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For the reaction 2A + 3B rarr product, A...

For the reaction 2A + 3B `rarr` product, A is in excess and on changing the concentration of B from `0.1` M to `0.4` M, rate becomes doubled. Thus, rate law is :

A

`(dx)/(dt)=k[A]^(2)[B]^(2)`

B

`(dx)/(dt)=k[A][B]`

C

`(dx)/(dt)=k[A]^(0)[B]^(2)`

D

`(dx)/(dt)=k[B]^(1//2)`

Text Solution

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The correct Answer is:
D
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