3 ~A arrow 2 ~B , rate of reaction +(d[B])/(d t) is equal to

Correct answer: (B) -2/3 d[A]. For reaction, 3 ~A arrow 2 ~B Rate of appearance = Rate of disappearance. +1/2 (d[B])/(d t)=-1/3 (d[A])/(d t) (d[B])/(d t)=-2/3 (d[A])/(d t)

Q11 / 180 · KCET 2016 · Chemistry

$ 3 \mathrm{~A} \rightarrow 2 \mathrm{~B} $, rate of reaction $ +\frac{d[B]}{d t} $ is equal to

A
$ -\frac{3 d[A]}{2 d t} $
Correct answer
B
$ -\frac{2}{3} d[A] $
C
$ +2 \frac{d[A]}{d t} $
D
$ -\frac{1}{3} \frac{d[A]}{d t} $

Solution

For reaction, $3 \mathrm{~A} \rightarrow 2 \mathrm{~B}$ Rate of appearance $=$ Rate of disappearance. $+\frac{1}{2} \frac{d[B]}{d t}=-\frac{1}{3} \frac{d[A]}{d t}$ $\frac{d[B]}{d t}=-\frac{2}{3} \frac{d[A]}{d t}$

Hence, the correct answer is option: 'B'.

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Solution

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