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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 of the earlier rate of the reaction will be as

A

`(m+n)`

B

`(n-m)`

C

`(2^(n-m))`

D

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

Text Solution

Verified by Experts

`"Rate"_(1) = k[A]^(n)[B]^(m), "Rate"_(2) = k[2A]^(n)[1//2B]^(m)`
`therefore ("Rate"_(2))/("Rate"_(1)) = (k[2A]^(n)[1//2B]^(m))/(k[A]^(n)[B]^(m)) = [2]^(n)[1//2]^(m) = 2^(n).2^(-m) = 2^(n-m)`
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