Class 12 Chemistry NCERT Solutions

~5 min read

Chemical Kinetics Class 12 NCERT Solutions

Every Chemical Kinetics NCERT question solved the way a coaching faculty would — define the formula, substitute, explain each step. Intext (3.1–3.5) and back-exercise working included, with final answers boxed.

Class:12Chapter:3 — Chemical KineticsCovers:Intext + Back Exercise

5 Key Formulas24 Practice MCQs

DWritten byDeep Narayan

Key Concept Summary

What is Chemical Kinetics in one line?

Chemical Kinetics is the study of reaction rates — how fast a reaction proceeds, the rate law and order, the integrated rate equations, half-life and how temperature/dilution affect the rate constant through the Arrhenius equation.

Intext questions — solved

Order of reaction and rate

Rate is always written positive; reactant terms carry a minus sign.
Order is the sum of experimentally determined exponents, not the stoichiometric coefficients.
Units of k depend on overall order: mol¹⁻ⁿ Lⁿ⁻¹ s⁻¹.

5Exercise questions

Step-by-step solution

1Average rate of reaction is the negative of the change in reactant concentration divided by the time interval:
2
$rate = - \frac{Δ [ R ]}{Δ t} = - \frac{[ R ] _{2} - [ R ] _{1}}{t _{2} - t _{1}}$
3Substituting the values $[R]_{1} = 0.03, [R]_{2} = 0.02, Δ t = 25 min$ :
4
$rate = - \frac{0.02 - 0.03}{25} = \frac{0.01}{25} = 4 \times 1 0^{- 4} mol L^{- 1} min^{- 1}$
5To express in seconds, $25 min = 25 \times 60 = 1500 s$ :
6
$rate = \frac{0.01}{1500} = 6.67 \times 1 0^{- 6} mol L^{- 1} s^{- 1}$

Final answer

4 \times 1 0^{- 4} mol L^{- 1} min^{- 1} = 6.67 \times 1 0^{- 6} mol L^{- 1} s^{- 1}

Step-by-step solution

1Rate of disappearance of A (per mole of A in the equation, coefficient 2):
2
$- \frac{1}{2} \frac{Δ [ A ]}{Δ t} = - \frac{1}{2} \times \frac{0.4 - 0.5}{10} = \frac{0.1}{20} = 5 \times 1 0^{- 3} mol L^{- 1} min^{- 1}$
3Rate of appearance of $A_{2} B$ is one-half the rate of disappearance of A (coefficient of $A_{2} B$ is 1):
4
$\frac{Δ [ A _{2} B ]}{Δ t} = 5 \times 1 0^{- 3} mol L^{- 1} min^{- 1}$

Final answer

Rate of reaction = 5×10⁻³ mol L⁻¹ min⁻¹; rate of appearance of A₂B = 5×10⁻³ mol L⁻¹ min⁻¹.

Step-by-step solution

1First-order integrated rate law with $[R] = \frac{[ R ] _{0}}{16}$ :
2
$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]} = \frac{2.303}{60} lo g 16$
3Substituting $lo g 16 = 1.204$ :
4
$t = \frac{2.303 \times 1.204}{60} = \frac{2.773}{60} \approx 4.6 \times 1 0^{- 2} s$

Final answer

t \approx 4.6 \times 1 0^{- 2} s

Step-by-step solution

1For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$ so:
2
$k = \frac{0.693}{t _{1/2}} = \frac{0.693}{69.3} = 1 \times 1 0^{- 2} s^{- 1}$

Final answer

k = 1 \times 1 0^{- 2} s^{- 1}

Step-by-step solution

1Original rate: $r_{1} = k [A]^{2} [B]$ . With $[A]^{'} = 2 [A]$ and $[B]^{'} = 3 [B]$ :
2
$r_{2} = k (2 [A])^{2} (3 [B]) = k \times 4 [A]^{2} \times 3 [B] = 12 k [A]^{2} [B] = 12 r_{1}$

Final answer

The rate increases by a factor of 12.

Back exercise — selected solved questions

Expressing rate and determining order

3Exercise questions

Step-by-step solution

1The order in each reactant is written as the exponent of its concentration in the rate law:
2
$rate = k [A]^{2} [B]$
3Overall order = sum of the exponents $2 + 1 = 3$ .

