Question

For the given reaction:
$4A\to 2B+C$
The initial concentration of $A$ is $2mol{L}^{-1}$ and it decreases to $1.5mol{L}^{-1}$ after $20min$.
Then the rate of formation of $B$ is:

• A. $1.25\times {10}^{-2}mol{L}^{-1}mi{n}^{-1}$
• B. $2.5\times {10}^{-1}mol{L}^{-1}mi{n}^{-1}$
• C. $2\times {10}^{-2}mol{L}^{-1}mi{n}^{-1}$
• D. $0.5\times {10}^{-3}mol{L}^{-1}mi{n}^{-1}$

Answer 1

The correct option is A $1.25\times {10}^{-2}mol{L}^{-1}mi{n}^{-1}$

According to the reaction:
$4A\to 2B+C$
$-\frac{1}{4}\times \frac{d\left[A\right]}{dt}=+\frac{1}{2}\times \frac{d\left[B\right]}{dt}$
Here,

$-\frac{d\left[A\right]}{dt}\text{is the rate of consumption of A}$

$+\frac{d\left[B\right]}{dt}\text{is the rate of formation of B}$

$\frac{d\left[B\right]}{dt}=\frac{1}{2}\times -\frac{d\left[A\right]}{dt}$

$\frac{d\left[B\right]}{dt}=\frac{1}{2}\times -\frac{\left[A\right]-\left[A_0\right]}{t-t_0}$

$\frac{d\left[B\right]}{dt}=\frac{1}{2}\times -\frac{1.5-2}{20-0}$
$\frac{d\left[B\right]}{dt}=1.25\times {10}^{-2}mol{L}^{-1}mi{n}^{-1}$