For a reaction $\dpi{100} \frac{1}{2}A\rightarrow 2B$ rate of disappearance of $\dpi{100} A$ is related to the rate of appearance of $\dpi{100} B$ by the expression Option 1) $-\frac{d\left [ A \right ]}{dt}=4\frac{d\left [ B \right ]}{dt}$ Option 2) $-\frac{d\left [ A \right ]}{dt}=\frac{1}{2}\frac{d\left [ B \right ]}{dt}$ Option 3) $-\frac{d\left [ A \right ]}{dt}=\frac{1}{4}\frac{d\left [ B \right ]}{dt}$ Option 4) $-\frac{d\left [ A \right ]}{dt}=\frac{d\left [ B \right ]}{dt}$

As we learnt in

Rates in presence of stoichiometry of reactants/products -

When stoichiometry coefficients of reactants/ products are not equal to one, the rate of disappearance of  & the rate of appearance of products is divided by their respective stoichiometric coefficients

- wherein

$e.g.\:2HI(g)\rightarrow H_{2}(g)+I_{2}(g)$

$r=\frac{-1}{2}.\frac{d}{dt}[HI]$

$=\frac{+d}{dt}[H2]=\frac{+ d}{dt}[I_{2}]$

$\frac{1}{2}A\rightarrow 2B$

$-\frac{2d\left [ A \right ]}{dt}=\frac{1}{2} \frac{d\left [ B \right ]}{dt}$

or

$-\frac{d\left [ A \right ]}{dt}=\frac{1}{4} \frac{d\left [ B \right ]}{dt}$

Option 1)

$-\frac{d\left [ A \right ]}{dt}=4\frac{d\left [ B \right ]}{dt}$

Incorrect Option

Option 2)

$-\frac{d\left [ A \right ]}{dt}=\frac{1}{2}\frac{d\left [ B \right ]}{dt}$

Incorrect Option

Option 3)

$-\frac{d\left [ A \right ]}{dt}=\frac{1}{4}\frac{d\left [ B \right ]}{dt}$

Correct Option

Option 4)

$-\frac{d\left [ A \right ]}{dt}=\frac{d\left [ B \right ]}{dt}$

Incorrect Option

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