Kinetics — Concepts, Formulas & Examples

Chemical kinetics studies the rates of chemical reactions and the factors affecting them. CBSE Class 12 and NEET test rate laws, half-life, Arrhenius equation and temperature effect. Expect two questions a year on this physical chemistry topic.

Core Concepts

Rate of reaction

Change in concentration per unit time. $r = -d[A]/dt$ (for reactant) or $+d[P]/dt$ (for product). Units — mol/L/s.

For a reaction $aA + bB \rightarrow cC + dD$ , the rate is expressed as:

r = -\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}

The $1/a$ , $1/b$ etc. ensure the rate is the same regardless of which species we measure. For example, in $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ :

r = -\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = \frac{1}{2}\frac{d[\text{NH}_3]}{dt}

Average rate vs instantaneous rate: Average rate is $\Delta[A]/\Delta t$ over a time interval. Instantaneous rate is the slope of the concentration-time curve at a specific moment ( $d[A]/dt$ ). Exam questions usually ask for instantaneous rate.

Factors affecting rate

Nature of reactants. Concentration (higher usually means faster). Temperature (higher almost always faster). Catalyst (lower activation energy). Surface area (for heterogeneous reactions).

Concentration: For most reactions, increasing reactant concentration increases the frequency of molecular collisions, speeding up the reaction. The exact relationship is given by the rate law.

Temperature: The Arrhenius equation quantifies this. A rough rule: for many reactions, rate doubles for every 10°C rise. The temperature coefficient $\mu = k_{T+10}/k_T \approx 2$ to 3 for most reactions.

Catalyst: Provides an alternative pathway with lower activation energy. Does not change thermodynamics ( $\Delta G$ ) — only changes how fast equilibrium is reached.

Order of reaction

The exponent of concentration in the rate law. Zero order — rate independent of concentration. First order — rate proportional to concentration. Second order — rate proportional to concentration squared. Determined experimentally, not from stoichiometry.

Rate law: $r = k[A]^m[B]^n$ , where $m$ and $n$ are the orders with respect to A and B. The overall order = $m + n$ .

Order	Rate law	Integrated form	Half-life	Units of $k$
Zero	$r = k$	$[A]_t = [A]_0 - kt$	$[A]_0/2k$	mol L $^{-1}$ s $^{-1}$
First	$r = k[A]$	$\ln[A]_t = \ln[A]_0 - kt$	$0.693/k$	s $^{-1}$
Second	$r = k[A]^2$	$1/[A]_t = 1/[A]_0 + kt$	$1/(k[A]_0)$	L mol $^{-1}$ s $^{-1}$

The units of $k$ change with order — this is a quick way to identify order from given data. If $k$ has units of s $^{-1}$ , the reaction is first order.

First order kinetics

$[A]_t = [A]_0 e^{-kt}$ . Half-life is constant — $t_{1/2} = 0.693/k$ . Radioactive decay is first order.

The integrated first-order rate law can also be written as:

\ln\frac{[A]_0}{[A]_t} = kt \quad \text{or} \quad \log\frac{[A]_0}{[A]_t} = \frac{kt}{2.303}

Key property: The time to reduce concentration by any fixed fraction is constant. If it takes 10 minutes for half the reactant to decompose, it takes another 10 minutes for half of the remaining to decompose, and so on.

After $n$ half-lives, the fraction remaining = $(1/2)^n$ . After 10 half-lives, only $\approx 0.1\%$ remains.

Arrhenius equation

$k = A e^{-E_a/RT}$ . Rate constant increases exponentially with temperature. Plotting $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ .

k = Ae^{-E_a/RT}

Where $A$ = pre-exponential factor (frequency factor), $E_a$ = activation energy (J/mol), $R$ = 8.314 J/(mol $\cdot$ K), $T$ = temperature in kelvin.

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

Use this when given rate constants at two different temperatures to find $E_a$ .

The pre-exponential factor $A$ represents the frequency of collisions with correct orientation. A high $A$ means collisions are frequent and well-oriented. $E_a$ is the energy barrier — only molecules with kinetic energy $\geq E_a$ can react.

Collision theory

For a reaction to occur, molecules must collide with (1) sufficient energy ( $\geq E_a$ ) and (2) proper orientation. The fraction of molecules with energy $\geq E_a$ at temperature $T$ is given by the Boltzmann factor: $e^{-E_a/RT}$ .

At room temperature (300 K), if $E_a = 50$ kJ/mol:

\text{Fraction} = e^{-50000/(8.314 \times 300)} = e^{-20} \approx 2 \times 10^{-9}

Only about 2 in a billion collisions have enough energy. Raising temperature to 310 K increases this fraction significantly — explaining why small temperature increases have large effects on rate.

Pseudo-first order reactions

When one reactant is in large excess, its concentration barely changes during the reaction. The rate law simplifies. For example, acid-catalysed hydrolysis of an ester in excess water:

r = k[\text{ester}][\text{H}_2\text{O}]

Since $[\text{H}_2\text{O}]$ is essentially constant: $r = k'[\text{ester}]$ where $k' = k[\text{H}_2\text{O}]$ .

This is a second-order reaction that behaves as first-order — hence “pseudo-first order.”

Key Formulas

t_{1/2} = \frac{0.693}{k}

Independent of initial concentration — the defining feature of first-order kinetics.

t_{1/2} = \frac{[A]_0}{2k}

Depends on initial concentration — the higher the initial concentration, the longer the half-life.

t_{1/2} = \frac{1}{k[A]_0}

Also depends on initial concentration, but inversely — higher concentration means shorter half-life.

