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Chemical Kinetics Formulas for JEE 2027
Chemical Kinetics formulas help students calculate reaction rates, determine reaction order, and understand how concentration and temperature affect the speed of a reaction. This section brings together key formulas for rate laws, integrated rate equations, half-life, and the Arrhenius equation for your 2027 JEE Main and JEE Advanced preparation. Knowing when to apply each formula, along with the correct units, is essential for solving numerical questions accurately.
While revising, pay particular attention to the differences between zero-order and first-order reactions, their graphs, and their half-life relationships. Use the formulas below alongside practice questions to strengthen your understanding and identify common calculation mistakes. For revision across other Chemistry chapters, explore the JEE Mains Chemistry Formula PDF, or refer to the JEE Formula Sheet PDF for formulas across Physics, Chemistry, and Mathematics.
Rate of a Reaction Formulas
Definition: Rate of Reaction
The change in concentration of a reactant or product per unit time. It tells us how quickly reactants are consumed or products are formed.
Rate Expressions
For a reaction $$A \rightarrow B$$:
- Average rate of disappearance of $$A$$: $$r_{\text{avg}} = -\dfrac{\Delta [A]}{\Delta t}$$
- Average rate of appearance of $$B$$: $$r_{\text{avg}} = +\dfrac{\Delta [B]}{\Delta t}$$
- Instantaneous rate: $$r = -\dfrac{d[A]}{dt} = +\dfrac{d[B]}{dt}$$
For a general reaction $$aA + bB \rightarrow cC + dD$$:
$$$r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$$$
Worked Example: Rate of Appearance
For $$2\text{NO}_2 \rightarrow 2\text{NO} + \text{O}_2$$, if the rate of disappearance of $$\text{NO}_2$$ is $$4.0 \times 10^{-3}$$ mol/L/s, find the rate of appearance of $$\text{O}_2$$.
$$r = -\dfrac{1}{2}\dfrac{d[\text{NO}_2]}{dt} = +\dfrac{d[\text{O}_2]}{dt}$$
$$r = \dfrac{1}{2} \times 4.0 \times 10^{-3} = 2.0 \times 10^{-3}$$ mol/L/s
Rate of appearance of O$$_2$$ = $$2.0 \times 10^{-3}$$ mol/L/s
Important Note
The "rate of reaction" $$r$$ is unique for a given reaction, but the rate of disappearance/appearance of individual species differs by stoichiometric coefficients. Always divide by the coefficient.
Rate Law and Order of Reaction Formulas
Definition: Rate Law
An equation of the form $$r = k[A]^m[B]^n$$, where $$k$$ is the rate constant and $$m$$, $$n$$ are the orders with respect to each reactant. It can only be determined experimentally.
Definition: Order of Reaction
The sum of the powers of concentration terms in the rate law. If $$r = k[A]^m[B]^n$$, then overall order $$= m + n$$.
Rate Law vs Molecularity
| Property | Order | Molecularity |
|---|---|---|
| Determined by | Experiment only | Reaction mechanism |
| Can be | Zero, fraction, integer | Only positive integer (1, 2, 3) |
| Applies to | Overall or elementary | Elementary steps only |
| Can be zero? | Yes | No |
| Can be fractional? | Yes | No |
Units of Rate Constant
For an $$n$$th-order reaction:
$$$\text{Units of } k = \text{mol}^{1-n} \, \text{L}^{n-1} \, \text{s}^{-1}$$$
| Order ($$n$$) | Units of $$k$$ |
|---|---|
| 0 | mol L$$^{-1}$$ s$$^{-1}$$ |
| 1 | s$$^{-1}$$ |
| 2 | L mol$$^{-1}$$ s$$^{-1}$$ |
| 3 | L$$^{2}$$ mol$$^{-2}$$ s$$^{-1}$$ |
Tip: A quick way to remember: for a first-order reaction, the units of $$k$$ are simply s$$^{-1}$$ (or min$$^{-1}$$). If you see $$k$$ given in s$$^{-1}$$, the reaction is first order.
Integrated Rate Laws Formulas
Zero-Order Reactions
Zero-Order Integrated Rate Law
Rate law: $$r = k$$
$$$[A] = [A]_0 - kt$$$
where $$[A]_0$$ = initial concentration, $$[A]$$ = concentration at time $$t$$.
Half-life:
$$$t_{1/2} = \frac{[A]_0}{2k}$$$
Graph: A plot of $$[A]$$ vs $$t$$ gives a straight line with slope $$= -k$$ and intercept $$= [A]_0$$.
Worked Example: Zero-Order Reaction
A zero-order reaction has $$k = 0.02$$ mol/L/min and $$[A]_0 = 0.5$$ mol/L. Find (a) the concentration after 10 min, and (b) the half-life.
