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. $5\times 10^{2}mol{L}^{-1}se{c}^{-1}$
• 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 :-
$2A\rightleftharpoons 4B+C$
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 $B=\frac{\text{ Increase in conc. of }B}{\text{ time }}$
$\begin{array}{l}=\frac{5\times 10^3}{10}mol\cdot L^{-1}se{c}^{-1}\\ =5\times 10^{2}mol\cdot L^{-1}se{c}^{-1}\end{array}$