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WBB • Class XII • Chemistry • Ch 8
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Chemical Kinetics

Chemical Kinetics is the central branch of physical chemistry dedicated to the quantitative investigation of chemical reaction velocities, the mechanistic sequences through which transformations proceed, and the energetic and environmental factors that dictate macroscopic reaction dynamics. While classical thermodynamics dictates whether a reaction is energetically feasible via negative Gibbs free energy changes, kinetics determines the actual timescale and molecular pathway required to achieve transformation. The discipline begins with the rigorous formulation of average and instantaneous reaction rates, structured by stoichiometric differential coefficients. The core analytical framework establishes the differential rate law, distinguishing between overall reaction order, an empirically determined quantity that may be zero, fractional, integer, or negative, and molecularity, a theoretical integer reflecting colliding species in an elementary step. Integration of differential rate laws yields exact mathematical expressions governing zero-order and first-order processes, elucidating the unique invariance of first-order half-lives with initial reactant concentration. Advanced kinetic analysis incorporates the Arrhenius equation, quantifying the exponential dependence of specific rate constants upon temperature through threshold activation energies and pre-exponential steric frequency factors. Transition state theory, activated complex formation, catalytic pathway lowering, simple collision models, and the natural first-order exponential kinetics of radioactive decay provide students with a unified theoretical foundation.

Why This Chapter Matters

Chemical kinetics governs virtually every natural and engineered transformation across industrial chemical synthesis, pharmacokinetics, atmospheric environmental monitoring, materials engineering, and renewable energy conversion. In industrial process engineering, the Haber-Bosch ammonia synthesis and contact sulfuric acid manufacturing rely on optimized temperature, pressure, and heterogeneous catalysts to achieve economically viable yields within acceptable operational time windows without destroying thermal equilibria. In medicinal pharmacology, kinetic profiling dictates drug delivery rates, bioavailability, metabolic clearance, and half-life dosages within the human bloodstream. In atmospheric science, understanding the reaction rates of chlorofluorocarbon radicals with ozone dictates global atmospheric remediation policies. Furthermore, nuclear geochronology and radiocarbon dating depend strictly on the invariant first-order radioactive decay kinetics of unstable isotopes like Carbon-14. In the WBCHSE Class 12 board examination, chemical kinetics represents a high-scoring foundational pillar comprising graphical order determinations, integrated rate mathematical derivations, half-life relationships, and multi-step Arrhenius activation energy calculations.

Chapter Roadmap & Progression

1 Module 1: Rate of Reaction, Stoichi...
2 Module 2: Rate Law, Rate Constant,...
3 Module 3: Integrated Rate Equations...
4 Module 4: Experimental Methods for...
5 Module 5: Temperature Dependence of...
6 Module 6: Collision Theory of React...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Rate of Reaction, Stoichiometric Expressions & Rate Factors

1.1 Definition and Mathematical Formulation of Reaction Rate

The rate of a chemical reaction is defined as the change in the molar concentration of any one of the reactants or products per unit time. Because reactant concentrations diminish while product concentrations accumulate as time elapses, a negative sign is appended to reactant terms to ensure the reaction rate is always a positive quantity:

  • Average Rate ($r_{\text{av}}$): Evaluated over a macroscopic finite time interval $\Delta t$:
    $$r_{\text{av}} = -\frac{\Delta [R]}{\Delta t} = +\frac{\Delta [P]}{\Delta t}$$
  • Instantaneous Rate ($r_{\text{inst}}$): The limiting value of the average rate as the time interval approaches infinitesimal zero ($\Delta t \to 0$). Graphically, it represents the slope of the tangent drawn to the concentration-versus-time curve at that specific instant $t$:
    $$r_{\text{inst}} = -\frac{d[R]}{dt} = +\frac{d[P]}{dt}$$
1.2 Stoichiometric Rate Relationships

When the stoichiometric coefficients of reactants and products are unequal, the individual rates of disappearance and appearance differ. For a general homogenous chemical reaction:

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

Example: For the synthesis of ammonia, $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$:
$$\text{Rate} = -\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = +\frac{1}{2}\frac{d[\text{NH}_3]}{dt}$$ Units of rate: $\text{mol L}^{-1}\text{ s}^{-1}$ or $\text{mol dm}^{-3}\text{ s}^{-1}$; for gaseous reactions, $\text{atm s}^{-1}$ or $\text{bar s}^{-1}$.

