Question

For an elementary reaction $2A+3B \to 4C+D$ the rate of appearance of $C$ at time $‘t’$ is $2.8\times10^{-3} \,mol \,L^{-1}S^{-1}$. Rate of disappearance of $B$ at ‘$t$’ will be

• A. $\frac{4}{3}(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• B. $\frac{3}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• C. $2(2.8\times10^{-3}) mol L^{-1}S^{-1}$
• D. $\frac{1}{4}(2.8\times10^{-3}) mol L^{-1}S^{-1}$

Answer 1

Answer: (B)

The given reaction is,

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

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

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

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