For a reaction `2Ararr3B,` if the rate of formation of B is `x"mol"//L,` the rate of consumption of A is

Options

A. `x`
B. `(3x)/(2)`
C. `3x`
D. `(2x)/(3)`

Answer

D

Explanation

For reaction, `2Ararr3B`
`-(1)/(2) (d[A])/(dt)=(1)/(3)(d[B])/(dt)`
`(d[B])/(dt)=xx"mol"//L`
`(d[A])/(dt)=-(2)/(3)rArr(d[B])/(dt)=-(2)/(3)x`
Hence, the rate of consumption of `A=[(-d[A])/(dt)=(2)/(3)x]`