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NEET CHEMISTRYEasy

In the reaction, 3A→2B3A \rightarrow 2B the rate of reaction +d[B]dt+\frac{d[B]}{dt} is equal to:

A

−32d[A]dt-\frac{3}{2}\frac{d[A]}{dt}

B

−23d[A]dt-\frac{2}{3}\frac{d[A]}{dt}

C

−13d[A]dt-\frac{1}{3}\frac{d[A]}{dt}

D

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

Step-by-Step Solution

For a general reaction aA→bBaA \rightarrow bB, the rate of the reaction is expressed by dividing the rate of disappearance of reactants or appearance of products by their respective stoichiometric coefficients: Rate=−1ad[A]dt=+1bd[B]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = +\frac{1}{b}\frac{d[B]}{dt} .

Applying this to the given reaction 3A→2B3A \rightarrow 2B: Rate=−13d[A]dt=+12d[B]dt\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt}

To find the value of +d[B]dt+\frac{d[B]}{dt}, rearrange the equation by multiplying both sides by 2: +d[B]dt=−23d[A]dt+\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}.

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