Chemical Kinetics — Rate Law, Order, Arrhenius Equation

How Fast Do Reactions Go?

Chemical kinetics answers one question: how fast does a reaction proceed? While thermodynamics tells us IF a reaction is feasible, kinetics tells us HOW FAST it happens. A diamond turning into graphite is thermodynamically favourable — but kinetically it takes billions of years.

This is a high-weightage chapter for CBSE Class 12 (5-7 marks) and JEE Main (1-2 questions).

graph TD
    A[Kinetics Problem] --> B{What's asked?}
    B -->|Find order| C[Use experimental data]
    C --> D{Method?}
    D -->|Initial rate method| E[Compare experiments]
    D -->|Graphical| F[Plot concentration vs time]
    D -->|Half-life| G["t½ independent of [A]? → 1st order"]
    B -->|Find rate constant| H[Integrated rate law]
    B -->|Temperature effect| I[Arrhenius equation]
    E --> J[Double conc, see rate change]
    H --> K{Order?}
    K -->|Zero| L["[A] = [A]₀ - kt"]
    K -->|First| M["ln[A] = ln[A]₀ - kt"]
    K -->|Second| N["1/[A] = 1/[A]₀ + kt"]

Essential Formulas

\text{Rate} = k[A]^m[B]^n

$m + n$ = overall order. Order is determined experimentally, NOT from stoichiometry.

Order	Rate Law	Integrated Form	Half-life	Units of $k$
0	$k$	$[A] = [A]_0 - kt$	$[A]_0/(2k)$	mol L $^{-1}$ s $^{-1}$
1	$k[A]$	$\ln[A] = \ln[A]_0 - kt$	$0.693/k$	s $^{-1}$
2	$k[A]^2$	$1/[A] = 1/[A]_0 + kt$	$1/(k[A]_0)$	L mol $^{-1}$ s $^{-1}$

k = Ae^{-E_a/RT}

\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

$E_a$ = activation energy, $A$ = pre-exponential factor.

For first-order reactions, the half-life is independent of initial concentration. This is a quick identifier — if a problem says “half-life is constant regardless of concentration,” it’s first order.

Solved Examples

Example 1 (Easy — CBSE)

A first-order reaction has $k = 0.01$ s $^{-1}$ . Find the half-life.

$t_{1/2} = 0.693/k = 0.693/0.01 = \mathbf{69.3 \text{ s}}$

Example 2 (Medium — JEE Main)

Rate data: when $[A]$ doubles (B constant), rate doubles. When $[B]$ doubles (A constant), rate quadruples. Find the rate law.

Doubling $[A]$ doubles rate: order w.r.t. $A$ = 1. Doubling $[B]$ quadruples rate: $2^n = 4$ , so $n = 2$ .

Rate $= k[A][B]^2$ . Overall order = 3.

Example 3 (Hard — JEE Main)

$k_1 = 2 \times 10^{-3}$ s $^{-1}$ at 300 K and $k_2 = 8 \times 10^{-3}$ s $^{-1}$ at 350 K. Find $E_a$ .

\ln\frac{8 \times 10^{-3}}{2 \times 10^{-3}} = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{350}\right)

\ln 4 = \frac{E_a}{8.314} \times \frac{50}{105000}

1.386 = \frac{E_a \times 4.76 \times 10^{-4}}{8.314}

E_a = \frac{1.386 \times 8.314}{4.76 \times 10^{-4}} \approx \mathbf{24.2 \text{ kJ/mol}}

Common Mistakes to Avoid

Mistake 1 — Assuming order = stoichiometric coefficient. For $2A \to B$ , the order is NOT necessarily 2. Order is always experimental.

Mistake 2 — Using $\log$ instead of $\ln$ (or vice versa). The first-order integrated form uses $\ln$ . Converting: $\ln x = 2.303 \log x$ .

Mistake 3 — Wrong units for $k$ . The units of $k$ depend on the order. Always check dimensional consistency.

Mistake 4 — Forgetting that $E_a$ is always positive. Activation energy is the barrier height — always a positive quantity.

Practice Questions

Q1. A zero-order reaction has $[A]_0 = 0.5$ M and $k = 0.01$ M/s. Find $t_{1/2}$ .

