In a reaction between A and B , the rate of the reaction becomes 1/4 th its initial rate if the concentration of B is doubled . Determine the order of the reaction with respect to B .

Explanation

Let `'alpha' and 'beta'` be the order of the reaction with respect to A and B respectively.
Rate of reaction , `(r)=k[A]^(alpha)[B]^(beta)` . [ k = rate constant If the initial concentrations of A and B are `[A]_(0) and [B]_(0)` respectively and the initial rate of reaction is `r_(0)` then ,
`r_(0)=k[A]_(0)^(alpha)[B]_(0)^(beta)" "...[1]`
When the concentration of B is doubled , the rate of the reaction becomes `(r_(0))/(4)` Therefore,
`(r_(0))/(4)=k[A]_(0)^(alpha)[2B]_(0)^(beta)=2^(beta)xxk[A]_(0)^(alpha)[B]_(0)^(beta)" "...[2]`
Comparing equations [1] and [2], we get `(1)/(4)=2^(beta)`
or , `(1)/(2^(2))=2^(beta) " or "2^(-2)=2^(beta) "i.e., "beta=-2`
Hence, order of the reaction with respect to B = - 2