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The rate law for a reaction between the ...

The rate law for a reaction between the substances 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 will be as:

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
`r=k[A]^(n)[B]^(m)`
`r.=k(2[A])^(n)(([B])/(2))^(m) = k[A]^(n)[B]^(m) xx 2^(n-m) rArr r.=2^(n-m)*r`
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