Chemical Kinetics

Physical Chemistry Weightage: 7 Marks CBSE Unit 3

“Hi there! Chemical Kinetics is one of the most scoring chapters in CBSE Class 12. It carries 7 marks (about 10% of the theory paper) and almost every year gives you a numerical worth 3 marks, plus 1-mark MCQs and assertion-reason questions that are pure concept. The chapter is small and logical: once you are clear on order vs molecularity, the two integrated rate laws and the Arrhenius equation, you can score full marks. Watch the red trap boxes; they are exactly where the board sets its questions.”

— SCORECHEM ACADEMIC TEAM

1. Rate of a Reaction

Rate = change in concentration of a reactant or product per unit time (unit: mol L⁻¹ s⁻¹; for gases with pressure, atm s⁻¹).

Rate falls as reactant is used up. Average rate is the slope of the chord between two times; instantaneous rate is the slope of the tangent at one instant (the chord shrinks to a point as Δt → 0). The negative sign makes the rate positive.

Average rate: $r_{av}=-\dfrac{\Delta[R]}{\Delta t}=+\dfrac{\Delta[P]}{\Delta t}$ over a finite interval.
Instantaneous rate: $r_{inst}=-\dfrac{d[R]}{dt}=+\dfrac{d[P]}{dt}$ , the slope of the tangent at one instant.
Reactant rate carries a minus sign only because $\Delta[R]$ is negative; the rate itself is always positive.

Rate with stoichiometry: divide each species' rate by its coefficient. For $aA+bB\rightarrow cC+dD$ :

\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}

Example (2024 board): for $X+2Y\rightarrow P$ : $-\dfrac{d[Y]}{dt}=2\dfrac{d[P]}{dt}$ .

⚠️ Board Exam Trap: Rate of the reaction ≠ rate of disappearance of one reactant. In $2\text{HI}\rightarrow\text{H}_2+\text{I}_2$, the rate of HI disappearance is twice the rate of H₂ formation. "Rate of reaction" means the coefficient-divided value ($-\tfrac12\,d[\text{HI}]/dt$).

2. Rate Law, Order & Molecularity

Rate law: for $aA+bB\rightarrow$ products, $\text{Rate}=k[A]^x[B]^y$ . The powers $x$ , $y$ come from experiment, not from the balanced equation. $k$ is the rate constant (rate when all concentrations are 1 M).

Order = sum of powers $=x+y$ . It can be 0, a fraction or even negative. Example: $\text{Rate}=k[A]^{1/2}[B]^{3/2}$ is order 2; $\text{Rate}=k[P]^{1/2}[Q]^{1}$ is order 1.5.

Units of k follow from $k=\dfrac{\text{rate}}{[\,]^n}$ : $(\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}$

Order	Rate law	Unit of k
0	Rate $=k$	mol L⁻¹ s⁻¹
1	Rate $=k[A]$	s⁻¹
2	Rate $=k[A]^2$	mol⁻¹ L s⁻¹
3	Rate $=k[A]^3$	mol⁻² L² s⁻¹

Molecularity = number of species that collide simultaneously in an elementary step: unimolecular (1), bimolecular (2), termolecular (3). Never zero or fractional; beyond 3 is practically impossible.

	Order	Molecularity
Nature	Experimental	Theoretical
Values	0, 1, 2, 3, fraction, negative	1, 2, 3 only
Applies to	Elementary and complex reactions	Elementary steps only

Complex reactions: several elementary steps; the slowest step is the rate-determining step and its molecularity decides the overall order. Intermediates (e.g. IO⁻ in the H₂O₂/I⁻ reaction) appear in the mechanism but not in the overall equation.

Determining order from initial-rate data (board 3-marker): compare two experiments where only one concentration changes.

Example (2024 board): 2NO + Br₂ → 2NOBr. Expt 1→2: [Br₂] tripled, rate tripled ⇒ order in Br₂ = 1. Expt 1→3: [NO] tripled, rate ×9 ⇒ order in NO = 2. Rate $=k[\text{NO}]^2[\text{Br}_2]$ ; $k=\dfrac{1.0\times10^{-3}}{(0.05)^2(0.05)}=8.0\ \text{mol}^{-2}\text{L}^2\text{s}^{-1}$ .

