JEE Main & Advanced

Chemical Kinetics

Chemical Kinetics for JEE Main & Advanced

1
Module 1

Rate, Order, and Integrated Rate Laws

Rate of Reaction, Rate Law, Order, MolecularityTopic 1

Rate of Reaction: Change in concentration of reactant or product per unit time.

For: $aA + bB \to cC + dD$: $$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$$

  • Reactants: $-$ (decreasing)
  • Products: $+$ (increasing)
  • Units: mol/L/s (M/s) or M/min

Average vs Instantaneous Rate:

  • Average: Over a finite time interval
  • Instantaneous: $dt \to 0$; tangent slope at a point

Rate Law (Differential): Relates rate to concentrations of reactants: $$\text{Rate} = k[A]^x[B]^y$$

  • $k$: rate constant; depends on T (not concentration)
  • $x, y$: orders w.r.t. A, B (determined experimentally)
  • Overall order = $x + y$

Order of Reaction:

  • Can be 0, 1, 2, 3, or fractional
  • Determined experimentally, NOT from stoichiometry
  • Cannot exceed 3 for elementary reactions

Examples:

  • $2N_2O_5 \to 4NO_2 + O_2$: rate = $k[N_2O_5]$ → first order
  • $2NO + Cl_2 \to 2NOCl$: rate = $k[NO]^2[Cl_2]$ → third order overall

Molecularity:

  • Number of reactant molecules in the elementary step
  • Always integer (1, 2, rarely 3)
  • Defined only for elementary reactions
  • e.g., $NH_4NO_2 \to N_2 + 2H_2O$ is unimolecular but kinetically can be first or zero order

Differences:

PropertyOrderMolecularity
DeterminedExperimentallyTheoretically
ValueCan be 0, fractionalAlways integer
Applies toOverall reaction (each step)Elementary step only
Can be 0YesNo

Units of Rate Constant:

  • Zero order: mol/L/s
  • First order: 1/s (or s⁻¹)
  • Second order: L/mol/s
  • nth order: $\text{(mol/L)}^{1-n} \cdot \text{s}^{-1}$
Worked Examples
1

For reaction $2A + B \to C$, rate $= k[A]^2[B]^0$. Find order overall and w.r.t. each species.

Show solution

Order w.r.t. A = 2; w.r.t. B = 0; overall order = 2.

Final Answer: Overall 2; A: 2; B: 0.

2

Find units of $k$ for first-order reaction.

Show solution

$\text{Rate} = k[A]$ $M/s = k \cdot M$ $k = 1/s = s^{-1}$.

Final Answer: $s^{-1}$ (or min⁻¹).

✎ Self-Check — 5 questions0 / 5
Q1.

Rate of reaction is expressed in:

Q2.

Order of reaction can be:

Q3.

Molecularity is:

Q4.

For zero order reaction, $k$ has units:

Q5.

Rate law is determined:

Integrated Rate Equations and Half-LifeTopic 2

Zero Order Reactions: Rate = $k$ (independent of concentration).

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

Plot of $[A]$ vs $t$: straight line, slope $= -k$, intercept $= [A]_0$.

Half-life ($t_{1/2}$): Time for $[A]$ to decrease to half: $$t_{1/2} = [A]_0/(2k)$$ Note: $t_{1/2}$ depends on initial concentration.

First Order Reactions: Rate = $k[A]$.

Integrated: $$\ln\frac{[A]_0}{[A]} = kt$$ $$[A] = [A]_0 e^{-kt}$$ $$\log[A] = \log[A]_0 - kt/2.303$$

Plot of $\log[A]$ vs $t$: straight line, slope $= -k/2.303$.

Half-life: $$t_{1/2} = \frac{0.693}{k}$$ Independent of initial concentration (characteristic of first order).

Second Order Reactions: Rate = $k[A]^2$.

Integrated: $$\frac{1}{[A]} - \frac{1}{[A]_0} = kt$$

Plot of $1/[A]$ vs $t$: straight line, slope $= k$.

Half-life: $t_{1/2} = 1/(k[A]_0)$ (depends on initial concentration).

