Question

For a reaction $\frac{1}{2} A \to 2B$ rate of disappearance of $A$ is related to rate of appearance of $B$ by the expression

• A. $ \frac{-d[A]}{dt }= 4 \frac{d[B]}{dt}$
• B. $ \frac{-d[A]}{dt }= \frac{1}{4} \frac{d[B]}{dt}$
• C. $\frac{-d[A]}{dt }= \frac{1}{2} \frac{d[B]}{dt}$
• D. $ \frac{-d[A]}{dt }= \frac{d[B]}{dt}$

Answer 1

Answer: (B)

For the given chemical equation, we have

$-\frac{1}{V_{A}} \frac{d[A]}{d t}=\frac{1}{V_{B}} \frac{d[B]}{d t}$

i.e. $-\frac{1}{(1 / 2)} \frac{d[A]}{d t}=\frac{1}{2} \frac{d[B]}{d t}$

Hence, $-\frac{d \mid A]}{d t}=\frac{1}{4} \frac{d[B]}{d t}$