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📑 In This Chapter Guide (Table of Contents)
1. Rate of Reaction, Order and Molecularity
| Feature | Order of Reaction | Molecularity of Reaction |
|---|---|---|
| Definition | Sum of powers of concentration terms of reactants in the experimentally determined rate law. | Number of colliding reacting species that collide simultaneously to bring about an elementary chemical change. |
| Nature of Value | Purely experimental property; can be zero, integer, or fractional. | Theoretical property; only positive integers (1, 2, 3); cannot be zero or fractional. |
| Applicability | Applicable to both elementary and complex multi-step reactions. | Meaningful only for elementary reactions (has no meaning for complex reactions). |
2. Integrated Rate Equations for Zero & First Order Reactions
A. Zero-Order Reaction (Rate = k[R]⁰ = k):
[R] = [R]₀ - k · t ⟹ k = ([R]₀ - [R]) / t
Half-Life (t½):
t½ = [R]₀ / (2k) (Directly proportional to initial reactant concentration).
Units of k:
mol · L⁻¹ · s⁻¹.
B. First-Order Reaction (Rate = k[R]¹):
k = (2.303 / t) · log₁₀ [ [R]₀ / [R] ]
Half-Life (t½):
t½ = (2.303 · log 2) / k = 0.693 / k
Notice: Half-life of a first-order reaction is completely INDEPENDENT of initial concentration [R]₀!
Units of k:
s⁻¹ (or min⁻¹).
3. Temperature Dependence & Arrhenius Equation
For most chemical reactions, reaction rate roughly doubles for every 10°C rise in temperature. Svante Arrhenius quantified this temperature dependence:
k = A · e^(-Ea / RT)
(where A = Arrhenius pre-exponential frequency factor, Ea = Activation Energy in J/mol, R = 8.314 J·K⁻¹·mol⁻¹)
Logarithmic Two-Temperature Form (For Exam Calculations):
log₁₀(k₂ / k₁) = (Ea / 2.303 R) · [ (T₂ - T₁) / (T₁ · T₂) ]
💡 Frequently Asked Questions (FAQ)
❓ Prove that the time required to complete 99.9% of a first-order reaction is 10 times its half-life (t½).
For 99.9% completion: [R] = [R]₀ - 0.999[R]₀ = 0.001[R]₀ = 10⁻³[R]₀. Using first-order formula: t_99.9% = (2.303/k) log([R]₀ / 10⁻³[R]₀) = (2.303/k) log(10³) = 3 × 2.303/k = 6.909/k. Half life is t½ = 0.693/k. Dividing: t_99.9% / t½ = (6.909/k) / (0.693/k) ≈ 10. Therefore, t_99.9% = 10 × t½. (Hence Proved)
❓ What is a Pseudo First Order reaction? Give an example.
A reaction which is truly higher order but behaves kinetically as a first-order reaction under specific conditions (usually when one reactant is in huge excess). Example: Acid-catalyzed hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O --[H⁺]--> CH₃COOH + C₂H₅OH. Because water is present in enormous excess, its concentration remains essentially constant, so Rate = k’[Ester].
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