Question

A gaseous hypothetical chemical equation $2A \rightleftharpoons \, \, 4B + C$ is carried out in a closed vessel. The concentration of B is found to increase by $5 \times 10^{-3} \, mol \, L^{-1}$ in 10 s. The rate of appearance of B is

• A. $5 \times 10^{2} mol L ^{-1} sec ^{-1}$
• B. $5 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$
• C. $6 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$
• D. $4 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$

Answer 1

Answer: (A)

The given reaction is :-
$ 2 A \rightleftharpoons 4 B + C $
Rate of reaction $=\frac{-1}{2} \frac{ d [ A ]}{ dt }=\frac{1}{4} \frac{ d [ B ]}{ dt }=\frac{ d [ C ]}{ dt }$
Now, Rate of appearance of $B=\frac{\text { Increase in conc. of } B}{\text { time }}$
$ \begin{array}{l} =\frac{5 \times 10^{3}}{10} mol \cdot L ^{-1} sec ^{-1} \\ =5 \times 10^{2} mol \cdot L ^{-1} sec ^{-1} \end{array} $