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Chemical Kinetics is the study of reaction rates and the step-by-step pathway by which reactants become products. For JEE you must master definitions, derive and apply integrated rate equations, interpret Arrhenius plots and avoid the very common graphical traps examiners set.
Chemical Kinetics JEE Notes: Important Concepts
Rate of reaction $$r$$: change in concentration of any species per unit time. For a general reaction $$aA+bB\rightarrow cC+dD$$
$$r=-\frac1a\frac{d[A]}{dt}=-\frac1b\frac{d[B]}{dt}=+\frac1c\frac{d[C]}{dt}=+\frac1d\frac{d[D]}{dt}$$
- Instantaneous rate: slope of the tangent to the concentration–time curve at that instant.
- Average rate: slope of the chord joining two points, useful for laboratory data tables.
- Rate law: experimentally determined relation $$r=k[A]^m[B]^n$$ where order $$m+n$$ need not match stoichiometry.
- Molecularity: number of particles colliding in an elementary step (always an integer ≤3). It is a theoretical concept, unlike order which is empirical.
- Factors affecting rate: concentration, temperature, nature & phase of reactants, presence of catalyst, surface area and radiation for photochemical reactions.
Collision and Transition-State views
The collision theory models bimolecular reactions by $$r=ZAB\,e^{-E_a/RT}$$ where $$ZAB$$ is the collision frequency. The better test model is the activated complex theory which introduces the Gibbs free energy of activation $$\Delta G^{\ddagger}$$ and explains catalytic action in terms of an alternative pathway with lower $$\Delta G^{\ddagger}$$.
Nomenclature pit-stop
| Term | Symbol | Usual unit (SI) |
|---|---|---|
| Rate constant | $$k$$ | Varies with order; for first order it is $$\text{s}^{-1}$$ |
| Activation energy | $$E_a$$ | J mol-1 |
| Half-life | $$t_{1/2}$$ | s (or min, hr as convenient) |
Rate Laws and Integrated Rate Equations
Zero, First and Second Order — derivations you must rehearse
| Order | Rate law | Integrated form | Half-life | Unit of $$k$$ |
|---|---|---|---|---|
| Zero | $$r=k$$ | $$[A]=[A]_0-kt$$ | $$t_{1/2}=\dfrac{[A]_0}{2k}$$ | mol L-1 s-1 |
| First | $$r=k[A]$$ | $$\ln\dfrac{[A]_0}{[A]}=kt$$ or $$[A]=[A]_0\,e^{-kt}$$ | $$t_{1/2}=\dfrac{\ln2}{k}$$ (independent of $$[A]_0$$) | s-1 |
| Second (same reactant) | $$r=k[A]^2$$ | $$\dfrac1{[A]}-\dfrac1{[A]_0}=kt$$ | $$t_{1/2}=\dfrac1{k[A]_0}$$ | L mol-1 s-1 |
Pseudo-first order: when one reactant is in large excess, e.g. hydrolysis of an ester in aqueous medium, the reaction behaves first order with effective rate constant $$k'=k[\text{H}_2\text{O}]$$.
$$t_{1/2}=\dfrac{\ln2}{k}\quad\text{and}\quad \ln k =\ln A-\dfrac{E_a}{RT}$$
Short cut graph tests
- Zero order: plot of $$[A]$$ vs $$t$$ is a straight line with slope $$-k$$.
- First order: plot of $$\ln[A]$$ vs $$t$$ is linear.
- Second order: plot of $$1/[A]$$ vs $$t$$ is linear.
Worked Example 1
The decomposition of $$N_2O$$ follows first order kinetics. Pressure falls from 600 torr to 200 torr in 230 s at 823 K. Calculate $$k$$ and $$t_{1/2}$$.
Rate law: $$\ln\dfrac{P_0}{P}=kt$$. $$\ln\dfrac{600}{200}=k(230) \Rightarrow \ln3=230k$$ $$k=\dfrac{1.099}{230}=4.78\times10^{-3}\,\text{s}^{-1}$$ Half-life: $$t_{1/2}=\dfrac{0.693}{k}=145\ \text{s}$$
Answer: $$k=4.78\times10^{-3}\,\text{s}^{-1},\;t_{1/2}=145\ \text{s}$$
After solving a couple of such variations, switch to the mixed difficulty filters under JEE Questions to cement your recognition speed.
Temperature Dependence and Arrhenius Equation
Arrhenius form
$$k=A\,e^{-E_a/RT}$$ where $$A$$ is the pre-exponential (frequency) factor.
Taking natural logs:
$$\ln k = \ln A -\dfrac{E_a}{R}\left(\dfrac1T\right)$$
- Slope $$=-E_a/R$$ when plotting $$\ln k$$ versus $$1/T$$.
- Rule of thumb: every 10 °C rise roughly doubles the rate for many reactions (origin: $$E_a\approx50\,$kJ mol-1$$).
