Identify the order of the two reactions whose rate constants are given below:
(i) \( k = 2.3 × 10^{-5}\, L\, mol\,^{-1} s^{-1} \)
(ii) \( k= 3 × 10^{-4} s^{-1} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (i) Second order, (ii) First order
The unit of a rate constant shows the order of the reaction. For rate \( = k[\mathrm{A}]^{n} \):
\( \displaystyle k = \frac{\text{rate}}{[\mathrm{A}]^{n}} = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{n}} = (\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)
This gives mol L−1 s−1 for zero order, s−1 for first order and L mol−1 s−1 for second order.
(i) \( k = 2.3 \times 10^{-5}\ \mathrm{L\,mol^{-1}\,s^{-1}} \): here \( (\mathrm{mol\,L^{-1}})^{1-n} = (\mathrm{mol\,L^{-1}})^{-1} \), so 1 − n = −1 and n = 2, a second order reaction.
(ii) \( k = 3 \times 10^{-4}\ \mathrm{s^{-1}} \): no concentration term, so 1 − n = 0 and n = 1, a first order reaction.
So (i) is second order and (ii) is first order. The numerical values of k play no part; only the units decide the order.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Order of a Reaction – …
