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The rate law for a reaction between A an...

The rate law for a reaction between `A` and `B` is given by rate `= k[A]^(n)[B]^(m)`. On doubling the concentration of `A` and halving the concentration of `B`, the ratio of the new rate to the earlier rate of the reaction becomes

A

`(1)/(2)^(m+n)`

B

`(m+n)`

C

`(n-m)`

D

`2^(n-m)`

Text Solution

Verified by Experts

The correct Answer is:
D
Rate `= k[A]^(n)[B]^(m)`
When `A' = 2A` and `B' = 0.5 B`
The new rate becomes Rate' `= [2A]^(n)[0.5]^(m)`
`(Rate')/(Rate) = (k2^(n)[A]^(n)(0.5)^(m)[B]^(m))/(k[A]^(n)[B]^(m))`
`= 2^(n)((1)/(2))^(m) = 2^((n-m))`
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