Final answer

rate = k [A]^{2} [B], overall order = 3

Step-by-step solution

1(a) For first order $k = \frac{2.303}{t} lo g \frac{[ X ] _{0}}{[ X ]}$ with $[X]_{0} = 0.10, [X] = 0.025, t = 40$ :
2
$k = \frac{2.303}{40} lo g \frac{0.10}{0.025} = \frac{2.303}{40} lo g 4 = \frac{2.303 \times 0.602}{40}$
3
$k = \frac{1.386}{40} \approx 3.47 \times 1 0^{- 2} min^{- 1}$
4(b) Half-life: $t_{1/2} = \frac{0.693}{k}$ :
5
$t_{1/2} = \frac{0.693}{3.47 \times 1 0 ^{- 2}} \approx 20 min$

Final answer

k = 3.47 \times 1 0^{- 2} min^{- 1}, t_{1/2} \approx 20 min

Step-by-step solution

1Arrhenius form for two temperatures: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ . With $\frac{k _{2}}{k _{1}} = 2$ :
2
$lo g 2 = \frac{E _{a}}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{310})$
3Compute the temperature bracket: $\frac{310 - 300}{300 \times 310} = \frac{10}{93000} = 1.075 \times 1 0^{- 4} K^{- 1}$ .
4
$E_{a} = \frac{2.303 \times 8.314 \times 0.3010}{1.075 \times 1 0 ^{- 4}} = \frac{5.765}{1.075 \times 1 0 ^{- 4}}$
5
$E_{a} \approx 5.36 \times 1 0^{4} J mol^{- 1} \approx 53.6 kJ mol^{- 1}$

Final answer

E_{a} \approx 53.6 kJ mol^{- 1}

How to attempt Kinetics numerically

Identify the order first — a constant half-life over time means first order.
For first order use the integrated form
$k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$
whenever you have two concentrations and a time.
Half-life is order-dependent:
$t_{1/2} = \frac{0.693}{k}$
(first order),
$t_{1/2} \propto \frac{1}{[ R ] _{0}}$
(second order).
For Arrhenius questions convert both temperatures to kelvin, use ln/k base 10 consistently, and double-check the sign of
$\frac{1}{T _{1}} - \frac{1}{T _{2}}$
.
Sanity-check units — a rate constant in s⁻¹ always indicates first order.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Average rate

rate = - \frac{1}{a} \frac{Δ [ A ]}{Δ t}

Positive rate; reactant terms negated and scaled by 1/coefficient.

Rate law

rate = k [A]^{x} [B]^{y}

Overall order = x + y (from experiment, not stoichiometry).

First-order integrated

k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}

Half-life (first order)

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

Arrhenius

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Rate of appearance vs disappearance: scale by the stoichiometric coefficient.
Order ≠ molecularity for complex (multi-step) reactions.
For first order, fraction remaining after n half-lives = (1/2)ⁿ.
A 10 °C rise roughly doubles a reaction's rate near room temperature.
Catalysts lower Ea but never change ΔH or K.

FAQ

Frequently asked questions

Why is the rate of a reaction always positive?

The rate is defined as the change in concentration of products divided by time (positive), or the negative of the change for reactants, so the magnitude comes out positive in every species.

How do I know a reaction is first order?

If the half-life stays constant as concentration decreases (t½ = 0.693/k) or a plot of ln[R] vs t is a straight line, the reaction is first order.

Do these solutions follow the current CBSE syllabus?

Yes — the chapters and exercise numbering follow the latest rationalised NCERT syllabus; units are kept SI-consistent and worked in the same notation the textbook uses.

Is there a solved PDF for Chemical Kinetics?

The full step-by-step solutions are free to read on this page, and the paired chapter notes are downloadable as a no-signup PDF from the notes page.

Next Chapters

Continue learning

Class 12 Chemistry NCERT Solutions (all chapters)Chemical Kinetics Class 12 Notes Class 12 Chemistry Notes

Interactive Quiz

Chapter MCQ practice test

Instant scoring with complete solutions — test your mastery in under 15 minutes.

Active Recall Practice

Chapter MCQ Mock Test

Evaluate how well you have retained the concepts, formulas, and reaction mechanisms from this chapter. Questions adhere strictly to latest CBSE, JEE & NEET trends.

15 questions (of 24)~23 minutesInstant Score & Solutions

Same solutions, live doubt-clearing help

Reading a solution is step one — getting a doubt resolved in real time is what clears it. ClassApna runs small-batch CBSE, JEE & NEET coaching with daily doubt sessions and mock tests.

Book a Free Demo Class Explore Plans & Batches

Small batches · 1-on-1 personal mentorship · Live online & offline centre