1. Chapter at a Glance
Thermodynamics tells us whether a reaction can happen; chemical kinetics tells us how fast it happens and by what path. This chapter covers the rate of reaction, the factors that affect it, rate laws, order and molecularity, integrated rate equations for zero- and first-order reactions, half-life, and the effect of temperature through the Arrhenius equation. NEET asks one or two questions every year, many of them on first-order kinetics and activation energy.
- Relating rates of disappearance and appearance using stoichiometry.
- Order from the rate law or from units of k; order vs molecularity.
- First-order integrated equation and half-life.
- Zero-order reactions and their half-life.
- Arrhenius equation; effect of catalysts; collision theory.
2. Rate of a 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.
- Average rate is the change over a time interval; instantaneous rate is the slope of the concentration–time curve at one moment.
- Units: mol L⁻¹ s⁻¹.
- Factors: concentration, temperature, catalyst, surface area, and the nature of the reactants.
3. Rate Law, Order and Molecularity
- x and y are found by experiment; they need not equal the coefficients in the balanced equation.
- Order can be zero, a whole number or a fraction.
- Molecularity is the number of species that collide in an elementary step. It is always a whole number (1, 2 or rarely 3), never zero.
- For a complex reaction, the slowest step (rate-determining step) decides the rate law.
- Pseudo-first-order reactions: one reactant is in large excess, so its concentration hardly changes. Examples: acid hydrolysis of ethyl acetate and inversion of cane sugar.
| Order | Units of k | NEET Note |
|---|---|---|
| 0 | mol L⁻¹ s⁻¹ | Rate independent of concentration |
| 1 | s⁻¹ | Radioactive decay is first order |
| 2 | L mol⁻¹ s⁻¹ | — |
| n | (mol L⁻¹)1−n s⁻¹ | General rule |
4. Integrated Rate Equations and Half-life
| Property | Zero order | First order |
|---|---|---|
| Integrated law | [A] = [A]₀ − kt | k = (2.303/t) log([A]₀/[A]) |
| Straight-line plot | [A] vs t (slope −k) | log[A] vs t (slope −k/2.303) |
| Half-life | t½ = [A]₀/2k | t½ = 0.693/k |
| Dependence of t½ on [A]₀ | Directly proportional | Independent |
| Examples | Decomposition of NH₃ on hot Pt; photochemical H₂ + Cl₂ | Decomposition of N₂O₅; radioactive decay |
- For first order: time for 75% completion = 2t½; 87.5% = 3t½; 99.9% ≈ 10t½.
- t₉₉ = 2 × t₉₀ for a first-order reaction.
- For a gas-phase first-order reaction measured by pressure: k = (2.303/t) log[p₀/(2p₀ − pt)] for A(g) → B(g) + C(g).
5. Temperature and the Arrhenius Equation
- For most reactions, the rate roughly doubles for every 10 K rise in temperature (temperature coefficient ≈ 2).
- Activation energy Ea is the minimum extra energy needed to form the activated complex. A higher Ea means a slower reaction and a stronger effect of temperature.
- A plot of ln k against 1/T is a straight line with slope −Ea/R.
- A catalyst provides a path with a lower Ea. It speeds up forward and backward reactions equally, does not change ΔH or ΔG or K, and is not consumed.
- Collision theory: rate = ZAB × e−Ea/RT × P, where P (the steric factor) accounts for proper orientation. Only collisions with enough energy and the right orientation are effective.
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Rate (aA → products) | rate = −(1/a)d[A]/dt |
| Rate law | rate = k[A]ˣ[B]ʸ |
| Units of k (order n) | (mol L⁻¹)1−n s⁻¹ |
| Zero order | [A] = [A]₀ − kt ; t½ = [A]₀/2k |
| First order | k = (2.303/t) log([A]₀/[A]) ; t½ = 0.693/k |
| Fraction left (first order) | [A]/[A]₀ = (½)t/t½ |
| Arrhenius | k = Ae−Ea/RT |
| Two temperatures | log(k₂/k₁) = Ea(T₂ − T₁)/(2.303RT₁T₂) |
| ln k vs 1/T | slope = −Ea/R |
7. Master Summary Table
| Concept | Key idea | NEET Note |
|---|---|---|
| Order | Experimental | Can be 0 or fractional |
| Molecularity | Elementary step only | Never zero or fractional |
| Zero order | t½ ∝ [A]₀ | Rate constant = rate |
| First order | t½ independent of [A]₀ | Radioactive decay |
| Pseudo first order | Excess reagent | Ester hydrolysis in acid |
| Arrhenius | k rises with T | About ×2 per 10 K |
| Catalyst | Lowers Ea | No change in ΔH or K |
8. Mind Map
- Average, instantaneous
- Stoichiometry
- Factors
- Order
- Molecularity
- Units of k
- Zero order
- First order
- Half-life
- Arrhenius
- Ea, catalyst
- Collision theory
9. Tips and Tricks
- Initial-rate method: double one concentration at a time; if the rate doubles, order = 1; quadruples, order = 2; unchanged, order = 0.
