Chapter 5. Kinetics

5.1 Reaction Rates

5.1.1

5.1.2

5.1.3

The instantaneous rate of disappearance of C4H9Cl at t=0 s is: −0.0004 M/s

Detailed explanation here.

PART A. -Δ[E]/Δt = 2Δ[G]/Δt

Detailed explanation here.

PART B. -Δ[F]/Δt = (3/4)Δ[G]/Δt

Detailed explanation here.

PART C. Δ[H₂]/Δt = (-1/2)Δ[E]/Δt

Detailed explanation here.

At a higher temperature, the rate of decomposition of H₂O₂ will increase. This is because, based on kinetic molecular theory, higher temperatures increase the average kinetic energy of molecules, resulting in more frequent and more effective collisions.

Detailed explanation here.

5.2 Introduction to Rate Law

5.2.1

5.2.2

5.2.3

5.2.4

5.2.5

PART A. This is done using a spectrophotometer or colorimeter (to measure absorbance), cuvettes (to hold the solution), and a timer. Plotting absorbance vs. time allows for rate determination and kinetic analysis.

Detailed explanation here.

PART B. Given that color intensity is directly proportional to crystal violet concentration, a 75% decrease in color intensity means the crystal violet concentration ([CV⁺]) is also reduced by 75%, leaving 25% of its initial concentration ([CV⁺]₀): [CV⁺] = 0.25[CV⁺]₀. The reaction rate is calculated by finding the change in concentration over the time (t) it takes for this 75% decrease: Rate = -Δ[CV⁺]/Δt = -([CV⁺] - [CV⁺]₀)/t. If absorbance data is available, Beer’s Law (A = εcl) can be used to determine concentration (c) from absorbance (A), molar absorptivity (ε), and path length (l), allowing for rate calculation using these concentration values.

Detailed explanation here.

PART C. The higher temperature (308 K) increases the rate of reaction due to more frequent and effective collisions as predicted by collision theory.

Detailed explanation here.

PART A. The overall order of the reaction is 4.

Detailed explanation here.

PART B. The rate increases by a factor of 2 when [P] is doubled.

Detailed explanation here.

PART C. The rate increases by a factor of 27 when [Q] is tripled.

Detailed explanation here.

PART A.

(i) The order of the reaction with respect to A is 1.

(ii) The order of the reaction with respect to B is 2.

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PART B. The overall order of the reaction is 3.

Detailed explanation here.

PART C. If the concentration of A is tripled, the rate of the reaction will increase by a factor of 3.

Detailed explanation here.

The units of the rate constant k are: L mol⁻¹ s⁻¹

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PART A. Order with respect to each reactant:

Br⁻ = 1
BrO₃⁻ = 1
H⁺ = 3

Detailed explanation here.

PART B. Rate Law Expression: Rate = k[Br⁻][BrO₃⁻][H⁺]³

Detailed explanation here.

PART C. Overall Order of Reaction: 5

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PART D. Rate Constant (k): k = 5.00 × 10⁴ L⁴ mol⁻⁴ s⁻¹

Detailed explanation here.

5.3 Concentration Changes over Time

5.3.1

5.3.2

5.3.3

5.3.4

5.3.5

5.3.6

This is a second order reaction.

Detailed explanation here.

PART A. The flask initially contained 0.0317 moles of X.

Detailed explanation here.

PART B. The reaction is first-order with respect to X.

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PART C. The rate law is: ln[X] = ln[X]₀ - 0.0197 min⁻¹ t

Detailed explanation here.

PART A. The reaction is verified to be second-order based on the linearity of the 1/[N₂O₅] vs. time plot. See link for the actual graph.

Detailed explanation here.

PART B. The rate constant k = 0.500 M⁻¹ s⁻¹.

Detailed explanation here.

PART C. At t = 180 s, the concentration of N₂O₅ is: [N₂O₅] = 0.0071 M

Detailed explanation here.

The slope of the graph for a second-order reaction is equal to the rate constant k. For this reaction, the rate constant is: k = 0.025 L/(mol⋅s)

Detailed explanation here.

