The rate constant for a reaction is \( 3.5\, mol^{-1}\, L\, s^{-1} \). The order of the reaction is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Second
The rate of a reaction is the change in concentration per unit time, so its unit is always mol L−1 s−1. For a reaction of order n:
\( \displaystyle \mathrm{Rate}=k[\mathrm{A}]^n,\quad k=\frac{\mathrm{Rate}}{[\mathrm{A}]^n} \)
\( \displaystyle \text{Unit of }k=\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^n}=(\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)
Putting in the common orders:
\( \displaystyle \text{Zero order }(n=0)\text{: }\mathrm{mol\,L^{-1}\,s^{-1}} \)
\( \displaystyle \text{First order }(n=1)\text{: }\mathrm{s^{-1}} \)
\( \displaystyle \text{Second order }(n=2)\text{: }\frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol^2\,L^{-2}}}=\mathrm{mol^{-1}\,L\,s^{-1}} \)
\( \displaystyle \text{Third order }(n=3)\text{: }\mathrm{mol^{-2}\,L^2\,s^{-1}} \)
The given rate constant, 3.5 mol−1 L s−1, has the unit mol−1 L s−1. So the reaction is of second order.
NCERT (2026-27 edition):
Order of a Reaction
| Reaction | Order | Units of rate constant |
|---|---|---|
| Zero order reaction | 0 | \( \displaystyle \frac{\mathrm{mol\,L^{-1}}}{\mathrm{s}}\times\frac{1}{(\mathrm{mol\,L^{-1}})^{0}}=\mathrm{mol\,L^{-1}s^{-1}} \) |
| First order reaction | 1 | \( \displaystyle \frac{\mathrm{mol\,L^{-1}}}{\mathrm{s}}\times\frac{1}{(\mathrm{mol\,L^{-1}})^{1}}=\mathrm{s^{-1}} \) |
| Second order reaction | 2 | \( \displaystyle \frac{\mathrm{mol\,L^{-1}}}{\mathrm{s}}\times\frac{1}{(\mathrm{mol\,L^{-1}})^{2}}=\mathrm{mol^{-1}L\,s^{-1}} \) |
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 24-05 S2 22-05 S1 | 15-05 S1 | – | – | – |