Question

For the reaction: $2A+B\to A_{2}B$ the $rate=k\left[A\right][B{]}^2$ with $k=2.0\times {10}^{-6}{\text{mol}}^{-2}{\text{L}}^2{\text{S}}^{-1}$. If the initial concentration of $A$ and $B$ are $0.1\text{mol}{\text{L}}^{-1}$ and $0.2{\text{mol L}}^{-1}$ respectively then which of the following is /are correct:

• A. Initial rate of reaction is $8\times {10}^{-9}{\text{mol L}}^{-1}{\text{S}}^{-1}$
• B. When $A$ is reduced to $0.06mol{L}^{-1}$ then rate becomes $3.89\times {10}^{-9}mol{L}^{-1}Se{c}^{-1}$
• C. When $A$ is reduced to $0.06mol{L}^{-1}$ then $B$ will be $0.16mol{L}^{-1}$
• D. All are correct

Answer 1

Answer: (A), (B)

When $A$ is reduced to $0.06mol{L}^{-1}$ then rate becomes $3.89\times {10}^{-9}mol{L}^{-1}Se{c}^{-1}$

$Rate=k\left[A\right][B{]}^2$
$\Rightarrow Initialrate=2\times {10}^{-6}\left[0.1\right][0.2{]}^2$
$=8\times {10}^{-9}\text{mol}{\text{L}}^{-1}{\text{sec}}^{-1}$
Now as $\left[A\right]$ is reduced to $0.06{\text{mol L}}^{-1}$ i.e. $0.04{\text{mol L}}^{-1}$ of $\left[A\right]$ is reacted
$\because \text{2 moles A react with 1 mole B}.$
$\therefore \text{0.04 moles of A reacts with 0.02 moles of B}$
$\therefore [B{]}_{left}=0.2-0.02=0.18$
$\therefore rate=2\times {10}^{-6}\left[0.06\right][0.18{]}^2=3.89\times {10}^{-9}{\text{mol}}^{-2}{\text{L}}^{-1}{\text{sec}}^{-1}$