Chemical Kinetics 20 Most Important Questions and Numericals

Chemical Kinetics is a crucial branch of physical chemistry that studies the speed of chemical reactions and the factors that affect them. It helps us understand how reaction rates change with concentration, temperature, catalysts, and surface area. 

This topic explains important concepts such as rate laws, order and molecularity of reactions, half-life, activation energy, and the Arrhenius equation. A strong grasp of both theory and numericals in Chemical Kinetics is essential for board exams and competitive exams like JEE and NEET. 

Practicing conceptual questions and calculation-based problems builds accuracy, speed, and a deeper understanding of reaction mechanisms. Students must practice the 20 most important questions and numericals to boost their score in the CBSE Class 12 Chemistry exam on 28 February.

Assertion-Reason: Order of Reaction

• Assertion (A): The order of the reaction can be zero or fractional.

• Reason (R): We cannot determine order from a balanced chemical equation.

Analysis: Both Assertion (A) and Reason (R) are correct. Reaction order is an experimental value, determined from the rate law expression, and can indeed be zero or fractional. A balanced chemical equation only indicates stoichiometry, not the reaction mechanism or rate-determining step; thus, it cannot predict the order. However, Reason (R) does not explain why the order can be zero or fractional; it only states why it cannot be determined stoichiometrically.

Conclusion: Both A and R are true, but R is not the correct explanation of A.

Assertion-Reason: Order vs. Molecularity

• Assertion (A): Order and molecularity are the same.

• Reason (R): Order is determined experimentally, and molecularity is the sum of the stoichiometric coefficients of the rate-determining elementary step.

Analysis: The assertion is false. Order and molecularity are generally different; they are only the same for elementary (single-step) reactions. Reason (R) is true, correctly defining order as experimentally determined and molecularity for an elementary step as the sum of reactant stoichiometric coefficients. Molecularity is not defined for complex reactions as a whole.

Conclusion: Assertion (A) is false, but Reason (R) is true.

Assertion-Reason: Effect of a Catalyst

• Assertion (A): The enthalpy of the reaction remains constant in the presence of a catalyst.

• Reason (R): The catalyst participates in the reaction, forms a different activated complex, and lowers the activation energy, but the difference in energy between the reactant and product remains the same.

Analysis: Both Assertion (A) and Reason (R) are true. A catalyst provides an alternate reaction pathway with a lower activation energy (Ea), forming a different activated complex, thus increasing the reaction rate. However, a catalyst does not change the overall enthalpy (ΔH) of a reaction because it does not alter the initial energy of reactants or the final energy of products. Reason (R) correctly explains the mechanism by which a catalyst acts and why ΔH remains constant.

Conclusion: Both A and R are true, and R is the correct explanation for A.

Calculating Average Rate from a Graph

Problem: Calculate the average rate of reaction from a concentration-time graph showing concentration decreasing from 5 M at 4 seconds to 1 M at 10 seconds.

Solution:

1. Identify Change: The concentration decreases, indicating a reactant.

2. Formula: Average Rate = - (C_final - C_initial) / (t_final - t_initial). A negative sign must be included for reactants to ensure a positive rate value.

3. Calculation:

• C_initial = 5 M, C_final = 1 M

• t_initial = 4 s, t_final = 10 s

• Average Rate = - (1 M - 5 M) / (10 s - 4 s) = - (-4 M) / 6 s = 4/6 M/s = 2/3 mol L⁻¹ s⁻¹.

Assertion-Reason: Collision Theory

• Assertion (A): All collisions of reactant molecules lead to product formation.

• Reason (R): Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.

Analysis: The assertion is false. According to collision theory, not all collisions are effective. Reason (R) is true, stating the two necessary conditions for an effective collision: molecules must possess sufficient energy (threshold energy) and collide with proper orientation.

Conclusion: Assertion (A) is false, but Reason (R) is true.

Assertion-Reason: Arrhenius Equation

• Assertion (A): Rate constants determined from the Arrhenius equation are fairly accurate for single-step as well as complex reactions.

• Reason (R): Reactant molecules undergo a chemical change irrespective of their orientation during the collision.

Analysis: Assertion (A) is true. The Arrhenius equation is a reliable model for determining rate constants across various reaction types. Reason (R) is false. Collision theory emphasizes that correct orientation is critical for product formation during a collision; incorrect orientation, even with sufficient energy, will not lead to a reaction.

Conclusion: Assertion (A) is true, but Reason (R) is false.

Assertion-Reason: Activation Energy

• Assertion (A): A reaction can have zero activation energy.

• Reason (R): The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold energy is called activation energy.

Analysis: Both Assertion (A) and Reason (R) are true. Some reactions, such as the combination of free radicals (e.g., CH₃• + CH₃• → C₂H₆), can have zero activation energy. Reason (R) provides the correct definition of activation energy (Ea). However, the definition doesn't explain why some reactions have zero Ea.

