Chemical Kinetics Practice Questions NEET: 8 Free MCQs

Q: Where can I practise Chemical Kinetics questions for NEET for free?

This article provides eight free, original, single-correct Chemical Kinetics MCQs with worked explanations. Attempt them before reading the solutions, and write the governing relation alongside each chosen option. If rate laws and reaction order are unfamiliar, study the relevant NCERT explanations first.

Q: Are these Chemical Kinetics questions NEET PYQs?

No, these are eight original practice questions, not previous-year NEET questions. They provide a limited diagnostic rather than a complete question bank or proof of chapter mastery. Use chapter-wise PYQs after repairing the conceptual errors the set reveals.

Q: How does doubling initial concentration affect half-life?

At fixed rate constant and temperature, doubling initial concentration doubles the half-life of a zero-order disappearance process. The half-life of a first-order disappearance process remains unchanged because it is independent of initial concentration.

Q: What should I do after getting Chemical Kinetics MCQs wrong?

Record the exact mistaken step, such as choosing the wrong reaction-order formula or omitting the common-logarithm conversion factor. Revise the relevant concept and reattempt the missed question with the answer key hidden. Once you can reproduce the method independently, attempt relevant PYQs and then mixed Chemistry practice.

Where should I start with free Chemical Kinetics practice?

Attempt these free Chemical Kinetics practice questions for NEET first if you have studied the chapter. Use NCERT to repair conceptual gaps, then chapter-wise PYQs to check how those concepts are tested.

If rate laws and reaction order are unfamiliar, read the relevant NCERT explanations before attempting the set. Below are eight original, free, single-correct MCQs with worked explanations, not PYQs or proof of chapter mastery.

How do NCERT exercises, chapter-wise PYQs and fresh MCQs compare?

Use NCERT to build understanding, PYQs to check actual exam phrasing, and fresh MCQs to test your method without relying on remembered answers. Choose your next resource by the error you need to correct.

Best use
NCERT examples and exercises: Understand definitions and connect equations to conditions.
Chapter-wise NEET PYQs: See how concepts have actually been examined.
Original MCQs here: Attempt an immediate diagnostic before choosing revision material.
Question format
NCERT examples and exercises: Worked examples and exercises, not uniformly single-correct MCQs.
Chapter-wise NEET PYQs: Actual examination questions.
Original MCQs here: Eight single-correct questions with four options each.
Explanation requirements
NCERT examples and exercises: Read the surrounding theory alongside solutions.
Chapter-wise NEET PYQs: Worked-solution quality depends on the source.
Original MCQs here: Each explanation gives the governing relation and a tempting error.
Principal limitation
NCERT examples and exercises: Understanding alone does not establish MCQ accuracy.
Chapter-wise NEET PYQs: Remembering an answer can conceal a weak method.
Original MCQs here: A limited diagnostic, not a complete question bank.
When to move on
NCERT examples and exercises: When you can explain why a relation applies.
Chapter-wise NEET PYQs: When you can reconstruct solutions without relying on remembered options.
Original MCQs here: After correcting and reattempting every missed question.

Which resource fits my current preparation problem?

Pick NCERT for unfamiliar concepts, fresh questions for suspected answer recall, and PYQs after concept repair. Your class helps determine the starting point, but your written solution should decide the next step.

Class 11 student studying ahead: Build comfort with concentration, stoichiometry and logarithms. Study the chapter before using PYQs as a test, otherwise unfamiliar notation can look like poor problem-solving.
Class 12 student who has just finished the chapter: Attempt this original set without looking at explanations. Repair errors through NCERT, then attempt relevant chapter-wise PYQs.
Dropper who recognises PYQ answers: Choose fresh questions and write the governing relation before selecting an option. A remembered answer is not evidence that you can reproduce the method.
Student who knows formulas but misses numericals: Identify reaction order, select the matching relation and check units. Correct the formula-selection error before collecting more PDFs.
Student who gets untimed questions right: Move to mixed Chemistry practice. A chapter label supplies a clue that disappears when kinetics sits beside equilibrium, electrochemistry and organic chemistry.

Can I attempt eight free Chemical Kinetics MCQs now?

Attempt all eight original questions below before reading the answer section. Each has exactly one correct option. Write your chosen letter and the relation you used, even when the question looks familiar.

