The following data were obtained during the first order thermal decomposition of N_2O_5 (g) at constant volume : 2N_2O_5(g) → 2N_2O_4(g) + O_2(g) S.No. Time/s Total Pressure/atm --- --- --- 1 0 0.5 2 100 0.625 Calculate rate constant. [ Given : log 2 = 0.3010, log 10 = 1 ]

ChemistryCBSE Class 12 · Board exam 2025, set 56/7/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2025: question 24

Set in the CBSE Class 12 board exam of 2025, paper 56/7/1, section C, question 24, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24 · 3 marks · Short answerChemical Kinetics

The following data were obtained during the first order thermal decomposition of $N_{2} O_{5}$ (g) at constant volume :
$2 N_{2} O_{5} (g) \to 2 N_{2} O_{4} (g) + O_{2} (g)$

S.No.	Time/s	Total Pressure/atm
1	0	0.5
2	100	0.625

Calculate rate constant.
[ Given : log 2 = 0.3010, log 10 = 1 ]

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/7/1.

Hints

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

Hint 1

Write the partial pressures of all gases at time t in terms of x.

Hint 2

Express the pressure of $N_{2} O_{5}$ using the total pressure.

Hint 3

Use $k = \frac{2.303}{t} lo g \frac{p _{i}}{p _{A}}$ .

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.

Let the initial pressure be $p_{i}$ and the decrease in $N_{2} O_{5}$ pressure be $2 x$ .
$2 N_{2} O_{5}$ $2 N_{2} O_{4}$ $O_{2}$
t = 0 $p_{i}$ 0 0
t $p_{i} - 2 x$ $2 x$ $x$
Total pressure: $p_{t} = p_{i} - 2 x + 2 x + x = p_{i} + x$ , so $x = p_{t} - p_{i}$ .
1 mark
Pressure of $N_{2} O_{5}$ at time t: $p_{A} = p_{i} - 2 (p_{t} - p_{i}) = 3 p_{i} - 2 p_{t}$
$p_{A} = 3 (0.5) - 2 (0.625) = 1.5 - 1.25 = 0.25$ atm
1 mark
For a first order reaction, $k = \frac{2.303}{t} lo g \frac{p _{i}}{p _{A}} = \frac{2.303}{100} lo g \frac{0.5}{0.25}$
$k = 0.02303 \times lo g 2 = 0.02303 \times 0.3010$
$k = 6.93 \times 1 0^{- 3} s^{- 1}$
1 mark

	$2 N_{2} O_{5}$	$2 N_{2} O_{4}$	$O_{2}$
t = 0	$p_{i}$	0	0
t	$p_{i} - 2 x$	$2 x$	$x$

Answer: k = 6.93 × 10⁻³ s⁻¹

Where the marks are

Relates total pressure to x: $p_{t} = p_{i} + x$ (1 mark)
$p_{A} = 3 p_{i} - 2 p_{t} = 0.25$ atm (1 mark)
First order expression giving $k = 6.93 \times 1 0^{- 3}$ s⁻¹ (1 mark)

Revise the chapter

Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)
Topic: Integrated Rate Equations and Half-Life

Practise this chapterBoard-style questions from Chemical Kinetics, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/7/1: every question, by section.

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CBSE Class 12 Chemistry 2025: question 24

Hints

Step-by-step solution

Revise the chapter

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