆) - Synthetic gases used in various industrial applications.

Environmental Impact:
- Increased greenhouse gas concentrations in the atmosphere lead to the greenhouse effect, trapping more heat and causing global warming.
- Global warming results in rising temperatures, melting of ice caps and glaciers, sea-level rise, changes in weather patterns, and ecosystem disruptions.
- It contributes to climate change, which can have severe consequences for the environment, agriculture, biodiversity, and human health.
- Greenhouse gases are the primary drivers of human-induced climate change, with CO₂ being the most significant contributor.

By reducing greenhouse gas emissions from human activities, such as transitioning to renewable energy sources, improving energy efficiency, and implementing sustainable practices, we can mitigate the impacts of climate change and work towards a more sustainable future for the planet.

Question 2

List all of the factors that may affect the rate of a chemical reaction.

Answer 2

Here are the key factors that can affect the rate of a chemical reaction:

• Temperature
• Concentration or pressure of reactants
• Physical state of reactants (solid, liquid, gas)
• Surface area of solid reactants
• Presence and concentration of a catalyst
• Nature of the reaction mechanism (elementary steps)
• Use of radiation or electricity

More detailed points:

• Temperature: Higher temperatures increase reaction rates due to more molecular collisions and greater fraction of molecules with sufficient energy.
• Reactant Concentration/Pressure: Higher concentrations or partial pressures of reactants lead to more frequent molecular collisions.
• Physical State: Reactions occur faster in liquid or gaseous states where molecules can move freely compared to solids.
• Surface Area: For heterogeneous reactions with solids, a larger surface area provides more sites for reaction.
• Catalysts: Catalysts provide an alternative lower energy pathway, increasing reaction rates without being consumed.
• Reaction Mechanism: More elementary steps or higher activation energies slow down overall reactions.
• Radiation/Electricity: Providing energy via light, microwaves or electricity can initiate or accelerate some reactions.

So in summary, factors that increase molecular collisions, provide alternative reaction pathways, or supply additional activation energy can enhance the rate of a chemical reaction.

Question 3

Explain how increasing concentration, and temperature, can all increase the rate of the reaction.

Answer 3

Increasing Concentration:
- In a chemical reaction, the reactant molecules must collide with each other to react.
- By increasing the concentration of reactants, there are more reactant molecules present per unit volume.
- This leads to more frequent collisions between the reactant molecules.
- More frequent collisions mean a higher probability that reactant molecules will collide with proper orientation and sufficient energy to overcome the activation energy barrier.
- Therefore, increasing reactant concentration increases the rate of the reaction.

Increasing Temperature:
- According to the Arrhenius equation, the rate constant (k) is exponentially dependent on temperature (k = Ae^(-Ea/RT)).
- As temperature increases, reactant molecules gain more kinetic energy from increased thermal motions.
- With higher kinetic energies, more molecules possess energy greater than or equal to the activation energy (Ea) required to react.
- This results in a higher fraction of productive molecular collisions that can overcome the energy barrier.
- Additionally, higher temperatures lead to more frequent molecular collisions due to faster molecular motion.
- Both of these factors - more molecules with sufficient energy and more frequent collisions - cause the reaction rate to increase significantly with rising temperature.

In summary, higher concentrations provide more molecular collisions per unit time, while higher temperatures provide more molecules with sufficient energy to react as well as more frequent collisions, collectively increasing the rate of the chemical reaction.

Question 4

Interpret graphical data to determine average rate, instantaneous rate, and initial rate.

Answer 4

Average Rate:
- Plot concentration or amount of reactant/product vs time
- Choose two different time points
- Calculate slope between those two points: Δ[Reactant]/Δt or Δ[Product]/Δt
- Slope represents average rate over that time interval

Instantaneous Rate:
- Instantaneous rate is the slope of the tangent line at a given point
- Find point of interest on concentration vs time graph
- Draw tangent line touching that single point
- Slope of the tangent gives instantaneous rate at that specific time

Initial Rate:
- Initial rate is the instantaneous rate at time = 0
- Find the first point on the curve as time approaches 0
- Draw tangent line at that initial point
- Slope of that tangent is the initial rate of reaction

Graphical Analysis:
- Instantaneous rates can vary over the course of the reaction
- Initial rate is instantaneous rate only at the start
- Average rates calculated over long time intervals represent overall behavior
- Curvature in graphs indicates changing instantaneous rates over time

So in essence, finding slopes of tangent lines gives instantaneous rates, while slopes between two points give average rates over that time period, with the initial rate being the instantaneous rate right at the start.

