Class 12 · Chemistry · Chapter 4
Chemical Kinetics
Master the science of reaction rates — from rate laws and integrated equations to the Arrhenius equation and collision theory. Aligned with NCERT for Boards, JEE & NEET.
📘 NCERT Based
🎓 Board Exam Ready
⚗️ JEE Advanced
💉 NEET Relevant
📝 Practice Problems Included
🧬 Why Study Chemical Kinetics?
Thermodynamics tells us whether a reaction will occur (feasibility). Chemical Kinetics tells us how fast a reaction will occur and by what mechanism. This branch of physical chemistry deals with the speed of chemical reactions and the factors that control them.
🔑 Key Insight: A reaction may be thermodynamically feasible yet kinetically very slow (e.g., rusting of iron at room temperature, conversion of diamond to graphite). Kinetics bridges the gap between theoretical possibility and practical reality.
Scope of Chemical Kinetics
- Measurement of the rate of a reaction
- Finding how rate depends on concentration, temperature, pressure, catalyst
- Proposing and verifying the reaction mechanism
- Industrial applications: designing reactors, optimizing yield and speed
📐 Rate of a Chemical Reaction
For a general reaction: A → B
The rate of reaction is defined as the change in concentration of a reactant or product per unit time.
Average Rate
Rate = −Δ[A]/Δt = +Δ[B]/Δt
(Units: mol L⁻¹ s⁻¹ or mol L⁻¹ min⁻¹)
⚠️ Sign Convention: For reactants, concentration decreases over time → negative sign is used. For products, concentration increases → positive sign. The rate is always a positive quantity.
Rate for a General Balanced Reaction
For: aA + bB → cC + dD
Rate = −(1/a)·d[A]/dt = −(1/b)·d[B]/dt = +(1/c)·d[C]/dt = +(1/d)·d[D]/dt
The stoichiometric coefficients ensure the rate is unique and consistent regardless of which species is used for measurement.
Instantaneous Rate vs. Average Rate
Average Rate
- Rate over a finite time interval
- Calculated from initial and final concentrations
- = Δ[concentration] / Δt
- Changes over the course of reaction
Instantaneous Rate
- Rate at a specific moment in time
- = −d[A]/dt (calculus limit)
- Obtained from the slope of tangent on [A] vs. t graph
- Most commonly referred to in kinetics
📊
Concentration vs. Time Graph
📖 NCERT Figure 4.1
Graph showing [HI] decreasing with time and [H₂], [I₂] increasing. Average rate = slope of chord; Instantaneous rate = slope of tangent.
NCERT Example: For 2HI(g) → H₂(g) + I₂(g), rate = −(1/2)·d[HI]/dt = +d[H₂]/dt = +d[I₂]/dt
📏 Units of Rate of Reaction
Rate = Δ[concentration]/Δt = mol L⁻¹ / time = mol L⁻¹ s⁻¹ (in SI)
| State of Reaction | Concentration Measure | Common Units of Rate |
|---|
| Solution phase | Molarity [mol/L] | mol L⁻¹ s⁻¹ or mol L⁻¹ min⁻¹ |
| Gas phase | Partial pressure (atm) | atm s⁻¹ or atm min⁻¹ |
| Heterogeneous | Surface area | mol m⁻² s⁻¹ |
Practice Problems — Rate of Reaction
Problem 1 · Board Level
For the reaction N₂ + 3H₂ → 2NH₃, if the rate of consumption of H₂ is 0.06 mol L⁻¹ s⁻¹, what is the rate of formation of NH₃ and the overall rate of reaction?
▶ Show Answer
Solution:
Rate equation: Rate = −d[N₂]/dt = −(1/3)d[H₂]/dt = +(1/2)d[NH₃]/dt
Given: −d[H₂]/dt = 0.06 mol L⁻¹ s⁻¹
Overall Rate = (1/3) × 0.06 = 0.02 mol L⁻¹ s⁻¹
Rate of formation of NH₃ = 2 × 0.02 = 0.04 mol L⁻¹ s⁻¹
Problem 2 · Board Level
For 2SO₂ + O₂ → 2SO₃, express the rate in terms of concentration changes of all species. If [SO₂] decreases at 0.04 mol L⁻¹ s⁻¹, find the rate of decrease of O₂ and rate of increase of SO₃.
▶ Show Answer
Solution:
Rate = −(1/2)d[SO₂]/dt = −d[O₂]/dt = +(1/2)d[SO₃]/dt
Rate of decrease of [SO₂] = 0.04 mol L⁻¹ s⁻¹
Overall rate = 0.04/2 = 0.02 mol L⁻¹ s⁻¹
Rate of decrease of O₂ = 0.02 mol L⁻¹ s⁻¹
Rate of increase of SO₃ = 2 × 0.02 = 0.04 mol L⁻¹ s⁻¹
Problem 3 · JEE Level
The concentration of a reactant A changes from 0.5 M to 0.3 M in 10 minutes and from 0.3 M to 0.25 M in the next 10 minutes. Comment on the rate of reaction with time.
▶ Show Answer
Solution:
Avg. rate (0–10 min) = (0.5 − 0.3)/10 = 0.02 mol L⁻¹ min⁻¹
Avg. rate (10–20 min) = (0.3 − 0.25)/10 = 0.005 mol L⁻¹ min⁻¹
Conclusion: The rate decreases with time as reactant concentration decreases. This is a characteristic of most reactions (rate ∝ [A]ⁿ).
Problem 4 · NEET Level
In a reaction, 2A → products, the concentration of A decreases from 0.5 mol/L to 0.4 mol/L in 10 minutes. The rate of appearance of products is:
▶ Show Answer
Rate of disappearance of A = 0.1/10 = 0.01 mol L⁻¹ min⁻¹
Since 2A → products, Rate = −(1/2)d[A]/dt
Overall Rate = 0.01/2 = 0.005 mol L⁻¹ min⁻¹
Rate of appearance of product = Overall rate = 0.005 mol L⁻¹ min⁻¹
⚡ Overview of Factors
🧪
Concentration
More molecules → more collisions → faster rate
🌡️
Temperature
Higher T → more kinetic energy → more effective collisions
⚗️
Catalyst
Lowers activation energy → alternate pathway → faster rate
📐
Surface Area
Smaller particles → more surface → faster heterogeneous reaction
💡
Light (Radiation)
Photochemical reactions: light provides energy to initiate reaction
📦
Pressure (Gases)
Higher pressure → higher concentration of gas → faster rate
🧪 1. Effect of Concentration
As a reaction proceeds, the concentration of reactants decreases, so the rate generally decreases with time. This is why a freshly prepared solution reacts faster than a dilute, aged one.