1.3 Factors Influencing Chemical Reaction Rates
Influencing Factor Observed Kinetic Impact Molecular / Physical Rationale
Concentration / Pressure Rate increases with concentration Higher particle density escalates the collision frequency per unit volume per second ($Z_{AB}$).
Temperature Rate doubles or triples for every $10^\circ\text{C}$ rise Significantly expands the fraction of reactant molecules possessing energy equal to or exceeding activation energy ($e^{-E_a/RT}$). Temperature coefficient $\mu = k_{T+10} / k_T \approx 2\text{--}3$.
Catalyst Dramatically accelerates reaction rate Provides an alternate reaction pathway characterized by a lower activation energy ($E_a' < E_a$), without shifting chemical equilibrium or altering $\Delta H$.
Surface Area (Heterogeneous) Finely divided solids react much faster Subdividing a solid exposes more active interfacial collision sites per unit mass to attacking fluid molecules.

Module 2: Rate Law, Rate Constant, Order of Reaction & Molecularity

2.1 The Rate Law and Specific Rate Constant ($k$)

The Rate Law is an experimentally deduced mathematical equation correlating the observed reaction rate with the molar concentrations of reactants raised to empirical powers:

$$\text{Rate} = k[A]^x[B]^y$$
  • The exponents $x$ and $y$ represent the orders of reaction with respect to reactants $A$ and $B$, respectively. They cannot be deduced from the stoichiometric coefficients of a balanced overall equation; they must be determined empirically.
  • Specific Reaction Rate Constant ($k$): The proportionality constant representing the reaction rate when the molar concentration of each reactant is unity ($[A] = [B] = 1\text{ mol L}^{-1}$).
  • Dimensional Units of Rate Constant: For an $n^{\text{th}}$-order reaction, the unit of $k$ depends strictly on $n$:
    $$\text{Unit of } k = \frac{\text{Rate}}{(\text{Concentration})^n} = \frac{\text{mol L}^{-1}\text{ s}^{-1}}{(\text{mol L}^{-1})^n} = (\text{mol L}^{-1})^{1-n}\text{ s}^{-1} = \text{mol}^{1-n}\text{ L}^{n-1}\text{ s}^{-1}$$
2.2 Order vs Molecularity of Reaction
Parameter Order of Reaction Molecularity of Reaction
Definition Sum of exponents of concentration terms ($x + y$) in the experimental rate law equation. Number of reacting species that must collide simultaneously to bring about an elementary chemical step.
Nature of Value Purely experimental quantity; can be zero, integer ($1, 2, 3$), fractional, or negative. Theoretical quantity; can only be a positive non-zero whole integer ($1, 2, 3$). Never zero, fractional, or $>3$.
Applicability Applicable to both elementary and complex (multi-step) chemical reactions. Meaningful strictly for elementary reactions; has no physical significance for complex overall reactions.
Pseudo-First-Order Reactions (ছদ্ম-একম আণবিক বিক্রিয়া):
Reactions that possess a molecularity of two or higher, but adhere mathematically to first-order kinetics because one reacting partner is maintained in massive stoichiometric excess such that its concentration remains virtually unchanged throughout the reaction.
১. Acid-catalyzed Hydrolysis of Ethyl Acetate:
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$$ Rate $= k'[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]$. Because $[\text{H}_2\text{O}] \gg [\text{ester}]$, we define $k = k'[\text{H}_2\text{O}]$, simplifying the rate law to Rate $= k[\text{CH}_3\text{COOC}_2\text{H}_5]$.
২. Inversion of Cane Sugar: $\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 (\text{glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{fructose})$.

Module 3: Integrated Rate Equations & Half-Life Kinetics (Zero & First Order)

3.1 Zero-Order Reaction Kinetics ($n = 0$)

A reaction is zero-order when its rate is completely independent of the concentration of reactants: $$-\frac{d[R]}{dt} = k[R]^0 = k \implies d[R] = -k\,dt$$ Integrating between boundaries $[R]_0$ at $t = 0$ to $[R]_t$ at time $t$:

$$[R]_t = [R]_0 - kt \iff k = \frac{[R]_0 - [R]_t}{t}$$
  • Units of $k$: $\text{mol L}^{-1}\text{ s}^{-1}$ (identical to the units of reaction rate).
  • Graphical Plot: Plotting $[R]_t$ versus $t$ gives a straight line with slope $= -k$ and y-intercept $= [R]_0$.
  • Half-Life ($t_{1/2}$): The time required for reactant concentration to decrease to half its initial value ($[R]_t = [R]_0 / 2$):
    $$\frac{[R]_0}{2} = [R]_0 - k t_{1/2} \implies t_{1/2} = \frac{[R]_0}{2k} \quad (t_{1/2} \propto [R]_0)$$
  • Total Time for Complete Consumption ($t_{100\%}$): $t_{100\%} = \frac{[R]_0}{k} = 2 \times t_{1/2}$.
  • Classical Examples: Thermal decomposition of $\text{NH}_3$ on hot platinum surface at high pressure ($2\text{NH}_3 \xrightarrow{\text{Pt}} \text{N}_2 + 3\text{H}_2$), photochemical chlorination of methane over water.
3.2 First-Order Reaction Kinetics ($n = 1$)