Calculating $$E_a$$ from two-temperature data
$$\ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$$
Worked Example 2
For a reaction, $$k$$ is $$1.5\times10^{-3}\,\text{s}^{-1}$$ at 300 K and $$4.5\times10^{-3}\,\text{s}^{-1}$$ at 310 K. Find $$E_a$$.
$$\ln\dfrac{4.5\times10^{-3}}{1.5\times10^{-3}}=\ln3=1.099$$ $$\dfrac{1}{T_1}-\dfrac{1}{T_2}=\dfrac{1}{300}-\dfrac{1}{310}=1.075\times10^{-4}$$ $$E_a=\dfrac{1.099R}{1.075\times10^{-4}}= \dfrac{1.099\times8.314}{1.075\times10^{-4}}$$ $$E_a\approx 85\ \text{kJ mol}^{-1}$$
$$E_a=8.5\times10^{4}\ \text{J mol}^{-1}$$
Multi-concept numericals on $$E_a$$ + equilibrium frequently appear in Paper 2 of JEE Advanced; scan the trend in JEE Advanced Previous Papers to benchmark difficulty.
Catalysis insight
Catalysts create an alternate pathway with a lower $$E_a$$, thereby increasing both forward and reverse rates equally. They do not shift equilibrium, a popular JEE distractor.
Experimental Methods and Graphs for JEE Numericals
Initial rate method
- Run several experiments varying one reactant’s initial concentration.
- Keep other conditions identical and measure slope of the tangent at $$t\to0$$.
- Form ratio of rates to deduce reaction orders.
Integrated (graphical) method
Plot data sequentially as $$[A]$$ vs $$t$$, $$\ln[A]$$ vs $$t$$ and $$1/[A]$$ vs $$t$$. The straightest line reveals the order. Remember to check $$R^2$$ values if regression is allowed in the question stem.
Half-life method
If successive half-lives increase in an arithmetic series, the reaction is first order. For a second-order reaction each half-life doubles, and for zero order each is shorter than the previous. This is a quick elimination hack when only time data are given.
Worked Example 3
A reactant concentration drops from 0.10 M to 0.025 M in 40 min with successive half-lives of 20 min. Identify the order and evaluate $$k$$.
Successive half-lives equal ⇒ first order. Two half-lives (0.10→0.05→0.025) in 40 min are reported, agreeing with the data.
For first order, $$k=\dfrac{\ln2}{t_{1/2}}=\dfrac{0.693}{20\,\text{min}}=3.47\times10^{-2}\,\text{min}^{-1}$$
First order, $$k=3.5\times10^{-2}\,\text{min}^{-1}$$
Rate-order identification by raw data plotting has been a Mains favourite. Cross-solve at least 30 past problems using the graph option filter inside JEE Mains Previous Papers before the final week.
Important Formulas and Results at a Glance
| Topic | Key Formula | Remarks |
|---|---|---|
| Average rate | $$\Delta[\text{X}]/\Delta t$$ | Sign convention important |
| Instantaneous rate | $$-\dfrac{d[A]}{dt}$$ | Derivative at a point |
| Zero order $$t_x$$ (time for $$[A]_0$$ to become $$[A]_0-x$$) | $$t_x=\dfrac{x}{k}$$ | Linear decrease |
| First order percentage completion | $$k=\dfrac{2.303}{t}\log\dfrac{100}{100-\%\,\text{completion}}$$ | Use common log when calculators are restricted |
| Second order different reactants | $$\dfrac{1}{b-a}\ln\dfrac{a[B]}{b[A]}=kt$$ | $$A+B\rightarrow P$$ with $$[A]_0=a,\,[B]_0=b$$ |
| Relationship between $$k$$ and $$t_{0.75}$$ for first order | $$t_{0.75}=2t_{1/2}$$ | General: $$t_{f}=\dfrac{\ln(1/(1-f))}{k}$$ |
| Arrhenius in log10 form | $$\log k =\log A -\dfrac{E_a}{2.303RT}$$ | Helpful when only log tables are allowed |
| Temperature coefficient $$\phi$$ | $$\phi=\dfrac{k_{T+10}}{k_T}$$ | Typically 2–3 |
Download all chemistry formula PDFs in one go from the JEE Formula Sheets hub; rewrite them once in your own hand for muscle memory.
JEE Important Points, Common Mistakes and Quick Revision
- Order vs Molecularity mix-up: order is experimentally found, can be fractional or zero; molecularity is theoretical and never fractional.
- Dimension check: many traps rely on you assigning wrong unit to $$k$$ — always derive the unit before plugging numbers.
- Graph reading errors: JEE frequently flips the axes; confirm which quantity is on the Y-axis before inferring slope sign.
- Activation energy unit slip: Convert kJ to J when using $$R=8.314\text{ J mol}^{-1}\text{K}^{-1}$$.
- Surface reactions in solids are often zero order because the rate is limited by surface sites, not bulk concentration.
- For first order decay no initial concentration data needed if you have two concentration-time points; use $$k=\dfrac{2.303}{t_2-t_1}\log\dfrac{[A]_1}{[A]_2}$$.
- Quick fire checklist 24 h before exam: definitions, unit of $$k$$ for each order, three integrated equations, Arrhenius slope sign, catalyst effect statement.
Once these bullets feel obvious, attempt 10 mixed mock problems inside your personalised dashboard from our JEE Mains Online Coaching course to lock the concepts.