Key Fact: The quantitative relationship between concentration and rate is given by the Rate Law (discussed in Section 3). Generally: Rate ∝ [A]m[B]n
Dependence on Nature of Reactants
The physical and chemical nature of reactants also affects rate:
- Ionic reactions: Very fast (milliseconds) — e.g., precipitation reactions
- Covalent bond breaking: Slower — e.g., organic reactions
- Physical state: Gases react faster than liquids; liquids react faster than solids
🌡️ 2. Effect of Temperature
Temperature has a dramatic effect on reaction rate. As a rule of thumb (van't Hoff rule):
Temperature Coefficient (μ)
μ = Rate at (T + 10°C) / Rate at T°C ≈ 2 to 3
i.e., Rate doubles or triples for every 10°C rise in temperature
Important for JEE/NEET: The precise quantitative relationship is given by the Arrhenius equation (Section 7). The van't Hoff rule is only an approximation.
Why does temperature increase rate?
- Higher temperature → higher kinetic energy of molecules
- More molecules have energy ≥ Activation energy (Ea)
- Frequency of effective collisions increases
- The fraction of molecules with energy ≥ Ea increases exponentially with temperature
⚗️ 3. Effect of Catalyst
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. It does this by providing an alternative reaction pathway with lower activation energy.
📉
Effect of Catalyst on Activation Energy
📖 NCERT Figure 4.9
Energy profile diagram showing the uncatalyzed pathway (higher Ea) and the catalyzed pathway (lower Ea). Both pathways have the same reactants, products, and overall ΔH.
Homogeneous Catalyst
- Same phase as reactants
- Example: H₂SO₄ (acid) in esterification
- Example: NO(g) in oxidation of SO₂(g)
- Forms intermediate with reactant
Heterogeneous Catalyst
- Different phase from reactants
- Example: Fe(s) in Haber process (N₂ + H₂)
- Example: Pt(s) in catalytic converters
- Adsorption on surface is key
Important Characteristics of a Catalyst:
- Lowers Ea (activation energy) → increases rate
- Does NOT change equilibrium constant (K) — only speeds up approach to equilibrium
- Does NOT appear in the overall balanced equation
- A small amount can catalyze a large amount of reaction
- Can be specific (selective) — acts on specific substrates
Practice Problems — Factors Affecting Rate
Problem 1 · Board Level
The rate of a reaction doubles when temperature is raised from 298 K to 308 K. Calculate the activation energy of the reaction. (R = 8.314 J mol⁻¹ K⁻¹)
▶ Show Answer
Using Arrhenius equation:
log(k₂/k₁) = Ea/2.303R × (T₂ − T₁)/(T₁ × T₂)
k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K
log(2) = Ea / (2.303 × 8.314) × (10)/(298 × 308)
0.3010 = Ea / 19.147 × (10/91784)
0.3010 = Ea × 5.689 × 10⁻⁶
Ea = 52,897 J mol⁻¹ ≈ 52.9 kJ mol⁻¹
Problem 2 · NEET Level
A catalyst lowers the activation energy of a reaction from 75 kJ/mol to 45 kJ/mol at 500 K. By what factor does the reaction rate increase? (R = 8.314 J/mol·K)
▶ Show Answer
Rate ratio:
k_cat / k_uncat = e^[(Ea(uncat) − Ea(cat))/RT] = e^[(75000 − 45000)/(8.314 × 500)]
= e^[30000/4157] = e^7.22 ≈ 1371
The catalyzed reaction is about 1371 times faster.
Problem 3 · JEE Level
Explain with reason why finely divided iron is a better catalyst than large iron chunks in the Haber process.
▶ Show Answer
Iron is a heterogeneous catalyst. The reaction occurs at the catalyst's surface. Finely divided iron has a much larger surface area per unit mass compared to large chunks. This provides more active sites for adsorption of N₂ and H₂ molecules, leading to more simultaneous reactions and a significantly higher overall rate.
⚖️ Rate Law (Rate Expression)
For a reaction: aA + bB → Products
The rate law is an experimentally determined expression:
Rate Law
Rate = k [A]ˣ [B]ʸ
where k = rate constant, x = order w.r.t. A, y = order w.r.t. B
⚠️ Critical Point: The orders x and y are experimentally determined and are NOT necessarily equal to the stoichiometric coefficients a and b. They can be zero, fractional, or even negative!
- x + y = overall order of the reaction
- The rate law can only be established from experiments, not from the balanced equation
- For elementary reactions, order = stoichiometry (exception to the rule above)
🔢 Order of Reaction
The order of reaction with respect to a particular reactant is the power to which the concentration of that reactant is raised in the experimentally determined rate law.
| Order | Rate Law | Example Reaction | Units of k |
|---|
| Zero Order |
Rate = k[A]⁰ = k |
Decomposition of NH₃ on Pt surface; photochemical H₂ + Cl₂ |
mol L⁻¹ s⁻¹ |
| First Order |
Rate = k[A]¹ |
Radioactive decay; hydrolysis of esters (dilute acid) |
s⁻¹ |
| Second Order |
Rate = k[A]² or k[A][B] |
NO₂ + CO → NO + CO₂; saponification |
L mol⁻¹ s⁻¹ |
| Third Order |
Rate = k[A]²[B] |
2NO + O₂ → 2NO₂; 2NO + Cl₂ → 2NOCl |
L² mol⁻² s⁻¹ |
| Fractional Order |
Rate = k[A]^(1/2)[B] |
Para-H₂ → ortho-H₂ (order = 3/2) |
Depends on order |
Units of Rate Constant k — General Formula
Units of k = (mol L⁻¹)^(1−n) × s⁻¹
where n = overall order of reaction
💡 Memory Trick
"Zero stays constant, first is simple, second needs two" — Zero order: rate = k (constant). First order: rate doubles when [A] doubles. Second order: rate quadruples when [A] doubles.
🔬 Rate Constant (k) — Specific Rate Constant
The rate constant k is the rate of reaction when all reactant concentrations are unity (1 mol/L). It is a characteristic of a reaction at a given temperature.