A reaction is first-order when its rate is directly proportional to the first power of reactant concentration: $$-\frac{d[R]}{dt} = k[R] \implies \frac{d[R]}{[R]} = -k\,dt$$ Integrating between limits $[R]_0$ at $t = 0$ to $[R]_t$ at time $t$:

$$\ln\left(\frac{[R]_0}{[R]_t}\right) = kt \iff k = \frac{2.303}{t}\log_{10}\left(\frac{[R]_0}{[R]_t}\right) = \frac{2.303}{t}\log_{10}\left(\frac{a}{a - x}\right)$$ $$[R]_t = [R]_0 \cdot e^{-kt} \quad (\text{Exponential decay form})$$
  • Units of $k$: $\text{time}^{-1}$ ($ ext{s}^{-1}, \text{min}^{-1}, \text{h}^{-1}$); completely free of concentration units.
  • Graphical Plots:
    1. $\log_{10}[R]_t$ vs $t$: Straight line with slope $= -\frac{k}{2.303}$ and intercept $= \log_{10}[R]_0$.
    2. $\log_{10}([R]_0 / [R]_t)$ vs $t$: Straight line passing through the origin with slope $= +\frac{k}{2.303}$.
  • Half-Life Period ($t_{1/2}$): Setting $[R]_t = [R]_0 / 2$:
    $$t_{1/2} = \frac{2.303}{k}\log_{10}(2) = \frac{2.303 \times 0.30103}{k} = \frac{0.693}{k}$$ Key Characteristic: The half-life of a first-order reaction is completely independent of the initial concentration $[R]_0$.
  • Critical Board Proofs:
    $t_{99.9\%} = \frac{2.303}{k}\log_{10}\left(\frac{100}{100 - 99.9}\right) = \frac{2.303}{k}\log_{10}(10^3) = \frac{2.303 \times 3}{k} = \frac{6.909}{k} \approx 10 \times t_{1/2}$.
    $t_{75\%} = 2 \times t_{1/2}$.
  • Gas-Phase First-Order Kinetics: For $A(g) \rightarrow B(g) + C(g)$ starting with initial pressure $P_i$ and reaching total pressure $P_t$ at time $t$:
    $$k = \frac{2.303}{t}\log_{10}\left(\frac{P_i}{2P_i - P_t}\right)$$

Module 4: Experimental Methods for Determining Reaction Order

4.1 Initial Rate Method (প্রাথমিক হার পদ্ধতি)

The most versatile laboratory technique for complex multi-reactant reactions. Systematic experiments are conducted where the initial concentration of one reactant is varied (e.g., doubled or tripled) while maintaining the initial concentrations of all other reactants constant:

$$r_1 = k[A]_1^x[B]_1^y, \quad r_2 = k[A]_2^x[B]_1^y \implies \frac{r_1}{r_2} = \left(\frac{[A]_1}{[A]_2}\right)^x$$

Taking logarithms yields $x = \frac{\log(r_1 / r_2)}{\log([A]_1 / [A]_2)}$. Repeating for reactant $B$ isolates exponent $y$.

4.2 Half-Life Method (অর্ধায়ু পদ্ধতি)

For any general $n^{\text{th}}$-order reaction, the half-life depends inversely on the initial concentration raised to power $(n - 1)$:

$$t_{1/2} \propto \frac{1}{[R]_0^{n-1}} \implies \frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{[R]_{0,2}}{[R]_{0,1}}\right)^{n-1}$$ $$n = 1 + \frac{\log\left[\frac{(t_{1/2})_1}{(t_{1/2})_2}\right]}{\log\left[\frac{[R]_{0,2}}{[R]_{0,1}}\right]}$$
4.3 Ostwald's Isolation Method (অসওয়াল্ডের পৃথকীকরণ পদ্ধতি)

When a reaction involves multiple reactants (e.g., $A + B + C \rightarrow \text{products}$), all reactants except one are taken in overwhelming excess. If $[B]$ and $[C]$ are huge relative to $[A]$, their concentrations remain virtually constant, reducing the effective rate law to $\text{Rate} = k'[A]^x$, allowing the isolation of order $x$. Repeating by isolating $B$ and $C$ in turn determines the complete order $n = x + y + z$.

Module 5: Temperature Dependence of Reaction Rates, Activation Energy & Arrhenius Theory

5.1 Concept of Activation Energy & Transition State Theory

Molecules must collide to react, but simple collision is insufficient. Reacting species must collide with a minimum threshold kinetic energy ($E_{\text{th}}$) and an appropriate spatial orientation.

