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

$-\frac{1}{V_A}\frac{d[A]}{dt}=\frac{1}{V_B}\frac{d[B]}{dt}$

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

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