Step 1: Understanding the Question:
We are given the experimental rate of a reaction, the theoretical rate law expression, and the instantaneous concentrations of the reactants. We must calculate the numerical value and units of the rate constant ($k$).
Step 2: Detailed Explanation:
The given rate law is:
$r = k [\text{A}] [\text{B}]^2$
We are provided with the following values:
Rate ($r$) = $3.6 \times 10^{-2} \text{ mol dm}^{-3}\text{s}^{-1}$
Concentration of A ($[\text{A}]$) = $0.2 \text{ M} = 0.2 \text{ mol dm}^{-3}$
Concentration of B ($[\text{B}]$) = $0.1 \text{ M} = 0.1 \text{ mol dm}^{-3}$
Rearrange the rate law to solve for the specific rate constant ($k$):
$k = \frac{r}{[\text{A}] [\text{B}]^2}$
Substitute the numerical values:
$k = \frac{3.6 \times 10^{-2}}{(0.2) \times (0.1)^2}$
$k = \frac{0.036}{0.2 \times 0.01}$
$k = \frac{0.036}{0.002}$
Multiply numerator and denominator by 1000 to simplify:
$k = \frac{36}{2} = 18$
Now, let's verify the complex units for a third-order reaction (order $1 + 2 = 3$):
Unit of $k = \frac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2}$
Unit of $k = \frac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol}^3 \text{dm}^{-9})}$
Unit of $k = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1}$.
The numerical value matches perfectly with the correct units.
Step 3: Final Answer:
The rate constant is $18\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$, matching option (a).