1.

The rate law for 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 cocentration of B the ratio of the new rate to the earlier rate of reaction will be :

Answer»

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

Solution :( C) Earlier rate rate =k `a^(n) B^(m)`
New rate rate = `k(2A)^(n) (b)/(2)`
`("rate")/("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