1.

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

Answer»

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

Solution :Earler RATE `= K a^(n) B^(m)`
New rate `= k(2A)^(n) (b)^(m)`
`("New rate")/("Earlier rate") = (2^(n) a^(n) b^(m) 2^(-m))/(a^(n)b^(m)) = 2^(n) . 2^(-m) = 2^(n-m)`


Discussion

No Comment Found

Related InterviewSolutions