Chemical Kinetics
Chemical Kinetics for JEE Main & Advanced
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:
| Property | Order | Molecularity |
|---|---|---|
| Determined | Experimentally | Theoretically |
| Value | Can be 0, fractional | Always integer |
| Applies to | Overall reaction (each step) | Elementary step only |
| Can be 0 | Yes | No |
Units of Rate Constant:
- Zero order: mol/L/s
- First order: 1/s (or s⁻¹)
- Second order: L/mol/s
- nth order: $\text{(mol/L)}^{1-n} \cdot \text{s}^{-1}$
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.
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).
Rate of reaction is expressed in:
Order of reaction can be:
Molecularity is:
For zero order reaction, $k$ has units:
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:
| Order | Rate eq | Integrated | Plot for straight line | Slope | $t_{1/2}$ |
|---|---|---|---|---|---|
| 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)
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.
$A \to B$ 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.
Half-life of first order reaction:
For first-order, plot of $\log[A]$ vs $t$:
Pseudo first order example:
Half-life of zero order reaction depends on:
Units of rate constant for second order:
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 $\Delta H$:
$\Delta H = E_a^{forward} - E_a^{reverse}$
$k$ 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.
$k = 2.0 \times 10^{-4}$ 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.
Arrhenius equation:
$E_a$ is:
Plot of $\log k$ vs $1/T$:
Rate $\approx$ doubles per:
Activation energy for catalyzed reaction:
Collision Theory, Mechanism, CatalysisTopic 2
Collision Theory: For reaction to occur:
- Reactants must collide
- Collision must have sufficient energy ($\geq E_a$) — energy condition
- 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:
| Type | Description | Example |
|---|---|---|
| Homogeneous | Same phase as reactants | $NO$ in lead chamber process |
| Heterogeneous | Different phase (usually solid catalyst, gas/liquid reactants) | $V_2O_5$ in contact process; Ni in hydrogenation |
| Enzyme | Biological catalyst (protein) | Amylase, pepsin |
Properties of Catalysts:
- Don't appear in net equation
- Provide alternative pathway with lower $E_a$
- Same effect on forward and reverse rates (doesn't shift equilibrium)
- Small amount enough
- Catalyst-specific to certain reactions
Adsorption Theory (Heterogeneous Catalysis):
- Reactants adsorb on catalyst surface
- Adsorbed species more reactive (weak bonds form/break)
- Reaction occurs on surface
- Products desorb
Examples of Catalysts in Industry:
| Reaction | Catalyst |
|---|---|
| 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 |
| Cracking | Zeolites, $SiO_2$-$Al_2O_3$ |
| Polymerization | Ziegler-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).
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.
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.
Collision theory requires:
Catalyst:
Enzymes are catalysts of:
In multi-step mechanism, rate = rate of:
Promoter is:
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