Question 5

Determine the rate of formation or disappearance of any reactant or product knowing only one rate.

Answer 5

1) Using the Balanced Chemical Equation:
- Write out the balanced chemical equation for the reaction.
- The coefficients represent the mole ratios of the reactants and products.

2) Rate of Formation = Rate of Disappearance:
- At any given time, the rate of formation of products equals the rate of disappearance of reactants.
- This is due to the law of conservation of mass.

3) Relating Rates Using Mole Ratios:
- If you know the rate for one species, use the mole ratios from the balanced equation.
- Multiply the known rate by the coefficients to get rates of other species.

For example, if the balanced equation is:
2A + B → 3C

And the rate of disappearance of A is known to be x M/s:
Rate of disappearance of A = -x M/s
Rate of disappearance of B = -x/2 M/s (Divide by 2 from mole ratio)
Rate of formation of C = 3x/2 M/s (Multiply by 3/2 from mole ratio)

4) Sign Conventions:
- Rates of disappearance are negative
- Rates of formation are positive

So in essence, by using the mole ratios from the balanced equation, you can calculate any reaction rate if you know just one rate of formation or disappearance.

Question 6

Explain the difference between rate and rate constant. Determine the overall order of a reaction from the rate constant units.

Answer 6

Rate vs Rate Constant:
- Rate: The change in concentration of reactants/products per unit time (e.g. M/s). Depends on concentrations.
- Rate Constant (k): Proportionality constant that does not depend on concentrations. Has units that incorporate reaction order.

Determining Overall Order from Rate Constant Units:
- The units of the rate constant reveal the overall order of the reaction.
- Compare the units to the generic rate constant units for different orders:

Zero-order: k has units of M/s
First-order: k has units of 1/s or s^-1
Second-order: k has units of 1/(M.s) or M^-1.s^-1
Third-order: k has units of 1/(M^2.s) or M^-2.s^-1

• The exponent on the molarity (M) units equals the overall order.
• For example:
  k = 2.5 x 10^-4 M^-1.s^-1 implies a second-order overall reaction
  k = 8.2 x 10^-3 s^-1 implies a first-order overall reaction

So in summary:
- Rate depends on concentrations, rate constant does not
- Rate constant units reveal overall order directly from the exponent on M
- This allows you to classify the overall reaction order from the rate constant

Question 7

Use initial rate data to determine the rate law and rate constant for a reaction.

Answer 7

• Measure initial rates at different concentrations of reactants
• Plot data (rate vs. concentration) and look for trends
• Order of reaction determined by powers needed to fit rate = k[A]^x[B]^y
  
  • First order in A if rate ∝ [A]
  • Second order in B if rate ∝ [B]^2
  • Zero order if rate is constant
• Once order known, plug in one data point to calculate rate constant k
• Rate law: Rate = k[A]^x[B]^y
• Units of k depend on overall order of reaction

Question 8

Use the integrated first-order rate law to calculate the amount of residual reactant or amount of product formed over a certain time.

Answer 8

• Applies to first-order reactions only (rate = k[A])
• Integrated rate law: ln([A]t/[A]0) = -kt
• To find [A]t at time t:
  
  • Rearrange to: [A]t = [A]0 e^(-kt)
  • Plug in [A]0, k, and t
• To find amount A remaining at time t:
  
  • [A]t x initial volume
• To find amount product P formed by time t:
  
  • Initial amount A - amount A remaining

Question 9

Use integrated rate laws to distinguish between 1st and 2nd order kinetics given the time-dependent concentration of a reactant.

Answer 9

• For first-order:
  
  • Plot ln[A] vs. t
  • Should give a straight line with slope = -k
• For second-order:
  
  • Plot 1/[A] vs. t
  • Should give a straight line with slope = k
• If neither gives a straight line, reaction is not first or second order
• Compare linear regression R^2 values
  
  • Higher R^2 indicates better fit to that order
• Can also calculate half-life:
  
  • t1/2 = ln(2)/k for first-order
  • t1/2 = 1/([A]0k) for second-order

Question 10

Do calculations with the Arrhenius equation and use collision theory to explain by what mechanism temperature can increase the rate

Answer 10

• Arrhenius equation: k = A * e^(-Ea/RT)
  
  • k is rate constant
  • A is pre-exponential factor
  • Ea is activation energy
  • R is gas constant
  • T is absolute temperature
• To calculate k at different temps:
  
  • Plug in Ea, A, R and T values
• To determine Ea graphically:
  
  • Plot ln(k) vs 1/T
  • Slope = -Ea/R
• Collision theory explains temperature dependence:
  
  • Higher T = faster molecular motion
  • More frequent collisions
  • More collisions have E > Ea
  • So more collisions are effective
• Temperature also affects orientation
  
  • At higher T, more molecules have proper orientation for reaction

Question 11

Define what a greenhouse gas is, give examples, and explain their environmental impact.

