If the rate law for a reaction is Rate = k[B]^(3/2), by what factor does the reaction rate increase if the concentration of B is doubled?

The given experimental rate law is:

$$\text{Rate}_1 = k[B]^{3/2}$$

where $[B]$ is the initial concentration of B.

If the concentration of B is doubled, the new concentration is $2[B]$. The new rate, $\text{Rate}_2$, will be:

$$\text{Rate}_2 = k(2[B])^{3/2}$$

$$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k \cdot 2^{3/2} \cdot [B]^{3/2}}{k[B]^{3/2}}$$

Cancel out $k$ and $[B]^{3/2}$:

$$\frac{\text{Rate}_2}{\text{Rate}_1} = 2^{3/2}$$

$$2^{3/2} = 2^{1} \cdot 2^{1/2} = 2\sqrt{2}$$

The reaction rate will increase by a factor of $2\sqrt{2}$.