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Study Guide: Chemistry Physical - How to Solve: Chemical Kinetics (Order, Rate Law, Arrhenius Equation, Half-Life, Zero/First/Second Order) – NEET UG Guide
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Chemistry Physical - How to Solve: Chemical Kinetics (Order, Rate Law, Arrhenius Equation, Half-Life, Zero/First/Second Order) – NEET UG Guide

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

⏱️ ~6 min read

How to Solve: Chemical Kinetics (Order, Rate Law, Arrhenius Equation, Half-Life, Zero/First/Second Order) – NEET UG Guide

Introduction

Mastering Chemical Kinetics unlocks 5-7 high-yield NEET questions—worth 20+ marks—on rate laws, half-life, and the Arrhenius equation. Miss this, and you lose easy marks to competitors. Let’s make sure you own this topic.

WHAT YOU NEED TO KNOW FIRST

  1. Basic algebra (solving for x, logarithms).
  2. Units of concentration (mol/L or M).
  3. Graph interpretation (slope, intercepts).

If you’re shaky on these, pause and review first.

KEY TERMS & FORMULAS

1. Rate of Reaction

  • Definition: Change in concentration of reactant/product per unit time.
  • Formula: [ \text{Rate} = -\frac{\Delta [A]}{\Delta t} \quad \text{(for reactant A)} ]
  • Δ[A]: Change in concentration (M).
  • Δt: Time interval (s).
  • Negative sign: Reactant concentration decreases over time.

2. Rate Law (Differential Rate Law)

  • Definition: Shows how rate depends on reactant concentrations.
  • Formula: [ \text{Rate} = k[A]^m[B]^n ]
  • k: Rate constant (units depend on order).
  • [A], [B]: Concentrations of reactants (M).
  • m, n: Orders with respect to A and B (determined experimentally).
  • MEMORISE THIS: Order ≠ stoichiometric coefficient.

3. Integrated Rate Laws (Zero, First, Second Order)

Order Integrated Rate Law Half-Life (t₁/₂) Linear Plot
Zero ([A]_t = [A]_0 - kt) (t_{1/2} = \frac{[A]_0}{2k}) ([A]_t) vs. (t) (slope = (-k))
First (\ln[A]_t = \ln[A]_0 - kt) (t_{1/2} = \frac{\ln 2}{k}) (\ln[A]_t) vs. (t) (slope = (-k))
Second (\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt) (t_{1/2} = \frac{1}{k[A]_0}) (\frac{1}{[A]_t}) vs. (t) (slope = (k))
  • MEMORISE ALL THREE HALF-LIFE FORMULAS.
  • Given on exam sheet: Sometimes, but not always—memorise them.

4. Arrhenius Equation

  • Definition: Relates rate constant (k) to temperature (T).
  • Formula: [ k = A e^{-E_a/RT} ]
  • k: Rate constant.
  • A: Pre-exponential factor (frequency of collisions).
  • Eₐ: Activation energy (J/mol).
  • R: Gas constant (8.314 J/mol·K).
  • T: Temperature (K).
  • Linear form (for graphing): [ \ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right) ]
  • MEMORISE THIS: Slope = (-E_a/R).

5. Units of Rate Constant (k)

Order Units of k
Zero M/s or mol L⁻¹ s⁻¹
First s⁻¹
Second M⁻¹ s⁻¹ or L mol⁻¹ s⁻¹
  • MEMORISE THIS TABLE.

STEP-BY-STEP METHOD

Step 1: Identify the Problem Type

Ask: "What am I solving for?" - Rate law? → Use experimental data to find k and order. - Half-life? → Use integrated rate law or half-life formula. - Arrhenius? → Use (k = A e^{-E_a/RT}) or linear form. - Order from data? → Compare initial rates at different concentrations.

Step 2: Write Down Given Data

  • List all given values (concentrations, times, k, T, etc.).
  • Convert units if needed (e.g., °C → K, kJ → J).

Step 3: Choose the Right Formula

  • Rate law? → (\text{Rate} = k[A]^m[B]^n).
  • Half-life? → Pick zero/first/second order formula.
  • Arrhenius? → (k = A e^{-E_a/RT}) or (\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)).

Step 4: Solve for the Unknown

  • Algebra first: Rearrange the formula.
  • Plug in numbers: Substitute given values.
  • Calculate: Use a calculator (logarithms, exponents).

Step 5: Check Units and Reasonableness

  • Units of k: Must match the order.
  • Half-life: Should decrease with higher k (first order) or depend on ([A]_0) (zero/second order).
  • Arrhenius: k should increase with T.

WORKED EXAMPLES

Example 1 – Basic: First-Order Half-Life

Problem: The half-life of a first-order reaction is 30 minutes. If the initial concentration is 0.8 M, what is the concentration after 90 minutes?