  • Threshold Energy ($E_{\text{th}}$): The minimum energy that colliding molecules must possess to produce an effective chemical collision resulting in product formation.
  • Activation Energy ($E_a$): The excess energy over and above the ground-state average kinetic energy of the reactants ($E_R$) that must be supplied to reach the transition state:
    $$E_a = E_{\text{th}} - E_R$$
  • Activated Complex (Transition State): An unstable, high-energy intermediate configuration where old reactant bonds are partially breaking while new product bonds are partially forming simultaneously.
5.2 The Arrhenius Equation

In 1889, Svante Arrhenius formulated the quantitative exponential relationship linking temperature and reaction rate constant:

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

where $A$ is the pre-exponential factor (frequency factor), $E_a$ is the activation energy in $\text{J mol}^{-1}$, $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$ is the universal gas constant, and $T$ is absolute temperature in Kelvin.
The term $e^{-E_a / RT}$ represents the Boltzmann fraction of molecules possessing kinetic energy equal to or greater than $E_a$.

5.3 Logarithmic Forms and Graphical Evaluation
$$\ln k = \ln A - \frac{E_a}{RT} \iff \log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$$
  • Arrhenius Plot: A plot of $\log_{10} k$ versus $1/T$ yields a straight line with: $$\text{Slope} = -\frac{E_a}{2.303 R} \implies E_a = -2.303 R \times \text{Slope}$$ $$\text{Y-Intercept} = \log_{10} A$$
  • Two-Temperature Form: If the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$ ($T_2 > T_1$):
    $$\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left[\frac{T_2 - T_1}{T_1 T_2}\right] = \frac{E_a}{2.303 R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$$
Catalytic Action on Energy Profile:
A catalyst accelerates a chemical reaction by providing an alternative transition state characterized by a significantly lower activation energy ($E_a' < E_a$). Because $E_a'$ appears in the exponent of $e^{-E_a/RT}$, a modest decrease in $E_a$ yields a massive exponential multiplication of the rate constant. Catalysts speed up forward and reverse reactions by the exact same factor, preserving $\Delta H$, $\Delta G$, and equilibrium constant $K_c$ completely unchanged.

Module 6: Collision Theory of Reactions & Radioactive Decay Kinetics

6.1 Collision Theory of Chemical Reactions

Proposed by Max Trautz and William Lewis, collision theory is based on the kinetic theory of gases. The rate of a bimolecular elementary reaction $A + B \rightarrow \text{products}$ is given by:

$$\text{Rate} = Z_{AB} \cdot e^{-E_a / RT} \quad \text{and factoring steric orientation } P: \quad k = P \cdot Z_{AB} \cdot e^{-E_a / RT}$$
  • Collision Frequency ($Z_{AB}$): The number of collisions per second per unit volume of the reaction mixture (typically $10^{25}\text{ to }10^{28}\text{ collisions s}^{-1}\text{ cm}^{-3}$).
  • Steric / Probability Factor ($P$): Accounts for the proper spatial orientation of colliding molecules during contact. Simple diatomic molecules have $P \approx 1$, while complex organic molecules have $P \ll 1$.
  • Limitations: Assumes reacting molecules are rigid hard spheres, completely ignoring vibrational-rotational modes and electronic structural distortions.
6.2 Kinetics of Radioactive Disintegration

All natural and artificial radioactive decay processes adhere strictly to first-order kinetics. The rate of nuclear disintegration at any instant is directly proportional to the total number of active nuclei ($N$) present:

$$-\frac{dN}{dt} = \lambda N \implies \lambda = \frac{2.303}{t}\log_{10}\left(\frac{N_0}{N_t}\right), \quad N_t = N_0 \cdot e^{-\lambda t}$$ $$t_{1/2} = \frac{0.693}{\lambda}, \quad \text{Average Life } (\tau) = \frac{1}{\lambda} = \frac{t_{1/2}}{0.693} \approx 1.443 \times t_{1/2}$$

where $\lambda$ is the decay constant (disintegration constant), $N_0$ is the initial radioactive nuclei population, and $N_t$ is the surviving nuclei population at elapsed time $t$.

