How do I find the order of a reaction from experimental data?

From the rate law: rate = k[A]^m[B]^n. Keep one concentration constant and vary the other. If doubling [A] doubles rate, it is first order in A.

Samacheer Kalvi 12th Chemistry Chapter 7 Exercise Solutions

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Chapter 7 — Chemical Kinetics Exercise Solutions

Chemical Kinetics is numerical-friendly. Always identify the order of reaction before selecting the formula.

Key Concepts Quick Recap

Term	Definition
Rate of reaction	Change in concentration per unit time; r = −(1/a)(d[A]/dt)
Rate law	Experimentally determined: rate = k[A]^m[B]^n
Order of reaction	Sum of powers m + n in the rate law; experimentally found
Molecularity	Number of molecules colliding in a single elementary step; theoretical
Rate constant (k)	Proportionality constant; depends on temperature, not concentration
Half-life (t½)	Time for [A] to fall to half its initial value
Arrhenius equation	k = Ae^(−Ea/RT); relates k to temperature
Activation energy (Ea)	Minimum energy needed for a reaction to occur
Frequency factor (A)	Rate of collisions; pre-exponential factor in Arrhenius equation
Pseudo-first order	Higher order reaction behaves as first order when one reactant is in large excess

Formula Recap

Concept	Formula
Zero order: k	k = ([A₀] − [A]) / t
First order: k	k = (2.303/t) log([A₀]/[A])
First order: t½	t½ = 0.693 / k
Second order: k	k = (1/t)(1/[A] − 1/[A₀])
Arrhenius	k = Ae^(−Ea/RT) or log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)
Units of k	Zero order: mol L⁻¹ s⁻¹; First order: s⁻¹; Second order: L mol⁻¹ s⁻¹

I. Choose the Best Answer (MCQs)

1. The rate of a first order reaction does NOT depend on: (a) Temperature (b) Initial concentration (c) Nature of reactant (d) Catalyst

Answer(b) Initial concentration Wait — the rate does depend on [A] for a first order reaction (rate = k[A]). But the half-life of a first order reaction is independent of initial concentration. This question tests that distinction: t½ = 0.693/k (no [A₀] term).

2. The unit of rate constant for a second order reaction is: (a) s⁻¹ (b) mol L⁻¹ s⁻¹ (c) L mol⁻¹ s⁻¹ (d) mol² L⁻² s⁻¹

Answer(c) L mol⁻¹ s⁻¹ From rate = k[A][B]: k = rate / [A][B] = (mol L⁻¹ s⁻¹) / (mol L⁻¹)² = L mol⁻¹ s⁻¹.

3. For a first order reaction, a plot of log[A] vs time gives: (a) A curve (b) A straight line with positive slope (c) A straight line with negative slope (d) A parabola

Answer(c) Straight line with negative slope First order: log[A] = log[A₀] − (k/2.303)t → y = mx + c form → slope = −k/2.303 (negative).

4. The Arrhenius equation k = Ae^(−Ea/RT) shows that k increases when: (a) Ea increases (b) T increases (c) A decreases (d) R increases

Answer(b) T increases As T increases, e^(−Ea/RT) → e^(−smaller value) → larger value → k increases. Higher temperature gives more molecules enough energy to cross the activation energy barrier.

5. Which of the following reactions is pseudo-first order? (a) H₂ + I₂ → 2HI (b) CH₃COOC₂H₅ + H₂O (excess) → acid hydrolysis (c) 2NO → N₂ + O₂ (d) 2H₂ + O₂ → 2H₂O

Answer(b) Acid hydrolysis of ester in excess water Rate = k[ester][H₂O], but [H₂O] is so large it barely changes → rate = k'[ester] → pseudo-first order.

6. Doubling the concentration of reactant A doubles the rate. Tripling [A] triples the rate. The order with respect to A is: (a) 0 (b) 1 (c) 2 (d) 3

Answer(b) 1 (First order) rate ∝ [A]^m. Doubling [A] → rate × 2 → 2^m = 2 → m = 1.

7. A catalyst increases the rate of a reaction by: (a) Increasing the activation energy (b) Decreasing the activation energy (c) Increasing the temperature (d) Changing the equilibrium constant

Answer(b) Decreasing the activation energy A catalyst provides an alternative reaction pathway with lower Ea. This allows more molecules to react per unit time. Kₑq is not affected by a catalyst.

II. Short Answer Questions

Q1. Distinguish between order and molecularity of a reaction.

Answer

Feature	Order	Molecularity
Nature	Experimental	Theoretical
Basis	Rate law (rate = k[A]^m[B]^n)	Number of molecules in elementary step
Value	0, 1, 2, 3 or fractional	Always a positive integer (1, 2, 3)
For complex reactions	Can be found from experiment	Applies only to elementary steps
Can be zero	Yes (zero order)	No (minimum 1)

Q2. State three factors that affect the rate of a chemical reaction.