125 seconds

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PART A. The decay constant is: k = 0.0866 day⁻¹

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PART B. After 16 days, 12.5 mg of I-131 remains.

Detailed explanation here.

PART C. It will take approximately 26.6 days for I-131 to decay to 10% of its original amount.

Detailed explanation here.

5.4 Elementary Reactions

5.4.1

5.4.2

PART A. Rate = k[O₃][NO]

Detailed explanation here.

PART B. The rate of reaction increases by a factor of 3.

Detailed explanation here.

PART C. The rate law can be directly inferred from the stoichiometry of this elementary reaction because the reaction occurs in a single step, and the stoichiometric coefficients represent the molecularity of the reaction.

Detailed explanation here.

A termolecular elementary step, where three molecules collide simultaneously, is highly improbable due to the extremely low probability of all three molecules meeting at the same time, with the correct orientation and sufficient energy to react. Most observed reactions proceed via unimolecular or bimolecular elementary steps because these involve fewer simultaneous collisions and are more kinetically feasible. Complex reactions, like the given reaction, typically occur in multiple elementary steps rather than a single termolecular step. Therefore, the proposed mechanism is unlikely, and a more reasonable explanation would involve breaking the reaction into simpler bimolecular steps.

Detailed explanation here.

5.5 Collision Model

5.5.1

5.5.2

PART A. Not all collisions between A₂ and BC result in a reaction because the molecules must have enough kinetic energy to overcome the activation energy barrier and collide with the correct orientation for new bonds to form.

Detailed explanation here.

PART B. Increasing temperature speeds up reactions by increasing molecular kinetic energy, leading to more collisions with sufficient energy to overcome the activation barrier and slightly higher collision frequency, collectively resulting in a faster reaction rate.

Detailed explanation here.

PART C. Improper molecular orientation decreases the reaction rate because it prevents the necessary atoms from aligning for bond formation, rendering collisions ineffective even if they have enough energy. Proper orientation is essential for successful reactions.

Detailed explanation here.

PART D. Increasing the temperature raises the average kinetic energy of molecules, increasing the fraction of collisions with energy above the activation energy and thereby boosting the reaction rate. According to the Arrhenius equation, the rate constant k increases exponentially with temperature, further accelerating the reaction.

Detailed explanation here.

PART A. Shading the Area: Shade the area under the curve to the right of the Ea line, representing molecules with sufficient energy to react.

Detailed explanation here.

PART B. Effect of Increased Temperature: Increasing the temperature shifts the curve to the right and flattens it slightly, increasing the shaded area to the right of Ea, as more molecules reach or exceed the activation energy.

Detailed explanation here.

PART C. Explanation of Temperature's Effect on Reaction Rate: Higher temperature increases the fraction of molecules with sufficient energy to overcome Ea, resulting in more frequent successful collisions and a higher reaction rate

Detailed explanation here.

5.6 Reaction Energy Profile

5.6.1

5.6.2

5.6.3

PART A. The reaction coordinate represents the progress of the reaction, showing the energy changes and molecular motions as reactants are converted into products. It tracks bond-breaking and bond-forming processes.

Detailed explanation here.

PART B. A slower-than-expected reaction could indicate a high activation energy, improper molecular orientation, or inefficient energy transfer, making it difficult for molecules to reach the transition state along the reaction coordinate.

Detailed explanation here.

PART A. These are the features of your diagram:

Draw the axes:

x-axis: Reaction Coordinate (progress of the reaction)
y-axis: Potential Energy (kJ/mol)

Plot the reactants and products:

Reactants (CH₄ + 2O₂): Place this point on the y-axis at a certain energy level. The exact value is not crucial for this rough sketch.
Products (CO₂ + 2H₂O): Place this point below the reactants on the y-axis at 200 kJ/mol. Since the products have lower energy, they are at a lower position on the diagram.