- If rate = k[A][B]², doubling both multiplies the rate by 2 × 4 = 8.
- For first order, count half-lives: 1/2, 1/4, 1/8, 1/16… of the reactant remains.
- 2.303R = 19.15 J K⁻¹ mol⁻¹ for Arrhenius calculations.
- Reading the order from the coefficients of an overall equation.
- Thinking a catalyst changes the equilibrium constant.
- Using t½ = 0.693/k for a zero-order reaction.
- Forgetting to divide by the stoichiometric coefficient when relating rates.
10. Important One-liners
- The rate constant depends on temperature, not on concentration.
- Molecularity of an elementary reaction cannot be zero.
- The half-life of a first-order reaction does not depend on the initial concentration.
- All radioactive decays follow first-order kinetics.
- A catalyst lowers activation energy for both forward and backward reactions equally.
- The units of k for a zero-order reaction are the same as the units of rate.
- The slope of ln k against 1/T is −Ea/R.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take R = 8.314 J K⁻¹ mol⁻¹ and log 2 = 0.301. Click to reveal each answer and working.
Average level
For 2N₂O₅ → 4NO₂ + O₂, N₂O₅ disappears at 0.02 mol L⁻¹ s⁻¹. Find the rate of formation of NO₂ and O₂, and the rate of the reaction.
Show answer & solution
Concept: rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt.
- Rate of reaction = 0.02/2 = 0.01 mol L⁻¹ s⁻¹.
- NO₂ forms at 4 × 0.01 = 0.04 mol L⁻¹ s⁻¹.
- O₂ forms at 1 × 0.01 = 0.01 mol L⁻¹ s⁻¹.
Answer: NO₂ 0.04; O₂ 0.01; reaction rate 0.01 mol L⁻¹ s⁻¹
The rate constant of a reaction is 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹. What is its order?
Show answer & solution
Concept: units of k = (mol L⁻¹)1−n s⁻¹.
- L mol⁻¹ s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ ⇒ 1 − n = −1.
- n = 2: second order.
Answer: second order
A first-order reaction has k = 0.0693 min⁻¹. Find its half-life and the time for 75% completion.
Show answer & solution
Concept: t½ = 0.693/k; 75% completion = 2 half-lives.
- t½ = 0.693/0.0693 = 10 min.
- 75% complete means 25% left = (½)², so 2 × 10 = 20 min.
Answer: 10 min; 20 min
A zero-order reaction starts at [A]₀ = 0.8 M with k = 0.02 mol L⁻¹ min⁻¹. Find its half-life and the time needed for the reaction to be complete.
Show answer & solution
Concept: [A] = [A]₀ − kt; t½ = [A]₀/2k.
- t½ = 0.8/(2 × 0.02) = 20 min.
- Complete when [A] = 0: t = 0.8/0.02 = 40 min.
Answer: 20 min; 40 min
The rate law of a reaction is rate = k[A][B]². By what factor does the rate change if [A] is doubled and [B] is tripled?
Show answer & solution
Concept: substitute the new concentrations.
- New rate = k(2[A])(3[B])² = 2 × 9 × k[A][B]².
- The rate becomes 18 times.
Answer: 18 times
Advanced level
A first-order reaction has k = 4.606 × 10⁻³ s⁻¹. Find the time for 90% and for 99% completion, and show that t₉₉ = 2t₉₀.
Show answer & solution
Concept: t = (2.303/k) log([A]₀/[A]).
- 90%: [A]₀/[A] = 10 ⇒ t₉₀ = 2.303/(4.606 × 10⁻³) × 1 = 500 s.
- 99%: [A]₀/[A] = 100 ⇒ t₉₉ = 500 × 2 = 1000 s.
- Since log 100 = 2 log 10, t₉₉ is always twice t₉₀ for first order.
Answer: 500 s; 1000 s
The rate constant of a reaction doubles when the temperature rises from 300 K to 310 K. Find its activation energy.
Show answer & solution
Concept: log(k₂/k₁) = Ea(T₂ − T₁)/(2.303RT₁T₂).
- log 2 = Ea × 10/(2.303 × 8.314 × 300 × 310).
- Ea = 0.301 × 2.303 × 8.314 × 93 000/10.
- Ea ≈ 53 600 J ≈ 53.6 kJ mol⁻¹.
Answer: about 53.6 kJ mol⁻¹
For A + B → products, initial rates were: [A] = 0.1, [B] = 0.1 → 2 × 10⁻³; [A] = 0.2, [B] = 0.1 → 4 × 10⁻³; [A] = 0.1, [B] = 0.2 → 8 × 10⁻³ mol L⁻¹ s⁻¹. Find the rate law and k.
Show answer & solution
Concept: change one concentration at a time.
- Doubling [A] doubles the rate ⇒ order in A = 1.
- Doubling [B] makes the rate 4 times ⇒ order in B = 2.