Worked Examples

Half-life depends only on $k$ , not on concentration. So any starting concentration takes the same time to halve. Used to date things — Carbon-14 has $t_{1/2}$ of 5730 years.

Arrhenius gives $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$ . Measure $k$ at two temperatures, solve for $E_a$ .

Given data for a reaction A → products:

$[A]_0$ (M)	Initial rate (M/s)
0.1	0.02
0.2	0.08
0.4	0.32

When $[A]$ doubles (0.1 → 0.2), rate quadruples (0.02 → 0.08). Rate $\propto [A]^2$ . This is second order.

Check: when $[A]$ doubles again (0.2 → 0.4), rate again quadruples (0.08 → 0.32). Confirmed second order.

For 90% completion, 10% remains: $[A]_t/[A]_0 = 0.1$ .

t = \frac{2.303}{k}\log\frac{[A]_0}{[A]_t} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}

Since $t_{1/2} = 0.693/k$ , we get $t_{90\%} = 2.303/k = 3.32 \times t_{1/2}$ .

So 90% completion takes about 3.3 half-lives. This is a frequently asked NEET numerical.

Given: $k_1 = 2 \times 10^{-3}$ s $^{-1}$ at 300 K, $k_2 = 8 \times 10^{-3}$ s $^{-1}$ at 320 K.

\ln\frac{8 \times 10^{-3}}{2 \times 10^{-3}} = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)

\ln 4 = \frac{E_a}{8.314} \times \frac{20}{96000}

1.386 = \frac{E_a}{8.314} \times 2.083 \times 10^{-4}

E_a = \frac{1.386 \times 8.314}{2.083 \times 10^{-4}} = 55,300\;\text{J/mol} \approx 55.3\;\text{kJ/mol}

Common Mistakes

Assuming order equals stoichiometry. They are unrelated; order is experimental.

Saying half-life of second order is constant. Only first order has constant half-life.

Confusing rate and rate constant. Rate changes with concentration; rate constant depends only on temperature.

Using Celsius in the Arrhenius equation. Temperature must be in kelvin. Forgetting to convert is one of the most common calculation errors in kinetics problems.

Writing that a catalyst shifts the equilibrium. It does not — it speeds up both forward and reverse reactions equally, reaching the same equilibrium faster.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2024 had a numerical on calculating $E_a$ from two temperatures. NEET 2023 tested first-order half-life. CBSE boards ask about factors affecting rate and the Arrhenius equation regularly. Kinetics gives 2-3 questions across JEE + NEET every year.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Memorise three formulas — first order integrated rate, half-life, Arrhenius. NEET questions cluster around these.

Practice Questions

Q1. A first-order reaction has $k = 0.01$ s $^{-1}$ . What is its half-life?

$t_{1/2} = 0.693/k = 0.693/0.01 = 69.3$ seconds. After 69.3 s, exactly half the reactant remains.

Q2. How do you determine the order of a reaction experimentally?

Method 1 (Initial rates): Vary the concentration of one reactant while keeping others constant. Measure the initial rate. If doubling concentration doubles rate → first order. If doubling concentration quadruples rate → second order. If rate is unchanged → zero order.

Method 2 (Integrated rate law): Plot $[A]$ vs $t$ (zero order gives straight line), $\ln[A]$ vs $t$ (first order gives straight line), or $1/[A]$ vs $t$ (second order gives straight line). The plot that gives a straight line indicates the order.

Q3. The rate constant of a reaction doubles when temperature increases from 300 K to 310 K. Calculate $E_a$ .

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

\ln 2 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{310}\right) = \frac{E_a}{8.314} \times \frac{10}{93000}

E_a = \frac{0.693 \times 8.314 \times 93000}{10} = 53,598\;\text{J/mol} \approx 53.6\;\text{kJ/mol}

Q4. What is a pseudo-first order reaction? Give an example.

A reaction that is actually higher order but behaves as first order because one reactant is in such large excess that its concentration does not change appreciably. Example: hydrolysis of ethyl acetate in excess water. $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$ . The rate depends on both ester and water concentrations, but water is the solvent (~55 M) and barely changes.

Q5. After how many half-lives will 99% of a first-order reactant be consumed?

1% remains: $[A]_t/[A]_0 = 0.01 = (1/2)^n$ . Taking log: $n = \log(100)/\log(2) = 2/0.301 = 6.64$ half-lives.

Alternatively: $t_{99\%} = 2.303 \times 2/k = 4.606/k = 6.65 \times t_{1/2}$ .

FAQs

What is the difference between rate and rate constant? Rate ( $r$ ) is how fast the concentration changes at a given moment — it depends on both $k$ and the current concentrations. Rate constant ( $k$ ) is a proportionality constant that depends only on temperature and the nature of the reaction. Rate changes as the reaction proceeds; $k$ stays the same throughout the reaction (at constant T).

Can order of a reaction be fractional or negative? Yes. Fractional orders (like 1.5) occur in complex reactions with multiple steps. Negative orders occur when a substance inhibits the reaction — increasing its concentration slows the reaction down. Order is purely experimental and can take any value.

Why does a catalyst not change the equilibrium position? A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount. Both rates increase equally, so the ratio $k_f/k_r$ (which determines the equilibrium constant) remains unchanged. The system reaches equilibrium faster, but at the same position.

What is activation energy physically? It is the minimum energy that colliding molecules must have for the collision to result in a reaction. Molecules must have enough energy to break existing bonds before new bonds can form. The higher the $E_a$ , the slower the reaction at a given temperature.

Kinetics connects chemistry to time. Understanding rate is what lets chemists design fast industrial processes and safe drugs.