These are original practice questions, not previous-year NEET questions. Use consistent concentration units and assume constant temperature unless stated otherwise.

Question 1. How is reaction rate related to disappearance rate?

At constant volume, consider the following reaction and disappearance rate. Assume the stoichiometric coefficients are distinct and neither is unity.

a A ⟶ b B, r_{A} = - \frac{d [A]}{d t}

Which expression gives the reaction rate?

A: & a r_{A} B: & \frac{r_{A}}{a} C: & r_{A} D: & \frac{r_{A}}{b}

Question 2. What is the overall reaction order?

The balanced overall reaction and experimentally determined rate law are:

A + B ⟶ P, r = k [A] [B]^{2}

Select the correct statement.

A: The order is two, from the sum of reactant coefficients.
B: The order is one, because only the first reactant matters.
C: The order is three, from the rate-law exponents.
D: The order cannot be determined without a mechanism.

Question 3. What are the units of the rate constant?

A reaction is second order overall. Concentration is measured in moles per litre and time in seconds.

A: & s^{- 1} B: & m o l L^{- 1} s^{- 1} C: & L^{2} m o l^{- 2} s^{- 1} D: & L m o l^{- 1} s^{- 1}

Question 4. Which graph identifies zero-order disappearance?

For disappearance of a reactant with unit stoichiometric coefficient, which plot is linear while the zero-order law remains valid?

A: & [A] versus t, slope - k B: & \ln [A] versus t, slope - k C: & [A] versus t, slope + k D: & 1 / [A] versus t, slope + k

Question 5. Which first-order expression is correct?

A first-order disappearance process starts at concentration: $[A]_{0} = C_{0}$

After elapsed time:

t = τ, [A] = C, 0 < C < C_{0}

Which expression gives its rate constant?

A: & \frac{1}{τ} \ln \frac{C}{C_{0}} B: & \frac{1}{τ} \ln \frac{C_{0}}{C} C: & \frac{1}{τ} \log_{10} \frac{C_{0}}{C} D: & \frac{C_{0} - C}{τ}

Question 6. How does doubling initial concentration change half-life?

Compare separate zero-order and first-order disappearance processes. Double the initial concentration in each, keeping its rate constant and temperature fixed.

A: Both half-lives double.
B: The zero-order half-life stays unchanged; the first-order half-life doubles.
C: The zero-order half-life doubles; the first-order half-life stays unchanged.
D: The zero-order half-life halves; the first-order half-life stays unchanged.

Question 7. When does pseudo-first-order behaviour occur?

The underlying rate law is: $r = k [A] [B]$

Reactant B is in such large excess that its concentration remains effectively constant during observation. Which statement is correct?

A: The observed rate varies linearly with A concentration; the apparent constant includes B concentration.
B: The observed rate varies with the square of A concentration because the underlying order is two.
C: The observed rate is independent of A concentration because B is constant.
D: The underlying reaction becomes an elementary unimolecular reaction.

Question 8. How does an Arrhenius slope give activation energy?

Over a range where the Arrhenius plot is linear, natural logarithm of the rate constant is plotted against reciprocal absolute temperature. The slope is negative: $m < 0$

Using the gas constant, which expression gives activation energy?

A: & m R B: & - 2.303 m R C: & - \frac{m}{R} D: & - m R

What are the correct answers and worked explanations?

The key is 1 B, 2 C, 3 D, 4 A, 5 B, 6 C, 7 A, 8 D. Check the governing relation as well as the letter. A correct choice reached through a wrong rule still needs repair.

1. B: Divide by the stoichiometric coefficient.

r = - \frac{1}{a} \frac{d [A]}{d t} = \frac{r_{A}}{a}

Reaction rate is normalised by the coefficient; disappearance rate is not. Option A reverses that normalisation by multiplying.

2. C: Add the experimental rate-law exponents. $n = 1 + 2 = 3$

Option A uses coefficients from the balanced equation rather than the given rate law. Overall reaction coefficients do not generally determine order; using coefficients as exponents requires an elementary-step rate law.

3. D: Divide rate units by concentration squared.

[k] = \frac{m o l L^{- 1} s^{- 1}}{(m o l L^{- 1})^{2}} = L m o l^{- 1} s^{- 1}

Option A is tempting because inverse seconds is familiar. It applies to a first-order constant, not a second-order constant.

