Q71 / 200 · NEET 2002 · Chemistry

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

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

Solution

For the reaction, $\begin{gathered} 3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\ \text { Rate }=-\frac{1}{3} \frac{d[\mathrm{~A}]}{d t}=+\frac{1}{2} \frac{d[\mathrm{~B}]}{d t} \\ \therefore+\frac{d[\mathrm{~B}]}{d t}=-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t} \end{gathered}$

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

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