For a reaction \frac{1}{2}A\rightarrow 2B rate of disappearance of A is related to the rate of appearance of 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}

 

Answers (1)

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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