Key Formulas, Reactions & Definitions

Differential and Stoichiometric Rate of Reaction
$$[\text{Rate}] = \text{mol L}^{-1}\text{ s}^{-1} \text{ or } \text{atm s}^{-1}$$
Divide rate of consumption/formation of each species by its balanced stoichiometric coefficient.
Units of Specific Rate Constant (k)
$$n=0 \implies \text{mol L}^{-1}\text{ s}^{-1}; \quad n=1 \implies \text{s}^{-1}$$
A rate constant with units of s^-1 always indicates an overall first-order reaction.
Integrated Rate Equation for Zero-Order Reactions
$$t_{100\%} = \frac{[R]_0}{k} = 2 \times t_{1/2}$$
Plot of [R] vs t gives a straight line with slope = -k and intercept = [R]0.
Integrated Rate Equation for First-Order Reactions
$$[R]_t = [R]_0 \cdot e^{-kt}$$
For gas-phase first-order: k = (2.303/t) * log[Pi / (2Pi - Pt)].
First-Order Half-Life and Fractional Life Relationships
$$t_{99\%} = 2 \times t_{90\%}, \quad t_{99.9\%} = 10 \times t_{1/2}$$
If doubling initial concentration leaves half-life unchanged, the reaction is strictly first-order.
Arrhenius Equation & Two-Temperature Expression
$$\text{Slope of } \log_{10} k \text{ vs } 1/T = -\frac{E_a}{2.303 R}$$
Ensure Ea is converted to Joules per mole (J/mol) before pairing with R = 8.314 J/(K mol).

Conceptual Solved Examples & Case Studies

Example 1
(a) What is a pseudo-first-order reaction? Give an example. (b) Why is molecularity never zero or fractional?
Step-by-Step Solution:
(a) Pseudo-First-Order Reaction (1 Mark):
A chemical reaction having a molecularity of two or more, but adhering mathematically to first-order kinetics because one of the reacting species is present in large stoichiometric excess so that its concentration remains practically constant throughout the reaction.
Example: Acid-catalyzed hydrolysis of ethyl acetate:
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$$ $$\text{Rate} = k[\text{CH}_3\text{COOC}_2\text{H}_5] \quad (\text{where } k = k'[\text{H}_2\text{O}])$$
(b) Why Molecularity Cannot Be Zero or Fractional (1 Mark):
Molecularity represents the actual physical count of discrete atoms, ions, or molecules that must collide simultaneously in an elementary step. Because collision cannot occur without at least one particle (molecularity cannot be zero) and fractions of molecules cannot participate in collisions (molecularity cannot be fractional), molecularity can only be a non-zero positive whole integer ($1, 2, 3$).
Example 2
For Rate = k[A][B]^2: find overall order, units of k, and rate factor when volume is reduced to 1/3.
Step-by-Step Solution:
(a) Overall Order of Reaction (1 Mark):
$$\text{Overall Order } n = x + y = 1 + 2 = \mathbf{3} \quad (\text{Third-order reaction})$$
(b) Units of Rate Constant $k$ (1 Mark):
$$\text{Unit of } k = (\text{mol L}^{-1})^{1-n}\text{ s}^{-1} = (\text{mol L}^{-1})^{1-3}\text{ s}^{-1} = (\text{mol L}^{-1})^{-2}\text{ s}^{-1} = \mathbf{\text{L}^2\text{ mol}^{-2}\text{ s}^{-1}}$$
(c) Effect of Volume Reduction (1 Mark):
Concentration is inversely proportional to volume ($C = n/V$). When volume is reduced to $1/3$, the concentration of every gaseous/solution species increases by a factor of 3:
$$[A]' = 3[A], \quad [B]' = 3[B]$$
New rate $r' = k[A]'[B]'^2 = k(3[A])(3[B])^2 = k(3[A])(9[B]^2) = 27 \cdot k[A][B]^2 = \mathbf{27 \cdot r}$
Answer: The reaction rate increases by a factor of 27.
Example 3
First-order reaction with k = 1.15 x 10^-3 s^-1. Calculate t_1/2 and time to reduce 5 g to 3 g.
Step-by-Step Solution:
(a) Calculation of Half-Life ($t_{1/2}$) (1.5 Marks):
For a first-order reaction:
$$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1.15 \times 10^{-3}\text{ s}^{-1}} = \mathbf{602.6\text{ seconds}} \quad (\text{or } 10.04\text{ minutes})$$
(b) Calculation of Time Elapsed ($t$) (1.5 Marks):
Initial amount $[R]_0 = 5.0\text{ g}$, Remaining amount $[R]_t = 3.0\text{ g}$.
Using the first-order integrated rate law:
$$t = \frac{2.303}{k}\log_{10}\left(\frac{[R]_0}{[R]_t}\right)$$
$$t = \frac{2.303}{1.15 \times 10^{-3}}\log_{10}\left(\frac{5.0}{3.0}\right) = \frac{2.303}{1.15 \times 10^{-3}}\log_{10}(1.667)$$
Since $\log_{10}(1.667) = 0.2219$:
$$t = \frac{2.303 \times 0.2219}{1.15 \times 10^{-3}} = \frac{0.5110}{1.15 \times 10^{-3}} = \mathbf{444.35\text{ seconds}} \quad (\text{or } 7.41\text{ minutes})$$
Answer: (a) $t_{1/2} = 602.6\text{ s}$; (b) Time required $= 444.4\text{ s}$.
Example 4
(a) Prove t_99.9% = 10 * t_1/2 for first order. (b) Find order wrt A and B from kinetic data table.
Step-by-Step Solution:
(a) Proof of $t_{99.9\%} = 10 \times t_{1/2}$ (2 Marks):
For a first-order reaction, $t = \frac{2.303}{k}\log_{10}\left(\frac{a}{a - x}\right)$.
When $99.9\%$ of the reaction is complete, $x = 0.999a \implies a - x = a - 0.999a = 0.001a = 10^{-3}a$.
$$t_{99.9\%} = \frac{2.303}{k}\log_{10}\left(\frac{a}{10^{-3}a}\right) = \frac{2.303}{k}\log_{10}(10^3) = \frac{2.303 \times 3}{k} = \frac{6.909}{k}$$
We know that the half-life is $t_{1/2} = \frac{0.693}{k}$.
Taking the ratio:
$$\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} = \frac{6.909}{0.693} = 9.9697 \approx \mathbf{10}$$
Hence proved: $t_{99.9\%} \approx 10 \times t_{1/2}$.

