Match the units of k for different reaction orders.
| List-I Order of reaction | List-II Units |
| (A) Zero order reaction | (I) mol−2 L2 s−1 |
| (B) First order reaction | (II) mol−1 L s−1 |
| (C) Second order reaction | (III) s−1 |
| (D) Third order reaction | (IV) mol L−1 s−1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
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) Zero order reaction — This matches (IV) mol L−1 s−1. With n = 0, k has the same unit as the rate itself.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{1}\,\mathrm{s^{-1}}=\mathrm{mol\,L^{-1}\,s^{-1}} \)
(B) First order reaction — This matches (III) s−1. With n = 1, the concentration unit cancels completely.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{0}\,\mathrm{s^{-1}}=\mathrm{s^{-1}} \)
(C) Second order reaction — This matches (II) mol−1 L s−1. 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) Third order reaction — This matches (I) mol−2 L2 s−1. 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 (C)–(II) appears only in Option (3).
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 | – |