Match List-I with List-II
|
List-I Order of reaction |
List-II Unit |
|
(A). Zero order reaction |
(I). $L^2\, mol^{-2}\, s^{-1}$ |
|
(B). First order reaction |
(II). $mol^{-1}\, L\, s^{-1}$ |
|
(C). Second order reaction |
(III). $mol\, L^{-1}\,s^{-1}$ |
|
(D). Third order reaction |
(IV). $s^{-1}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
|
List-I Order of reaction |
List-II Unit |
|
(A). Zero order reaction |
(III). $mol\, L^{-1}\,s^{-1}$ |
|
(B). First order reaction |
(IV). $s^{-1}$ |
|
(C). Second order reaction |
(II). $mol^{-1}\, L\, s^{-1}$ |
|
(D). Third order reaction |
(I). $L^2\, mol^{-2}\, s^{-1}$ |
To solve this, we can use the general formula for the unit of the rate constant ($k$) for a reaction of $n^{th}$ order:
$\text{Unit of } k = (mol \ L^{-1})^{1-n} \ s^{-1}$
Or, simplified:
$\text{Unit of } k = mol^{1-n} \ L^{n-1} \ s^{-1}$
Applying the formula to each order:
(A) Zero Order ($n=0$): Substituting $n=0$ into the formula: $mol^{1-0} \ L^{0-1} \ s^{-1} = \mathbf{mol \ L^{-1} \ s^{-1}}$ Matches with (III).
(B) First Order ($n=1$): Substituting $n=1$ into the formula: $mol^{1-1} \ L^{1-1} \ s^{-1} = mol^{0} \ L^{0} \ s^{-1} = \mathbf{s^{-1}}$ Matches with (IV).
(C) Second Order ($n=2$): Substituting $n=2$ into the formula: $mol^{1-2} \ L^{2-1} \ s^{-1} = \mathbf{mol^{-1} \ L \ s^{-1}}$ Matches with (II).
(D) Third Order ($n=3$): Substituting $n=3$ into the formula: $mol^{1-3} \ L^{3-1} \ s^{-1} = \mathbf{mol^{-2} \ L^2 \ s^{-1}}$ Matches with (I).