Pseudo-first-order reaction: a higher-order reaction that behaves as first order because one reactant is in large excess (its concentration is practically constant). Examples: acid hydrolysis of ethyl acetate (water in excess) and inversion of cane sugar.

⚠️ Board Exam Trap: Do not read order from the balanced equation. $\text{H}_2+\text{Br}_2\rightarrow2\text{HBr}$ looks bimolecular but is a chain reaction (complex), so its molecularity is meaningless (2023 AR question). Similarly, $\text{H}_2+\text{Cl}_2\xrightarrow{h\nu}2\text{HCl}$ is zero order (rate depends on light intensity), and the reaction $\text{KClO}_3+6\text{FeSO}_4+3\text{H}_2\text{SO}_4\rightarrow\ldots$ is second order although its equation has 10 reactant species.

⚠️ Board Exam Trap: Volume and concentration changes. If the container volume of a gaseous elementary reaction is reduced to 1/3, every concentration triples. For $2A+B\rightarrow2C$ (rate $=k[A]^2[B]$) the rate becomes $3^3=27$ times (2026 board). The order does not change: it is a property of the rate law, not of the vessel.

3. Integrated Rate Laws & Half-Life

	Zero order	First order
Rate law	$-\dfrac{d[R]}{dt}=k$	$-\dfrac{d[R]}{dt}=k[R]$
Integrated	$k=\dfrac{[R]_0-[R]}{t}$	$k=\dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$
Straight-line plot	$[R]$ vs $t$	$\ln[R]$ vs $t$
Slope	$-k$	$-k$ (or $-k/2.303$ for log)
Half-life $t_{1/2}$	$\dfrac{[R]_0}{2k}$	$\dfrac{0.693}{k}$
$t_{1/2}$ depends on $[R]_0$ ?	Yes (proportional)	No

Straight-line tests: [R] vs t is linear for zero order; ln [R] vs t is linear for first order. Both have slope −k. Half-life is constant for first order but proportional to [R]₀ for zero order.

Where they occur: zero order: decomposition of NH₃ on hot platinum at high pressure (metal surface saturated), HI on gold, many enzyme-catalysed reactions. First order: radioactive decay, decomposition of N₂O₅, hydrogenation of ethene.

Useful first-order results:

Two-point form: $k=\dfrac{2.303}{t_2-t_1}\log\dfrac{[R]_1}{[R]_2}$
$t_{99.9\%}=10\,t_{1/2}$ and $t_{99\%}=2\,t_{90\%}$ (both are NCERT/board proofs).
Fraction left after $n$ half-lives $=(1/2)^n$ .
Worked example: 30% decomposition in 40 min: $k=\dfrac{2.303}{40}\log\dfrac{100}{70}=8.9\times10^{-3}\,\text{min}^{-1}$ , so $t_{1/2}=0.693/k\approx77.7$ min.
Gas-phase $A(g)\rightarrow B(g)+C(g)$ at constant volume: $k=\dfrac{2.303}{t}\log\dfrac{p_i}{2p_i-p_t}$ .

⚠️ Board Exam Trap: The curve that shows first order. A plot of $t_{1/2}$ against $[R]_0$ is a horizontal line for first order (2026 MCQ). Also, use $\ln$ vs $\log$ carefully: with $\log$, slope is $-k/2.303$. Half-life of first order needs no concentration at all.

4. Temperature & the Arrhenius Equation

For most reactions, a 10 K rise nearly doubles (or triples) the rate constant. Reason: the number of molecules with energy $\ge E_a$ increases sharply.

Maxwell–Boltzmann distribution. Raising the temperature by 10° flattens the curve and shifts the peak right; the area beyond E_a (fraction e^−E_a/RT) roughly doubles, so the rate roughly doubles. Total area stays 1.

k=A\,e^{-E_a/RT}\qquad\ln k=\ln A-\frac{E_a}{RT}\qquad\log k=\log A-\frac{E_a}{2.303\,RT}

$A$ = Arrhenius (frequency / pre-exponential) factor, related to collision frequency and orientation.
$e^{-E_a/RT}$ = fraction of molecules with energy equal to or greater than $E_a$ (2026 MCQ).
$E_a$ = minimum extra energy the reactants need to reach the activated complex (threshold energy minus average energy of reactants).

