Chemical kinetics is the branch of physical chemistry that studies the rates of chemical reactions and the factors that control them. For NEET 2026, it consistently contributes 2–3 questions per year, making it one of the most rewarding topics per hour of revision.

Rate of Reaction

For a general reaction $aA + bB \rightarrow cC + dD$, the rate is defined so that it is always positive:

$\text{rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$

Factors that increase the reaction rate include: higher concentration, higher temperature, increased pressure (for gaseous reactions), use of a catalyst, greater surface area, and more reactive nature of reactants. A 10 °C rise in temperature typically doubles the rate.

Rate Law and Order

The rate law is determined experimentally: $\text{rate} = k[A]^m[B]^n$. The exponents $m$ and $n$ are the orders with respect to A and B, and the overall order is $m + n$. These exponents are NOT stoichiometric coefficients — they must be measured in the laboratory.

Units of the rate constant $k$ are $(mol\,L^{-1})^{1-n}\,s^{-1}$, giving:

• Zero order: $mol\,L^{-1}\,s^{-1}$
• First order: $s^{-1}$
• Second order: $L\,mol^{-1}\,s^{-1}$

Order vs Molecularity

Order is experimental; it can be zero, fractional (e.g., 1.5), or any integer, and can change with conditions. Molecularity is the number of molecules that participate in an elementary step; it is always a positive integer (1, 2, or 3), never fractional or zero, and cannot change. Molecularity applies only to elementary (single-step) reactions, while order applies to both elementary and complex reactions. The rate-determining step (slowest step) in a multi-step mechanism controls the overall rate law.

Zero-Order Kinetics

Integrated rate equation: $[A] = [A]_0 - kt$

The plot of $[A]$ vs $t$ is a straight line with slope $= -k$.

$t_{1/2} = \frac{[A]_0}{2k} \quad \text{(depends on initial concentration)}$

Classic example: decomposition of $\text{NH}_3$ on a platinum surface at high pressure.

First-Order Kinetics

Integrated rate equation:

$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$

Half-life derivation: At $t_{1/2}$, $[A] = [A]_0/2$. Therefore $k \cdot t_{1/2} = \ln 2 = 0.693$, giving:

$\boxed{t_{1/2} = \frac{0.693}{k}} \quad \text{(independent of initial concentration)}$

Percentage-completion shortcuts (first order):

• 50% complete → $1 \times t_{1/2}$
• 75% complete → $2 \times t_{1/2}$
• 87.5% complete → $3 \times t_{1/2}$
• 93.75% complete → $4 \times t_{1/2}$

Examples: radioactive decay, $\text{H}_2\text{O}_2$ decomposition, $\text{N}_2\text{O}_5$ decomposition.

Worked example: $k = 6.93 \times 10^{-3}\,s^{-1}$

$t_{1/2} = \frac{0.693}{6.93 \times 10^{-3}} = 100\,s$

Time for 75% completion ($[A] = 0.25[A]_0$):

$t = \frac{2.303}{k}\log 4 = \frac{2.303}{6.93 \times 10^{-3}} \times 0.602 = 200\,s = 2 \times t_{1/2}$

Pseudo First-Order Reactions

When one reactant is present in large excess, its concentration remains essentially constant throughout the reaction, and the rate law simplifies to apparent first-order behaviour. Example: acid-catalysed hydrolysis of ethyl acetate in excess water:

$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$

True rate $= k[\text{ester}][\text{H}_2\text{O}]$, but since $[\text{H}_2\text{O}]$ is constant, pseudo-rate $= k'[\text{ester}]$ where $k' = k[\text{H}_2\text{O}]$.

Arrhenius Equation

$k = A\,e^{-E_a/RT}$

where $A$ is the pre-exponential (frequency) factor, $E_a$ is the activation energy, $R = 8.314\,J\,mol^{-1}\,K^{-1}$, and $T$ is absolute temperature.

Taking logarithms: $\ln k = \ln A - \dfrac{E_a}{RT}$, so a plot of $\ln k$ vs $1/T$ is a straight line with slope $= -E_a/R$.

Two-temperature form:

$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Worked example: $k_{300} = 2.5 \times 10^{-3}\,s^{-1}$; $k_{310} = 5.0 \times 10^{-3}\,s^{-1}$

$\log 2 = \frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300} - \frac{1}{310}\right) \implies E_a \approx 53.6\,kJ\,mol^{-1}$

Activation Energy and Catalysis

$E_a(\text{forward}) - E_a(\text{backward}) = \Delta H$

A catalyst provides an alternative reaction pathway with lower $E_a$. Crucially, a catalyst does not alter $\Delta H$ or the equilibrium constant $K$; it speeds up both forward and backward reactions equally.