For a reaction (1)/(2)A rarr 2B, rate of...

For a reaction `(1)/(2)A rarr 2B`, rate of disappearance of `'A'` is related to the rate of apperance of `'B'` by the expression:

A

`-(d[A])/(dt) =(1)/(2) (d[B])/(dt)`

B

`-(d[A])/(dt) =(1)/(4) (d[B])/(dt)`

C

`-(d[A])/(dt)=(d[B])/(dt)`

D

`-(d[A])/(dt)=4(d[B])/(dt)`

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The correct Answer is:

B

`(1)/(2)A to 2B`
`(-2d[A])/(dt)= + (d[B])/(2dt) rArr (-d[A])/(dt)=(1)/(4)(d[B])/(dt)`

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