Conclusion: Both A and R are true, but R is not the correct explanation of A.

Thermodynamic vs. Kinetic Feasibility

Question: "Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction." Explain with an example.

Explanation:

1. Thermodynamic Feasibility depends on the Gibbs Free Energy change (ΔG). A negative ΔG indicates a spontaneous (thermodynamically feasible) reaction.

2. Kinetic Feasibility relates to the reaction rate, which is governed by factors like activation energy. A reaction can be thermodynamically feasible but kinetically very slow due to high activation energy.
   Example: The conversion of diamond to graphite (Diamond → Graphite) has a negative ΔG, making it thermodynamically spontaneous. However, it possesses an extremely high activation energy, making its rate negligibly slow under normal conditions. Thus, it is not kinetically viable on a practical timescale.

Assertion-Reason: Zero-Order Reaction

• Assertion (A): For a zero-order reaction, the unit of the rate constant and the rate of reaction are the same.

• Reason (R): The rate of reaction for a zero-order reaction is independent of the concentration of the reactant.

Analysis: Both Assertion (A) and Reason (R) are true, and R is the correct explanation for A. For a zero-order reaction, the rate law is Rate = k[R]⁰, which simplifies to Rate = k. This means the rate of reaction is equal to the rate constant and is independent of reactant concentration. Consequently, the units of rate and rate constant must be identical (e.g., mol L⁻¹ s⁻¹).

Conclusion: Both A and R are true, and R is the correct explanation for A.

Numerical: Activation Energy Calculation (Two-Temperature Arrhenius)

Problem: A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate the activation energy (Ea). (Given: R = 8.314 J K⁻¹ mol⁻¹)

Solution Steps:

1. Half-life (t₁/₂): For a first-order reaction, t₁/₂ = 0.693 / k.

• k₁ = 0.693 / 30 min (at 300 K)

• k₂ = 0.693 / 10 min (at 320 K)

• Ratio k₂/k₁ = (0.693/10) / (0.693/30) = 3.

2. Arrhenius Equation (logarithmic form): log(k₂/k₁) = (Ea / 2.303R) * [(T₂ - T₁) / (T₁ * T₂)]

3. Substitute and Solve:

• log(3) = (Ea / (2.303 * 8.314)) * [(320 - 300) / (300 * 320)]

• 0.4771 = (Ea / 19.147) * [20 / 96000]

• Ea ≈ 43,848 J/mol or 43.8 kJ/mol.

Case-Based Study: General Rate Concepts

• A. Average Rate: Change in concentration of a reactant or product over a large, appreciable interval of time.

• B. Factors affecting rate: Concentration of reactants and Temperature.

• C. Zero-order reaction rate: The rate (Rate = k) is independent of the initial concentration of reactants and remains constant.

• OR Question: Order Calculation: For Rate = k[P]¹/²[Q]¹, Order = 0.5 + 1 = 1.5.

• OR Question: Pseudo-First Order Reaction: A reaction that appears first-order under specific conditions (e.g., one reactant in large excess), though its actual molecularity is higher. Example: Hydrolysis of an ester in excess water.

Case-Based Study: Factors Affecting Reaction Rate

• A. Effect of Temperature on Rate Constant (k): The rate constant (k) increases with a rise in temperature. It approximately doubles for every 10°C (or 10 K) rise.

• B. Order Calculation: For 2A + B → C with Rate = k[A]²[B]⁰.⁵, the order is 2 + 0.5 = 2.5.

• C. Order vs. Molecularity for Complex Reactions:

• Molecularity is defined only for elementary reactions (number of colliding species).

• Order is an experimentally determined quantity derived from the rate law, often reflecting the slowest (rate-determining) step of a complex reaction.

Numerical: First-Order Kinetics

Problem: A first-order reaction has a rate constant k = 2 x 10⁻³ s⁻¹. How long will it take for 6 g of this reactant to reduce to 2 g?

Solution:

1. Integrated Rate Law (First-Order): t = (2.303 / k) * log(A₀ / A)

2. Substitute Values:

• A₀ = 6 g, A = 2 g, k = 2 x 10⁻³ s⁻¹

• t = (2.303 / (2 x 10⁻³)) * log(6 / 2)

• t = (2.303 / (2 x 10⁻³)) * log(3)

• t = (2.303 / (2 x 10⁻³)) * 0.4771

• t ≈ 549.4 s

Expressing Reaction Rate

Problem: For the reaction N₂ + 3H₂ → 2NH₃, identify the correct relationship for the rate of reaction.

Solution: The general rate expression relates the change in concentration of reactants and products, divided by their stoichiometric coefficients. Reactants have a negative sign, products a positive sign.

