Chemical Kinetics: CBSE Class 12 NCERT Guide

Core Concepts in Chemical Kinetics

Rate of Reaction

The change in concentration of any one of the reactants or products per unit time. For a reaction R → P, Rate = -Δ[R]/Δt = +Δ[P]/Δt. The negative sign for reactants indicates their concentration decreases over time.

Rate Law

An expression that relates the rate of a reaction to the concentration of the reactants raised to some power. For a reaction aA + bB → Products, Rate = k[A]^x[B]^y, where 'x' and 'y' are determined experimentally.

Order of Reaction

The sum of the powers of the concentration terms in the experimentally determined rate law. In the above example, the overall order is x + y. The order can be 0, 1, 2, or even a fraction.

Molecularity

The number of reacting species (atoms, ions, or molecules) that must collide simultaneously to bring about a chemical reaction in an elementary step. It is a theoretical concept and is always a positive integer (1, 2, or 3).

Activation Energy (Ea)

The minimum extra amount of energy that must be supplied to the reactant molecules to enable them to cross the energy barrier and convert into products. It's the energy required to form the activated complex.

Half-Life (t₁/₂)

The time required for the concentration of a reactant to be reduced to one-half of its initial value. It's a useful measure of reaction speed, especially for first-order reactions where it is constant.

Factors Affecting the Rate of a Reaction

Several factors can influence how fast a reaction proceeds. Understanding them is key to controlling chemical processes.

1. Concentration of Reactants: According to collision theory, for a reaction to occur, reactant molecules must collide. Increasing the concentration of reactants increases the number of particles per unit volume. This leads to more frequent collisions, and therefore, a higher reaction rate. This is why the rate law directly relates rate to concentration.

1. Temperature: Increasing the temperature almost always increases the reaction rate. Why? At higher temperatures, molecules have greater kinetic energy. This has a two-fold effect: they collide more frequently, and more importantly, a larger fraction of these collisions have energy equal to or greater than the activation energy (Ea). The Arrhenius equation quantitatively describes this relationship.

1. Presence of a Catalyst: A catalyst is a substance that increases the rate of a reaction without being consumed itself. It does this by providing an alternative reaction pathway with a lower activation energy. With a smaller energy barrier to overcome, more reactant molecules can successfully convert to products upon collision, speeding up the reaction significantly.

1. Surface Area of Reactants: For reactions involving a solid reactant (heterogeneous reactions), the rate depends on the surface area. Grinding a solid into a powder increases its surface area, exposing more particles to the other reactants. This increases the frequency of collisions at the surface, thus increasing the reaction rate. A classic example is a powdered sugar dissolving faster than a sugar cube.

Worked Examples: Integrated Rate Equations

• Example 1: First-Order Reaction A first-order reaction is found to have a rate constant, k = 5.5 × 10⁻¹⁴ s⁻¹. Find the half-life of the reaction. Step 1: Identify the formula. For a first-order reaction, the relationship between the rate constant (k) and half-life (t₁/₂) is given by the formula: t₁/₂ = 0.693 / k Step 2: Substitute the given value of k. We are given k = 5.5 × 10⁻¹⁴ s⁻¹. Step 3: Calculate the half-life. t₁/₂ = 0.693 / (5.5 × 10⁻¹⁴ s⁻¹) t₁/₂ = (0.693 / 5.5) × 10¹⁴ s t₁/₂ ≈ 0.126 × 10¹⁴ s t₁/₂ = 1.26 × 10¹³ s Final Answer: The half-life of the reaction is 1.26 × 10¹³ s.
• Example 2: Zero-Order Reaction The decomposition of a substance is a zero-order reaction. The initial concentration of the substance is 1.5 M and after 120 seconds, the concentration is 0.75 M. Calculate the rate constant (k). Step 1: Identify the integrated rate law for a zero-order reaction. The formula is: [R] = -kt + [R]₀ Where [R] is the final concentration, [R]₀ is the initial concentration, k is the rate constant, and t is the time. Step 2: Rearrange the formula to solve for k. kt = [R]₀ - [R] k = ([R]₀ - [R]) / t Step 3: Substitute the given values. [R]₀ = 1.5 M [R] = 0.75 M t = 120 s k = (1.5 M - 0.75 M) / 120 s k = 0.75 M / 120 s k = 0.00625 M s⁻¹ Step 4: Express the answer in scientific notation and state the units. k = 6.25 × 10⁻³ mol L⁻¹ s⁻¹ Final Answer: The rate constant for this zero-order reaction is 6.25 × 10⁻³ mol L⁻¹ s⁻¹.

Exam Traps and Key Points

Pay close attention to these points to avoid common mistakes in your exams:

• Units of Rate Constant (k): The units of 'k' depend on the overall order of the reaction. For a reaction of order 'n', the units are (concentration)¹⁻ⁿ (time)⁻¹.
• Zero order: mol L⁻¹ s⁻¹
• First order: s⁻¹
• Second order: L mol⁻¹ s⁻¹

Forgetting or mixing up these units is a very common way to lose marks.