4. A: Concentration falls linearly with time.

The concentration itself is the vertical-axis variable for zero order. The slope is negative because the reactant is being consumed.

A zero-order concentration-versus-time graph with horizontal axis labelled t, vertical axis labelled [A], a straight descending line starting at [A]₀ and ending at zero at t = [A]₀/k, and the line labelled slope = −k with no extension into negative concentration.

$[A] = [A]_{0} - k t$ Option B belongs to first-order disappearance. A straight line is useful only when you check its axes.

5. B: Use the initial-to-remaining concentration ratio.

\ln \frac{C_{0}}{C} = k τ, k = \frac{1}{τ} \ln \frac{C_{0}}{C}

This ratio exceeds one, giving a positive constant. Option C omits the conversion factor of 2.303 required when using common logarithms.

6. C: Only the zero-order half-life doubles.

t_{1 / 2, zero} = \frac{[A]_{0}}{2 k}, t_{1 / 2, first} = \frac{\ln 2}{k}

Initial concentration appears only in the zero-order expression. Option A incorrectly transfers that dependence to first order.

7. A: Absorb the effectively constant concentration into an apparent constant.

r = k [A] [B]_{excess} = k_{obs} [A], k_{obs} = k [B]_{excess}

The observed dependence on A is first order. Option B confuses underlying overall order with the dependence measured under excess-reactant conditions.

8. D: Use the natural-logarithm slope.

\ln k = \ln A_{A r r} - \frac{E_{a}}{R T}, m = - \frac{E_{a}}{R}, E_{a} = - m R

The negative slope gives positive activation energy. Option B would apply to a slope from a common-logarithm plot, not this natural-logarithm plot.

What should I revise after each wrong answer?

Use the question number to identify the concept, then record the exact mistaken step. “Weak Chemical Kinetics” is too broad to guide revision; “used the wrong logarithm base” names something you can correct.

Questions 1–3: Revisit NCERT explanations of reaction-rate definition, empirical rate laws and dimensions. Write what each coefficient or exponent means before attempting more MCQs.
Questions 4–6: Review integrated equations, graph axes and half-life dependence. Name the reaction order before choosing a formula, then check whether initial concentration should appear.
Question 7: Separate the underlying rate law from the observed dependence when an excess reactant remains effectively constant. The condition changes the apparent expression, not the underlying law.
Question 8: Check Arrhenius interpretation, slope sign and logarithm base. For a related worked example, use Chemical Kinetics NEET 2024: Activation Energy Solution.

Use this error-log format. Complete the final entry after reattempting.

Question: 5.
Mistaken step: Used a common logarithm without its conversion factor.
Corrected rule: Match the coefficient to the logarithm base.
Reattempt result: Record whether you selected the right option and independently reproduced the relation.

Do not convert an eight-question score into a readiness band. Record corrected methods instead.

How should I move from chapter questions to a full NEET paper?

Correct the concept, reattempt missed questions without the key, attempt relevant PYQs, then practise within a mixed paper. Chapter-specific practice isolates a weakness; mixed practice checks whether you can recognise the right method without a chapter label.

The current paper has 180 compulsory questions in 180 minutes. That is an overall average of one minute per question, not a compulsory time limit for every Chemical Kinetics problem.

Marking gives 4 marks for a correct answer, deducts 1 for a wrong answer, and gives 0 for an unattempted answer. Reviewing avoidable errors therefore tells you more than counting attempts alone.

Once your corrections hold, NEET JEEnius AI offers 30 free full-length mock tests a month: each has 180 questions, 720 marks and 180 minutes, comprising 45 Physics, 45 Chemistry and 90 Biology questions, with a scored per-subject breakdown. These are full-length mocks, not chapter-specific practice. Before moving to one, reattempt your missed questions here with the answer key hidden.

Next step: full-length mock tests on NEET JEEnius AI and sit a full 180-question, 720-mark, 180-minute paper, 45 Physics, 45 Chemistry and 90 Biology, and get a scored per-subject breakdown (30 free a month).

Frequently asked questions

Where can I practise Chemical Kinetics questions for NEET for free?