(b) Determination of Rate Law by Initial Rate Method (2 Marks):
Let Rate $= k[A]^x[B]^y$.
From Exp 2 and Exp 3 (where $[A]$ is held constant at $0.3\text{ M}$):
$$\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \left(\frac{0.4}{0.2}\right)^y \implies 4 = 2^y \implies 2^2 = 2^y \implies \mathbf{y = 2}$$
From Exp 1 and Exp 2:
$$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{7.2 \times 10^{-2}}{6.0 \times 10^{-3}} = \left(\frac{0.3}{0.1}\right)^x \left(\frac{0.2}{0.1}\right)^2$$
$$12 = (3)^x \times (2)^2 = (3)^x \times 4 \implies (3)^x = \frac{12}{4} = 3 \implies \mathbf{x = 1}$$
Rate Law: $\mathbf{\text{Rate} = k[A][B]^2}$ (Order with respect to $A$ is 1, with respect to $B$ is 2).
Example 5
(a) Reaction rate triples from 293 K to 313 K; calculate Ea. (b) Find k for gas phase decomposition of N2O5.
Step-by-Step Solution:
(a) Calculation of Activation Energy ($E_a$) (2.5 Marks):
Given: $T_1 = 293\text{ K}$, $T_2 = 313\text{ K}$, $k_2 / k_1 = 3$, $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$.
Using the two-temperature Arrhenius equation:
$$\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left[\frac{T_2 - T_1}{T_1 T_2}\right]$$
$$\log_{10}(3) = \frac{E_a}{2.303 \times 8.314}\left[\frac{313 - 293}{293 \times 313}\right]$$
$$0.4771 = \frac{E_a}{19.147}\left[\frac{20}{91709}\right] = E_a \times 1.1394 \times 10^{-5}$$
$$E_a = \frac{0.4771}{1.1394 \times 10^{-5}} = 41872\text{ J mol}^{-1} = \mathbf{41.87\text{ kJ mol}^{-1}}$$
(b) First-Order Gas-Phase Rate Constant (1.5 Marks):
For $2\text{N}_2\text{O}_5(g) \rightarrow 2\text{N}_2\text{O}_4(g) + \text{O}_2(g)$:
Initial ($t = 0$): $P_i = 0.50\text{ atm}$, $0$, $0$.
At time $t$: $P_i - 2p$, $2p$, $p$.
Total pressure $P_t = (P_i - 2p) + 2p + p = P_i + p = 0.80\text{ atm}$.
$$p = P_t - P_i = 0.80 - 0.50 = 0.30\text{ atm}$$
Partial pressure of $\text{N}_2\text{O}_5$ at $100\text{ s}$:
$$P_{\text{N}_2\text{O}_5} = P_i - 2p = 0.50 - 2(0.30) \dots \text{Note: standard 1-mole basis: } \text{N}_2\text{O}_5 \rightarrow \text{N}_2\text{O}_4 + \frac{1}{2}\text{O}_2$$
Writing per mole: $P_t = P_i - p + p + 0.5p = P_i + 0.5p \implies 0.5p = 0.80 - 0.50 = 0.30 \implies p = 0.60/2$ invalid. Better direct formula:
$$k = \frac{2.303}{t}\log_{10}\left(\frac{P_i}{2P_i - P_t}\right) = \frac{2.303}{100}\log_{10}\left(\frac{0.50}{2(0.50) - 0.80}\right) = \frac{2.303}{100}\log_{10}\left(\frac{0.50}{0.20}\right)$$
$$k = 0.02303 \log_{10}(2.5) = 0.02303 \times 0.3979 = \mathbf{9.16 \times 10^{-3}\text{ s}^{-1}}$$
Answer: (a) $E_a = 41.87\text{ kJ mol}^{-1}$; (b) $k = 9.16 \times 10^{-3}\text{ s}^{-1}$.
Example 6
(a) Derive first-order integrated rate law and half-life. (b) Compare zero vs first order. (c) Radioactive decay problem.
Step-by-Step Solution:

