If 2A + 3D -> products is first-order in A and second-order in D, the rate law is: rate =

The general rate law for a reaction involving reactants A and D is given by:

$$\text{rate} = k[A]^m[D]^n$$

where $k$ is the rate constant, $[A]$ and $[D]$ are the concentrations of reactants A and D, and $m$ and $n$ are the reaction orders with respect to A and D, respectively.

The problem states that the reaction is "first-order in A" and "second-order in D".
This means:

• The order with respect to A, $m$, is 1.
• The order with respect to D, $n$, is 2.

Substitute $m=1$ and $n=2$ into the rate law equation:

$$\text{rate} = k[A]^1[D]^2$$

This simplifies to:

$$\text{rate} = k[A][D]^2$$

The derived rate law is $k[A][D]^2$.
Let's check the given options:

(a) $k[A][D]^2$

(b) $k[A]^2[D]$

(c) $k[A]^2[D]^2$

(d) $k[A][D]$

The derived rate law matches option (a).

The final answer is $\boxed{\text{(a)} k[A][D]^2}$.