$t_{1/2} = [A]_0/(2k) = 0.5/0.02 = 25$ s.

Q2. In a first-order reaction, 75% of reactant decomposes in 100 min. Find $k$ .

$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{100}\log\frac{100}{25} = \frac{2.303 \times 0.602}{100} = 0.01387$ min $^{-1}$ .

Q3. By what factor does rate increase if temperature rises from 300 K to 310 K? (Assume rate doubles every 10 K.)

Factor = $2^1 = 2$ . This is the rough “temperature coefficient” rule.

Q4. A reaction has $E_a = 50$ kJ/mol. Find the ratio $k_{310}/k_{300}$ . ( $R = 8.314$ J/mol K)

$\ln(k_2/k_1) = \frac{50000}{8.314}(\frac{1}{300} - \frac{1}{310}) = 6014.4 \times 1.075 \times 10^{-4} = 0.646$ . $k_2/k_1 = e^{0.646} \approx 1.91$ .

FAQs

What is the difference between order and molecularity?

Order is the sum of concentration powers in the rate law (experimental). Molecularity is the number of molecules involved in an elementary step (theoretical). Molecularity is always a positive integer; order can be 0, fractional, or negative.

Can a reaction be zero order?

Yes. In zero-order reactions, rate is independent of concentration. Example: decomposition of NH $_3$ on a hot platinum surface (surface is saturated).

What is a catalyst and how does it affect kinetics?

A catalyst provides an alternative reaction pathway with lower activation energy. It speeds up the reaction without being consumed. It does not change the equilibrium position — only how fast equilibrium is reached.

What does the Arrhenius plot look like?

Plot $\ln k$ vs $1/T$ . It gives a straight line with slope $= -E_a/R$ and intercept $= \ln A$ .

Advanced Concepts

Pseudo-first-order reactions

Some reactions are second order overall but appear first order experimentally. This happens when one reactant is in large excess, so its concentration barely changes during the reaction. The rate law then depends only on the other reactant’s concentration.

Classic example: Hydrolysis of ethyl acetate in excess water.

\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}

True rate law: $r = k[\text{ester}][\text{H}_2\text{O}]$

Since water is in vast excess, $[\text{H}_2\text{O}] \approx$ constant. So $r = k'[\text{ester}]$ where $k' = k[\text{H}_2\text{O}]$ .

JEE Main 2023 asked: “The hydrolysis of sucrose is an example of which type of reaction?” Answer: pseudo-first-order.

Collision theory — why not every collision gives a product

For a reaction to occur, molecules must:

Collide — collision frequency depends on concentration and temperature
Have sufficient energy — kinetic energy must exceed $E_a$ (activation energy)
Have proper orientation — the reactive parts must face each other

The rate constant in collision theory:

k = Z \cdot e^{-E_a/RT} \cdot p

where $Z$ is the collision frequency, $e^{-E_a/RT}$ is the fraction of collisions with enough energy, and $p$ is the steric factor (accounts for orientation, always < 1).

Temperature dependence — the “10-degree rule”

A rough rule of thumb: rate doubles for every 10 K rise in temperature (temperature coefficient $\approx 2$ ). This comes from the Arrhenius equation — a small increase in $T$ significantly increases the fraction $e^{-E_a/RT}$ .

\frac{k_{T+10}}{k_T} = 2 \text{ to } 3 \text{ (approximate)}

This rule works only as an estimate. The exact ratio depends on $E_a$ through the Arrhenius equation.

Graphical identification of order

Order	Plot that gives a straight line	Slope	Intercept
Zero	$[A]$ vs $t$	$-k$	$[A]_0$
First	$\ln[A]$ vs $t$	$-k$	$\ln[A]_0$
Second	$1/[A]$ vs $t$	$k$	$1/[A]_0$

This table is the quickest way to determine order from experimental data. If plotting $\ln[A]$ vs $t$ gives a straight line, the reaction is first order. JEE loves this.

Worked Examples — Additional

Given: When $[A]$ is doubled (B constant), rate doubles. When $[B]$ is tripled (A constant), rate increases 9 times.

