A gas-phase reaction has a highly specific rate constant unit of . Kinetically, plotting versus time — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

The units of $k$ ($\text{L}^2 \text{ mol}^{-2} \text{ s}^{-1}$) identify it as a third-order reaction ($1-n = -2 \Rightarrow n=3$). For a 3rd order reaction, the integrated equation is $2kt = \frac{1}{(a-x)^2} - \frac{1}{a^2}$. Rearranging into $y = mx + c$ format gives $\frac{1}{(a-x)^2} = 2kt + \frac{1}{a^2}$. The slope $m$ is exactly $2k$.