- k is independent of concentration — depends only on temperature
- k increases with temperature (Arrhenius equation)
- k depends on catalyst — different values for catalyzed and uncatalyzed reactions
- k is specific to the reaction and mechanism
- When k is large → reaction is fast; when small → reaction is slow
Molecularity vs. Order
Molecularity
- Number of molecules/ions/atoms in an elementary step
- Always a whole number (1, 2, or 3)
- Defined for elementary steps only
- Theoretical concept — cannot be zero or fractional
- Determined from the mechanism
Order
- Power of concentration term in the rate law
- Can be 0, 1, 2, fractional, or even negative
- Defined for overall reaction
- Experimental quantity
- Determined from experiments
Relationship: For a complex (multi-step) reaction, the order is determined by the rate-determining step (the slowest elementary step). The molecularity of the RDS equals the order of the overall reaction.
🔍 Determination of Order — Method of Initial Rates
In this method, experiments are designed so that all concentrations are varied one at a time while others are kept constant. By comparing initial rates, we determine orders.
Example Procedure
- Run Experiment 1: [A] = a₁, [B] = b₁ → Rate = R₁
- Run Experiment 2: [A] = 2a₁, [B] = b₁ → Rate = R₂
- If R₂ = 2R₁ → first order in A; if R₂ = 4R₁ → second order in A
- Similarly vary [B] to find order in B
Finding Order from Two Experiments
R₁ = k[A₁]ˣ[B₁]ʸ
R₂ = k[A₂]ˣ[B₂]ʸ
─────────────────
R₂/R₁ = ([A₂]/[A₁])ˣ × ([B₂]/[B₁])ʸ
→ Solve for x and y
Practice Problems — Rate Law & Order
Problem 1 · Board Level
For the reaction: A + B → Products, the following data is obtained:
| Exp | [A] (mol/L) | [B] (mol/L) | Rate (mol/L·s) |
|---|
| 1 | 0.1 | 0.1 | 2×10⁻² |
| 2 | 0.2 | 0.1 | 4×10⁻² |
| 3 | 0.1 | 0.2 | 8×10⁻² |
Find the rate law, order w.r.t each reactant, and the value of k.
▶ Show Answer
Step 1: Order w.r.t. A (Exp 1 vs 2, [B] constant)
4×10⁻² / 2×10⁻² = (0.2/0.1)ˣ → 2 = 2ˣ → x = 1 (First order in A)
Step 2: Order w.r.t. B (Exp 1 vs 3, [A] constant)
8×10⁻² / 2×10⁻² = (0.2/0.1)ʸ → 4 = 2ʸ → y = 2 (Second order in B)
Rate Law: Rate = k[A][B]² ; Overall order = 3
Step 3: Finding k (using Exp 1)
2×10⁻² = k × 0.1 × (0.1)² = k × 0.001
k = 20 L² mol⁻² s⁻¹
Problem 2 · JEE Level
For a reaction A → B + C, it is found that the rate of the reaction doubles when the concentration of A is increased four times. What is the order of the reaction?
▶ Show Answer
Rate = k[A]ⁿ
2R/R = (4[A]/[A])ⁿ → 2 = 4ⁿ = (2²)ⁿ = 2²ⁿ
Therefore 2²ⁿ = 2¹ → 2n = 1 → n = 1/2 (Half order or 0.5 order)
Problem 3 · NEET Level
What are the units of rate constant for a zero order, first order, and second order reaction? Derive using dimensional analysis.
▶ Show Answer
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 = (mol L⁻¹ s⁻¹)/(mol L⁻¹) = s⁻¹
Second order (n=2): k = (mol L⁻¹ s⁻¹)/(mol L⁻¹)² = L mol⁻¹ s⁻¹
General formula: Units of k = (mol L⁻¹)^(1−n) s⁻¹
Problem 4 · Board Level
Distinguish between order and molecularity of a reaction. Give one example each of a reaction with order = molecularity and a reaction where they differ.
▶ Show Answer
Order = Molecularity: For elementary reactions, they are equal.
Example: H₂ + I₂ → 2HI (molecularity = 2, order = 2 — both second order)
Order ≠ Molecularity: For complex (multi-step) reactions.
Example: H₂ + Br₂ → 2HBr — Molecularity of overall reaction is 2 but experimental order = 3/2 (fractional). The mechanism involves multiple steps.
📊 Zero Order Reaction
In a zero order reaction, the rate is independent of the concentration of the reactant.
Rate Law:
Rate = k[A]⁰ = k (constant)
Derivation of Integrated Rate Equation:
- Rate = −d[A]/dt = k
- d[A] = −k·dt
- Integrating: ∫d[A] = −k∫dt → [A] = −kt + C
- At t = 0: [A] = [A]₀ → C = [A]₀
- [A]₀ − [A]t = kt OR [A]t = [A]₀ − kt
Integrated Rate Law — Zero Order
[A]t = [A]₀ − kt
OR k = ([A]₀ − [A]t) / t
Units of k: mol L⁻¹ s⁻¹
Key Features of Zero Order Reaction
- [A] vs. t graph: Straight line with slope = −k
- Half-life (t½): t½ = [A]₀ / 2k → depends on initial concentration
- Examples: Photochemical reactions (H₂ + Cl₂), enzyme-catalyzed reactions at saturation, decomposition of NH₃ on Pt surface
📉
Zero Order Reaction Graphs
📖 NCERT Figure 4.4
Left: [R] vs t — straight line with negative slope (−k). Right: Rate vs [R] — horizontal line (rate independent of concentration).
📈 First Order Reaction — Most Important
In a first order reaction, the rate is directly proportional to the concentration of one reactant.
Rate Law:
Rate = k[A]¹ = k[A]
Complete Derivation:
- Rate = −d[A]/dt = k[A]
- Rearranging: d[A]/[A] = −k·dt
- Integrating both sides: ∫d[A]/[A] = −k∫dt
- ln[A] = −kt + C
- At t = 0: ln[A]₀ = C
- Subtracting: ln[A]t − ln[A]₀ = −kt
- Therefore: ln([A]₀/[A]t) = kt
Integrated Rate Law — First Order (★ Most Important ★)
ln[A]t = ln[A]₀ − kt
────────────────────────────────────
OR: k = (2.303/t) × log([A]₀/[A]t)
────────────────────────────────────
[A]t = [A]₀ × e^(−kt)
Units of k: s⁻¹ or min⁻¹ (time⁻¹)
In Terms of Pressure (for gas phase reactions):
k = (2.303/t) × log(P₀/Pt)
where P₀ = initial partial pressure, Pt = pressure at time t
Key Features of First Order Reaction
- [A] vs. t: Exponential decay curve
- ln[A] vs. t: Straight line, slope = −k (used to find k experimentally)
- log[A] vs. t: Straight line, slope = −k/2.303
- Half-life: t½ = 0.693/k — independent of initial concentration ⭐
- Concentration after n half-lives: [A]n = [A]₀ × (1/2)ⁿ
📊
First Order Reaction — [A] vs t and log[A] vs t
📖 NCERT Figure 4.5 & 4.6
Left: Exponential decay of [R] vs time. Right: Linear plot of log[R] vs time — slope = −k/2.303, intercept = log[R]₀.
