ChemistryCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2025-26: question 33(A)

From the official CBSE sample question paper for 2025-26, section E, question 33(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(A) · 5 marks · Long answerChemical Kinetics

I. The rate of a reaction triples when the temperature changes from 298 K to 318 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. (Given R = 8.314 $J K^{- 1} mo l^{- 1}$ , log 3 = 0.4771) (3 marks)
II. Identify the order of reaction and write its integrated rate equation mentioning what each term in the equation represents. (2 marks)

Figure in the paper: Graph of concentration of R (y-axis) versus time (x-axis): a straight line falling linearly from an initial value $[R]_{0}$ on the y-axis down to the right, labelled 'k = -slope'. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Use the form of the Arrhenius equation that compares rate constants at two temperatures.

Hint 2

Tripling of rate means k2/k1 = 3; check carefully what T2 − T1 equals.

Hint 3

For II, which order gives a linear decrease of concentration with time?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

I. Here $T_{1} = 298$ K, $T_{2} = 318$ K and $\frac{k _{2}}{k _{1}} = 3$ . Using the Arrhenius equation: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$
½ mark
Substituting: $0.4771 = \frac{E _{a}}{2.303 \times 8.314} (\frac{318 - 298}{298 \times 318}) = \frac{E _{a}}{19.147} \times \frac{20}{94764}$
½ mark
Rearranging: $E_{a} = \frac{0.4771 \times 19.147 \times 94764}{20} = 9.135 \times 4738.2$
1 mark
$E_{a} \approx 43284 J mo l^{- 1} = 43.28 kJ mo l^{- 1}$
1 mark
II. The graph of [R] against t is a straight line with a negative slope. This means concentration falls at a constant rate, independent of [R], so the reaction is of zero order.
1 mark
Integrated rate equation: $[R] = [R]_{0} - k t$ (or $k = \frac{[ R ] _{0} - [ R ]}{t}$ ). Here $[R]_{0}$ is the initial concentration of the reactant, $[R]$ is its concentration at time $t$ , and $k$ is the rate constant (equal to $-$ slope of the line).
1 mark

Answer: I. Ea ≈ 43.28 kJ mol⁻¹; II. Zero order, [R] = [R]₀ − kt

Where the marks are

Writes the Arrhenius two-temperature equation (½ mark)
Substitutes the correct values, including k2/k1 = 3 and T2 − T1 = 20 K (½ mark)
Rearranges and simplifies correctly (1 mark)
Gets Ea ≈ 43.3 kJ mol⁻¹ (½ mark) with the correct unit (½ mark)
Identifies the reaction as zero order from the straight-line [R] vs t graph (1 mark)
Writes [R] = [R]₀ − kt and explains each term (1 mark)

Revise the chapter

Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)
Topic: Arrhenius Equation and Activation Energy

Practise this chapterBoard-style questions from Chemical Kinetics, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Chemistry sample paper 2025-26: question 33(A)

Hints

Step-by-step solution

Revise the chapter

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