Match List-I with List-II. (Concentration is in mol L−1 and time in seconds.)
| List-I (Rate law) | List-II (Units of the rate constant, k) |
|---|---|
| (A) Rate = k[A]1/2[B]3/2 | (I) s−1 |
| (B) Rate = k[A]3/2[B]−1 | (II) mol−1/2 L1/2 s−1 |
| (C) Rate = k[A][B]1/2 | (III) mol−1 L s−1 |
| (D) Rate = k[A]2[B]−1 | (IV) mol1/2 L−1/2 s−1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-II, D-I
First add the exponents in the rate law to get the order of the reaction, n. Then the units of k follow from k = Rate/(concentration)n:
\( \displaystyle k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{n}} = (\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)
(A) Rate = k[A]1/2[B]3/2 — This matches (III) mol−1 L s−1. n = 1/2 + 3/2 = 2, so 1 − n = −1, the unit of a second order rate constant.
(B) Rate = k[A]3/2[B]−1 — This matches (IV) mol1/2 L−1/2 s−1. n = 3/2 + (−1) = 1/2, so 1 − n = 1/2 and k has the unit (mol L−1)1/2 s−1.
(C) Rate = k[A][B]1/2 — This matches (II) mol−1/2 L1/2 s−1. n = 1 + 1/2 = 3/2, so 1 − n = −1/2 and k has the unit (mol L−1)−1/2 s−1.
(D) Rate = k[A]2[B]−1 — This matches (I) s−1. n = 2 + (−1) = 1, so 1 − n = 0 and only s−1 is left, as for a first order reaction.
NCERT (2026-27 edition):
3.2.3 Order of a Reaction – “Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively.” … “Order of a reaction can be 0, 1, 2, 3 and even a fraction.”
Example 3.3 – “(a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 … So order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order.”
Units of rate constant – “Taking SI units of concentration, mol L–1 and time, s, the units of k for different reaction order are listed in Table 3.3”
