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Chemistry Question on Chemical Kinetics

For a reaction 12A→2B\frac{1}{2} A \to 2B rate of disappearance of AA is related to rate of appearance of BB by the expression

A

−d[A]dt=4d[B]dt{ \frac{-d[A]}{dt }= 4 \frac{d[B]}{dt}}

B

−d[A]dt=14d[B]dt{ \frac{-d[A]}{dt }= \frac{1}{4} \frac{d[B]}{dt}}

C

−d[A]dt=12d[B]dt{ \frac{-d[A]}{dt }= \frac{1}{2} \frac{d[B]}{dt}}

D

−d[A]dt=d[B]dt{ \frac{-d[A]}{dt }= \frac{d[B]}{dt}}

Answer

−d[A]dt=14d[B]dt{ \frac{-d[A]}{dt }= \frac{1}{4} \frac{d[B]}{dt}}

Explanation

Solution

For the given chemical equation, we have

−1VAd[A]dt=1VBd[B]dt-\frac{1}{V_{A}} \frac{d[A]}{d t}=\frac{1}{V_{B}} \frac{d[B]}{d t}

i.e. −1(1/2)d[A]dt=12d[B]dt-\frac{1}{(1 / 2)} \frac{d[A]}{d t}=\frac{1}{2} \frac{d[B]}{d t}

Hence, −d∣A]dt=14d[B]dt-\frac{d \mid A]}{d t}=\frac{1}{4} \frac{d[B]}{d t}