Match the LIST-I with LIST-II.
| LIST-I Units | LIST-II Order of reaction | ||
| A. | s−1 | I. | Zero |
| B. | mol L−1 s−1 | II. | First |
| C. | mol−1 L s−1 | III. | Third |
| D. | mol−2 L2 s−1 | IV. | Second |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-I, C-IV, D-III
The unit of rate is always mol L−1 s−1. For a reaction of order n, Rate = k[A]n, so:
\( \displaystyle \text{Unit of }k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{n}}=(\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)
A. s−1 — This matches II. First. With n = 1, the concentration unit cancels completely.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{0}\,\mathrm{s^{-1}}=\mathrm{s^{-1}} \)
B. mol L−1 s−1 — This matches I. Zero. With n = 0, k has the same unit as the rate itself.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{1}\,\mathrm{s^{-1}}=\mathrm{mol\,L^{-1}\,s^{-1}} \)
C. mol−1 L s−1 — This matches IV. Second. With n = 2, one concentration unit is left in the denominator.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{-1}\,\mathrm{s^{-1}}=\mathrm{mol^{-1}\,L\,s^{-1}} \)
D. mol−2 L2 s−1 — This matches III. Third. With n = 3, two concentration units are left in the denominator.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{-2}\,\mathrm{s^{-1}}=\mathrm{mol^{-2}\,L^{2}\,s^{-1}} \)
Shortcut: The pair A–II appears only in Option (2).
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 07-06 S2 26-05 S2 21-05 S2 | 30-05 S1 | – | 05-06 S3 27-05 S3 | – |