Question

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^-1 L² sec^-1
Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is A

The units of the rate constant \(k\) depend on the order of the reaction. The general form for the rate of a reaction is:

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

Where:

\(\text{Rate}\) has units of concentration per time (mol/L·sec),

\([A] \)is the concentration of the reactant in mol/L,

\(n\) is the order of the reaction,

\(k\) is the rate constant.

The units of the rate constant change based on the order of the reaction:

Zero-order reaction \((n=0)\):

Rate = \(k\)

Units of \(k\) = mol/L·sec (since Rate = mol/L·sec).

So, Unit of \(k\) for zero-order = mol/L·sec.

First-order reaction \((n=1)\):

\(Rate = k⋅[A]\)

Units of k = sec⁻¹ (since concentration has units of mol/L).

So, Unit of k for first-order = sec⁻¹.

Second-order reaction \((n=2)\):

\(Rate = k \cdot [A]^2\)

Units of k = mol⁻¹ L sec⁻¹ (since concentration squared is (mol/L)²).

So, Unit of k for second-order = mol⁻¹ L sec⁻¹.

Third-order reaction \((n=3)\):

\(Rate = k \cdot [A]^3\)

Units of k = mol⁻² L² sec⁻¹ (since concentration cubed is (mol/L)³).

So, Unit of k for third-order = mol⁻² L² sec⁻¹.

Now, let's examine the units of the rate constant for the given options:

• (A) sec⁻¹ corresponds to a first-order reaction.
• (B) mol L⁻¹ sec⁻¹ corresponds to a second-order reaction.
• (C) mol⁻¹ L sec⁻¹ corresponds to a third-order reaction.
• (D) mol⁻² L² sec⁻¹ corresponds to a fourth-order reaction.

Increasing order of the order of reaction based on the units:

• (C) (third-order) < (A) (first-order) < (B) (second-order) < (D) (fourth-order).

Thus, the correct order is Option A: (C) < (A) < (B) < (D).