Summary Table:

OrderRate eqIntegratedPlot for straight lineSlope$t_{1/2}$
0$\text{Rate} = k$$[A] = [A]_0 - kt$$[A]$ vs $t$$-k$$[A]_0/(2k)$
1$\text{Rate} = k[A]$$\ln([A]_0/[A]) = kt$$\log[A]$ vs $t$$-k/2.303$$0.693/k$
2$\text{Rate} = k[A]^2$$1/[A] - 1/[A]_0 = kt$$1/[A]$ vs $t$$k$$1/(k[A]_0)$

Pseudo-Order Reactions: Reactions with multiple reactants behave as lower order when one reactant is in large excess. e.g., hydrolysis of ester in water: $CH_3COOC_2H_5 + H_2O \to CH_3COOH + C_2H_5OH$

  • True rate = $k[ester][H_2O]$ (2nd order)
  • With excess water, $[H_2O]$ ≈ constant: rate = $k'[ester]$ (pseudo first order)
Worked Examples
1

First-order reaction; rate constant $k = 0.04$ min⁻¹. Find half-life.

Show solution

$t_{1/2} = 0.693/k = 0.693/0.04 = 17.3$ min.

Final Answer: $t_{1/2} = 17.3$ min.

2

$A \to B$ is first order; $A_0 = 0.5$ M. After $30$ min, $[A] = 0.1$ M. Find $k$ and $t_{1/2}$.

Show solution

$\ln([A]_0/[A]) = kt$ $\ln(0.5/0.1) = k \times 30$ $\ln 5 = 30k$ $1.609 = 30k \implies k = 0.0537$ min⁻¹. $t_{1/2} = 0.693/0.0537 = 12.9$ min.

Final Answer: $k = 0.054$ min⁻¹; $t_{1/2} = 12.9$ min.

✎ Self-Check — 5 questions0 / 5
Q1.

Half-life of first order reaction:

Q2.

For first-order, plot of $\log[A]$ vs $t$:

Q3.

Pseudo first order example:

Q4.

Half-life of zero order reaction depends on:

Q5.

Units of rate constant for second order:

2
Module 2

Temperature, Arrhenius, and Mechanism

Arrhenius Equation and Activation EnergyTopic 1

Temperature Effect on Rate:

  • Rate increases with $T$
  • Empirical rule: rate ~ doubles per $10°$C rise (rough)

Arrhenius Equation: $$k = A e^{-E_a/RT}$$

  • $A$: pre-exponential (frequency) factor; related to collision frequency
  • $E_a$: activation energy
  • $R$: gas constant ($8.314$ J/mol·K)
  • $T$: absolute temperature

Logarithmic Form: $$\ln k = \ln A - E_a/(RT)$$ $$\log k = \log A - E_a/(2.303 RT)$$

Plot of $\log k$ vs $1/T$: straight line, slope $= -E_a/(2.303R)$, intercept $= \log A$.

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

Activation Energy ($E_a$): Minimum energy reactants must have for successful reaction. Energy difference between reactants and transition state (activated complex).

Threshold energy: Total energy of activated complex. Activation energy = Threshold energy − Energy of reactants.

Energy Profile Diagram:

  • Endothermic reaction: products at higher energy
  • Exothermic reaction: products at lower energy
  • $E_a$ (forward) and $E_a$ (reverse) related to $\Delta H$:

$\Delta H = E_a^{forward} - E_a^{reverse}$

Worked Examples
1

$k$ doubles when T changes from $300$ K to $310$ K. Find $E_a$.

Show solution

$\log(k_2/k_1) = (E_a/2.303R)(T_2-T_1)/(T_1T_2)$ $\log 2 = (E_a/2.303 \times 8.314)(10/93000)$ $0.301 = E_a \times 5.62 \times 10^{-5}$ $E_a = 0.301/5.62 \times 10^{-5} = 5356 \times 10 = 53,560$ J $\approx 53.6$ kJ/mol.

Final Answer: $E_a \approx 53.6$ kJ/mol.

2

$k = 2.0 \times 10^{-4}$ s⁻¹ at $300$ K; $E_a = 100$ kJ/mol. Find $k$ at $310$ K.

Show solution

$\log(k_2/k_1) = (100000/2.303 \times 8.314)(10/93000) = 0.564$ $k_2/k_1 = 10^{0.564} = 3.66$ $k_2 = 3.66 \times 2 \times 10^{-4} = 7.3 \times 10^{-4}$ s⁻¹.

