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 initial rate of the reaction is :
$Rate=k\left[A\right][B{]}^2$

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

When [A] is reduced from $0.1mol{L}^{-1}$to $0.06mo{l}^{-1}$, the concentration of A reacted$=(0.1-0.06)mol{L}^{-1}=0.04mol{L}^{-1}$

Therefore, concentration of B reacted$=\frac{1}{2}\times 0.04mol{L}^{-1}=0.02mol{L}^{-1}$

Then, concentration of B available, $\left[B\right]=(0.2-0.02)mol{L}^{-1}=0.18mol{L}^{-1}$

After [A] is reduced to $0.06mol{L}^{-1}$, the rate of the reaction is given by,

$Rate=k\left[A\right][B{]}^2=(2.0\times {10}^{-6}mo{l}^{-2}{L}^2{s}^{-1}\left)\right(0.06mol{L}^{-1}\left)\right(0.18mol{L}^{-1}{)}^2=3.89\times {10}^{-9}mol{L}^{-1}{s}^{-1}$