Answer 11

A greenhouse gas is gases in the earth’s atmosphere that trap heat
They remain in the atmosphere for years, something that can be determined by the rate of reaction in the atmosphere. The worst GHG is SF6 but it is not common, while CO2 is the most common
New refrigerant used to reduce this

Question 12

List all of the factors that may affect the rate of a chemical reaction.

Answer 12

A rate is a measure of how some value varies w/ time
the rate of reaction is the change in amount of reactant or product over time
Factors that affect the rate:
Chemical properties of reactants (bond strength, size, ect.)
Physical states of reactants (SA of reaction)
Chem rxn between 2 chemicals require contact when they’re in different states
Temperature of Reactants
RXN faster @ high temps
Concentrations of reactants
Increase rate with increased concentration
Presence of a catalyst
Catalyst increases the rate

Question 13

Explain how increasing concentration, and temperature, can all increase the rate of the reaction.

Answer 13

Increasing the concentrations of reactants will increase the rate of reaction
Increasing the temperature increases the average amount of molecules that are capable of reacting, so it increases the rate of reaction
As a reaction proceeds, the rate decreases due to less reactant concentrations

Question 14

Interpret graphical data to determine average rate, instantaneous rate, and initial rate.

Answer 14

Instantaneous rate: use limits or estimate the slope of the tangent line
For example: aA → bB: rate = (1/a)(ΔA/Δt)=(1/b)(ΔB/Δt)
Note: add a negative sign when there is a difference between reactants and products
EX: 2NH3 → N2 +3H2
-(1/2)(Δ[NH2]/Δt)=(1/1)(Δ[N2]/Δt)

Question 15

Determine the rate of formation or disappearance of any reactant or product knowing only one rate.

Answer 15

Each rate is connected by the stoich—1/coefficient (Δ[concentration]/Δt)
rate=k[A]^m[B]^n
k=rate constant
[A]=concentration of reactant A
[B]=concentration of reactant B
m/n=order of reaction

Question 16

Explain the difference between rate and rate constant. Determine the overall order of a reaction from the rate constant units.

Answer 16

The rate constant is the proportionality constant relating the rate of the reaction to the concentrations of reactants.
The order of the reaction can be determined by ths

Question 17

Use initial rate data to determine the rate law and rate constant for a reaction.

Answer 17

Use the concentration and rate data to determine it
Overall order: order of reactant 1 plus order of reactant 5

Question 18

In SN1 (unimolecular nucleophilic substitution) reactions, the lowest unoccupied molecular orbital (LUMO) plays a crucial role in determining the stereochemistry of the product.

Answer 18

1. Rate-determining step:
   In SN1 reactions, the rate-determining step is the heterolytic cleavage of the carbon-leaving group bond, forming a planar carbocation intermediate.
2. Carbocation stability:
   The stability of the carbocation intermediate is a key factor in determining the feasibility of the SN1 reaction. More stable carbocations have lower activation energies, making the reaction more favorable.
3. Orbital interactions:
   The stability of the carbocation intermediate is influenced by the interaction between the vacant p-orbital of the carbocation and the adjacent sigma (σ) bonds or lone pairs.
4. LUMO involvement:
   The LUMO is the vacant p-orbital of the carbocation intermediate. This orbital accepts electron density from the adjacent σ bonds or lone pairs, leading to delocalization and stabilization of the carbocation.
5. Stereochemistry:
   The stereochemistry of the SN1 reaction is determined by the approach of the nucleophile to the planar carbocation intermediate. The nucleophile can attack from either face of the planar carbocation, leading to a racemic mixture of products (unless the substrate or the product is chiral).
6. LUMO geometry:
   The geometry of the LUMO (the vacant p-orbital) dictates the preferred trajectory of the incoming nucleophile. The nucleophile tends to approach along the axis of the LUMO, minimizing orbital repulsions and maximizing orbital overlap.
7. Steric effects:
   While the LUMO geometry influences the trajectory of the nucleophilic attack, steric factors can also play a role. Bulky substituents near the carbocation center can hinder the approach of the nucleophile, leading to preferential attack from the less hindered side.