Important Examples of First Order Reactions:
- Radioactive decay — all radioactive processes are first order
- Decomposition of N₂O₅: N₂O₅ → 2NO₂ + ½O₂
- Hydrolysis of sucrose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (pseudo-first order)
- Inversion of cane sugar
🔄 Second Order Reaction
For a second order reaction with one reactant: Rate = k[A]²
Derivation:
- −d[A]/dt = k[A]²
- d[A]/[A]² = −k·dt
- Integrating: −1/[A] = −kt + C
- At t = 0: C = −1/[A]₀
- 1/[A]t − 1/[A]₀ = kt
Integrated Rate Law — Second Order
1/[A]t = 1/[A]₀ + kt
k = (1/t) × (1/[A]t − 1/[A]₀)
Units of k: L mol⁻¹ s⁻¹
- 1/[A] vs. t: Straight line with slope = k
- Half-life: t½ = 1/(k[A]₀) — depends on initial concentration
📋 Comparison of Integrated Rate Equations
| Property |
Zero Order |
First Order |
Second Order |
| Rate Law |
k |
k[A] |
k[A]² |
| Integrated Equation |
[A]t = [A]₀ − kt |
ln[A]t = ln[A]₀ − kt |
1/[A]t = 1/[A]₀ + kt |
| Linear Graph |
[A] vs t |
ln[A] vs t |
1/[A] vs t |
| Slope |
−k |
−k |
+k |
| Half-life (t½) |
[A]₀/2k |
0.693/k |
1/(k[A]₀) |
| Units of k |
mol L⁻¹ s⁻¹ |
s⁻¹ |
L mol⁻¹ s⁻¹ |
Practice Problems — Integrated Rate Equations
Problem 1 · Board Level
The decomposition of N₂O₅ in CCl₄ solution at 318 K is a first order reaction. If the initial concentration is 2.0 × 10⁻² M and k = 6.2 × 10⁻⁴ s⁻¹, what will be the concentration after 900 s?
▶ Show Answer
ln[A]t = ln[A]₀ − kt
[A]₀ = 2.0 × 10⁻² M, k = 6.2 × 10⁻⁴ s⁻¹, t = 900 s
ln[A]t = ln(2.0 × 10⁻²) − (6.2 × 10⁻⁴)(900)
ln[A]t = −3.912 − 0.558 = −4.47
[A]t = e^(−4.47) = 1.146 × 10⁻² M ≈ 1.15 × 10⁻² M
Problem 2 · Board Level (NCERT Type)
A first order reaction has a rate constant of 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
▶ Show Answer
t = (2.303/k) × log([A]₀/[A]t)
= (2.303 / 1.15 × 10⁻³) × log(5/3)
= 2002.6 × log(1.667)
= 2002.6 × 0.2218
= 444.2 s ≈ 444 s
Problem 3 · JEE Level
In a first order reaction, the concentration of the reactant decreases to 1/8 of its initial value in 96 minutes. Find the half-life of the reaction.
▶ Show Answer
If [A]t = [A]₀/8, then [A]₀/[A]t = 8 = 2³
This means the reaction has gone through 3 half-lives.
3 × t½ = 96 minutes
t½ = 32 minutes
Verification: k = 0.693/32 = 0.02166 min⁻¹
t = (2.303/0.02166) × log(8) = 106.3 × 0.903 = 96 min ✓
Problem 4 · NEET Level
Show that the time required for 99.9% completion of a first order reaction is approximately 10 times the time required for 50% completion (half-life).
▶ Show Answer
t99.9% = (2.303/k) × log(100/0.1) = (2.303/k) × log(1000) = (2.303 × 3)/k
t50% = (2.303/k) × log(100/50) = (2.303/k) × log(2) = (2.303 × 0.3010)/k
Ratio: t99.9% / t50% = (2.303 × 3) / (2.303 × 0.3010) = 3/0.3010 ≈ 9.97 ≈ 10 ✓
Problem 5 · JEE Level
The rate constant of a zero-order reaction is 2 × 10⁻² mol L⁻¹ s⁻¹. If the initial concentration of the reactant is 0.10 mol/L, how long will the reaction take to complete?
▶ Show Answer
For zero order: [A]t = [A]₀ − kt
Reaction completes when [A]t = 0
t = [A]₀ / k = 0.10 / (2 × 10⁻²) = 5 seconds
Note: For a zero order reaction, the reaction completes in a finite time (unlike first order which is theoretically infinite).
⏱️ Definition and Derivation
The half-life (t½) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value.
Half-Life for First Order Reaction (★ Most Important ★)
At t = t½: [A]t = [A]₀/2
- From: k = (2.303/t) × log([A]₀/[A]t)
- k = (2.303/t½) × log([A]₀/[A]₀/2)
- k = (2.303/t½) × log(2)
- k = (2.303/t½) × 0.3010
- k = 0.693/t½
Half-Life — First Order Reaction
t½ = 0.693 / k
══════════════════════════
t½ is INDEPENDENT of initial concentration!
⭐ JEE/NEET Key Fact: The half-life of a first order reaction is constant — it does NOT depend on the initial concentration. This is a unique and important property. Radioactive decay is first order, and its half-life is a fundamental property of each isotope.
| Order | Half-Life Formula | Depends on [A]₀? |
|---|
| Zero | t½ = [A]₀ / 2k | ✅ Yes — directly proportional |
| First | t½ = 0.693 / k | ❌ No — independent of [A]₀ |
| Second | t½ = 1 / (k[A]₀) | ✅ Yes — inversely proportional |
Concentration After n Half-Lives (First Order)
[A]n = [A]₀ × (1/2)ⁿ
Amount remaining = (1/2)ⁿ × Initial amount
% remaining = (1/2)ⁿ × 100
☢️ Radioactive Decay — Application of First Order Kinetics
All radioactive decay processes follow first order kinetics:
N(t) = N₀ × e^(−λt)
where λ = decay constant = 0.693/t½
t½ (radioactive) = 0.693/λ
Radiocarbon Dating: ¹⁴C decays by first order kinetics with t½ = 5730 years. By measuring the ratio of ¹⁴C to ¹²C in an artifact, scientists can determine its age. This is a direct application of integrated rate equations.
