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3A t0 2B ,rate of reaction + (d[B])/(dt)...

`3A t0 2B` ,rate of reaction `+ (d[B])/(dt)`is equal to

A

`-(3)/(2) (d[A])/(dt)`

B

`-(2)/(3)(d[B])/(dt)`

C

`+2(d[A])/(dt)`

D

`-(1)/(3)(d[A])/(dt)`

Text Solution

Verified by Experts

The correct Answer is:
b
`:-(1)/(3)(d[A])/(dt)=+(d[B])/(dt)`
`-(2)/(3)(d[A])/(dt)=+(d[B])/(dt)`
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