JEE Main Chemistry: Chemical Kinetics High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)
Published: October 22, 2026
Author: NEET & JEE STUDY HUB
🎯 Curriculum & Exam Overview
Chemical Kinetics examines the rates of chemical reactions and the molecular mechanisms that govern them under varying conditions. It is a high-weightage topic because it bridges theoretical thermodynamics with practical industrial process efficiency, frequently appearing in both numerical and conceptual formats.
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January Session 1 Sprint
JEE Main 2027 Engineering Focus: High-Yield Problem Solving & Negative Marking Defense
- Session 1 Registration Timeline: Registration for JEE Main 2027 Session 1 opens in November on jeemain.nta.nic.in. January exam performance sets your prime percentile baseline.
- Negative Marking Safeguard (-1): Eliminate blind guesswork. Prioritize question selection, dimensional checks, and options elimination to maximize net marks.
- Multi-Concept Problem Linkages: JEE questions integrate multiple domains (e.g. Calculus with Mechanics, or Thermodynamics with Equilibrium) into single composite problems.
- Timed Shift PYQ Practice: Solve at least 15-20 recent shift questions (2021-2026) under a strict 2.5-minute timer per numerical/MCQ.
⚡ JEE Main Exam Blueprint & Problem-Solving Guidelines
📊 Expected Weightage
Expected 4 to 8 marks in CBSE board exam (typically 1-2 MCQs + 1 Short/Long Answer or Case Study).
✍️ Examiner & Scoring Insight
Examiners award partial step marks for formula definition, intermediate substitution, correct SI units, and neat labeled diagrams.
1. 📚 Quick Revision Notes & Core Theory
1
Rate of Reaction: The change in concentration of a reactant or product per unit time, defined by the stoichiometry of the balanced chemical equation.
2
Integrated Rate Equations: Mathematical expressions relating the concentration of reactants to time for zero-order and first-order reactions.
3
Arrhenius Equation: The temperature dependence of the rate constant, highlighting the significance of activation energy (Ea) and frequency factor (A).
⚡ High-Yield Formula & Reaction Sheet
- Integrated Rate Equation (1st Order): ln[R] = -kt + ln[R]₀; where [R] is concentration at time t, k is rate constant, and [R]₀ is initial concentration.
- Arrhenius Equation: k = Ae^(-Ea/RT); where k is rate constant, A is pre-exponential factor, Ea is activation energy, R is gas constant, and T is temperature in Kelvin.
⚠️ Common Board Trap Alert
Confusing the units of the rate constant (k) for different orders of reaction; examiners often provide a rate constant with specific units to implicitly test if the student identifies the reaction order correctly.
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3. 📝 Interactive Chapter DPP & Practice Questions
Question 1 • MCQ
🔥 MHT CET (2026)
4 Mark(s)
In a reaction,
$2\text{N}_2\text{O}_{5(g)} \rightarrow 4\,\text{NO}_{2(g)} + \text{O}_{2(g)}$
$\text{N}_2\text{O}_5$ disappears at a rate of $0.06$ mol dm$^{-3}$ s$^{-1}$
Calculate rate of formation of $\text{O}_{2(g)}$ ?
(A) $0.02$ mol dm$^{-3}$ s$^{-1}$
(B) $0.03$ mol dm$^{-3}$ s$^{-1}$
(C) $0.04$ mol dm$^{-3}$ s$^{-1}$
(D) $0.06$ mol dm$^{-3}$ s$^{-1}$
💡 View Step-by-Step Solution & Answer Key
Correct Answer: (b)
Explanation: For the reaction,
$ 2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g)+\text{O}_2(g) $
The stoichiometric ratio between $\text{N}_2\text{O}_5$ consumed and $\text{O}_2$ formed is
$ 2:1 $
So, the rate of formation of $\text{O}_2$ is half the rate of disappearance of $\text{N}_2\text{O}_5$:
$ \text{Rate of formation of }\text{O}_2 =\frac{1}{2}\times 0.06 =0.03\ \text{mol dm}^{-3}\text{s}^{-1} $
Hence, the correct answer is **Option B:**
$ \boxed{0.03\ \text{mol dm}^{-3}\text{s}^{-1}} $
Question 2 • MCQ
4 Mark(s)
$\begin{matrix}A s p e r t h e f o l l o w i n g c o n s e c u t i v e r e a c t i o n s t h e r . d . s i s\end{matrix}$
$\begin{matrix}\left(P\right) \overset{k = 2 \times 1 0^{- 5} s^{- 1}}{\rightarrow} \left(Q\right) \overset{k = 8 \times 1 0^{- 6} s^{- 1}}{\rightarrow} \left(R\right) \overset{k = 3 \times 1 0^{- 3} s^{- 1}}{\rightarrow} S \\ \\ \\ \end{matrix}$
(A) P → Q
(B) Q → R
(C) R → S
(D) P → S
💡 View Step-by-Step Solution & Answer Key
Correct Answer: b
Explanation: Q → R
Slowest step (with lowest rate constant) is the rate determining step (r.d.s) i.e., Q→R.
