Class 12 Chemistry · Original EZT practice
Chemical Kinetics (Class 12 Chapter 3): summary and practice
Chemical kinetics measures how fast reactions go and what controls that speed: the rate law r = k[A]^x[B]^y, the order of a reaction, integrated equations for zero and first order, and half-lives that behave very differently in the two cases. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Rate of a reaction
Average rate is the change in concentration over a time interval; instantaneous rate is the slope at one moment, written r = -d[R]/dt for a reactant. Stoichiometry scales the rate: each species changes at its own coefficient's pace.
2. Rate law, order and molecularity
The rate law r = k[A]^x[B]^y is found experimentally; the order is x + y and may be fractional or zero. Molecularity counts the molecules in an elementary step - it is theoretical, whole-numbered, and not always equal to the order.
3. Integrated rate equations
Zero order: [R] = [R]0 - kt, a straight fall with time. First order: k = (2.303/t) log([R]0/[R]), so the same fraction reacts in equal times. The 2.303 factor converts natural log to base-10 log.
4. Half-life
Zero order: t1/2 = [R]0/2k - it depends on the starting concentration. First order: t1/2 = 0.693/k - constant, no matter how much you start with. That difference identifies the order from data.
5. Temperature, catalysts and collision theory
The Arrhenius equation k = A e^(-Ea/RT) explains why rates jump with temperature; the chapter notes a rough doubling near every 10 K rise. Section 3.4.1 adds that a catalyst lowers the activation energy, and section 3.5 explains rates through collisions with enough energy and proper orientation.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute an average rate
Choose the method: Average rate = change in concentration over change in time.
EZT practice question
For R -> P, the concentration of R falls from 0.03 mol/L to 0.02 mol/L in 25 minutes. Find the average rate.
Show the worked solution
- Average rate = -delta[R]/delta t = -(0.02 - 0.03)/25.
- = 0.01/25.
Answer: 4 x 10^-4 mol/L/min.
Check: The reactant disappears, so the negative sign turns the change positive.
Watch for this: Report the rate as a positive number - the minus sign is bookkeeping for reactants.
Now try it yourself
[R] falls from 0.04 to 0.01 mol/L in 10 minutes. Average rate?
Check your answer
3 x 10^-3 mol/L/min.
0.03/10 = 3 x 10^-3 mol/L/min.
Read a rate law
Choose the method: Order is the sum of the powers in the experimentally found rate law.
EZT practice question
A reaction has rate = k[A]^2[B]. What is the overall order, and what happens to the rate if [A] is doubled at fixed [B]?
Show the worked solution
- Order = 2 + 1 = 3.
- Doubling [A] multiplies the rate by 2^2 = 4.
Answer: Overall order is 3; the rate becomes 4 times.
Check: Only the doubled species matters - [B] stays put.
Watch for this: Do not read the order off the balanced equation's coefficients - the rate law comes from experiment.
Now try it yourself
For rate = k[A], what happens when [A] is tripled?
Check your answer
The rate triples.
First power means the rate tracks [A] directly.
Find a first-order half-life
Choose the method: t1/2 = 0.693/k - independent of the starting amount.
EZT practice question
A first order reaction has rate constant k = 1.15 x 10^-3 per second. Find its half-life.
Show the worked solution
- t1/2 = 0.693/k.
- t1/2 = 0.693/(1.15 x 10^-3).
Answer: t1/2 is about 603 s.
Check: 0.693/0.00115 = 602.6, which rounds to 603 - matches the chapter's own example value.
Watch for this: Keep units straight: k in per second gives t1/2 in seconds.
Now try it yourself
A first order reaction has t1/2 = 20 min. Find k.
Check your answer
k is about 0.0347 per minute.
k = 0.693/20 = 0.0347 per minute.
Track decay over half-lives
Choose the method: Each half-life halves what remains.
EZT practice question
A first order reaction has k = 0.0231 per minute. Find t1/2, the time to fall to 25 percent, and the fraction left after 3 half-lives.
Show the worked solution
- t1/2 = 0.693/0.0231 = 30 min.
- 25 percent means two halvings: 2 x 30 = 60 min.