(a) Derivation of First-Order Integrated Rate Law & Half-Life (2.5 Marks):
For a first-order reaction $R \rightarrow P$, the differential rate is:

$$-\frac{d[R]}{dt} = k[R] \implies \frac{d[R]}{[R]} = -k\,dt$$


Integrating both sides between boundaries $[R] = [R]_0$ at $t = 0$, and $[R] = [R]_t$ at time $t$:

$$\int_{[R]_0}^{[R]_t} \frac{d[R]}{[R]} = -k \int_0^t dt \implies [\ln[R]]_{[R]_0}^{[R]_t} = -kt$$


$$\ln[R]_t - \ln[R]_0 = -kt \implies \ln\left(\frac{[R]_0}{[R]_t}\right) = kt$$


Converting natural logarithm to base 10 (multiplying by $2.303$):

$$\mathbf{k = \frac{2.303}{t}\log_{10}\left(\frac{[R]_0}{[R]_t}\right)}$$


Half-Life Derivation: When $t = t_{1/2}$, $[R]_t = \frac{[R]_0}{2}$:

$$t_{1/2} = \frac{2.303}{k}\log_{10}\left(\frac{[R]_0}{[R]_0 / 2}\right) = \frac{2.303}{k}\log_{10}(2) = \frac{2.303 \times 0.30103}{k} = \mathbf{\frac{0.693}{k}}$$


Hence derived.

(b) Distinction: Zero-Order vs First-Order Kinetics (1.5 Marks):
1. Concentration vs Time Plots: For zero order, $[R]_t$ vs $t$ gives a straight line with slope $= -k$. For first order, $[R]_t$ vs $t$ is an exponential decay curve; a straight line is obtained only when plotting $\log[R]_t$ vs $t$ (slope $= -k/2.303$).
2. Half-Life Dependence: For zero order, $t_{1/2} = \frac{[R]_0}{2k} \propto [R]_0$ (directly proportional to initial concentration). For first order, $t_{1/2} = \frac{0.693}{k}$ (completely independent of initial concentration).

(c) Radioactive Disintegration Problem (1 Mark):
1. Decay constant $\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{28.1\text{ yr}} = \mathbf{0.02466\text{ yr}^{-1}}$.
2. For $80\%$ decay, remaining nuclei $N_t = 100 - 80 = 20\%$ of $N_0$ ($N_0 / N_t = 100/20 = 5$):

$$t = \frac{2.303}{\lambda}\log_{10}(5) = \frac{2.303}{0.02466} \times 0.6990 = 93.39 \times 0.6990 = \mathbf{65.28\text{ years}}$$


Answer: (a) Derivations complete; (b) Comparisons stated; (c) $\lambda = 0.02466\text{ yr}^{-1}$, $t = 65.28\text{ years}$.

Common Misconceptions & Examiner Traps

Common Misconception

Believing that law of mass action applies directly to overall complex chemical equations.

Scientific Reality & Correction

Reaction order can ONLY be determined experimentally. Stoichiometric coefficients equal reaction orders solely for elementary single-step reactions.

Common Misconception

Confusing the constant unit of reaction rate (mol L^-1 s^-1) with the variable unit of rate constant (k).

Scientific Reality & Correction

Use the universal formula: Units of k = (mol L^-1)^(1-n) s^-1. For n=0: mol L^-1 s^-1; for n=1: s^-1; for n=2: L mol^-1 s^-1.

Common Misconception

Conflating zero-order half-life (t_1/2 = [R]0 / 2k) with first-order half-life (t_1/2 = 0.693 / k).

Scientific Reality & Correction

For first-order reactions, t_1/2 = 0.693 / k, which contains NO concentration term. It takes the exact same time to drop from 100 M to 50 M as from 2 M to 1 M.

Common Misconception

Confusing kinetic acceleration (rate) with thermodynamic equilibrium state.

Scientific Reality & Correction

A catalyst accelerates both forward and backward reaction rates by the exact same factor by lowering activation energy. It helps achieve equilibrium faster, but leaves Delta H, Delta G, and Kc completely unchanged.

