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Chemistry Question on Chemical Kinetics

For a reaction 3A→2B3A \to 2B, rate of reaction +d[B]dt\frac{+d[B]}{dt} is equal to

A

−32d[A]dt -\frac{3}{2}\frac{d\left[A\right]}{dt}

B

−23d[A]dt -\frac{2}{3}\frac{d\left[A\right]}{dt}

C

−13d[A]dt -\frac{1}{3}\frac{d\left[A\right]}{dt}

D

+2d[A]dt +2 \frac{d\left[A\right]}{dt}

Answer

−23d[A]dt -\frac{2}{3}\frac{d\left[A\right]}{dt}

Explanation

Solution

3A→2B3A \to2B Rate =−13d[A]dt=+12d[B]dt=-\frac{1}{3} \frac{d \left[A\right]}{dt}=+\frac{1}{2}\frac{d\left[B\right]}{dt} or, +d[B]dt=−23d[A]dt+\frac{d\left[B\right]}{dt}=-\frac{2}{3} \frac{d \left[A\right]}{dt}