In a reaction between A and B, the initial rate of reaction `(r_0)` was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?

Explanation

Rare law states that rate`=k[A]^(x) [B]^(y)`
`("Rate")_(1) = k [0.20]^(x) [0.30]^(y) = 5.07 xx 10^(-5)" "...(i)`
`("Rate")_2 = k [0.20]^(x) [0.10]^(y) = 5.07 xx 10^(-5)" "...(ii)`
`("Rate")_(3) = k [ 0.40]^(x) [0.05]^(y) = 1.43 xx 10^(-4)" "...(iii)`
On dividing Eq. (i) by Eq. (ii), we get
`(("Rate")_2)/(("Rate")_2) = (k[0.20]^(x) [0.30]^(y) )/(k[0.20]^(x) [ 0.10]^(y)) = (5.07 xx 10^(-5) )/(5.07 xx 10^(-5) )= 1`
`[3]^(y) = [3]^(0) rArr y = 0`
Now, on dividing Eq. (iii) by Eq. (ii), we get
`(("Rate")_3)/(("Rate")_2) = (k (0.40)^(x) (0.05)^(0) )/( k(0.20)^(x) (0.10)^(0) ) = (1.43 xx 10^(-4) )/( 5.07 xx 10^(-5) )`
`2.8205 = 2^(x) ((1)/(2))^(0) =2^(x)`
`log 2.8205 = x log 2 rArr x = (0.4503)/(0.3010) = 1.5`
Thus, order of the reaction with respect to A and B are 1.5 and 0, respectively.