Chemistry Physical Chemistry → Class 12 Chemical Kinetics

Chemical Kinetics & Reaction Rates: Rate Laws, Arrhenius Equation & Half-Life Kinetics

By Dr. Evelyn Reed, Molecular Chemistry Specialist • 12 min read • Published: September 2026

Master zero and first-order integrated rate laws, half-life formulas, activation energy graphical analysis, and collision theory without rote memorization.

Chemical kinetics investigates the speeds at which chemical transformations occur, the mechanistic pathways of elementary steps, and the energetic barriers controlling reactivity.

Unlike thermodynamics (which determines feasibility via $\Delta G < 0$), kinetics determines the reaction timeline. The rate law expresses reaction rate as a function of reactant molar concentrations raised to empirical powers called partial reaction orders.

For a first-order reaction (such as radioactive decay or pharmacological clearance), rate is directly proportional to concentration. The integrated rate law displays exponential decay, and the half-life $t_{1/2} = \frac{0.693}{k}$ is completely independent of initial reactant concentration.

Temperature dramatically accelerates reaction rates. Svante Arrhenius formulated the governing exponential relationship: $k = A e^{-E_a / RT}$, where $E_a$ is the threshold activation energy required for reactant collisions to yield products.

Key Conceptual Takeaways

  • Order of reaction is an experimental quantity that can be zero, fractional, or negative; molecularity is an integer theoretical count of colliding species.
  • First-order half-life is strictly constant and independent of initial concentration ($t_{1/2} = \frac{0.693}{k}$).
  • The slope of a $\ln k$ vs $\frac{1}{T}$ Arrhenius plot equals $-\frac{E_a}{R}$, allowing direct calculation of activation energy.

1. Integrated Rate Laws and Half-Life Relationships

For a zero-order reaction: $-\frac{d[A]}{dt} = k \implies [A]_t = [A]_0 - kt$. The reaction completes in finite time $t_{\text{comp}} = \frac{[A]_0}{k}$, and its half-life is directly proportional to initial concentration: $t_{1/2} = \frac{[A]_0}{2k}$.

For a first-order reaction: $-\frac{d[A]}{dt} = k[A] \implies \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt \implies [A]_t = [A]_0 e^{-kt}$. Converting to base-10 logarithms yields $k = \frac{2.303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right)$.

First-Order Integrated Rate Law and Constant Half-Life
k = \frac{2.303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right) \quad \text{and} \quad t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}
k: Rate constant (units: $\text{s}^{-1}$ or $\text{min}^{-1}$ for 1st order)
[A]_0, [A]_t: Initial reactant concentration and concentration remaining at time $t$
t_{1/2}: Half-life period required for 50% concentration reduction
Worked Problem:

A first-order reaction is 75% completed in 32 minutes. Calculate its rate constant $k$ and the time required for 99.9% completion.

Solution:

When 75% is completed, remaining concentration $[A]_t = 0.25 [A]_0 = \frac{[A]_0}{4}$. Two half-lives have elapsed ($100\% \to 50\% \to 25\%$). Therefore: $2 \times t_{1/2} = 32\text{ min} \implies t_{1/2} = 16\text{ minutes}$. Rate constant $k = \frac{0.693}{16} = 0.0433\text{ min}^{-1}$. For 99.9% completion: $[A]_t = 0.1\% [A]_0 = \frac{[A]_0}{1000} = \frac{[A]_0}{2^{10}}$, which corresponds to 10 half-lives: $t_{99.9\%} = 10 \times t_{1/2} = 10 \times 16 = 160\text{ minutes}$.

Rule: For any first-order reaction: $t_{99.9\%} \approx 10 \times t_{1/2}$ and $t_{99\%} \approx 2 \times t_{90\%}$.

2. Temperature Dependence and the Arrhenius Activation Energy

Most chemical reactions double or triple in speed for every 10 °C rise in temperature. Arrhenius related rate constant $k$ to absolute temperature $T$ through the activation energy barrier $E_a$.

Taking natural logarithm: $\ln k = \ln A - \left(\frac{E_a}{R}\right)\left(\frac{1}{T}\right)$. Comparing two different temperatures $T_1$ and $T_2$ yields the two-point Arrhenius equation.

