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  • For the given reaction: 4A  2B + C The initial concentration of A is <img alt="2

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

Option: 1

1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}


Option: 2

2.5 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}


Option: 3

2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}


Option: 4

0.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}


Answers (1)

best_answer

According to the reaction:
4A \rightarrow 2B+C 
\mathrm{ -\frac{1}{4} \times \frac{d[A]}{d t}=+\frac{1}{2} \times \frac{d[B]}{d t}}

Here,

\mathrm{-\frac{d[A]}{d t} } is the rate of consumption of A

\mathrm{+\frac{d[B]}{d t} } is the rate of formation of B

\mathrm{ \frac{d[B]}{d t}=\frac{1}{2} \times-\frac{d[A]}{d t} }

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

\mathrm{\frac{d[B]}{d t}=\frac{1}{2} \times-\frac{1.5-2}{20-0} }

\mathrm{\frac{d[B]}{d t}=1.25 \times 10^{-2} \mathrm{~mol} ~L^{-1} \mathrm{~min}^{-1} }

 

Posted by

Devendra Khairwa

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