Arrhenius plot: ln k against 1/T is a straight line with negative slope −E_a/R and intercept ln A. Steeper line = higher activation energy = rate more sensitive to temperature.

Two-temperature form (most asked numerical):

\log\frac{k_2}{k_1}=\frac{E_a}{2.303\,R}\left[\frac{T_2-T_1}{T_1T_2}\right]

Worked example (2025 board): first-order reaction 50% complete in 20 min at 300 K and in 5 min at 350 K. Since $t_{1/2}\propto1/k$ : $k_2/k_1=20/5=4$ .

0.602=\frac{E_a}{2.303\times8.314}\times\frac{50}{300\times350}\ \Rightarrow\ E_a\approx24.2\text{ kJ mol}^{-1}

Shortcut: "rate doubles for 10 K rise at 298 K" gives $E_a\approx52.9$ kJ mol⁻¹ (NCERT 4.8).

⚠️ Board Exam Trap: Slope and intercept of the Arrhenius plot. For $\log k$ vs $1/T$, slope $=-E_a/(2.303R)$ and intercept $=\log A$ (2024 MCQ). For $\ln k$, slope is $-E_a/R$. Also: temperatures must be in kelvin, $E_a$ in joules when $R=8.314$.

5. Catalyst

A catalyst speeds up a reaction by forming an intermediate and giving an alternative path with lower $E_a$ . It is not consumed.

A catalyst offers an alternative path with a lower E_a (green dashed). Reactant and product levels are the same, so ΔH, ΔG and the equilibrium constant do not change; the catalyst speeds forward and backward reactions equally.

A catalyst changes: $E_a$ , the rate constant $k$ , the time to reach equilibrium.
A catalyst does NOT change: $\Delta H$ , $\Delta G$ , $\Delta S$ , the equilibrium constant, or whether a reaction is spontaneous (2023 MCQ: only $E_a$ is affected). It cannot start a non-spontaneous reaction.

6. Collision Theory

Reactant molecules must collide, but only effective collisions give products. A collision is effective if it has (i) energy $\ge$ threshold energy and (ii) proper orientation.

\text{Rate}=P\,Z_{AB}\,e^{-E_a/RT},\qquad k=P\,Z_{AB}\,e^{-E_a/RT}

$Z_{AB}$ = collision frequency (number of collisions per second per unit volume).
$P$ = probability (steric) factor for correct orientation.
Threshold energy = $E_a$ + average energy of reactants.

⚠️ Board Exam Trap: "Collision" does not mean "reaction". Only a tiny fraction of collisions is effective. Always mention both energy and orientation (e.g. CH₃Br + OH⁻: OH⁻ must attack the carbon from the side opposite to Br).

7. Formula Sheet & Last-Minute Checklist

Concept	Formula
Rate (general)	$-\frac{1}{a}\frac{d[A]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}$
Order	$x+y$ from $\text{Rate}=k[A]^x[B]^y$
Zero order	$k=\dfrac{[R]_0-[R]}{t}$ ; $t_{1/2}=\dfrac{[R]_0}{2k}$
First order	$k=\dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}$ ; $t_{1/2}=\dfrac{0.693}{k}$
Units of $k$	$(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$
Arrhenius	$k=Ae^{-E_a/RT}$
Two-temperature	$\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\dfrac{T_2-T_1}{T_1T_2}$
Fraction with $E\ge E_a$	$e^{-E_a/RT}$

Before the exam, check:

Can I find the order from initial-rate data by changing one concentration at a time?
Do I know the unit of $k$ for order 0, 1, 2 (and can I derive it)?
Can I state three differences between order and molecularity?
Can I derive $t_{1/2}$ for zero and first order in two lines?
Do I use kelvin and joules in the Arrhenius numerical?