What are the units of the rate constant kf given the rate law Rate = -kf CA1 CB1/2 and the units for Rate, CA, and CB?

Step 4: Determine the units of the rate constant $k_f$.
The rate law is $Rate = -k_f C_A^m C_B^n$.
From previous steps, we have $m=1$ and $n=\frac{1}{2}$.
So, the rate law is $Rate = -k_f C_A^1 C_B^{1/2}$.

The units for each term are:

• $Rate$: $\frac{\text{mol}}{\text{L} \cdot \text{s}}$
• $C_A$: $\frac{\text{mol}}{\text{L}}$
• $C_B$: $\frac{\text{mol}}{\text{L}}$

Substitute these units into the rate law:

$$\frac{\text{mol}}{\text{L} \cdot \text{s}} = \text{Units}(k_f) \cdot \left(\frac{\text{mol}}{\text{L}}\right)^1 \cdot \left(\frac{\text{mol}}{\text{L}}\right)^{1/2}$$

Combine the concentration terms:

$$\frac{\text{mol}}{\text{L} \cdot \text{s}} = \text{Units}(k_f) \cdot \left(\frac{\text{mol}}{\text{L}}\right)^{1 + 1/2}$$ $$\frac{\text{mol}}{\text{L} \cdot \text{s}} = \text{Units}(k_f) \cdot \left(\frac{\text{mol}}{\text{L}}\right)^{3/2}$$

Now, solve for the units of $k_f$:

$$\text{Units}(k_f) = \frac{\frac{\text{mol}}{\text{L} \cdot \text{s}}}{\left(\frac{\text{mol}}{\text{L}}\right)^{3/2}}$$ $$\text{Units}(k_f) = \frac{\text{mol}}{\text{L} \cdot \text{s}} \cdot \frac{\text{L}^{3/2}}{\text{mol}^{3/2}}$$ $$\text{Units}(k_f) = \text{mol}^{1 - 3/2} \cdot \text{L}^{-1 + 3/2} \cdot \text{s}^{-1}$$ $$\text{Units}(k_f) = \text{mol}^{-1/2} \cdot \text{L}^{1/2} \cdot \text{s}^{-1}$$

The units of the rate constant $k_f$ are $\boxed{\text{mol}^{-1/2} \text{ L}^{1/2} \text{ s}^{-1}}$.