Practice Problems — Half Life
Problem 1 · Board Level
The half-life of a first order reaction is 60 minutes. How much time will be required for 80% completion of this reaction?
▶ Show Answer
t½ = 60 min → k = 0.693/60 = 0.01155 min⁻¹
80% complete → [A]t = 20% of [A]₀ = 0.2[A]₀
t = (2.303/k) × log([A]₀/[A]t) = (2.303/0.01155) × log(100/20)
= 199.4 × log(5) = 199.4 × 0.699 = 139.4 minutes
Problem 2 · NEET Level
A radioactive isotope has a half-life of 10 years. In how many years will only 6.25% of the original amount remain?
▶ Show Answer
6.25% = 6.25/100 = 1/16 = (1/2)⁴
So 4 half-lives are required.
Time = 4 × 10 = 40 years
Problem 3 · JEE Level
A second order reaction has k = 0.5 L mol⁻¹ s⁻¹. If initial concentration is 1 mol/L, find the half-life. How does t½ change if initial concentration is doubled?
▶ Show Answer
For second order: t½ = 1/(k[A]₀)
t½ = 1/(0.5 × 1) = 2 seconds
If [A]₀ is doubled to 2 mol/L:
t½ = 1/(0.5 × 2) = 1 second → t½ is halved
(For 2nd order, t½ ∝ 1/[A]₀)
🎭 Concept of Pseudo-First Order
Some reactions appear to be of one order but are actually of a higher order. This happens when one reactant is present in such a large excess that its concentration remains effectively constant throughout the reaction.
Classic Example: Hydrolysis of Cane Sugar
C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
(Sucrose) (Water) (Glucose) (Fructose)
True rate law: Rate = k[C₁₂H₂₂O₁₁][H₂O] → Second order
But water is present as solvent in large excess (~55.5 mol/L). Its concentration barely changes during the reaction.
Rate = k[C₁₂H₂₂O₁₁][H₂O] = k[H₂O] × [C₁₂H₂₂O₁₁] = k' × [C₁₂H₂₂O₁₁]
where k' = k[H₂O] = pseudo-first order rate constant
Definition: A pseudo-first order reaction is a second (or higher) order reaction that behaves as a first order reaction because one of the reactants is present in large excess (or is a catalyst/solvent) and maintains a constant concentration.
Other Examples of Pseudo-First Order Reactions
- Acid hydrolysis of ester: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (H₂O in excess)
- Hydrolysis of methyl acetate in dilute acid solution
- Inversion of cane sugar (classical pseudo-first order)
- Many enzymatic reactions at low substrate concentration
💡 Key Point
Pseudo-first order rate constant k' = k × [excess reactant]. So k' has units of s⁻¹ (first order), while the true k has units appropriate for the true order. The reaction is mechanistically second order but kinetically observed as first order.
Practice Problems — Pseudo-First Order
Problem 1 · Board Level
Hydrolysis of ethyl acetate in the presence of dilute H₂SO₄ follows pseudo-first order kinetics. If k' = 1.98 × 10⁻³ s⁻¹ and [H₂O] = 55.5 mol/L, calculate the true second order rate constant k.
▶ Show Answer
k' = k × [H₂O]
k = k' / [H₂O] = 1.98 × 10⁻³ / 55.5 = 3.57 × 10⁻⁵ L mol⁻¹ s⁻¹
Problem 2 · JEE Level
For the reaction A + B → products, the rate = k[A][B]. In an experiment, [B]₀ = 100 × [A]₀. Write the effective rate law and explain why it becomes pseudo-first order.
▶ Show Answer
Since [B]₀ >> [A]₀ (100× excess), [B] remains nearly constant throughout the reaction.
[B] ≈ [B]₀ = constant
Rate = k[A][B] ≈ k[B]₀ × [A] = k'[A]
where k' = k[B]₀ is the pseudo-first order rate constant.
The reaction now appears to be first order in [A] only, even though it is actually second order overall. This is pseudo-first order kinetics.
🌡️ Activation Energy and Energy Barrier
For a reaction to occur, the reacting molecules must possess a minimum energy — called the Activation Energy (Ea). This is the energy required to break bonds in the reactants so that new bonds can form in the products.
⛰️
Energy Profile Diagram (Reaction Coordinate Diagram)
📖 NCERT Figure 4.9
Reaction coordinate diagram showing: Reactants → Transition State (peak, energy = Ea) → Products. The Ea is the energy difference between the transition state and reactants. For an exothermic reaction, products are lower in energy; for endothermic, products are higher.
Threshold Energy & Transition State
- Threshold energy: The minimum energy required for reactants to transform into products
- Activation energy (Ea): Threshold energy − Average kinetic energy of reactants
- Transition State: The unstable, high-energy intermediate formed at the peak of the energy barrier. It cannot be isolated.
- Activated complex: Another name for the transition state
High Activation Energy (Ea)
- Few molecules have enough energy
- Slow reaction rate
- Rate more sensitive to temperature
- Example: combustion in absence of catalyst
Low Activation Energy (Ea)
- Many molecules have enough energy
- Fast reaction rate
- Rate less sensitive to temperature
- Example: ionic reactions in solution
🧮 Arrhenius Equation — The Core Formula
In 1889, Swedish chemist Svante Arrhenius proposed a quantitative relationship between temperature and rate constant:
Arrhenius Equation (★ Most Important ★)
k = A × e^(−Ea/RT)
━━━━━━━━━━━━━━━━━━━━━━━━━━
where:
k = rate constant
A = frequency factor (pre-exponential factor / Arrhenius factor)
Ea = activation energy (J mol⁻¹)
R = universal gas constant = 8.314 J mol⁻¹ K⁻¹
T = absolute temperature (Kelvin)
The Frequency Factor (A)
- Represents the frequency of collisions with the correct orientation
- Also called the pre-exponential factor or Arrhenius factor
- Has the same units as k (depends on order of reaction)
- Temperature independent (approximately)
- Related to collision frequency Z and steric factor p: A = p × Z
Taking Logarithm — Useful Forms
Logarithmic Form
ln k = ln A − Ea/RT
━━━━━━━━━━━━━━━━━━━━━━
log k = log A − Ea/(2.303 RT)
━━━━━━━━━━━━━━━━━━━━━━
Graph: log k vs. 1/T → Slope = −Ea/2.303R
📉
Arrhenius Plot: log k vs. 1/T
📖 NCERT Figure 4.10
A straight line graph of log k (y-axis) vs 1/T (x-axis). Slope = −Ea/2.303R (negative, so line goes from upper left to lower right). The y-intercept = log A. From this graph, Ea and A can be calculated experimentally.
