## Chemistry 9th Edition

Published by Cengage Learning

# Chapter 12 - Chemical Kinetics - Exercises: 28

#### Answer

$\frac{L^\frac{1}{2}}{mol^\frac{1}{2}\times s}$

#### Work Step by Step

Let the concentration of $Cl_2$ be $A$ and the concentration of $CHCl_3$ be $B$. Therefore, we are given $Rate=kA^\frac{1}{2}B$. Because the rate is always in $\frac{mol}{L\times s}$ and the concentrations are always in $\frac{mol}{L}$, our equation becomes $\frac{mol}{L\times s}=k(\frac{mol}{L})^\frac{1}{2}(\frac{mol}{L})$ or $\frac{mol}{L\times s}=k(\frac{mol}{L})^\frac{3}{2}$. Solving for $k$ gives us $k=\frac{L^\frac{1}{2}}{mol^\frac{1}{2}\times s}$.

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