Match List-I with List-II
|
List-I (Order of reaction) |
List-II (Unit of rate constant, k) |
|
(A) nth order reaction |
(I) \( (mol\, L^{-1})^{1-n}\, s^{-1} \) |
|
(B) zero order reaction |
(II) \( mol\, L^{-1}\, s^{-1} \) |
|
(C) first order reaction |
(III) \( mol^{-1}\, L\, s^{-1} \) |
|
(D) second order reaction |
(IV) \( s^{-1} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
The unit of the rate constant follows from the rate law. For a reaction of order n, rate \( = k[\mathrm{A}]^{n} \), where the rate has the unit mol L−1 s−1 and the concentration mol L−1:
\( \displaystyle k = \frac{\text{rate}}{[\mathrm{A}]^{n}} \)
\( \displaystyle \text{unit of } k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{n}} = (\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \)
(A) nth order reaction — This matches (I) \((mol\, L^{-1})^{1-n}\, s^{-1}\). This is the general result derived above.
(B) zero order reaction — This matches (II) \(mol\, L^{-1}\, s^{-1}\). With n = 0 the concentration term drops out, so k has the same unit as the rate.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{1-0}\,\mathrm{s^{-1}} = \mathrm{mol\,L^{-1}\,s^{-1}} \)
(C) first order reaction — This matches (IV) \(s^{-1}\). With n = 1 the concentration units cancel, so k has the unit of reciprocal time.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{1-1}\,\mathrm{s^{-1}} = \mathrm{s^{-1}} \)
(D) second order reaction — This matches (III) \(mol^{-1}\, L\, s^{-1}\). With n = 2 one concentration unit is left in the denominator.
\( \displaystyle (\mathrm{mol\,L^{-1}})^{1-2}\,\mathrm{s^{-1}} = \mathrm{mol^{-1}\,L\,s^{-1}} \)
Shortcut: The best-known unit, \( \mathrm{s^{-1}} \) for a first order reaction, gives (C)-(IV), which appears only in Option (3).
NCERT (2026-27 edition):
Order of a Reaction – …