- After 3 half-lives, (1/2)^3 = 1/8 remains.
Answer: t1/2 = 30 min; 25 percent remains after 60 min; 12.5 percent remains after 90 min.
Check: 100 -> 50 -> 25 -> 12.5: three halvings in 90 minutes.
Watch for this: Half-life counts halvings of the CURRENT amount, not fixed chunks of the start.
Now try it yourself
80 g of a first order reactant sits for 3 half-lives. How much is left?
Check your answer
10 g.
80 -> 40 -> 20 -> 10 g.
Work a zero-order reaction
Choose the method: [R] = [R]0 - kt and t1/2 = [R]0/2k.
EZT practice question
A zero order reaction starts with [A]0 = 0.5 mol/L and k = 0.02 mol/L/min. Find [A] after 10 minutes and the half-life.
Show the worked solution
- [A] = 0.5 - 0.02 x 10 = 0.5 - 0.2.
- t1/2 = 0.5/(2 x 0.02) = 0.5/0.04.
Answer: [A] = 0.3 mol/L after 10 min; t1/2 = 12.5 min.
Check: In 12.5 min the reaction eats 0.02 x 12.5 = 0.25 mol/L - exactly half of 0.5.
Watch for this: Zero-order half-life shrinks as the reaction proceeds - it is not constant like first order.
Now try it yourself
Zero order with [A]0 = 0.4 mol/L and k = 0.01 mol/L/min: when is the reactant exhausted?
Check your answer
After 40 min.
[A] hits zero when t = 0.4/0.01 = 40 min.
Pick the right units for k
Choose the method: Units of k = (mol/L)^(1-order) per unit time.
EZT practice question
What are the units of the rate constant for zero, first and second order reactions?
Show the worked solution
- Rate always carries mol/L/time.
- Zero order: k = rate, so mol/L/time.
- First order: k = rate/[A], so per unit time. Second order: k = rate/[A]^2, so L/mol/time.
Answer: Zero order: mol L^-1 time^-1; first order: time^-1; second order: L mol^-1 time^-1.
Check: Multiply k by the concentration powers - you must land back on mol/L/time every time.
Watch for this: Quoting per-second units for every order is wrong - the unit shifts with the order.
Now try it yourself
A third order rate constant has what units?
Check your answer
L^2 mol^-2 time^-1.
(mol/L)^(1-3) = (mol/L)^-2 per unit time.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Rate law form?
Show answer and reason
r = k[A]^x[B]^y, found by experiment.
The exponents need not match the balanced equation.
2. Order of a reaction?
Show answer and reason
The sum of powers in the rate law.
It can be zero or fractional; molecularity cannot.
3. First-order half-life?
Show answer and reason
t1/2 = 0.693/k.
Constant - it does not depend on [R]0.
4. Why the 2.303 in the first-order equation?
Show answer and reason
It converts natural log to base-10 log.
k = (2.303/t) log([R]0/[R]).
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A first order reaction has t1/2 = 20 min. Find its rate constant.
- 80 g of a first order reactant is left for 3 half-lives. How much remains?
- For rate = k[A], what happens to the rate when [A] is tripled?
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Questions about this practice page
- How are order and molecularity different?
- Order is the sum of powers in the experimentally measured rate law - it can be zero or fractional. Molecularity counts the molecules colliding in one elementary step - it is always a whole number and comes from theory, not measurement.
- What are the units of the rate constant?
- They depend on the order: mol/L/time for zero order, per unit time for first order, L/mol/time for second order. Check by multiplying k by the concentration powers - the rate must come out in mol/L/time.
- How does half-life differ between zero and first order?
- First order: t1/2 = 0.693/k, a constant whatever the starting amount. Zero order: t1/2 = [R]0/2k, which shrinks as the reactant gets used up.
- Does this chapter include collision theory?
- Yes - the Reprint 2026-27 PDF runs to section 3.5, covering the Arrhenius equation, the catalyst's effect in 3.4.1, and collision theory in 3.5: reactions need collisions with enough energy and the right orientation.
- Are these the NCERT exercise questions for Chemical Kinetics (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.