• Order vs. Molecularity: Do not use these terms interchangeably. Order is an experimental quantity determined from the rate law and can be zero or fractional. Molecularity is a theoretical concept for elementary reactions and is always a whole number.
• Rate Law from Stoichiometry: Never determine the rate law from the stoichiometric coefficients of a balanced chemical equation unless you are explicitly told the reaction is an elementary step. The powers in the rate law (x and y) must be found experimentally.
• Graph Interpretation: Be very comfortable with the graphical representations for different orders.
• Zero Order: A plot of [R] vs. time is a straight line with slope = -k.
• First Order: A plot of ln[R] vs. time is a straight line with slope = -k.
• Second Order: A plot of 1/[R] vs. time is a straight line with slope = k.

Practice Questions with Solutions

• Q: For the reaction 2A + B → C, the rate of disappearance of A is 4 × 10⁻⁴ mol L⁻¹ s⁻¹. What is the rate of formation of C and the rate of disappearance of B? A: Step 1: Write the relationship between the rates of reaction for each species based on stoichiometry. Rate = -(1/2) d[A]/dt = -d[B]/dt = +d[C]/dt Step 2: Calculate the rate of formation of C. Rate of formation of C, d[C]/dt = (1/2) (rate of disappearance of A) d[C]/dt = (1/2) (4 × 10⁻⁴ mol L⁻¹ s⁻¹) = 2 × 10⁻⁴ mol L⁻¹ s⁻¹. Step 3: Calculate the rate of disappearance of B. Rate of disappearance of B, -d[B]/dt = (1/2) (rate of disappearance of A) -d[B]/dt = (1/2) (4 × 10⁻⁴ mol L⁻¹ s⁻¹) = 2 × 10⁻⁴ mol L⁻¹ s⁻¹. Final answer: The rate of formation of C is 2 × 10⁻⁴ mol L⁻¹ s⁻¹ and the rate of disappearance of B is 2 × 10⁻⁴ mol L⁻¹ s⁻¹.
• Q: A reaction has a rate law: Rate = k[A]¹/²[B]². What is the overall order of the reaction? What happens to the rate if the concentration of B is tripled? A: Step 1: Determine the overall order. The order of a reaction is the sum of the powers of the concentration terms in the rate law. Overall order = (order with respect to A) + (order with respect to B) = 1/2 + 2 = 2.5. Step 2: Analyze the effect of changing the concentration of B. The rate is proportional to [B]². Let the initial rate be R₁ = k[A]¹/²[B]². If [B] is tripled, the new concentration is 3[B]. The new rate, R₂ = k[A]¹/²(3[B])² = k[A]¹/² 9[B]² = 9 (k[A]¹/²[B]²) = 9 * R₁. Final answer: The overall order of the reaction is 2.5. If the concentration of B is tripled, the reaction rate will increase by a factor of 9.
• Q: A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life (t₁/₂). (Given: log(10/7) ≈ 0.155) A: Step 1: Use the integrated rate law for a first-order reaction. k = (2.303/t) log([R]₀/[R]) Step 2: Determine the values for the formula. If 30% is decomposed, 70% remains. So, if [R]₀ = 100, then [R] = 70. The ratio [R]₀/[R] = 100/70 = 10/7. Time, t = 40 minutes. Step 3: Calculate the rate constant, k. k = (2.303 / 40 min) log(10/7) k = (2.303 / 40) 0.155 min⁻¹ k ≈ 0.00892 min⁻¹ Step 4: Calculate the half-life using k. t₁/₂ = 0.693 / k t₁/₂ = 0.693 / 0.00892 min⁻¹ t₁/₂ ≈ 77.7 minutes. Final answer: The half-life of the reaction is approximately 77.7 minutes.
• Q: How does a catalyst increase the rate of a chemical reaction? Explain with reference to activation energy. A: Step 1: Define a catalyst's primary function. A catalyst is a substance that accelerates a chemical reaction without being consumed in the overall process. Step 2: Explain the mechanism involving activation energy. A catalyst provides an alternative reaction pathway or mechanism. This new pathway has a lower activation energy (Ea) compared to the uncatalyzed reaction. Step 3: Relate lower activation energy to reaction rate. According to the Arrhenius equation and collision theory, a lower activation energy means that a larger fraction of reactant molecules will possess the minimum energy required for a successful collision at a given temperature. This leads to a higher number of effective collisions per unit time, thus increasing the reaction rate. Final answer: A catalyst increases the reaction rate by providing an alternative pathway with a lower activation energy, which allows more reactant molecules to overcome the energy barrier and form products.

Frequently Asked Questions