For a reaction that is second order overall, what are the units of the rate constant k?

• A. s^{-1}
• B. mol L s^{-1}
• C. mol L^{-1} s^{-1}
• D. mol^{-1} L s^{-1}

Answer: (D)

Verified step by step guidance

1. Recall the general rate law for a reaction: \(\text{rate} = k [\text{A}]^m [\text{B}]^n\), where \(m + n\) is the overall order of the reaction.
2. For a second-order reaction, the overall order is 2, so the rate law can be written as \(\text{rate} = k [\text{A}]^2\) or \(k [\text{A}][\text{B}]\).
3. The units of rate are always concentration per time, typically \(\mathrm{mol\,L^{-1}\,s^{-1}}\).
4. The units of concentration are \(\mathrm{mol\,L^{-1}}\), so for a second-order reaction, the units of \(k\) must satisfy: \(\mathrm{mol\,L^{-1}\,s^{-1}} = k \times (\mathrm{mol\,L^{-1}})^2\).
5. Rearranging to solve for the units of \(k\), we get: \(k = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})^2} = \mathrm{mol^{-1}\,L\,s^{-1}}\).