Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

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:

Options:

(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)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Explanation:

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).