Match List-I with List-II
|
List-I Order of reaction |
List-II Unit |
|
(A). Zero order reaction |
(I). \( L^2\, mol^{-2}\, s^{-1} \) |
|
(B). First order reaction |
(II). \( mol^{-1}\, L\, s^{-1} \) |
|
(C). Second order reaction |
(III). \( mol\, L^{-1}\,s^{-1} \) |
|
(D). Third order reaction |
(IV). \( s^{-1} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (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 (III) 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 (IV) 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) L2 mol−2 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 (A)–(III) appears only in Option (4).
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 | – |