Chemical Kinetics for NEET UG

Kinetics is about how fast reactions go and what controls their speed. Rate laws, half-lives and the Arrhenius equation are the main tools.

ChemistryClass 1274 MCQs3 lessons13 flashcards3 memory tricks

Practise Chemical Kinetics free

What you'll learn

  1. Rate, rate law and order
    Write a rate law, find the order of a reaction and the units of k.
  2. First-order reactions
    Use the integrated first-order equation to find time, concentration or k.
  3. Temperature, activation energy and catalysts
    Use the Arrhenius equation and explain how catalysts speed up reactions.

Key formulas

Integrated rate lawk = (2.303/t) log([A]₀/[A])
Half-lifet½ = 0.693/k
Time for 99.9% completion≈ 10 half-lives
Arrhenius equationk = A e^(−Ea/RT)
Two temperatureslog(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)

Memory tricks

Temperature and rateFood spoils faster in May than in December.

For many reactions, a 10 °C rise roughly doubles the rate (Arrhenius: k = A e^(−Ea/RT)). A fridge slows spoilage.

What a catalyst doesA catalyst digs a tunnel through the mountain instead of climbing over it. The start and the end stay the same.

A catalyst provides a path with lower activation energy. It does not change ΔH, ΔG or K.

First-order half-lifeHostel mess: half the rotis vanish every 5 minutes, whether there were 100 or 1,000.

For a first-order reaction t½ = 0.693/k, independent of starting concentration. Units of k: s⁻¹ (first order), mol L⁻¹ s⁻¹ (zero), L mol⁻¹ s⁻¹ (second).

Try 3 NEET-style questions

Q1. For the decomposition CH₃CHO(g) → CH₄(g) + CO(g), the experimentally found rate law is rate = k[CH₃CHO]^(3/2). The order of the reaction is

  1. 1
  2. 2
  3. 1.5
  4. 3
Show answer

Answer: (3) 1.5

The order is the exponent of the concentration term in the experimental rate law, here 3/2 = 1.5. Orders can be fractional.

Q2. The rate law of a reaction is rate = k[A]^(1/2)[B]². The overall order of the reaction is

  1. 2.0
  2. 3.0
  3. 2.5
  4. 1.0
Show answer

Answer: (3) 2.5

Overall order = ½ + 2 = 2.5.

Q3. A first-order reaction has k = 2.303 × 10⁻² s⁻¹. The time for the reactant concentration to fall to one-tenth of its initial value is

  1. 10 s
  2. 100 s
  3. 1000 s
  4. 230 s
Show answer

Answer: (2) 100 s

t = (2.303/k) log([A]₀/[A]) = (2.303/2.303 × 10⁻²) × log 10 = 100 × 1 = 100 s.

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