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Question: If \(3A \rightarrow 2B\) then the rate of reaction of \(+ \frac{d(B)}{dt}\) is equal to...

If 3A→2B3A \rightarrow 2B then the rate of reaction of +d(B)dt+ \frac{d(B)}{dt} is equal to

A

+2d(A)dt+ 2\frac{d(A)}{dt}

B

−13d(A)dt- \frac{1}{3}\frac{d(A)}{dt}

C

−23d(A)dt- \frac{2}{3}\frac{d(A)}{dt}

D

−32d(A)dt- \frac{3}{2}\frac{d(A)}{dt}

Answer

−23d(A)dt- \frac{2}{3}\frac{d(A)}{dt}

Explanation

Solution

3A→2B3A \rightarrow 2B; Rate

=−13d[A]dt=12d[B]dt= - \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = \frac{1}{2}\frac{d\lbrack B\rbrack}{dt}; ∴+d[B]dt=−23d[A]dt\therefore + \frac{d\lbrack B\rbrack}{dt} = - \frac{2}{3}\frac{d\lbrack A\rbrack}{dt}