1.

Reaction rate between two substance A and B is expressed as following, rate=k[A]^(n)[B]^(m) If the concentration of A is doubled and concentration of B is made half of initial concentration, the ratio of the new rate to the earlier rate will be,

Answer»

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

Solution :`R_(1)=K[A]^(n)[B]^(m)`
`R_(2)=k[2A]^(n)[(1)/(2)B]^(m)`
`THEREFORE (R_(2))/(R_(1))=(k[2A]^(n)[(1)/(2)B]^(m))/(k[A]^(n)B^(m))=2^(n)((1)/(2))^(m)=2^(n)*2^(-m)=2^(n-m)`


Discussion

No Comment Found

Related InterviewSolutions