Final Answer: $k_2 = 7.3 \times 10^{-4}$ s⁻¹.

✎ Self-Check — 5 questions0 / 5
Q1.

Arrhenius equation:

Q2.

$E_a$ is:

Q3.

Plot of $\log k$ vs $1/T$:

Q4.

Rate $\approx$ doubles per:

Q5.

Activation energy for catalyzed reaction:

Collision Theory, Mechanism, CatalysisTopic 2

Collision Theory: For reaction to occur:

  1. Reactants must collide
  2. Collision must have sufficient energy ($\geq E_a$) — energy condition
  3. Collision must be in correct orientation — orientation condition

Rate $\propto$ (collision frequency) × (fraction with $E \geq E_a$) × (steric/orientation factor): $$\text{Rate} = Z \cdot e^{-E_a/RT} \cdot P$$

Arrhenius factor $A = ZP$ where $P$ is orientation/steric factor.

Reaction Mechanism:

  • Sequence of elementary steps that combine to give overall reaction
  • Rate-determining step (RDS): Slowest step; determines overall rate

For multi-step reactions, rate law is derived from RDS.

Example: $NO_2 + CO \to NO + CO_2$. Mechanism: Step 1 (slow): $NO_2 + NO_2 \to NO_3 + NO$ Step 2 (fast): $NO_3 + CO \to NO_2 + CO_2$

Rate law = $k[NO_2]^2$ (from RDS).

Pre-equilibrium Approximation: For a fast equilibrium followed by slow step, use $K_{eq}$ to eliminate intermediates.

Catalysis: Substance that increases rate without being consumed.

Types:

TypeDescriptionExample
HomogeneousSame phase as reactants$NO$ in lead chamber process
HeterogeneousDifferent phase (usually solid catalyst, gas/liquid reactants)$V_2O_5$ in contact process; Ni in hydrogenation
EnzymeBiological catalyst (protein)Amylase, pepsin

Properties of Catalysts:

  1. Don't appear in net equation
  2. Provide alternative pathway with lower $E_a$
  3. Same effect on forward and reverse rates (doesn't shift equilibrium)
  4. Small amount enough
  5. Catalyst-specific to certain reactions

Adsorption Theory (Heterogeneous Catalysis):

  1. Reactants adsorb on catalyst surface
  2. Adsorbed species more reactive (weak bonds form/break)
  3. Reaction occurs on surface
  4. Products desorb

Examples of Catalysts in Industry:

ReactionCatalyst
Haber process: $N_2 + 3H_2 \to 2NH_3$Fe (with $K_2O$, $Al_2O_3$ promoters)
Contact process: $SO_2 + O_2 \to SO_3$$V_2O_5$ (Pt earlier)
Ostwald process: $NH_3 \to NO$Pt gauze
Hydrogenation: $C=C \to C-C$Ni, Pt, Pd
CrackingZeolites, $SiO_2$-$Al_2O_3$
PolymerizationZiegler-Natta ($TiCl_4$ + $Al(C_2H_5)_3$)

Promoters: Increase activity of catalyst (e.g., $K_2O$, $Al_2O_3$ promote $Fe$ in Haber).

Poisons: Decrease activity (e.g., $As$, $S$ poison Pt).

Worked Examples
1

Catalyst increases rate of reaction. Why doesn't it shift equilibrium?

Show solution

A catalyst provides an alternate path with lower $E_a$ for both forward and reverse reactions equally. It increases both rates by the same factor. At equilibrium, rates are still equal (faster but equal). So equilibrium position ($K_c$) is unchanged. Only the time to reach equilibrium decreases.

Final Answer: Lowers $E_a$ of both directions equally; $K$ unchanged.

2

For mechanism: $A \xrightarrow{k_1, \text{slow}} I \xrightarrow{k_2, \text{fast}} P$. What is the rate law?

Show solution

The slow step (RDS) is the formation of intermediate I from A. Therefore: Rate = $k_1[A]$ (first order in A, depends only on RDS).

Final Answer: Rate = $k_1[A]$; first order.

✎ Self-Check — 5 questions0 / 5
Q1.

Collision theory requires:

Q2.

Catalyst:

Q3.

Enzymes are catalysts of:

Q4.

In multi-step mechanism, rate = rate of:

Q5.

Promoter is:

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