Match List-I with List-II
|
List-I Order of reaction |
List-II Expressions/Unit |
|
(A) Unit of zero order |
(I) \( kt= [R]_0-[R]_t \) |
|
(B) Unit of First order |
(II) \( mol\, L^{-1} s^{-1} \) |
|
(C) Integrated Rate Law for zero order |
(III) \( s^{-1} \) |
|
(D) Integrated Rate Law for First order |
(IV) \( [R]= [R]_0e^{-kt} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
For a reaction of order n, rate \( = k[\mathrm{R}]^{n} \), so the unit of k is \( (\mathrm{mol\,L^{-1}})^{1-n}\,\mathrm{s^{-1}} \). The integrated rate law comes from integrating the rate law and links the concentration with time.
(A) Unit of zero order — This matches (II) mol L−1 s−1. With n = 0, k = rate, so its unit is that of rate:
\( \displaystyle k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^{0}} = \mathrm{mol\,L^{-1}\,s^{-1}} \)
(B) Unit of First order — This matches (III) s−1. With n = 1 the concentration units cancel:
\( \displaystyle k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol\,L^{-1}}} = \mathrm{s^{-1}} \)
(C) Integrated Rate Law for zero order — This matches (I) kt = [R]0 − [R]t. Integrating \( -d[\mathrm{R}]/dt = k \) from 0 to t gives \( [\mathrm{R}] = [\mathrm{R}]_0 - kt \): the concentration falls linearly.
(D) Integrated Rate Law for First order — This matches (IV) [R] = [R]0e−kt. Integrating \( -d[\mathrm{R}]/[\mathrm{R}] = k\,dt \) gives \( \ln([\mathrm{R}]_0/[\mathrm{R}]) = kt \), and taking the antilog gives the exponential form.
Shortcut: (A)-(II) and (B)-(III) appear in Options (2) and (4); (C)-(I) then leaves only Option (4).
NCERT (2026-27 edition):
Half-Life of a Reaction – …