Rate $= k[A]^m[B]^n$

Doubling $[A]$ doubles rate: $2^m = 2$ , so $m = 1$ . Tripling $[B]$ gives $3^n = 9$ , so $n = 2$ .

Rate $= k[A][B]^2$ . Overall order = 3.

For first-order: $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]}$

99% complete means $[A] = 0.01[A]_0$ .

$t_{99\%} = \frac{2.303}{k}\log\frac{1}{0.01} = \frac{2.303 \times 2}{k} = \frac{4.606}{k}$

Compare with $t_{1/2} = 0.693/k$ . So $t_{99\%} = \frac{4.606}{0.693} \times t_{1/2} \approx 6.65 \times t_{1/2}$ .

It takes about 6.65 half-lives for 99% completion. This is a commonly asked JEE numerical.

Additional Practice Questions

Q5. The half-life of a first-order reaction is 20 minutes. How long does it take for 75% of the reactant to decompose?

75% decomposed means 25% remains. $[A] = [A]_0/4 = [A]_0 \times (1/2)^2$ . So 2 half-lives are needed. $t = 2 \times 20 = 40$ minutes.

Q6. For a reaction, $E_a$ (forward) = 180 kJ/mol and $\Delta H = -20$ kJ/mol. Find $E_a$ (reverse).

$\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$ . So $E_a(\text{reverse}) = 180 - (-20) = 200$ kJ/mol.

Q7. A zero-order reaction has $k = 0.02$ mol/L/s. If initial concentration is 1 M, how long until complete consumption?

For zero order: $[A] = [A]_0 - kt$ . Complete consumption: $0 = 1 - 0.02t$ . $t = 50$ s. Note: $t_{1/2} = [A]_0/(2k) = 1/(0.04) = 25$ s — exactly half the total time, as expected for zero order.

Q8. Why does a catalyst not change the equilibrium position?

A catalyst lowers the activation energy for both forward and reverse reactions equally. Both rates increase by the same factor. The ratio of forward to reverse rate constants ( $K_{eq} = k_f/k_r$ ) remains unchanged. Equilibrium is reached faster but at the same position.

Exam Weightage

Exam	Typical weight	Key topics tested
CBSE Class 12	5–7 marks	Integrated rate law, half-life, Arrhenius equation
JEE Main	1–2 questions	Order determination, graphical methods, Arrhenius
NEET	1 question	First-order kinetics, half-life, rate law basics

The single most tested question type: “A first-order reaction has $t_{1/2} = X$ min. Find time for Y% completion.” Use $t = \frac{2.303}{k}\log\frac{100}{100-Y}$ with $k = 0.693/t_{1/2}$ .

Photochemical reactions

Some reactions are initiated by light rather than heat. In photochemical reactions, the rate depends on the intensity of light, not temperature. The Arrhenius equation does not apply. Examples: photosynthesis, photography (AgBr + light), ozone decomposition by UV.

Chain reactions

A chain reaction has three stages: initiation (free radical formation), propagation (radicals react and regenerate), and termination (radicals combine and disappear). Nuclear fission is a chain reaction where each fission event releases neutrons that trigger more fissions.

In organic chemistry, free radical halogenation (Cl $_2$ + light $\rightarrow$ 2Cl $\cdot$ ) follows this pattern. The chain length (number of propagation steps per initiation) determines the overall rate.

Transition state theory

The transition state (or activated complex) is the highest-energy arrangement along the reaction coordinate. It is not an intermediate — it cannot be isolated. It exists for only about $10^{-13}$ seconds.

The activation energy $E_a$ is the energy difference between reactants and the transition state. A catalyst lowers $E_a$ by providing an alternative pathway with a lower-energy transition state.

\Delta H = E_a(\text{forward}) - E_a(\text{reverse})

For an exothermic reaction ( $\Delta H < 0$ ): $E_a(\text{forward}) < E_a(\text{reverse})$ .

For an endothermic reaction ( $\Delta H > 0$ ): $E_a(\text{forward}) > E_a(\text{reverse})$ .

This relationship is tested directly in JEE — given any two of $\Delta H$ , $E_a$ (forward), $E_a$ (reverse), find the third.