Calculating Ea from Two Temperatures
Two-Temperature Arrhenius Equation (★ Used in All Calculations ★)
log(k₂/k₁) = Ea/2.303R × (T₂ − T₁)/(T₁ × T₂)
━━━━━━━━━━━━━━━━━━━━━━
OR: ln(k₂/k₁) = −Ea/R × (1/T₂ − 1/T₁)
⚠️ Important Sign Note: When T₂ > T₁ (temperature increases), k₂ > k₁ (rate increases). Since Ea is always positive, the formula gives a positive result when T₂ > T₁, consistent with the increase in k. Always check signs carefully!
Physical Meaning of e^(−Ea/RT)
The factor e^(−Ea/RT) represents the fraction of molecules that have kinetic energy ≥ Ea at temperature T. This fraction increases exponentially with temperature.
Fraction of molecules with energy ≥ Ea = e^(−Ea/RT)
As T increases → −Ea/RT becomes less negative → e^(−Ea/RT) increases → more reactive molecules
📊 Maxwell-Boltzmann Distribution
The Maxwell-Boltzmann distribution describes how molecular energies are distributed in a gas at a given temperature.
📊
Maxwell-Boltzmann Energy Distribution
📖 NCERT Figure 4.8
Bell-shaped curve showing number of molecules (y-axis) vs kinetic energy (x-axis). The peak represents the most probable energy. A vertical line at Ea shows the threshold energy. The shaded area to the right of Ea represents the fraction of molecules with enough energy to react. At higher temperature T₂ > T₁, the curve flattens and shifts right, with a larger shaded area — explaining the increased rate.
- The curve is asymmetric — skewed to the right (not symmetric)
- At higher temperature, the peak shifts to higher energy and the curve broadens
- The area under the curve (shaded portion) beyond Ea increases significantly with small increases in T
- This exponential increase in reactive fraction explains why rate roughly doubles per 10°C rise
Practice Problems — Arrhenius Equation
Problem 1 · Board Level (NCERT)
The rate constant of a reaction at 500 K and 700 K are 0.02 s⁻¹ and 0.07 s⁻¹ respectively. Calculate the value of Ea.
▶ Show Answer
log(k₂/k₁) = Ea/2.303R × (T₂−T₁)/(T₁T₂)
log(0.07/0.02) = Ea/(2.303 × 8.314) × (700−500)/(500×700)
log(3.5) = Ea/19.147 × (200/350000)
0.5441 = Ea × 2.971 × 10⁻⁵
Ea = 0.5441 / (2.971 × 10⁻⁵)
Ea = 18,313 J mol⁻¹ ≈ 18.3 kJ mol⁻¹
Problem 2 · Board Level
Calculate the temperature at which the rate constant of a reaction is 1.5 times the rate constant at 298 K. Ea = 50 kJ mol⁻¹.
▶ Show Answer
log(k₂/k₁) = Ea/2.303R × (T₂−T₁)/(T₁T₂)
log(1.5) = (50000/(2.303 × 8.314)) × (T₂ − 298)/(298 × T₂)
0.1761 = 2613.3 × (T₂ − 298)/(298 × T₂)
Solving: 0.1761 × 298 × T₂ = 2613.3 × (T₂ − 298)
52.48 T₂ = 2613.3 T₂ − 778,364
778,364 = 2560.8 T₂
T₂ ≈ 304 K (approximately 31°C)
Problem 3 · JEE Level
The activation energy of a reaction is 100 kJ/mol. What fraction of molecules at 500 K would have energy ≥ Ea? (R = 8.314 J/mol·K)
▶ Show Answer
Fraction = e^(−Ea/RT) = e^(−100000/(8.314 × 500)) = e^(−24.05)
= e^(−24.05) ≈ 3.7 × 10⁻¹¹
This is an extremely small fraction — only ~37 molecules per trillion have enough energy! This explains why high-Ea reactions are very slow.
Problem 4 · NEET Level
A catalyst reduces the activation energy of a reaction from 100 kJ/mol to 50 kJ/mol at 300 K. By what factor is the rate increased?
▶ Show Answer
Rate ratio = k_cat/k = e^(−Ea_cat/RT) / e^(−Ea/RT) = e^[(Ea − Ea_cat)/RT]
= e^[(100000 − 50000)/(8.314 × 300)]
= e^[50000/2494] = e^20.05
= ≈ 5 × 10⁸
The rate is increased by a factor of about 500 million — demonstrating why catalysts are so powerful!
Problem 5 · JEE Advanced Level
For a reaction, log k = −(5000/T) + 6. Find: (a) Activation energy, (b) Frequency factor A, (c) Rate constant at 500 K.
▶ Show Answer
Compare with: log k = log A − Ea/(2.303RT) = log A − (Ea/2.303R) × (1/T)
(a) Ea: Slope = −Ea/2.303R = −5000
Ea = 5000 × 2.303 × 8.314 = 95,710 J/mol ≈ 95.7 kJ/mol
(b) A: Intercept = log A = 6 → A = 10⁶ (in appropriate units)
(c) At T = 500 K:
log k = −5000/500 + 6 = −10 + 6 = −4
k = 10⁻⁴ (in appropriate units)
💥 Postulates of Collision Theory
Collision theory, developed by Max Trautz and William Lewis (1916–1918), provides a molecular explanation for the rate of chemical reactions.
Main Postulates:
- For a reaction to occur, the reactant molecules must collide with each other.
- Not all collisions lead to reaction. Only those collisions in which the molecules have sufficient kinetic energy (≥ Ea) lead to reaction. These are called effective collisions.
- The reacting molecules must also have the proper orientation at the time of collision for the reaction to occur.
🎯
Effective vs Ineffective Collisions — Orientation Effect
📖 NCERT Figure 4.11
Diagram showing: (a) Effective collision — molecules approach with correct orientation, reaction occurs, products formed. (b) Ineffective collision — molecules bounce off without reaction due to wrong orientation, even if energy is sufficient. Example: NO + Cl₂ → NOCl + Cl.