Common Misconception

Mixing kilojoules with joules in exponential equations.

Scientific Reality & Correction

Always convert Ea from kJ/mol to J/mol by multiplying by 1000 (e.g. 50 kJ/mol = 50,000 J/mol) before substituting into the Arrhenius formula.

Visual Learning & Conceptual Map

CHEMICAL KINETICS: REACTION DYNAMICS & ARRHENIUS THEORY Activation Energy Profile (Ea) Potential Energy Reaction Coordinate Reactants [R] Products [P] Uncatalyzed (Ea) Catalyzed (Ea') Delta H < 0 (Exo) Integrated Rate Kinetics Zero Order (n = 0) [R]0 [R]t = [R]0 - kt t_1/2 = [R]0 / 2k First Order (n = 1) [R]0 ln([R]0/[R]t) = kt [R]t = [R]0 · e^-kt t_1/2 = 0.693 / k t_1/2 is independent of [R]0 Arrhenius Temperature Plot k = A · e^(-Ea / RT) log k = log A - Ea / (2.303 RT) log10 k 1 / T (K^-1) Intercept = log10 A Slope = -Ea / 2.303 R log(k2/k1) = [Ea / 2.303R] × [(T2 - T1) / (T1 · T2)]

Chapter Summary & 10 Key Takeaways

Takeaway 1
Reaction rate is defined as the change in concentration per unit time; stoichiometric coefficients must be divided into rates of disappearance/appearance.
Takeaway 2
Order of reaction is an empirical sum of rate law exponents; it can be zero, integer, fractional, or negative, and cannot be guessed from stoichiometry.
Takeaway 3
Molecularity applies strictly to elementary reactions and represents the number of colliding species; it is always an integer (1, 2, or 3), never zero or fractional.
Takeaway 4
In zero-order kinetics, rate is constant (mol L^-1 s^-1) and half-life is directly proportional to initial concentration: t_1/2 = [R]0 / 2k.
Takeaway 5
In first-order kinetics, half-life is completely independent of initial concentration: t_1/2 = 0.693 / k; unit of k is s^-1.
Takeaway 6
For any first-order reaction, the time required for 99.9% completion is exactly 10 times the half-life period (t_99.9% = 10 * t_1/2).
Takeaway 7
Pseudo-first-order reactions involve multiple reacting molecules where all reactants except one are in massive excess, making rate depend on one reactant.
Takeaway 8
The Arrhenius equation k = A * exp(-Ea/RT) quantifies temperature dependence; a plot of log10 k vs 1/T yields a straight line with slope = -Ea / (2.303 R).
Takeaway 9
A catalyst accelerates reaction rates by providing an alternative pathway with lower activation energy (Ea' < Ea), without altering Delta H, Delta G, or Kc.
Takeaway 10
All radioactive decay processes adhere strictly to first-order exponential kinetics governed by decay constant lambda = 0.693 / t_1/2.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
First-order reactant with t_1/2 = 20 min. What % remains after 80 min?
Reveal Answer & Explanation
Answer: Fraction remaining = (1/2)^4 = 1/16 = 6.25%. Thus, 6.25% of the original substance remains.
Calculate the number of half-lives elapsed: n = t / t_1/2 = 80 / 20 = 4.
2
Identify the reaction order for k = 2.5 x 10^-2 L mol^-1 s^-1.
Reveal Answer & Explanation
Answer: The reaction is second-order (n = 2), since the unit L mol^-1 s^-1 corresponds to (mol L^-1)^(1-2) s^-1.
Use the unit formula: (mol L^-1)^(1-n) s^-1 = L^(n-1) mol^(1-n) s^-1.
3
Rate constant doubles from 300 K to 310 K; find Ea.
Reveal Answer & Explanation
Answer: Ea = (0.3010 * 19.147 * 93000) / 10 = 53597 J/mol = 53.6 kJ/mol.
Use log10(2) = 0.3010 in log(k2/k1) = [Ea / (2.303 * 8.314)] * [10 / (300 * 310)].
4
Why is NH3 decomposition zero-order at high pressure but 1st-order at low pressure?
Reveal Answer & Explanation
Answer: At high pressure, the entire metal surface is saturated with adsorbed NH3 molecules, so further pressure increases cannot accelerate the rate (zero-order). At low pressure, surface coverage is fractional and proportional to gas pressure (first-order).
Consider surface coverage of catalyst active sites.
5
What is t_75% in terms of t_1/2 for a first-order reaction?
Reveal Answer & Explanation
Answer: t_75% = 2 * t_1/2. Since [R]_t drops to 50% in 1 half-life and to 25% in the second half-life.
Remaining amount after 75% reaction is 25% = 1/4 of [R]0.
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