Two-Point Arrhenius Equation for Temperature Dependence
\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)
k_1, k_2: Reaction rate constants at absolute temperatures $T_1$ and $T_2$ (in Kelvin)
E_a: Activation energy of the reaction (in Joules/mol)
R: Universal gas constant ($8.314\text{ J}/(\text{mol}\cdot\text{K})$)
Worked Problem:

The rate of a reaction quadruples when temperature increases from 300 K to 320 K. Calculate the activation energy $E_a$ (use $R = 8.314\text{ J}/(\text{mol}\cdot\text{K})$ and $\log 4 = 0.602$).

Solution:

$\log_{10}\left(\frac{k_2}{k_1}\right) = \log_{10}(4) = 0.602$. Using the Arrhenius equation: $0.602 = \frac{E_a}{2.303 \times 8.314} \left(\frac{320 - 300}{300 \times 320}\right) = \frac{E_a}{19.147} \times \left(\frac{20}{96000}\right)$. Solving for $E_a$: $E_a = 0.602 \times 19.147 \times \left(\frac{96000}{20}\right) = 0.602 \times 19.147 \times 4800 = 55{,}327\text{ J/mol} = 55.33\text{ kJ/mol}$.

Rule: A higher activation energy $E_a$ signifies a reaction that is dramatically more sensitive to temperature increases.

Common Misconceptions & Examination Traps

Trap: Using balanced stoichiometric coefficients to write the rate law directly.
Why it is wrong: Stoichiometric coefficients determine molecularity of elementary steps, but reaction order is an empirical macroscopic quantity that must be determined experimentally.
First-Principles Approach: Never write $\text{Rate} = k[A]^a [B]^b$ from balanced equations unless the problem explicitly states the reaction is an elementary single-step process.
Trap: Confusing units of rate constant k across different orders.
Why it is wrong: Units of $k$ change with order $n$ according to: $(\text{mol/L})^{1-n} \cdot \text{s}^{-1}$. For 0-order: $\text{mol}/(\text{L}\cdot\text{s})$. For 1st-order: $\text{s}^{-1}$. For 2nd-order: $\text{L}/(\text{mol}\cdot\text{s})$.
First-Principles Approach: Inspect the units of $k$ in numerical problems to immediately deduce the reaction order without reading the prompt text.

Interactive Practice Checkpoints

Question 1 (Easy)

For a reaction A -> B, the rate doubles when the concentration of A is quadrupled. The order of the reaction with respect to A is:

A. 1
B. 0.5 (half-order) ✔ (Correct)
C. 2
D. 0 (zero-order)
Explanation: Rate $r = k[A]^n$. When $[A]$ becomes $4[A]$, $r$ becomes $2r$: $2 = 4^n \implies 2^1 = (2^2)^n = 2^{2n} \implies 2n = 1 \implies n = 1/2$.

Previous-Year Exam Questions (PYQ Vault)

NEET 2024 +4 Marks

For a first-order reaction, the time required for 99% completion is approximately how many times the time required for 90% completion?

Verified Answer: Option B (2 times)
$t_{99\%} = \frac{2.303}{k} \log_{10}\left(\frac{100}{1}\right) = \frac{2.303}{k} \times 2$. $t_{90\%} = \frac{2.303}{k} \log_{10}\left(\frac{100}{10}\right) = \frac{2.303}{k} \times 1$. $\text{Ratio} = \frac{t_{99\%}}{t_{90\%}} = \frac{2}{1} = 2$.

2-Minute High-Yield Exam Revision

Governing Equation:
k = \frac{2.303}{t}\log_{10}\left(\frac{[A]_0}{[A]_t}\right) \quad \Big| \quad t_{1/2} = \frac{0.693}{k} \quad \Big| \quad \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)
Key Recall Checkpoints:
  • Zero-order: constant rate, $t_{1/2} \propto [A]_0$, linear plot of $[A]$ vs $t$.
  • First-order: exponential decay, $t_{1/2}$ independent of $[A]_0$, linear plot of $\ln[A]$ vs $t$.
  • Catalyst lowers activation energy $E_a$ equally for both forward and reverse reactions without shifting equilibrium constant $K$.
Academic Peer Review Certification
Reviewed by Dr. Evelyn Reed, Ph.D.
Associate Professor of Physical and Computational Chemistry • Royal Society of Chemistry Academic Network
Verification Date: September 2026

Frequently Asked Questions

Can a reaction have a negative or zero order?

Yes. A zero order means the rate is completely independent of reactant concentration (e.g., enzyme-catalyzed reactions at saturation or decomposition of $\text{NH}_3$ on hot platinum). A negative order occurs when a reactant or product inhibits the reaction rate.

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