Question

The unit of rate constant of a second order reaction is

• A. mol\(^{-1}\) L sec\(^{-1}\)
• B. mol\(^{-1}\) L\(^{-1}\) sec\(^{-1}\)
• C. mol\(^{-1}\) L sec
• D. mol L sec\(^{-1}\)

Hint:

For second-order reactions, the unit of the rate constant is always mol\(^{-1}\) L sec\(^{-1}\), ensuring that the rate law maintains dimensional consistency.

Answer 1

The Correct Option is A

Step 1: Understanding the order of reaction.
For a second-order reaction, the rate law is given as: \[ \text{Rate} = k[A]^2 \] where \(k\) is the rate constant and \( [A] \) is the concentration of the reactant. The unit of rate is mol L\(^{-1}\) sec\(^{-1}\), and the unit of concentration is mol L\(^{-1}\). To maintain the dimensional consistency, the unit of \(k\) will be: \[ k = \frac{\text{Rate}}{[A]^2} = \frac{\text{mol L}^{-1} \text{sec}^{-1}}{(\text{mol L}^{-1})^2} = \text{mol}^{-1} \text{L sec}^{-1} \] Step 2: Conclusion.
The correct answer is (A) because the unit of the rate constant for a second-order reaction is mol\(^{-1}\) L sec\(^{-1}\).