In this guide
Thermodynamics tells you whether a reaction can happen. Kinetics tells you how fast it does. The two are independent: diamond turning into graphite is favoured by ΔG, yet it is far too slow to see. Kinetics is about rates, the rate law, and how temperature and catalysts change the speed.
NEET questions here come in a handful of shapes: units of k for a given order, half-life and fraction remaining for a first-order reaction, order from an initial-rates table, the relation between rates of disappearance and formation, and activation energy from rate constants at two temperatures. The first-order formulas carry most of the numericals.
Rate of reaction
For aA + bB → cC + dD, the rate is the same number whichever species you track, once you divide by the coefficient:
Rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt
Reactants have a minus sign because their concentrations fall. The average rate is taken over a time interval; the instantaneous rate is the slope of the concentration–time curve at one moment.
Rate depends on concentration, temperature, pressure (for gases), catalysts and, for heterogeneous reactions, surface area.
Rate law, order and molecularity
The rate law is found by experiment, not from the balanced equation:
Rate = k[A]ˣ[B]ʸ
The order is x + y. It can be zero, a whole number or a fraction. The rate constant k is the rate when every concentration is 1 mol L⁻¹.
| Feature | Order | Molecularity |
|---|---|---|
| Definition | Sum of powers in the experimental rate law | Number of species colliding in one elementary step |
| Found by | Experiment | Theory (the mechanism) |
| Values | 0, fractional or whole number | Whole number only: 1, 2 or rarely 3 |
| Applies to | Elementary and complex reactions | Elementary steps only |
Complex reactions happen in several elementary steps. The slowest step, the rate-determining step, fixes the overall rate. For a complex reaction, order and molecularity need not match, and molecularity has no meaning for the overall reaction.
Units of k follow from the order n: (mol L⁻¹)¹⁻ⁿ s⁻¹.
| Order | Units of k |
|---|---|
| Zero | mol L⁻¹ s⁻¹ |
| First | s⁻¹ |
| Second | L mol⁻¹ s⁻¹ |
| Third | L² mol⁻² s⁻¹ |
Integrated rate laws and half-life
| Feature | Zero order | First order |
|---|---|---|
| Integrated rate law | [A] = [A]₀ − kt | k = (2.303/t) log([A]₀/[A]) |
| Straight-line plot | [A] against t, slope −k | ln[A] against t, slope −k; or log[A] against t, slope −k/2.303 |
| Half-life | t½ = [A]₀/2k | t½ = 0.693/k |
| Half-life depends on [A]₀? | Yes, directly proportional | No, constant |
| Examples | Decomposition of NH₃ on a hot platinum surface at high pressure | Decomposition of N₂O₅; hydrogenation of ethene; radioactive decay |
Useful first-order shortcuts:
- After n half-lives, the fraction left is (½)ⁿ. After 3 half-lives, 1/8 remains.
- 75% completion takes 2 half-lives; 87.5% takes 3.
- 99% completion takes about 6.64 half-lives, and 99.9% about 10 half-lives.
Pseudo first-order reactions, such as the acid hydrolysis of ethyl acetate, look first order because one reactant (water) is in such large excess that its concentration hardly changes.
Temperature and the Arrhenius equation
For many reactions, the rate roughly doubles or triples for a 10 °C rise. The ratio k(T + 10)/k(T) is called the temperature coefficient.
The Arrhenius equation explains why:
k = A exp(−Ea/RT), or ln k = ln A − Ea/RT
- A is the pre-exponential (frequency) factor.
- Ea is the activation energy, the minimum extra energy the reactants need to form the activated complex.
- exp(−Ea/RT) is the fraction of molecules with energy of at least Ea.
- A plot of ln k against 1/T is a straight line with slope −Ea/R.
For two temperatures:
log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
A rise in temperature increases the fraction of molecules with energy above Ea sharply, which is why a small temperature change has a large effect on rate.
Catalysts provide an alternative path with a lower Ea. They speed up the forward and backward reactions equally, so they do not change ΔH, ΔG or the equilibrium constant. They only help equilibrium arrive sooner.
