Question

For the reaction:
$2A+B\to A_{2}B$, the rate $=k\left[A\right][B{]}^2$ with $k=2.0\times {10}^{-6}mo{l}^{-2}{L}^2{s}^{-1}$.

Calculate the initial rate of the reaction when $\left[A\right]=0.1mol{L}^{-1},\left[B\right]=0.2mol{L}^{-1}$. Calculate the rate of reaction after $\left[A\right]$ is reduced to $0.06mol{L}^{-1}$.

Answer 1

The reaction is $2A+B\to A_{2}B$.

The rate law expression is as follows:
Rate $=k\left[A\right][B{]}^2$

The rate constant $k=2.0\times {10}^{-6}mo{l}^{-2}{L}^2{s}^{-1}$.

When $\left[A\right]=0.1mol{L}^{-1},\left[B\right]=0.2mol{L}^{-1}$,

The initial rate of the reaction is,

Rate$=2.0\times {10}^{-6}\times \left(0.1\right)\times (0.2{)}^2=8.0\times {10}^{-9}mol{L}^{-1}{s}^{-1}$

After $\left[A\right]$ is reduced to $0.06mol{L}^{-1}$, the rate of reaction is as follows:

Rate $=2.0\times {10}^{-6}\times 0.06\times (0.18{)}^2=3.89\times {10}^{-9}mol{L}^{-1}{s}^{-1}$

[Note: A reacted $=0.1-0.06=0.04$
B reacted $=0.02$
$\left[B\right]=0.2-0.02=0.18mol{L}^{-1}$]