In summary, the LUMO of the carbocation intermediate in SN1 reactions is crucial for determining the stereochemistry of the product. The geometry of the LUMO guides the trajectory of the incoming nucleophile, while steric effects can further influence the stereochemical outcome. Understanding the role of the LUMO is essential for predicting and rationalizing the stereochemistry of SN1 reactions.

Question 19

key feature of Sn2 reaction

Answer 19

proceeds via a carbocation

Question 20

Carbocations, also known as carbon cations or carbenium ions, are reactive intermediates in organic chemistry that contain a trivalent carbon atom bearing a positive charge.

Answer 20

1. Structure:
   
   • Carbocations have a trigonal planar geometry with an empty p-orbital perpendicular to the plane.
   • The positive charge is delocalized through resonance stabilization, which can involve adjacent σ bonds or π bonds.
2. Formation:
   
   • Carbocations are typically formed by the heterolytic cleavage of a carbon-leaving group bond, where the carbon atom retains the bonding electrons.
   • Common precursors include alkyl halides, alcohols, ethers, and alkenes.
3. Stability:
   
   • The stability of carbocations is influenced by several factors, including inductive effects, hyperconjugation, and resonance stabilization.
   • Tertiary carbocations (3°) are more stable than secondary (2°), which are more stable than primary (1°) carbocations.
   • Resonance stabilization through delocalization of the positive charge is a significant factor in increasing carbocation stability.
4. Reactivity:
   
   • Carbocations are electrophilic species and can undergo nucleophilic addition reactions.
   • The rate of reaction depends on the stability of the carbocation, with more stable carbocations being less reactive.
   • Nucleophiles can attack carbocations from either face, leading to stereochemical consequences.
5. Rearrangements:
   
   • Carbocations can undergo rearrangements, such as hydride shifts, alkyl shifts, and Wagner-Meerwein rearrangements, to form more stable carbocations or products.
6. Reaction mechanisms:
   
   • Carbocations are involved in various reaction mechanisms, including SN1 (unimolecular nucleophilic substitution), E1 (unimolecular elimination), and certain rearrangement reactions.
   • Understanding the formation, stability, and reactivity of carbocations is crucial for predicting and rationalizing reaction pathways and stereochemical outcomes.

Carbocations play a central role in many organic transformations, and their study has led to a deeper understanding of reaction mechanisms, stereochemistry, and the principles governing the stability and reactivity of organic compounds.

Question 21

No carbocation intermediate in SN2 reaction

Answer 21

LUMO in Sn2 is the sigma anti bonding orbital

Here are the key points regarding the role of the LUMO in SN2 reactions:

1. Rate-determining step:
   In an SN2 reaction, the rate-determining step is the backside attack of the nucleophile on the carbon atom bearing the leaving group.
2. Orbital interaction:
   The nucleophile, which is a species with a high electron density, interacts with the σ* orbital of the carbon-leaving group bond, which is the LUMO.
3. LUMO geometry:
   The σ* orbital has a cylindrically symmetrical geometry, aligning with the internuclear axis of the carbon-leaving group bond.
4. Nucleophilic trajectory:
   The nucleophile approaches along the axis of the σ* orbital, maximizing the orbital overlap and facilitating the backside attack.
5. Orbital overlap:
   The interaction between the nucleophile’s lone pair or π-bonding electrons and the σ* orbital weakens and eventually breaks the carbon-leaving group bond, leading to the formation of the new carbon-nucleophile bond.
6. Stereochemistry:
   The backside attack of the nucleophile on the σ* orbital leads to an inversion of configuration at the carbon center, resulting in a stereochemical outcome consistent with an SN2 mechanism.
7. Steric effects:
   Steric hindrance around the carbon center can influence the trajectory of the nucleophilic attack and the orbital overlap with the LUMO, affecting the reaction kinetics and stereochemistry.

In contrast to SN1 reactions, where the LUMO is the vacant p-orbital of a planar carbocation, the LUMO in SN2 reactions is the σ* orbital of the carbon-leaving group bond. Understanding this orbital interaction is crucial for rationalizing the stereochemical outcome and the kinetics of SN2 reactions.

Question 22

Transition states with more strained or destabilized structures, weaker bonds, or less resonance stabilization are likely to have higher energies.