Collision theory
Collision theory, for bimolecular gaseous reactions, says molecules react only when they collide with enough energy (at least Ea) and the right orientation:
Rate = P × Z_AB × exp(−Ea/RT)
Z_AB is the collision frequency and P is the probability (steric) factor, which accounts for orientation. The syllabus asks for the idea only, with no derivation.
Worked NEET-style questions
Example 1: relating rates
For 2N₂O₅ → 4NO₂ + O₂, the rate of formation of O₂ is x mol L⁻¹ s⁻¹. Find the rates for N₂O₅ and NO₂.
- −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt = x.
- N₂O₅ disappears at 2x and NO₂ forms at 4x.
Example 2: first-order half-life from data
A first-order reaction falls from 0.8 M to 0.1 M in 60 minutes. Find t½ and k.
- 0.8 → 0.4 → 0.2 → 0.1 is 3 half-lives.
- t½ = 60/3 = 20 min.
- k = 0.693/20 ≈ 0.0347 min⁻¹.
Example 3: order from initial rates
| [A] (mol L⁻¹) | [B] (mol L⁻¹) | Initial rate (mol L⁻¹ s⁻¹) |
|---|---|---|
| 0.1 | 0.1 | 2 × 10⁻³ |
| 0.2 | 0.1 | 8 × 10⁻³ |
| 0.1 | 0.2 | 4 × 10⁻³ |
- Doubling [A] with [B] fixed multiplies the rate by 4, so the order in A is 2.
- Doubling [B] with [A] fixed doubles the rate, so the order in B is 1.
- Rate = k[A]²[B]; overall order 3.
- k = 2 × 10⁻³/(0.1² × 0.1) = 2 × 10⁻³/10⁻³ = 2 L² mol⁻² s⁻¹.
Example 4: activation energy
The rate constant of a reaction doubles when the temperature rises from 300 K to 310 K. Find Ea. (R = 8.314 J K⁻¹ mol⁻¹, log 2 = 0.301)
- log 2 = (Ea/2.303R) × (10/(300 × 310)).
- Ea = 2.303 × 8.314 × 0.301 × 9300 ≈ 53,600 J mol⁻¹ = 53.6 kJ mol⁻¹.
Practice MCQs
- The unit of k for a second-order reaction is: (a) L mol⁻¹ s⁻¹ (b) mol L⁻¹ s⁻¹ (c) s⁻¹ (d) L² mol⁻² s⁻¹
- The half-life of a zero-order reaction is: (a) independent of [A]₀ (b) proportional to [A]₀ (c) inversely proportional to [A]₀ (d) proportional to [A]₀²
- A first-order reaction has t½ = 10 min. The fraction left after 40 min is: (a) 1/4 (b) 1/8 (c) 1/16 (d) 1/32
- A reaction is of order one-half in A and 2 in B. Its overall order is: (a) 1 (b) 2 (c) 2.5 (d) 3
- A catalyst changes: (a) ΔH (b) K (c) ΔG (d) Ea
- Molecularity cannot be: (a) 1 (b) 2 (c) 3 (d) ½
- For a first-order reaction, a plot of log[A] against t has slope: (a) k (b) −k/2.303 (c) −k (d) −2.303k
- The time for 99.9% completion of a first-order reaction is about: (a) 2 t½ (b) 5 t½ (c) 10 t½ (d) 100 t½
Answers
- (a) (mol L⁻¹)¹⁻² s⁻¹ = L mol⁻¹ s⁻¹.
- (b) t½ = [A]₀/2k.
- (c) 40 min is 4 half-lives: (½)⁴ = 1/16.
- (c) ½ + 2 = 2.5.
- (d) It gives a lower-energy path; ΔH, ΔG and K stay the same.
- (d) Molecularity counts molecules, so it is a whole number.
- (b) log[A] = log[A]₀ − kt/2.303.
- (c) t = (2.303/k) log 1000 = 6.909/k, and 6.909/0.693 ≈ 10.
What to do next
- Fill in the zero- versus first-order table from memory, including the plots.
- Solve ten first-order questions using half-life counting before reaching for logarithms.
- Practise five initial-rates tables and five two-temperature Arrhenius questions.
- Revisit chemical thermodynamics to keep "spontaneous" and "fast" separate, and see the NEET chemistry plan for where this chapter fits.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the National Testing Agency website .
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