We need to identify which of the following reactions has a rate constant with units of \( \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \):
The unit \( \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) corresponds to a second-order reaction with respect to concentration.
Typically, acid catalysed hydrolysis reactions are first order with respect to the ester and first order with respect to the acid catalyst. Thus, the rate constant would have units of \( \text{s}^{-1} \) or \( \text{L} \, \text{mol}^{-1} \, \text{s}^{-1} \).
This reaction is a second-order reaction with respect to \( \text{Cl}_2 \). Thus, the rate constant would have units of \( \text{L} \, \text{mol}^{-1} \, \text{s}^{-1} \).
This reaction is a second-order reaction with respect to \( \text{NO} \) and first order with respect to \( \text{O}_2 \). Thus, the rate constant would have units of \( \text{L}^2 \, \text{mol}^{-2} \, \text{s}^{-1} \).
The decomposition of \( \text{HI} \) on a surface is typically a first-order reaction. However, if it occurs on a surface, the rate constant is often expressed in terms of surface area, not concentration. This could potentially lead to a rate constant with units of \( \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) if the surface area effects are normalized to concentration units.
Based on the analysis, the most likely answer is (D) Decomposition of \( \text{HI} \) on the surface of Gold, as it could potentially have a rate constant with units of \( \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) under certain conditions.