For a reaction `1//2 A rarr 2B` rate of disappearance of A is related to rate of appearance of B by the expression

Options

A. `(-d[A])/(dt)=4(d[B])/(dt)`
B. `(-d[A])/(dt) = (1)/(4)(d[B])/(dt)`
C. `(-d[A])/(dt)=1/2(d[B])/(dt)`
D. `(-d[A])/(dt)=(d[B])/(dt)`

Answer

B

Explanation

For the given chemical equation, we have
`(1)/(V_A)(d[A])/(dt) = (1)/(V_B)(d[B])/(dt)`
i.e `-(1)/(1//2)(d[A])/(dt) = (1)/(2)(d[B])/(dt)`
Hence, `-(d[A])/(dt) = 1/4 (d[B])/(dt)`