The rate law for the reaction $$ \mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \operatorname{HCl}(g)+\mathrm{CCl}_{4}(g) $$ is $$ \text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right] $$ What are the units for \(k\), assuming time in seconds and concentration in \(\mathrm{mol} / \mathrm{L}\) ?

The rate law equation is given as: Rate = \(k[\mathrm{Cl}_{2}]^{\frac{1}{2}}[\mathrm{CHCl}_{3}]\) The units of concentration are mol/L, and the units of time are seconds.

Replace the terms "Rate," "Cl2" and "CHCl3" with the concentration unit (mol/L) and "time" with the unit for time (s). The equation will now be: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)^{\frac{1}{2}} \left(\frac{\mathrm{mol}}{\mathrm{L}}\right)\)

Now, we need to solve the equation to find the units of "k". First, multiply the exponents of the concentration units: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)^{\frac{1}{2}} \left(\frac{\mathrm{mol}}{\mathrm{L}}\right) = k\left(\frac{\mathrm{mol}^{\frac{1}{2}}}{\mathrm{L}^{\frac{1}{2}}}\right)\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)\) Now, we simplify by multiplying the concentration units: \( \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} = k\left(\frac{\mathrm{mol}^{\frac{3}{2}}}{\mathrm{L}^{\frac{3}{2}}}\right)\) Finally, solve for the units of "k": \( k = \frac{\mathrm{mol}}{\mathrm{L} \cdot\mathrm{s}} \times\frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{3}{2}}} = \frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{1}{2}} \cdot \mathrm{s}} \)

The units for the rate constant "k" in the given reaction are: \(\frac{\mathrm{L}^{\frac{3}{2}}}{\mathrm{mol}^{\frac{1}{2}} \cdot \mathrm{s}}\).