Question

In the reaction, $A+2B⟶6C+2D$, if the initial rate $-\frac{d\left[A\right]}{dt}$ at t=0 is $2.6\times {10}^{-2}M{sec}^{-1}$,

what will be the value of $-\frac{d\left[B\right]}{dt}$ at $t=0$?

• A. $8.5\times {10}^{-2}M{sec}^{-1}$
• B. $2.5\times {10}^{-2}M{sec}^{-1}$
• C. $5.2\times {10}^{-2}M{sec}^{-1}$
• D. $7.5\times {10}^{-2}M{sec}^{-1}$

Answer 1

Answer: (C)

$5.2\times {10}^{-2}M{sec}^{-1}$

$A+2B\to 6C+2D$

By rate law,

Rate, $r=\frac{-d\left[A\right]}{dt}=\frac{-1}{2}\frac{d\left[B\right]}{dt}=+\frac{1}{6}\frac{d\left[C\right]}{dt}=\frac{1}{2}\frac{d\left[D\right]}{dt}$ ...(1)

Given that,

$\frac{-d\left[A\right]}{dt}=2.6\times {10}^{-2}Mse{c}^{-1}$

$2.6\times {10}^{-2}=\frac{1}{2}\frac{d\left[B\right]}{dt}$

$⟹\frac{-d\left[B\right]}{dt}=2\times 2.6\times {10}^{-2}$

$=5.2\times {10}^{-2}M{s}^{-1}$.

So, the correct option is ( C )