Answer

1. Concentration: Higher concentration → more collisions per second → faster rate. Rate = k[A]^m[B]^n. 2. Temperature: Higher T → more molecules have Ea energy → faster rate. Rate roughly doubles per 10 °C rise. 3. Catalyst: Lowers Ea by providing an alternative pathway → increases rate without being consumed.

Other factors: surface area (for solids), nature of reactants, light (for photochemical reactions).

Q3. Explain with an example how the order of a reaction is determined experimentally.

Answer

Consider the reaction: A + B → products

Experiment 1: [A] = 0.1, [B] = 0.1, rate = r₁ Experiment 2: [A] = 0.2, [B] = 0.1, rate = 2r₁ (rate doubled when [A] doubled → first order in A) Experiment 3: [A] = 0.1, [B] = 0.2, rate = 4r₁ (rate quadrupled when [B] doubled → second order in B)

Rate law: rate = k[A][B]² Overall order = 1 + 2 = 3 (third order)

Q4. What is activation energy? How does the Arrhenius equation relate it to rate constant?

Answer

Activation energy (Ea) is the minimum amount of energy that reacting molecules must possess to form the activated complex and undergo reaction.

Arrhenius equation: k = Ae^(−Ea/RT)

Taking logarithm: ln k = ln A − Ea/RT

Or: log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)

A plot of log k vs 1/T is a straight line with slope = −Ea/2.303R → this allows Ea to be calculated experimentally.

Worked Examples

Example 1 — Finding k for a first order reaction

A first-order reaction decreases [A] from 0.8 M to 0.2 M in 20 min. Find k.

k = (2.303/t) × log([A₀]/[A])
  = (2.303/20) × log(0.8/0.2)
  = 0.1152 × log(4)
  = 0.1152 × 0.6021
  = 0.0693 min⁻¹

Example 2 — Half-life of a first order reaction

A first-order reaction has k = 0.0231 min⁻¹. Find its half-life.

t½ = 0.693 / k = 0.693 / 0.0231 = 30 min

Note: t½ is independent of initial concentration — this is a unique feature
of first order reactions.

Example 3 — Activation energy using Arrhenius equation

A reaction has k₁ = 0.1 s⁻¹ at 300 K and k₂ = 0.4 s⁻¹ at 320 K. Find Ea.

log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂)

log(0.4/0.1) = (Ea/(2.303 × 8.314)) × (1/300 − 1/320)

log 4 = (Ea/19.14) × (0.003333 − 0.003125)

0.6021 = (Ea/19.14) × (2.08×10⁻⁴)

Ea = 0.6021 × 19.14 / 2.08×10⁻⁴
   = 55,432 J/mol ≈ 55.4 kJ/mol

Example 4 — Units of rate constant

From rate = k[A]ⁿ:
  k = rate / [A]ⁿ = (mol L⁻¹ s⁻¹) / (mol L⁻¹)ⁿ

Zero order  (n=0): k = mol L⁻¹ s⁻¹
First order (n=1): k = s⁻¹
Second order(n=2): k = L mol⁻¹ s⁻¹
nth order:         k = L^(n-1) mol^(1-n) s⁻¹

Example 5 — Integrated rate law: zero order

A zero-order reaction has k = 0.05 mol L⁻¹ min⁻¹. If [A₀] = 1 M, when does [A] = 0.5 M?

[A] = [A₀] − kt
0.5 = 1.0 − 0.05 × t
0.05t = 0.5
t = 10 min

t½ (zero order) = [A₀] / 2k = 1.0 / (2 × 0.05) = 10 min
(Unlike first order, t½ of zero order depends on [A₀])

Numerical Pattern

1. Identify reaction order from given data or question. 2. Write the correct integrated rate equation. 3. Substitute known values (concentrations, time). 4. Keep log base correct (2.303 × log₁₀ = ln). 5. Write unit of rate constant based on order. 6. Check that the answer is dimensionally correct.

Common Mistakes

Confusing order and molecularity (order is experimental; molecularity is for elementary steps).
Missing the unit of k — always write it.
Using ln formula when question expects log₁₀ (2.303 factor must be included).
Not converting time units (hours → minutes → seconds) consistently.
Saying catalyst changes equilibrium constant — it does not; Keq is unchanged.

Exam Checklist

Order of reaction is clearly identified.
Correct formula (zero/first/second order) is used.
Unit of k is stated.
Activation energy units are J/mol or kJ/mol.
Final numerical answer has proper significant figures.

Frequently Asked Questions

k = (2.303/t) log([A₀]/[A]) and t½ = 0.693/k are the key formulas. The half-life of a first order reaction is independent of initial concentration.