🔢 Mathematical Expression from Collision Theory
The rate of a bimolecular reaction is given by:
Collision Theory Rate Equation
k = PZ_{AB} × e^(−Ea/RT)
━━━━━━━━━━━━━━━━━━━━━━━━━━
where:
P = steric factor (probability factor / orientation factor) [0 < P ≤ 1]
Z_{AB} = collision frequency (number of collisions per second per unit volume)
e^(−Ea/RT) = fraction of molecules with energy ≥ Ea
Comparison with Arrhenius: k = A × e^(−Ea/RT)
Therefore: A = P × Z_AB
The Arrhenius frequency factor A is the product of the steric factor (P) and the collision frequency (Z_AB).
Steric Factor (P)
- P accounts for the fraction of collisions with the correct orientation
- P is always ≤ 1 (typically much less than 1 for complex molecules)
- For simple atoms: P ≈ 1 (any orientation works)
- For complex molecules with specific reactive sites: P << 1
- P explains why observed rates are often much lower than predicted from collision frequency alone
⚖️ Limitations of Collision Theory
Limitations:
- Treats molecules as hard spheres — ignores their structure
- Cannot predict the value of P (steric factor) — it is fitted experimentally
- Not applicable to reactions in solution (solvent effects ignored)
- For many reactions, predicted rate (using Z_AB alone) is much higher than observed
More advanced theories (Transition State Theory / Activated Complex Theory) overcome some of these limitations by considering the structure of the transition state.
🆚 Collision Theory vs. Transition State Theory
Collision Theory
- Treats molecules as rigid spheres
- Rate depends on Z_AB, P, and Ea
- Simple and intuitive
- Cannot account for complex molecular shapes
- Good for simple gas-phase reactions
Transition State Theory
- Considers the activated complex (transition state)
- Rate depends on equilibrium constant for TS formation
- Accounts for molecular geometry
- More rigorous and general
- Applicable to reactions in solution
Practice Problems — Collision Theory
Problem 1 · Board Level
What do you understand by the term 'effective collision'? Why is every collision not effective?
▶ Show Answer
An effective collision is a collision between reactant molecules that actually results in the formation of products.
Not every collision is effective because:
1. Energy criterion: The colliding molecules must have kinetic energy ≥ activation energy (Ea). Most molecules at room temperature have energy below Ea.
2. Orientation criterion: The molecules must collide with the correct mutual orientation so that the reactive bonds are properly aligned for breaking and formation of new bonds.
Both conditions must be satisfied simultaneously for an effective collision.
Problem 2 · JEE Level
At 27°C, the rate constant of a reaction is 1.7 × 10⁻³ L mol⁻¹ s⁻¹. If the steric factor P = 0.001 and Ea = 60 kJ/mol, calculate the collision frequency Z_AB. (R = 8.314 J/mol·K)
▶ Show Answer
T = 27 + 273 = 300 K
k = P × Z_AB × e^(−Ea/RT)
e^(−Ea/RT) = e^(−60000/(8.314 × 300)) = e^(−24.05) = 3.7 × 10⁻¹¹
Z_AB = k / (P × e^(−Ea/RT)) = 1.7 × 10⁻³ / (0.001 × 3.7 × 10⁻¹¹)
Z_AB = 1.7 × 10⁻³ / 3.7 × 10⁻¹⁴ = 4.6 × 10¹⁰ L mol⁻¹ s⁻¹
📌 Master Formula Sheet
Rate of Reaction
For aA + bB → cC + dD:
Rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt
Integrated Rate Laws
Zero Order: [A]t = [A]₀ − kt | t½ = [A]₀/2k
First Order: k = (2.303/t) log([A]₀/[A]t) | t½ = 0.693/k
Second Order: 1/[A]t − 1/[A]₀ = kt | t½ = 1/(k[A]₀)
Arrhenius Equation
k = Ae^(−Ea/RT)
log k = log A − Ea/(2.303RT)
log(k₂/k₁) = Ea/2.303R × (T₂−T₁)/(T₁T₂)
Slope of log k vs 1/T = −Ea/2.303R
Units of Rate Constant
General: Units of k = (mol L⁻¹)^(1−n) × s⁻¹
n=0: mol L⁻¹ s⁻¹ | n=1: s⁻¹ | n=2: L mol⁻¹ s⁻¹ | n=3: L² mol⁻² s⁻¹
⭐ Important Points to Remember
- t½ of first order is independent of initial concentration — this is unique and very important
- Order is experimental; molecularity is theoretical for elementary steps only
- Rate constant k increases with temperature — quantified by Arrhenius equation
- Catalyst lowers Ea but NOT ΔH or equilibrium position
- Pseudo-first order: one reactant in large excess keeps concentration constant
- log([A]₀/[A]t) vs t → straight line only for first order
- A and k have same units (since e^(−Ea/RT) is dimensionless)
- t(99%) = 6.64t½; t(99.9%) = 9.97t½ ≈ 10t½ for first order
- Radioactive decay is always first order with constant t½
- Rate law must be determined experimentally — never from balanced equation alone
🧠 Key Graphs Summary
| Order | Straight Line Graph | Slope | Y-Intercept |
|---|
| Zero | [A] vs t | −k | [A]₀ |
| First | log[A] vs t | −k/2.303 | log[A]₀ |
| First | ln[A] vs t | −k | ln[A]₀ |
| Second | 1/[A] vs t | +k | 1/[A]₀ |
| Arrhenius | log k vs 1/T | −Ea/2.303R | log A |
🎯 Common Mistakes to Avoid
- ❌ Do NOT equate order with stoichiometric coefficient unless the reaction is elementary
- ❌ Do NOT confuse rate with rate constant — rate depends on concentration; k does not
- ❌ Do NOT forget that t½ for first order is constant; for zero and second order it depends on concentration
- ❌ Do NOT use °C in the Arrhenius equation — always convert to Kelvin!
- ❌ Do NOT say catalyst changes equilibrium — it only speeds up approach to equilibrium
- ❌ In the two-temperature Arrhenius formula, be careful: T₁ < T₂ gives k₁ < k₂
📖 Selected NCERT Solutions
NCERT Q4.4 · Board Level
The initial concentration of N₂O₅ in the following first order reaction N₂O₅(g) → 2NO₂(g) + ½O₂(g) was 1.24 × 10⁻² mol L⁻¹ at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10⁻² mol L⁻¹. Calculate the rate constant of the reaction at 318 K.