(a) $$[A] = 0.5 - (0.02)(10) = 0.5 - 0.2 = 0.3$$ mol/L
(b) $$t_{1/2} = \dfrac{[A]_0}{2k} = \dfrac{0.5}{2 \times 0.02} = \dfrac{0.5}{0.04} = 12.5$$ min
First-Order Reactions
First-Order Integrated Rate Law
Rate law: $$r = k[A]$$
$$$\ln[A] = \ln[A]_0 - kt \quad \text{or equivalently} \quad [A] = [A]_0 \, e^{-kt}$$$
Rearranged:
$$$k = \frac{1}{t} \ln\frac{[A]_0}{[A]} = \frac{2.303}{t} \log\frac{[A]_0}{[A]}$$$
Half-life:
$$$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$$$
Graphs:
- $$\ln[A]$$ vs $$t$$ — straight line, slope $$= -k$$
- $$\log[A]$$ vs $$t$$ — straight line, slope $$= -k/2.303$$
- $$[A]$$ vs $$t$$ — exponential decay curve
Important Note
The half-life of a first-order reaction is independent of initial concentration. This is a unique and defining property — if the half-life does not change when you change the starting concentration, the reaction is first order.
Worked Example: First-Order Rate Constant and Half-Life
A first-order reaction has $$k = 0.0693$$ min$$^{-1}$$. Find (a) the half-life, (b) the time to complete 75% of the reaction.
(a) $$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0693} = 10$$ min
(b) If 75% is complete, 25% remains. So $$[A] = 0.25[A]_0$$.
$$t = \dfrac{2.303}{0.0693} \log\dfrac{[A]_0}{0.25[A]_0} = \dfrac{2.303}{0.0693} \log 4 = 33.24 \times 0.6021 = 20$$ min
Shortcut: 75% completion means 2 half-lives ($$50\% \rightarrow 75\%$$), so $$t = 2 \times 10 = 20$$ min. ✓
Worked Example: Fraction Remaining After Multiple Half-Lives
The half-life of a first-order reaction is 30 s. What fraction of the reactant remains after 90 s?
Number of half-lives $$= \dfrac{90}{30} = 3$$
Fraction remaining $$= \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$$
1/8 of the reactant remains (i.e., 87.5% has reacted).
Tip: For first-order reactions, after $$n$$ half-lives the fraction remaining is $$(1/2)^n$$. This is a very fast shortcut for JEE — no need to use logarithms if the time is a whole multiple of $$t_{1/2}$$.
Second-Order Reactions
Second-Order Integrated Rate Law
For $$r = k[A]^2$$, integrating gives:
$$$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$$
Half-life:
$$$t_{1/2} = \frac{1}{k[A]_0}$$$
Graph: A plot of $$\dfrac{1}{[A]}$$ vs $$t$$ gives a straight line with slope $$= k$$.
Worked Example: Second-Order Reaction
A second-order reaction has $$k = 0.5$$ L mol$$^{-1}$$ s$$^{-1}$$ and $$[A]_0 = 0.1$$ mol/L. Find the half-life and the concentration after 10 s.
Half-life: $$t_{1/2} = \dfrac{1}{k[A]_0} = \dfrac{1}{0.5 \times 0.1} = 20$$ s
Concentration after 10 s:
$$\dfrac{1}{[A]} = \dfrac{1}{0.1} + 0.5 \times 10 = 10 + 5 = 15$$
$$[A] = \dfrac{1}{15} = 0.067$$ mol/L
Summary of Integrated Rate Laws
Comparison Table — Orders 0, 1, and 2
| Property | Zero Order | First Order | Second Order |
|---|---|---|---|
| Rate law | $$r = k$$ | $$r = k[A]$$ | $$r = k[A]^2$$ |
| Integrated law | $$[A] = [A]_0 - kt$$ | $$\ln[A] = \ln[A]_0 - kt$$ | $$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$ |
| Linear plot | $$[A]$$ vs $$t$$ | $$\ln[A]$$ vs $$t$$ | $$\frac{1}{[A]}$$ vs $$t$$ |
| Slope | $$-k$$ | $$-k$$ | $$+k$$ |
| Half-life | $$\frac{[A]_0}{2k}$$ | $$\frac{0.693}{k}$$ | $$\frac{1}{k[A]_0}$$ |
| $$t_{1/2}$$ depends on $$[A]_0$$? | Yes ($$\propto [A]_0$$) | No | Yes ($$\propto 1/[A]_0$$) |
| Units of $$k$$ | mol L$$^{-1}$$ s$$^{-1}$$ | s$$^{-1}$$ | L mol$$^{-1}$$ s$$^{-1}$$ |
Tip: For an $$n$$th-order reaction, $$t_{1/2} \propto [A]_0^{1-n}$$. This is a master formula: plug in $$n = 0, 1, 2$$ to recover each case.