- rate = k[A][B]²; k = 2 × 10⁻³/(0.1 × 0.01) = 2 L² mol⁻² s⁻¹ (third order overall).
Answer: rate = k[A][B]²; k = 2 L² mol⁻² s⁻¹
In a first-order reaction, the concentration falls from 0.6 M to 0.15 M in 40 minutes. Find the half-life, k, and the concentration after 60 minutes.
Show answer & solution
Concept: 0.6 → 0.3 → 0.15 is two half-lives.
- 2t½ = 40 min ⇒ t½ = 20 min.
- k = 0.693/20 ≈ 0.0347 min⁻¹.
- 60 min = 3 half-lives ⇒ [A] = 0.6/8 = 0.075 M.
Answer: 20 min; about 0.035 min⁻¹; 0.075 M
A reaction has k = 2 × 10⁻³ s⁻¹ at 300 K and an activation energy of 50 kJ mol⁻¹. Find k at 350 K.
Show answer & solution
Concept: Arrhenius equation at two temperatures.
- log(k₂/k₁) = 50 000/(2.303 × 8.314) × (1/300 − 1/350).
- = 2611 × 4.76 × 10⁻⁴ ≈ 1.24 ⇒ k₂/k₁ ≈ 17.5.
- k₂ ≈ 17.5 × 2 × 10⁻³ ≈ 3.5 × 10⁻² s⁻¹.
Answer: about 3.5 × 10⁻² s⁻¹
12. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
The unit of the rate constant for a first-order reaction is:
- mol L⁻¹ s⁻¹
- s⁻¹
- L mol⁻¹ s⁻¹
- L² mol⁻² s⁻¹
Show solution
For n = 1, (mol L⁻¹)⁰ s⁻¹ = s⁻¹. The other units belong to zero-, second- and third-order reactions.
Answer: (B)
A catalyst increases the rate of a reaction by:
- Increasing the activation energy
- Decreasing the activation energy
- Increasing ΔH
- Shifting the equilibrium
Show solution
A catalyst provides an alternative path with a lower Ea, so more collisions succeed. It does not change ΔH or the position of equilibrium.
Answer: (B)
For a zero-order reaction, the half-life is:
- Independent of initial concentration
- Directly proportional to initial concentration
- Inversely proportional to initial concentration
- Proportional to the square of initial concentration
Show solution
t½ = [A]₀/2k, so it is proportional to [A]₀. Independence from [A]₀ is the property of first-order reactions, and inverse proportionality belongs to second-order reactions.
Answer: (B)
The acid hydrolysis of ethyl acetate is an example of a:
- Zero-order reaction
- Pseudo-first-order reaction
- Second-order reaction in practice
- Third-order reaction
Show solution
Water is present in large excess, so its concentration stays almost constant. The rate then depends only on [ester]. Strictly the reaction is bimolecular, but it behaves as first order.
Answer: (B)
Which statement about molecularity is correct?
- It can be zero
- It can be fractional
- It applies only to elementary reactions
- It is always equal to the order
Show solution
Molecularity counts the species colliding in one elementary step, so it must be a positive whole number. It equals the order only for elementary reactions.
Answer: (C)
For a first-order reaction, a plot that gives a straight line is:
- [A] vs t
- log[A] vs t
- 1/[A] vs t
- [A]² vs t
Show solution
log[A] = log[A]₀ − kt/2.303 is linear in t. A plot of [A] against t is linear for zero order, and 1/[A] against t is linear for second order.
Answer: (B)
A first-order reaction is 20% complete in 10 minutes. The time for it to be 75% complete is about (log 2 = 0.301, log 1.25 = 0.097):
- 31 min
- 62 min
- 40 min
- 50 min
Show solution
k = (2.303/10) log(1/0.8) = 0.2303 × 0.097 ≈ 0.0223 min⁻¹. For 75%, t = (2.303/k) log 4 = 2.303 × 0.602/0.0223 ≈ 62 min. (C) and (D) come from assuming a linear change.
Answer: (B)
The slope of a graph of ln k against 1/T for a reaction is −6000 K. The activation energy is about:
- 6 kJ mol⁻¹
- 49.9 kJ mol⁻¹
- 115 kJ mol⁻¹
- 721 J mol⁻¹
Show solution
Slope = −Ea/R, so Ea = 6000 × 8.314 ≈ 49 900 J ≈ 49.9 kJ mol⁻¹. (C) wrongly multiplies by 2.303 as well, which is needed only when the graph uses log k instead of ln k.
Answer: (B)
13. Quick Revision Summary
- Rate = −(1/a)d[A]/dt; rate = k[A]ˣ[B]ʸ (experimental).
- Units of k: (mol L⁻¹)1−n s⁻¹; molecularity only for elementary steps.
- Zero order: [A] = [A]₀ − kt, t½ = [A]₀/2k.
- First order: k = (2.303/t) log([A]₀/[A]); t½ = 0.693/k; t₉₉ = 2t₉₀.
- Arrhenius k = Ae−Ea/RT; catalyst lowers Ea only.