Solution:
1. Identify: First-order half-life problem.
2. Given: - (t_{1/2} = 30) min. - ([A]0 = 0.8) M. - (t = 90) min.
3. Formula: [ t ]
4. } = \frac{\ln 2}{k} \implies k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{30} = 0.0231 \text{ min}^{-1Integrated rate law: [ \ln[A]_t = \ln[A]_0 - kt ] [ \ln[A]_t = \ln(0.8) - (0.0231)(90) ] [ \ln[A]_t = -0.223 - 2.079 = -2.302 ] [ [A]_t = e^{-2.302} = 0.1 \text{ M} ]
5. Check: 90 min = 3 half-lives → (0.8 \times (1/2)^3 = 0.1) M. Correct!

What we did and why: - Used first-order half-life to find k. - Applied integrated rate law to find concentration after time t. - Verified with half-life logic (3 half-lives = 1/8th concentration).

Example 2 – Medium: Determining Order from Data

Problem: For the reaction (A \rightarrow B), the following data were obtained:

Experiment [A] (M) Initial Rate (M/s)
1 0.1 0.02
2 0.2 0.08
3 0.4 0.32

Determine the order of the reaction with respect to A and the rate constant k.

Solution:
1. Identify: Rate law problem (find order and k).
2. Given: Rate data at different [A].
3. Compare experiments: - Exp 1 → Exp 2: [A] doubles (0.1 → 0.2), rate quadruples (0.02 → 0.08). [ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{0.08}{0.02} = 4 = \left( \frac{0.2}{0.1} \right)^m \implies 4 = 2^m \implies m = 2 ] - Exp 2 → Exp 3: [A] doubles (0.2 → 0.4), rate quadruples (0.08 → 0.32). [ \frac{0.32}{0.08} = 4 = 2^m \implies m = 2 ]
4. Order: Second order (m = 2).
5. Rate law: (\text{Rate} = k[A]^2).
6. Find k: [ 0.02 = k(0.1)^2 \implies k = \frac{0.02}{0.01} = 2 \text{ M}^{-1} \text{s}^{-1} ]
7. Check units: Second order → k should be M⁻¹ s⁻¹. Correct!

What we did and why: - Compared rate changes with concentration changes to find order. - Used one experiment to solve for k. - Verified k units match the order.

Example 3 – Exam-Style: Arrhenius Equation

Problem: The rate constant of a reaction doubles when the temperature is increased from 20°C to 30°C. Calculate the activation energy (Eₐ) of the reaction. (R = 8.314 J/mol·K)

Solution:
1. Identify: Arrhenius problem (find Eₐ).
2. Given: - (k_2 = 2k_1). - (T_1 = 20°C = 293) K. - (T_2 = 30°C = 303) K. - R = 8.314 J/mol·K.
3. Arrhenius equation (two-point form): [ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) ]
4. Substitute: [ \ln(2) = \frac{E_a}{8.314} \left( \frac{1}{293} - \frac{1}{303} \right) ] [ 0.693 = \frac{E_a}{8.314} \left( 0.00341 - 0.00330 \right) ] [ 0.693 = \frac{E_a}{8.314} (0.00011) ] [ E_a = \frac{0.693 \times 8.314}{0.00011} = 52,890 \text{ J/mol} = 52.9 \text{ kJ/mol} ]
5. Check: Eₐ is positive and reasonable (typical range: 40-200 kJ/mol).

What we did and why: - Used the two-point Arrhenius equation to relate k and T. - Converted °C to K (critical step!). - Solved for Eₐ and checked units.

COMMON MISTAKES

MISTAKE WHY IT HAPPENS CORRECT APPROACH
Confusing order with stoichiometry Assuming coefficients = orders. Orders are experimental, not from balanced equation.
Ignoring units of k Forgetting k units depend on order. Memorise the table: Zero (M/s), First (s⁻¹), Second (M⁻¹ s⁻¹).
Using °C instead of K in Arrhenius Temperature must be in Kelvin. Always convert °C → K (+273).
Misapplying half-life formulas Using first-order half-life for zero/second order. Memorise all three half-life formulas.
Forgetting negative sign in integrated rate laws Sign errors in (\ln[A]_t = \ln[A]_0 - kt). Rate decreases → negative slope.

EXAM TRAPS

TRAP HOW TO SPOT IT HOW TO AVOID IT
Disguised half-life question Asks for "time to reach 25% concentration" instead of half-life. Recognise 25% = 2 half-lives (first order).
Arrhenius with log base 10 Gives (\log k) instead of (\ln k). Convert: (\ln k = 2.303 \log k).
Rate law with multiple reactants Gives data for [A] and [B] but asks for order in A only. Hold [B] constant and compare [A] changes.

1-MINUTE RECAP

"Listen up—this is your last-minute kinetics cheat sheet. Memorise these three things:
1. Rate law: (\text{Rate} = k[A]^m[B]^n). Order ≠ coefficient. Use data to find m and n.
2. Half-life: - Zero: (t_{1/2} = [A]0/2k). - First: (t = \ln 2/k) (constant). - Second: (t_{1/2} = 1/k[A]_0).
3. Arrhenius: (\ln k = \ln A - E_a/RT). Slope = (-E_a/R). Always use Kelvin!

For problems: - Rate law? Compare experiments. - Half-life? Pick the right formula. - Arrhenius? Convert °C to K, use (\ln k) form.

Common traps? Units of k, sign errors, and disguised half-life questions. Now go crush those 20 marks!


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