Question

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

• A. $5\times {10}^2\text{mol}{\text{L}}^{\text{- 1}}\text{se}{\text{c}}^{\text{- 1}}$
• B. $5\times {10}^5\text{mol}{\text{L}}^{\text{- 1}}\text{se}{\text{c}}^{\text{- 1}}$
• C. $6\times {10}^{-5}\text{mol}{\text{L}}^{\text{- 1}}\text{se}{\text{c}}^{\text{- 1}}$
• D. $4\times {10}^{-4}\text{mol}{\text{L}}^{\text{- 1}}\text{se}{\text{c}}^{\text{- 1}}$

Answer 1

The correct option is A $5\times {10}^2\text{mol}{\text{L}}^{\text{- 1}}\text{se}{\text{c}}^{\text{- 1}}$

The given reaction is :-

$2A\rightleftharpoons 4B+C$

Rate of reaction $=\frac{-1}{2}\frac{d\left[A\right]}{dt}=\frac{1}{4}\frac{d\left[B\right]}{dt}=\frac{d\left[C\right]}{dt}$

Now, Rate of appearance of B $=\frac{Increaseinconc.ofB}{time}$

$=\frac{5\times {10}^3}{10}mol.L^{-1}se{c}^{-1}$

$=5\times {10}^{2}mol.L^{-1}{sec}^{-1}$