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$3 \mathrm{~A} \rightarrow 2 \mathrm{~B}$, then the rate of reaction $\frac{+d[\mathrm{~B}]}{d t}$ is equal to:
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Verified Answer
The correct answer is:
$-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}$
For the reaction,
$\begin{gathered}
3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\
\text { Rate }=-\frac{1}{3} \frac{d[\mathrm{~A}]}{d t}=+\frac{1}{2} \frac{d[\mathrm{~B}]}{d t} \\
\therefore+\frac{d[\mathrm{~B}]}{d t}=-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}
\end{gathered}$
$\begin{gathered}
3 \mathrm{~A} \rightarrow 2 \mathrm{~B} \\
\text { Rate }=-\frac{1}{3} \frac{d[\mathrm{~A}]}{d t}=+\frac{1}{2} \frac{d[\mathrm{~B}]}{d t} \\
\therefore+\frac{d[\mathrm{~B}]}{d t}=-\frac{2}{3} \frac{d[\mathrm{~A}]}{d t}
\end{gathered}$
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