▶ Show Solution
k = (2.303/t) × log([A]₀/[A]t)
[A]₀ = 1.24 × 10⁻² M, [A]t = 0.20 × 10⁻² M, t = 60 min
k = (2.303/60) × log(1.24 × 10⁻² / 0.20 × 10⁻²)
k = (2.303/60) × log(6.2)
k = 0.03838 × 0.7924
k = 3.04 × 10⁻² min⁻¹
NCERT Q4.8 · Board Level
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
▶ Show Solution
t½ = 5730 years → k = 0.693/5730 = 1.209 × 10⁻⁴ year⁻¹
[A]₀ = 100%, [A]t = 80% → [A]₀/[A]t = 100/80 = 1.25
t = (2.303/k) × log([A]₀/[A]t)
t = (2.303 / 1.209 × 10⁻⁴) × log(1.25)
t = 19049 × 0.09691
t ≈ 1845 years
NCERT Q4.13 · Board Level
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming it does not change with temperature.
▶ Show Solution
k₂/k₁ = 4, T₁ = 293 K, T₂ = 313 K
log(k₂/k₁) = Ea/2.303R × (T₂−T₁)/(T₁T₂)
log(4) = Ea/(2.303 × 8.314) × (313−293)/(293 × 313)
0.6021 = Ea/19.147 × 20/91709
0.6021 = Ea × 1.140 × 10⁻⁵
Ea = 0.6021 / (1.140 × 10⁻⁵)
Ea = 52,816 J mol⁻¹ ≈ 52.82 kJ mol⁻¹
NCERT Q4.15 · Board Level
The rate constant for the first order decomposition of H₂O₂ is given by: log k = 14.34 − 1.25 × 10⁴ K / T. Calculate Ea for the reaction. Also find the temperature at which the half-life of H₂O₂ is 256 minutes.
▶ Show Solution
Compare: log k = log A − Ea/(2.303R) × (1/T)
∴ Ea/2.303R = 1.25 × 10⁴
Ea = 1.25 × 10⁴ × 2.303 × 8.314 = 239,339 J mol⁻¹ ≈ 239.34 kJ mol⁻¹
For t½ = 256 min → k = 0.693/256 = 2.707 × 10⁻³ min⁻¹
log(2.707 × 10⁻³) = 14.34 − 1.25 × 10⁴/T
−2.567 = 14.34 − 1.25 × 10⁴/T
1.25 × 10⁴/T = 14.34 + 2.567 = 16.907
T = 1.25 × 10⁴ / 16.907
T = 739.5 K
Mixed Challenge Problems (Board + JEE + NEET)
Challenge 1 · JEE Main Level
For a reaction A + B → C + D, the rate = k[A]½[B]². If concentrations of both A and B are doubled simultaneously, by what factor does the rate change?
▶ Show Answer
New rate = k[2A]^(1/2)[2B]² = k × 2^(1/2) × [A]^(1/2) × 4 × [B]²
= 4√2 × k[A]^(1/2)[B]² = 4√2 × Rate
Factor of increase = 4√2 ≈ 5.66
Challenge 2 · JEE Level
At temperature T, a compound AB₂ dissociates according to: 2AB₂ ⇌ 2AB + B₂. This is a first order reaction. If the initial pressure of AB₂ is P₀ and total pressure at time t is P, find the rate constant k in terms of P₀, P, and t.
▶ Show Answer
Let x = decrease in pressure of AB₂
2AB₂ → 2AB + B₂
At t: P(AB₂) = P₀ − x; P(AB) = x; P(B₂) = x/2
Total P = (P₀ − x) + x + x/2 = P₀ + x/2
∴ x = 2(P − P₀) → P(AB₂) = P₀ − 2(P − P₀) = 3P₀ − 2P
k = (2.303/t) × log(P₀/P_{AB₂}) = (2.303/t) × log[P₀/(3P₀ − 2P)]
Challenge 3 · NEET Level
A first order gas phase reaction A(g) → 2B(g) + C(g) starts with initial pressure P₀ = 100 kPa. After 10 min, total pressure = 175 kPa. Calculate k.
▶ Show Answer
A → 2B + C
At t=0: P₀ = 100, 0, 0. Total = 100 kPa
At t=10: (100−x), 2x, x. Total = 100 + 2x = 175
2x = 75 → x = 37.5 kPa
P(A) at t=10 min = 100 − 37.5 = 62.5 kPa
k = (2.303/10) × log(100/62.5) = 0.2303 × log(1.6) = 0.2303 × 0.2041
k = 0.0470 min⁻¹
Challenge 4 · JEE Advanced Level
The following data is obtained for the decomposition of H₂O₂ in solution at 300 K:
| t (min) | 0 | 10 | 20 |
|---|
| [H₂O₂] (mol/L) | 0.50 | 0.25 | 0.125 |
Determine: (a) order of reaction, (b) rate constant, (c) rate at t = 20 min.
▶ Show Answer
(a) Order: Concentration halves every 10 minutes: 0.50 → 0.25 → 0.125
t½ is constant = 10 minutes → First Order Reaction
(b) Rate constant:
k = 0.693/t½ = 0.693/10 = 0.0693 min⁻¹
(c) Rate at t = 20 min:
[H₂O₂] at t = 20 min = 0.125 mol/L
Rate = k[H₂O₂] = 0.0693 × 0.125 = 8.66 × 10⁻³ mol L⁻¹ min⁻¹
Challenge 5 · Board Level
What is a pseudo-first order reaction? Give two examples. Why is it important to understand such reactions?
▶ Show Answer
A pseudo-first order reaction is a reaction that is actually of a higher order but appears to be first order because one reactant is in such large excess that its concentration remains effectively constant.
Examples:
1. Hydrolysis of cane sugar: C₁₂H₂₂O₁₁ + H₂O → glucose + fructose (H₂O in excess as solvent)
2. Acid hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Importance:
• Simplifies rate equations — complex reactions can be studied using simple first-order mathematics
• Allows measurement of rate constants more easily
• Important in biochemistry — enzyme kinetics at low substrate concentrations
• Essential in pharmacokinetics (drug absorption/elimination kinetics)
🎯 Chapter Complete!
You have covered all topics of Chemical Kinetics — from the basics of reaction rate to the Arrhenius equation and collision theory. Practice the problems above, memorize the key formulas, and pay special attention to first-order reactions and the Arrhenius equation as they are the most frequently tested topics in both Board exams and JEE/NEET.
Reference: NCERT Chemistry Part-I, Class XII, Chapter 4 — Chemical Kinetics