Question 3 • MCQ
4 Mark(s)
$\begin{matrix}O b s e r v e t h e f o l l o w i n g r e a c t i o n : \\ A \left(g\right) + 3 B \left(g\right) \rightarrow 2 C \left(g\right) \\ T h e r a t e o f t h i s r e a c t i o n \left\{\frac{- d \left[A\right]}{d t}\right\} i s 3 \times 1 0^{- 3} m o l l i t r e^{- 1} m i n^{- 1} . \\ W h a t i s t h e v a l u e o f \frac{- d \left[B\right]}{dt} i n m o l l i t r e^{- 1} m i n^{- 1} ?\end{matrix}$
(A) 3 × 10$^{–3}$
(B) 9 × 10$^{–3}$
(C) 10$^{–3}$
(D) 1.5 × 10$^{–3}$
💡 View Step-by-Step Solution & Answer Key
Correct Answer: b
Explanation: 9 × 10$^{–3}$
$\begin{matrix}\frac{ d \left[A\right]}{d t} = \frac{1}{3} \frac{d \left[B\right]}{d t} \\ \frac{ d \left[B\right]}{d t} = 3 \left\{\frac{ d \left[A\right]}{d t}\right\} = 3 3 1 0^{ 3} \\ = 9 1 0^{ 3} m o l l i t r e^{ 1} m i n^{ 1} .\end{matrix}$
Question 4 • MCQ
4 Mark(s)
The ratio of activities of two radio nuclides X and Y in a mixture at time t = 0 was found to be 4 : 1. After two hours, the ratio of activities become 1 : 1. If the t$_{1/2}$ of radio nuclide X is 20 min then t$_{1/2}$ [in minutes] of radio nuclide Y is:
💡 View Step-by-Step Solution & Answer Key
Correct Answer: c
Explanation: $\begin{matrix}\frac{A_{0 \left(X\right)}}{A_{0 \left(Y\right)}} = \frac{4}{1} ; \frac{A_{x}}{A_{y}} = 1 , \\ \lambda_{y} - \lambda_{x} = \frac{1}{t} = \left(\frac{\left(A_{0}\right)_{y}}{\left(A_{0}\right)_{x}} \times \frac{A_{x}}{A_{y}}\right) \\ \left(\lambda_{y} - \lambda_{x}\right) t = ln \left(\frac{1}{4}\right) ; \left(\right. t_{1 / 2} \left.\right)_{y} = 30 min\end{matrix}$
Question 5 • MCQ
🔥 MHT CET (2026)
4 Mark(s)
Which of the following is an example of fractional order reaction ?
(A) $2\text{H}_2\text{O}_2\text{(g)} \longrightarrow 2\text{H}_2\text{O}\text{(l)} + \text{O}_2\text{(g)}$
(B) $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \longrightarrow 2\text{HI}\text{(g)}$
(C) $\text{CH}_3\text{CHO}\text{(g)} \longrightarrow \text{CH}_4\text{(g)} + \text{CO}\text{(g)}$
(D) $2\text{NO}\text{(g)} + 2\text{H}_2\text{(g)} \longrightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O}\text{(l)}$
💡 View Step-by-Step Solution & Answer Key
Correct Answer: (c)
Explanation: The correct option is **Option C**.
$ \text{CH}_3\text{CHO(g)} \longrightarrow \text{CH}_4\text{(g)} + \text{CO(g)} $
This reaction is a well-known example of a **fractional order reaction**.
For the decomposition of acetaldehyde, the rate law is found experimentally as:
$ \text{Rate} = k[\text{CH}_3\text{CHO}]^{3/2} $
So, the order of the reaction is:
$ \frac{3}{2} $
Since $\frac{3}{2}$ is a fractional value, this is a **fractional order reaction**.
The other options have integral orders:
Decomposition of $\text{H}_2\text{O}_2$ is generally first order.
$\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}$ is second order.
Reaction between $\text{NO}$ and $\text{H}_2$ has an integral order.
So, the answer is:
$ \boxed{\text{Option C}} $
4. ❓ Frequently Asked Questions (Exam Strategy)
❓ How many questions appear from JEE Main Chemistry: Chemical Kinetics High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27) in JEE Main Session 1?
Typically 1 to 2 questions (4 to 8 marks) appear from JEE Main Chemistry: Chemical Kinetics High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27), frequently featuring combined multi-topic concepts.
❓ What is the best approach to avoid negative marks in JEE Main Chemistry: Chemical Kinetics High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)?
Master standard shortcut formulas, verify boundary states (e.g. t=0, infinity), and skip uncertain numerical value questions.
❓ What is the difference between average rate and instantaneous rate?
Average rate is measured over a long time interval, whereas instantaneous rate is the rate at a specific moment in time. Instantaneous rate is determined by the slope of the tangent to the concentration-time curve at that point.
❓ Why does the rate of reaction generally increase with temperature?
Temperature increase raises the kinetic energy of reactant molecules, leading to a higher fraction of collisions exceeding the activation energy. According to the collision theory, more successful collisions per unit time significantly increase the rate constant.
❓ Are all first-order reactions radioactive decay?
No, while radioactive decay is a classic example of a first-order process, many chemical reactions like the decomposition of N₂O₅ also follow first-order kinetics. The defining trait is that the rate is directly proportional to the first power of the concentration of a single reactant.
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