Rate = -d[N₂]/dt = -(1/3)d[H₂]/dt = +(1/2)d[NH₃]/dt

The correct option would align with this full expression, for instance, -(1/3)d[H₂]/dt = +(1/2)d[NH₃]/dt. It's crucial to correctly apply the negative sign for reactants and the inverse of the stoichiometric coefficient for all species.

Practice Problem: Calculating Activation Energy (Arrhenius Equation)

Problem: A rate constant for a first-order reaction is given by log k = [some value] - (1.0 x 10⁴) / T. Calculate the activation energy (Ea).

Solution:

1. Standard Arrhenius Equation (logarithmic form): log k = log A - Ea / (2.303 * R * T)

2. Compare: Equate the 1/T terms from the given equation and the standard form:
   Ea / (2.303 * R * T) = (1.0 x 10⁴) / T

3. Solve for Ea:

• Ea = (1.0 x 10⁴) * 2.303 * R

• Using R = 8.314 J/(mol·K): Ea = (1.0 x 10⁴) * 2.303 * 8.314

• Ea ≈ 19.14 x 10⁴ J/mol or 191.4 kJ/mol.

Practice Problem: Graphical Representation of a Zero-Order Reaction

• Problem 1: Slope of [R] vs. Time Plot: For a zero-order reaction, the integrated rate law is [R]t = -kt + [R]₀. Comparing this to y = mx + c, the slope m is -k (the negative of the rate constant).

• Problem 2: Identifying the Correct Graph: A zero-order reaction is represented by a straight line with a negative (downward) slope when plotting reactant concentration [R] versus time t.

Practice Problem: Half-Life Calculations

• Problem 1: First-Order Reaction: Given t₁/₂ = 10 s for a first-order reaction, find k.

• Formula: k = 0.693 / t₁/₂

• Calculation: k = 0.693 / 10 s = 0.0693 s⁻¹.

• Problem 2: Zero-Order Reaction Half-Life: What happens to the half-life period of a zero-order reaction if the initial concentration is doubled?

• Formula: t₁/₂ = [A]₀ / 2k.

• Analysis: The half-life (t₁/₂) is directly proportional to the initial concentration ([A]₀).

• Conclusion: If [A]₀ is doubled, t₁/₂ will also be doubled.

Practice Problem: Identifying Reaction Order from Completion Times

Problem: A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction?

Solution: For first-order reactions, a crucial relationship is t₇₅% = 2 * t₅₀% (Memory Tip: If the time for 75% completion is exactly double the time for 50% completion, it's first-order).

Given t₅₀% = 2 hours and t₇₅% = 4 hours. Since 4 hours = 2 * (2 hours), the reaction follows this relationship.

Conclusion: The reaction is first-order.

Practice Problem: Applying the t₇₅% vs t₅₀% Relation

Problem: For a first-order reaction, if 75% of the reaction is completed in 32 minutes, calculate the time taken for 50% completion (t₅₀%).

Solution: Using the first-order relation: t₇₅% = 2 * t₅₀%.

Given t₇₅% = 32 minutes.

32 minutes = 2 * t₅₀%

t₅₀% = 32 / 2 = 16 minutes.

Practice Problem: Gas-Phase First-Order Reaction Formula

For a gas-phase first-order reaction A(g) → B(g) + C(g), the correct formula for the rate constant (k) is:

k = (2.303 / t) * log [ Pᵢ / (2Pᵢ - Pₜ) ]

where Pᵢ is the initial pressure and Pₜ is the total pressure at time t.

Summary of Important Topics in Chemical Kinetics

For comprehensive preparation in Chemical Kinetics, focus on these critical areas:

1. Rate of Reaction: Understand how to write rate expressions, including the use of stoichiometric coefficients and sign conventions (negative for reactants, positive for products).

2. Arrhenius Equation: Master the equation, its logarithmic form, and its graphical representation. Be able to calculate activation energy (Ea) from given data or by comparing equations.

3. Collision Theory and Catalysis: Comprehend the conditions for effective collisions (sufficient energy and proper orientation) and how a catalyst lowers activation energy by providing an alternate pathway, without affecting overall enthalpy (ΔH).

4. Integrated Rate Laws: Know the derivations and applications of integrated rate laws for zero-order and first-order reactions.

5. Key Reaction Time Relations: Be familiar with relationships like t₇₅% = 2 * t₅₀% for first-order reactions and how half-life varies with concentration for different orders.

6. Graphical Representations: Interpret and identify graphs illustrating zero-order and first-order reactions (e.g., concentration vs. time, log[R] vs. time, 1/T vs. log k).

7. Pseudo-First Order Reactions: Understand their definition and be able to provide 2-3 common examples.

8. Assertion-Reason Questions: Practice analyzing statements related to reaction order, molecularity, catalyst effects, and Arrhenius theory.