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

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

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

Q33(B) · 5 marks · Long answerChemical Kinetics

I. Consider the following first order thermal decomposition of $S O_{2} C l_{2}$ at a constant volume
$S O_{2} C l_{2} (g) \to S O_{2} (g) + C l_{2} (g)$
If the total pressure of the gases is found to be 200 torr after 10 seconds and 300 torr upon the complete decomposition of $S O_{2} C l_{2}$ . Calculate the rate constant. (Given log 3 = 0.4771, log 2= 0.3010) (3 marks)
II. For a bimolecular elementary reaction A + B → Products. Write the expression for the rate of reaction relating temperature and activation energy for the reaction and also mention what each term represents in the equation. (2 marks)

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

One mole of SO2Cl2 gives two moles of gas, so the final pressure is twice the initial pressure.

Hint 2

Express the total pressure at 10 s as Pi + x and solve for x.

Hint 3

For II, think of collision frequency multiplied by the fraction of effective-energy molecules.

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. Let $P_{i}$ be the initial pressure of $S O_{2} C l_{2}$ and let $x$ torr of it decompose in 10 s.
$S O_{2} C l_{2}$ $S O_{2}$ $C l_{2}$
$t = 0$ $P_{i}$ 0 0
$t = 10$ s $P_{i} - x$ $x$ $x$
completion 0 $P_{i}$ $P_{i}$
½ mark
On complete decomposition, $P_{t o t a l} = 2 P_{i} = 300$ torr, so $P_{i} = 150$ torr.
½ mark
After 10 s, $P_{t o t a l} = P_{i} + x = 200$ torr, so $x = 200 - 150 = 50$ torr and $P_{i} - x = 100$ torr.
½ mark
First-order rate equation in terms of pressure: $k = \frac{2.303}{t} lo g \frac{P _{i}}{P _{i} - x}$
½ mark
Substituting: $k = \frac{2.303}{10} lo g \frac{150}{100} = 0.2303 (lo g 3 - lo g 2) = 0.2303 \times (0.4771 - 0.3010) = 0.2303 \times 0.1761$
½ mark
$k \approx 0.0406 s^{- 1}$ (about $4.06 \times 1 0^{- 2} s^{- 1}$ )
½ mark
II. By collision theory, for $A + B \to$ Products: $Rate = Z_{A B} e^{- E_{a} / R T}$
1 mark
In this expression:
- $Z_{A B}$ is the collision frequency, the number of collisions between A and B per second per unit volume.
- $e^{- E_{a} / R T}$ is the fraction of molecules with energy equal to or greater than $E_{a}$ .
- $E_{a}$ is the activation energy, $R$ is the gas constant and $T$ is the absolute temperature.
1 mark

	$S O_{2} C l_{2}$	$S O_{2}$	$C l_{2}$
$t = 0$	$P_{i}$	0	0
$t = 10$ s	$P_{i} - x$	$x$	$x$
completion	0	$P_{i}$	$P_{i}$

Answer: I. k ≈ 0.040 s⁻¹; II. Rate = Z_AB e^(−Ea/RT)

Where the marks are

Sets up the pressure table for reactant and products (½ mark)
Finds Pi = 150 torr from the final pressure (½ mark)
Finds x = 50 torr and Pi − x = 100 torr (½ mark)
Writes the first-order equation in terms of pressure (½ mark)
Substitutes values and uses the logs correctly (½ mark)
Obtains k ≈ 0.040 s⁻¹ with unit (½ mark)
Writes Rate = Z_AB e^(−Ea/RT) (1 mark)
Explains Z_AB and the exponential term correctly (1 mark)

Revise the chapter

Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)

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(B)

Hints

Step-by-step solution

Revise the chapter

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