Determination of Order of Reaction Formulas
Initial Rates Method
If $$r = k[A]^m[B]^n$$, then comparing two experiments where only $$[A]$$ changes (keeping $$[B]$$ constant):
$$$\frac{r_2}{r_1} = \left(\frac{[A]_2}{[A]_1}\right)^m$$$
Solve for $$m$$ by taking logarithms or by inspection.
Worked Example: Finding Order from Initial Rates
Given the data below, find the order with respect to $$A$$ and $$B$$.
| Expt | $$[A]$$ (M) | $$[B]$$ (M) | Rate (M/s) |
|---|---|---|---|
| 1 | 0.1 | 0.1 | $$2.0 \times 10^{-3}$$ |
| 2 | 0.2 | 0.1 | $$4.0 \times 10^{-3}$$ |
| 3 | 0.1 | 0.2 | $$8.0 \times 10^{-3}$$ |
Order w.r.t. $$A$$: Compare Expts 1 and 2 ($$[B]$$ is constant):
$$\dfrac{4.0 \times 10^{-3}}{2.0 \times 10^{-3}} = \left(\dfrac{0.2}{0.1}\right)^m \Rightarrow 2 = 2^m \Rightarrow m = 1$$
Order w.r.t. $$B$$: Compare Expts 1 and 3 ($$[A]$$ is constant):
$$\dfrac{8.0 \times 10^{-3}}{2.0 \times 10^{-3}} = \left(\dfrac{0.2}{0.1}\right)^n \Rightarrow 4 = 2^n \Rightarrow n = 2$$
Rate law: $$r = k[A]^1[B]^2$$, overall order $$= 1 + 2 = 3$$
Graphical and Half-Life Methods
Graphical Determination of Order
- If $$[A]$$ vs $$t$$ is linear → zero order
- If $$\ln[A]$$ vs $$t$$ is linear → first order
- If $$1/[A]$$ vs $$t$$ is linear → second order
Half-Life Method
For an $$n$$th-order reaction: $$t_{1/2} \propto [A]_0^{1-n}$$
If two experiments give half-lives $$t_1$$ and $$t_2$$ with initial concentrations $$[A]_1$$ and $$[A]_2$$:
$$$\frac{t_1}{t_2} = \left(\frac{[A]_2}{[A]_1}\right)^{n-1}$$$
Worked Example: Half-Life Method
A reaction has $$t_{1/2} = 100$$ s when $$[A]_0 = 0.1$$ M and $$t_{1/2} = 50$$ s when $$[A]_0 = 0.2$$ M. Find the order.
$$\dfrac{100}{50} = \left(\dfrac{0.2}{0.1}\right)^{n-1} \Rightarrow 2 = 2^{n-1} \Rightarrow n - 1 = 1 \Rightarrow n = 2$$
The reaction is second order.
Pseudo-First-Order Reactions Formulas
Pseudo-First-Order Reactions
For $$A + B \rightarrow$$ Products, if $$[B] \gg [A]$$:
Actual rate law: $$r = k[A][B]$$
Since $$[B] \approx \text{constant}$$, let $$k' = k[B]$$:
$$$r = k'[A] \quad \text{(pseudo-first-order)}$$$
$$k'$$ is called the pseudo-first-order rate constant.
Classic example: Hydrolysis of ethyl acetate in excess water:
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$$
Worked Example: Pseudo-First-Order Reaction
The rate constant for acid hydrolysis of an ester is $$k = 1.0 \times 10^{-2}$$ L mol$$^{-1}$$ s$$^{-1}$$. If $$[\text{H}_2\text{O}] = 55$$ M (constant), find the pseudo-first-order rate constant and half-life.
$$k' = k[\text{H}_2\text{O}] = 1.0 \times 10^{-2} \times 55 = 0.55$$ s$$^{-1}$$
$$t_{1/2} = \dfrac{0.693}{k'} = \dfrac{0.693}{0.55} = 1.26$$ s
Arrhenius Equation Formulas
Definition: Activation Energy ($$E_a$$)
The minimum energy that reactant molecules must have for a successful (productive) collision to occur. It is the "energy barrier" that must be crossed for reactants to become products.
