For the reaction: $2 A + B \rightarrow A_2 B$,the rate $= k[A][B]^2$ with $k = 2.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$. Calculate the initial rate of the reaction when $[A] = 0.1 \ mol \ L^{-1}$ and $[B] = 0.2 \ mol \ L^{-1}$. Calculate the rate of reaction after $[A]$ is reduced to $0.06 \ mol \ L^{-1}$.

The initial rate of the reaction is:
Rate $= k[A][B]^2$
$= (2.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1})(0.1 \ mol \ L^{-1})(0.2 \ mol \ L^{-1})^2$
$= 8.0 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$
When $[A]$ is reduced from $0.1 \ mol \ L^{-1}$ to $0.06 \ mol \ L^{-1}$,the concentration of $A$ reacted $= (0.1 - 0.06) \ mol \ L^{-1} = 0.04 \ mol \ L^{-1}$.
According to the stoichiometry $2 A + B \rightarrow A_2 B$,the concentration of $B$ reacted $= \frac{1}{2} \times [A]_{\text{reacted}} = \frac{1}{2} \times 0.04 \ mol \ L^{-1} = 0.02 \ mol \ L^{-1}$.
Then,the remaining concentration of $B$ is $[B] = (0.2 - 0.02) \ mol \ L^{-1} = 0.18 \ mol \ L^{-1}$.
After $[A]$ is reduced to $0.06 \ mol \ L^{-1}$,the rate of the reaction is:
Rate $= k[A][B]^2$
$= (2.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1})(0.06 \ mol \ L^{-1})(0.18 \ mol \ L^{-1})^2$
$= 3.888 \times 10^{-9} \ mol \ L^{-1} \ s^{-1} \approx 3.89 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$.