Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (A)-(I), (B)-(II), (C)-(IV), (D)-(III) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) |
(A)-(III), (B)-(IV), (C)-(II), (D)-(I) |
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
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). |