In the following reaction: xA to yB,log left[ | vedclass

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(B) For the reaction $xA \longrightarrow yB$,the rate of reaction $(ROR)$ is given by:$ROR = -\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt

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$2 : 1$

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(B) For the reaction $xA \longrightarrow yB$,the rate of reaction $(ROR)$ is given by:$ROR = -\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$Rearranging the terms,we get:$-\frac{d[A]}{dt} = \frac{x}{y} \frac{d[B]}{dt}$Taking the logarithm on both sides:$\log \left[ -\frac{d[A]}{dt} \right] = \log \left( \frac{x}{y} \cdot \frac{d[B]}{dt} \right)$Using the property $\log(ab) = \log a + \log b$:$\log \left[ -\frac{d[A]}{dt} \right] = \log \frac{x}{y} + \log \left[ \frac{d[B]}{dt} \right]$ ...... $(i)$Given in the question:$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$ ...... $(ii)$Comparing equation $(i)$ and $(ii)$:$\log \frac{x}{y} = \log 2$Therefore,$\frac{x}{y} = 2$,which implies $x : y = 2 : 1$.

In the following reaction: $xA \to yB$,$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$,where the negative sign indicates the rate of disappearance of the reactant. Thus,$x : y$ is:

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