Question

Write the unit of (i) second order reaction and (ii) zero order reaction.

Hint:

Quick rule: Unit of rate constant = $(\text{mol L}^{-1})^{1-n}\text{ s}^{-1}$. For first order ($n=1$), unit becomes s$^{-1}$.

Answer 1

Concept:
The units of the rate constant (\(k\)) depend on the overall order of the reaction (\(n\)). According to the rate law: \[ \text{Rate} = k[\text{Concentration}]^n \] Since the rate has units of mol L\(^{-1}\) s\(^{-1}\), the unit of \(k\) can be determined as: \[ \text{Unit of } k = (\text{mol L}^{-1})^{1-n}\text{ s}^{-1} \]
Step 1: Second-order reaction (\(n=2\)).
Substitute \(n = 2\) into the equation for the units of \(k\): \[ (\text{mol L}^{-1})^{1-2}\text{ s}^{-1} = (\text{mol L}^{-1})^{-1}\text{ s}^{-1} \] Simplifying: \[ = \text{L mol}^{-1}\text{ s}^{-1} \] Thus, for a second-order reaction, the units of the rate constant (\(k\)) are \(\mathbf{L mol^{-1} s^{-1}}\).
Step 2: Zero-order reaction (\(n=0\)).
Substitute \(n = 0\) into the equation for the units of \(k\): \[ (\text{mol L}^{-1})^{1-0}\text{ s}^{-1} = \text{mol L}^{-1}\text{ s}^{-1} \] For zero-order reactions, the unit of the rate constant is the same as the unit of the rate, which is \(\mathbf{mol L^{-1} s^{-1}}\).
Step 3: Final answers.

• For a second-order reaction: \(\mathbf{L mol^{-1} s^{-1}}\)
• For a zero-order reaction: \(\mathbf{mol L^{-1} s^{-1}}\)