Question

For an elementary reaction $2A+3B\to 4C+D$ the rate of appearance of $C$ at time $‘t’$ is $2.8\times 10^{-3}mol{L}^{-1}{S}^{-1}$. Rate of disappearance of $B$ at ‘$t$’ will be

• A. $\frac{4}{3}(2.8\times 10^{-3})mol{L}^{-1}{S}^{-1}$
• B. $\frac{3}{4}(2.8\times 10^{-3})mol{L}^{-1}{S}^{-1}$
• C. $2(2.8\times 10^{-3})mol{L}^{-1}{S}^{-1}$
• D. $\frac{1}{4}(2.8\times 10^{-3})mol{L}^{-1}{S}^{-1}$

Answer 1

Answer: (B)

The given reaction is,

$2A+3B\to 4C+D$

So, $-\frac{1}{3}\frac{d[B]}{dt}=\frac{1}{4}\frac{d[c]}{dt}$

$\Rightarrow -\frac{d[B]}{dt}=\frac{3}{4}\frac{d[C]}{dt}$

$=\frac{3}{4}\left(2.8\times 10^{-3}\right)mol{L}^{-1}{S}^{-1}$