Arrhenius Equation
$$$k = A \, e^{-E_a / RT}$$$
where:
- $$k$$ = rate constant
- $$A$$ = pre-exponential factor (frequency factor)
- $$E_a$$ = activation energy (J/mol)
- $$R$$ = gas constant $$= 8.314$$ J mol$$^{-1}$$ K$$^{-1}$$
- $$T$$ = absolute temperature (K)
Logarithmic form:
$$$\ln k = \ln A - \frac{E_a}{RT} \quad \text{or} \quad \log k = \log A - \frac{E_a}{2.303\,RT}$$$
Graph: A plot of $$\ln k$$ vs $$1/T$$ gives a straight line with slope $$= -E_a/R$$ and intercept $$= \ln A$$.
Two-Temperature Form of Arrhenius Equation
If rate constants are known at two temperatures $$T_1$$ and $$T_2$$:
$$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$$
Worked Example: Finding Activation Energy
$$k_1 = 2.0 \times 10^{-2}$$ s$$^{-1}$$ at 300 K and $$k_2 = 8.0 \times 10^{-2}$$ s$$^{-1}$$ at 320 K. Find $$E_a$$.
$$\log\dfrac{8.0 \times 10^{-2}}{2.0 \times 10^{-2}} = \dfrac{E_a}{2.303 \times 8.314}\left(\dfrac{1}{300} - \dfrac{1}{320}\right)$$
$$\log 4 = \dfrac{E_a}{19.147} \times \dfrac{20}{96000}$$
$$0.6021 = \dfrac{E_a}{19.147} \times 2.083 \times 10^{-4}$$
$$E_a = \dfrac{0.6021 \times 19.147}{2.083 \times 10^{-4}} \approx 55,350$$ J/mol $$\approx 55.4$$ kJ/mol
Important Note
General rule of thumb: for many reactions, the rate constant roughly doubles for every 10°C rise in temperature. This is sometimes given as the temperature coefficient $$= k_{T+10}/k_T \approx 2$$ to $$3$$.
Collision Theory Formulas
Collision Theory Rate Expression
$$$k = p \, Z_{AB} \, e^{-E_a / RT}$$$
where:
- $$Z_{AB}$$ = collision frequency (number of collisions per unit time per unit volume)
- $$p$$ = steric factor (fraction of collisions with correct orientation, $$0 < p \leq 1$$)
- $$e^{-E_a/RT}$$ = fraction of collisions with energy $$\geq E_a$$ (Boltzmann factor)
Comparing with Arrhenius: $$A = p \, Z_{AB}$$ (the pre-exponential factor).
Three conditions for a successful collision:
- Collision must occur
- Sufficient energy (at least $$E_a$$)
- Proper orientation of reactive parts
Effect of Catalyst: Formulas
Catalyst and Activation Energy
- Without catalyst: rate constant $$k = A \, e^{-E_a/RT}$$
- With catalyst: rate constant $$k' = A' \, e^{-E_a'/RT}$$ where $$E_a' < E_a$$
- Since $$E_a' < E_a$$, we have $$k' > k$$ (reaction is faster)
A catalyst lowers $$E_a$$ for both the forward and reverse reactions by the same amount.
Important Note
A catalyst does not change $$\Delta H$$ (enthalpy change) or the equilibrium constant $$K$$. It only changes the rate at which equilibrium is achieved. Many JEE questions test this conceptual point.
Worked Example: Effect of Catalyst on Rate
The activation energy is 75 kJ/mol without a catalyst and 50 kJ/mol with a catalyst at 300 K. By what factor does the rate increase?
$$\dfrac{k_{\text{cat}}}{k_{\text{uncat}}} = e^{(E_a - E_a')/RT} = e^{(75000 - 50000)/(8.314 \times 300)} = e^{25000/2494.2} = e^{10.02} \approx 22,400$$
The catalyst increases the rate by a factor of about 22,400.
Summary of Key Formulas: Quick Reference for JEE Mains
Chemical Kinetics: Key Formulas
| Formula | Description |
|---|---|
| $$r = k[A]^m[B]^n$$ | Rate law |
| $$[A] = [A]_0 - kt$$ | Zero-order integrated law |
| $$\ln[A] = \ln[A]_0 - kt$$ | First-order integrated law |
| $$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$$ | First-order $$k$$ formula |
| $$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$$ | Second-order integrated law |
| $$t_{1/2} = \frac{[A]_0}{2k}$$ | Zero-order half-life |
| $$t_{1/2} = \frac{0.693}{k}$$ | First-order half-life |
| $$t_{1/2} = \frac{1}{k[A]_0}$$ | Second-order half-life |
| $$k = A \, e^{-E_a/RT}$$ | Arrhenius equation |
| $$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ | Two-temperature Arrhenius |
Tip: For JEE Mains, the most frequently tested topics are: (1) first-order integrated rate law and half-life calculations, (2) determining order from initial rates data, and (3) Arrhenius equation problems with two temperatures. Master these three and you will handle most kinetics questions.
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