Arrange the following rate constant units in increasing order of their order of reaction:

(A) $sec^{-1}$
(B) $mol\, L^{-1} sec^{-1}$
(C) $mol^{-1} L sec^{-1}$
(D) $mol^{-2} L^2 sec^{-1}$

Choose the correct answer from the options given Below:

(B)<(A)<(C)<(D)

The correct answer is Option (3) → (B)<(A)<(C)<(D).

To arrange the rate constant units in increasing order of their order of reaction, let's first understand the relationship between the units of the rate constant (\(k\)) and the reaction order.

General Form of Rate Law:

The rate of a reaction can be expressed as:

\(\text{Rate} = k [A]^m [B]^n\)

Where:

\( k \) = rate constant

\( m + n \) = overall order of the reaction

Units of Rate Constant for Different Reaction Orders:

For a zero-order reaction, the units of \(k\) are \( \text{mol} \cdot \text{L}^{-1} \cdot \text{sec}^{-1} \).

For a first-order reaction, the units of \(k\) are \( \text{sec}^{-1} \).

For a second-order reaction, the units of \(k\) are \( \text{mol}^{-1} \cdot \text{L} \cdot \text{sec}^{-1} \).

For a third-order reaction, the units of \(k\) are \( \text{mol}^{-2} \cdot \text{L}^2 \cdot \text{sec}^{-1} \).

Analyzing the Units for Each Option:

(A) sec⁻¹:

This corresponds to a first-order reaction because its units match the rate constant for a first-order reaction.

(B) mol L⁻¹ sec⁻¹:

This corresponds to a zero-order reaction because these are the units of the rate constant for a zero-order reaction.

(C) mol⁻¹ L sec⁻¹:

This corresponds to a second-order reaction because these are the units for the rate constant of a second-order reaction.

(D) mol⁻² L² sec⁻¹:

This corresponds to a third-order reaction because these are the units of the rate constant for a third-order reaction.

Increasing Order of the Reaction Order:

Now, we can arrange the rate constants in increasing order of their reaction order:

\( B \) (zero-order) < \( A \) (first-order) < \( C \) (second-order) < \( D \) (third-order)

The correct order is: 3. (B) < (A) < (C) < (D)