Answer the following. a. Radioactive decay follows first-order kinetics. The initial amount of two radioactive elements X and Y is 1 gm each. What will be the ratio of X and Y after two days if their half-lives are 12 hours and 16 hours respectively? b. The hypothetical reaction P + Q → R is half order w.r.t 'P' and zero order w.r.t 'Q'. What is the unit of rate constant for this reaction?

ChemistryCBSE Class 12 · Sample paper, session 2023-24 · 2 marks · Very short answer

CBSE Class 12 Chemistry sample paper 2023-24: question 17

From the official CBSE sample question paper for 2023-24, section B, question 17, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q17 · 2 marks · Very short answerChemical Kinetics

Answer the following.

a.
Radioactive decay follows first-order kinetics. The initial amount of two radioactive elements X and Y is 1 gm each. What will be the ratio of X and Y after two days if their half-lives are 12 hours and 16 hours respectively?
0 marks
b.
The hypothetical reaction $P + Q \to R$ is half order w.r.t 'P' and zero order w.r.t 'Q'. What is the unit of rate constant for this reaction?
0 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

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

Hint 1

For first-order decay, the amount left after n half-lives is $(1/2)^{n}$ of the initial amount.

Hint 2

Convert 2 days into hours and divide by each half-life to get n.

Hint 3

For (b), write the rate law and divide the units of rate by the units of $[P]^{1/2}$ .

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.

(a) Number of half-lives in 2 days = 48 h: for X, $n = \frac{48}{12} = 4$ , so X left $= 1 \times (\frac{1}{2})^{4} = \frac{1}{16}$ g.
½ mark
For Y, $n = \frac{48}{16} = 3$ , so Y left $= 1 \times (\frac{1}{2})^{3} = \frac{1}{8}$ g. Ratio X : Y $= \frac{1}{16} : \frac{1}{8} = 1 : 2$ .
½ mark
(b) Rate $= k [P]^{1/2} [Q]^{0}$ , so $k = \frac{rate}{[ P ] ^{1/2}} = \frac{mol L ^{- 1} s ^{- 1}}{( mol L ^{- 1} ) ^{1/2}} = mol^{1/2} L^{- 1/2} s^{- 1}$ .
1 mark

Answer: (a) X : Y = 1 : 2; (b) mol^{1/2} L^{-1/2} s^{-1}

Where the marks are

Finds 4 half-lives for X giving 1/16 g left (½ mark)
Finds 3 half-lives for Y giving 1/8 g left and ratio 1 : 2 (½ mark)
Derives unit of k as mol^{1/2} L^{-1/2} s^{-1} from rate = k[P]^{1/2} (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 paperSample paper 2023-24: every question, by section.

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CBSE Class 12 Chemistry sample paper 2023-24: question 17

Hints

Step-by-step solution

Revise the chapter

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