Draw the activation energy barriers:

Forward Reaction: Draw a curve from the reactants upwards to a peak. This peak represents the transition state. The height of this peak above the reactants is the activation energy for the forward reaction (150 kJ/mol).
Reverse Reaction: Draw a curve from the products upwards to a peak. This peak represents the transition state for the reverse reaction. The height of this peak above the products is the activation energy for the reverse reaction (300 kJ/mol).

Detailed explanation here.

PART B.

ΔH = Energy of Products - Energy of Reactants

ΔH = 200 kJ/mol - Energy of Reactants

Since the products have lower energy than the reactants, ΔH will be negative. This confirms that the reaction is exothermic.

Detailed explanation here.

The activation energy directly determines the number of collisions that have enough energy to overcome the energy barrier and proceed to form products.

Detailed explanation here.

5.7 Introduction to Reaction Mechanism

5.7.1

5.7.2

5.7.3

5.7.4

Overall Reaction: 2SO₂(g) + O₂(g) → 2SO₃(g); NO is an intermediate; NO₂is a catalyst.

Detailed explanation here.

Both mechanisms (Mechanism 1 and Mechanism 2) are valid because: (1) Both add up to the overall reaction and (2) Both involve plausible elementary steps.

Detailed explanation here.

PART A. The overall balanced chemical equation is: 2N₂O(g) → 2N₂(g) + O₂(g)

Detailed explanation here.

PART B. The rate law for the reaction is: Rate = k[N₂O]

Detailed explanation here.

PART C. Intermediate: O(g) Catalyst: None

Detailed explanation here.

PART A. Mechanism 1 is supported by the experimental evidence.

Detailed explanation here.

PART B. Intermediates like D or E serve as evidence for a proposed mechanism by linking the individual steps to the overall reaction. Their detection supports the mechanism, while their absence casts doubt on its validity.

Detailed explanation here.

PART C. Distinguishing between D and E is crucial because the presence of one and absence of the other validates one mechanism and rules out the other, providing clarity about the true reaction pathway.

Detailed explanation here.

5.8 Reaction Mechanism and Rate Law

5.8.1

Valid Mechanism: Mechanism 1 is valid because its rate law matches the experimental rate law (Rate = k[H₂][NO]²). Mechanism 2 is invalid because its rate law (Rate = k[H₂][NO]) does not match the experimental rate law.

Detailed explanation here.

5.9 Steady-State Approximation

5.9.1

The rate law for the reaction is: Rate = k[A]; The overall reaction is: A → B + D + E

Detailed explanation here.

5.10 Multistep Reaction Energy Profile

5.10.1

Diagram Characteristics:

Reactants: Start at a low energy level.
First step: A small activation energy leads to the first intermediate.
Second step (rate-determining): A large activation energy leads to the second intermediate.
Third step: A moderate activation energy leads to the products, which are at a higher energy level than the reactants.

Detailed explanation here.

5.11 Catalysis

5.11.1

5.11.2

5.11.3

True statements:

C) Catalysts can speed up chemical reactions by providing an alternate pathway for the reaction that requires less energy.
F) Catalysts can speed up chemical reactions by forming intermediates and introducing new elementary steps that require less energy.
H) Catalysts are regenerated within a chemical reaction.

Detailed explanation here.

PART A. The overall reaction is: A+2B→2E

Detailed explanation here.

PART B. A is not a catalyst because it is consumed in the reaction and not regenerated. The correct catalyst is B.

Detailed explanation here.

PART C. The intermediates are C and D.

Detailed explanation here.

PART D. The rate law is: Rate = k[A][B]

Detailed explanation here.

PART A. The palladium catalyst lowers the activation energy by adsorbing H₂ and C₂H₄ onto its surface, weakening bonds in the reactants and stabilizing intermediates, thus providing an alternate pathway with lower energy.

Detailed explanation here.

PART B. The intermediates are: Pd-H and Pd-C₂H₄

Detailed explanation here.

PART C. Adsorption facilitates the reaction by weakening bonds in the reactants, bringing them into proximity for interaction, and providing a surface for efficient bond-breaking and bond-making processes.

Detailed explanation here.