The following initial rate data were obtained for the reaction : 2NO(g) + Br_2(g) → 2NOBr(g) Expt. No. [NO]/mol L^-1 [Br_2]/mol L^-1 Initial Rate (mol L^-1 s^-1) --- --- --- --- 1 0·05 0·05 1·0×10^-3 2 0·05 0·15 3·0×10^-3 3 0·15 0·05 9·0×10^-3 (a) What is the order with respect to NO and Br_2 in the reaction ? (b) Calculate the rate constant (k). (c) Determine the rate of reaction when concentration of NO and Br_2 are 0·4 M and 0·2 M, respectively.

ChemistryCBSE Class 12 · Board exam 2024, set 56/1/1 · 3 marks · Short answer

CBSE Class 12 Chemistry 2024: question 23

Set in the CBSE Class 12 board exam of 2024, paper 56/1/1, section C, question 23, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q23 · 3 marks · Short answerChemical Kinetics

The following initial rate data were obtained for the reaction :
$2 NO (g) + B r_{2} (g) \to 2 NOBr (g)$

Expt. No.	[NO]/mol L $^{- 1}$	[Br $_{2}$ ]/mol L $^{- 1}$	Initial Rate (mol L $^{- 1}$ s $^{- 1}$ )
1	0·05	0·05	$1 \cdot 0 \times 1 0^{- 3}$
2	0·05	0·15	$3 \cdot 0 \times 1 0^{- 3}$
3	0·15	0·05	$9 \cdot 0 \times 1 0^{- 3}$

(a)
What is the order with respect to NO and $B r_{2}$ in the reaction ?
1 mark
(b)
Calculate the rate constant (k).
1 mark
(c)
Determine the rate of reaction when concentration of NO and $B r_{2}$ are 0·4 M and 0·2 M, respectively.
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/1/1.

Hints

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

Hint 1

Pick two experiments where only one concentration changes.

Hint 2

Tripling [NO] makes the rate 9 times — what power of 3 is 9? Then substitute any experiment to find k.

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 rate $= k [NO]^{p} [B r_{2}]^{q}$ . Expt 1 ÷ Expt 2: $\frac{1.0 \times 1 0 ^{- 3}}{3.0 \times 1 0 ^{- 3}} = (\frac{0.05}{0.15})^{q} \Rightarrow \frac{1}{3} = (\frac{1}{3})^{q} \Rightarrow q = 1$ . Expt 1 ÷ Expt 3: $\frac{1}{9} = (\frac{1}{3})^{p} \Rightarrow p = 2$ .
(a) Order with respect to NO = 2; order with respect to Br₂ = 1.
1 mark
(b) From Expt 1: $1.0 \times 1 0^{- 3} = k (0.05)^{2} (0.05)$ , so $k = \frac{1.0 \times 1 0 ^{- 3}}{1.25 \times 1 0 ^{- 4}} = 8 L^{2} mo l^{- 2} s^{- 1}$
1 mark
(c) Rate $= 8 \times (0.4)^{2} \times (0.2) = 8 \times 0.16 \times 0.2 = 0.256 = 2.56 \times 1 0^{- 1} mol L^{- 1} s^{- 1}$
1 mark

Answer: (a) 2 w.r.t. NO, 1 w.r.t. Br₂; (b) k = 8 L² mol⁻² s⁻¹; (c) 0.256 mol L⁻¹ s⁻¹

Where the marks are

Orders: NO = 2, Br₂ = 1 from comparing experiments (1 mark)
k = 8 L² mol⁻² s⁻¹ (1 mark)
Rate = 2.56 × 10⁻¹ mol L⁻¹ s⁻¹ (1 mark)

Revise the chapter

Chapter: Chemical Kinetics (notes, formulas, where marks are lost and every past question)
Topic: Rate Law, Order and Molecularity

Practise this chapterBoard-style questions from Chemical Kinetics, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 56/1/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 23

Hints

Step-by-step solution

Revise the chapter

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