In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given below:
`{:(A//M,0.20,,0.20,0.40),(B//M,0.30,,0.10,0.05),(r_(0)//M s^(-1),5.07xx10^(-5),,5.07xx10^(-5),7.6xx10^(-5)):}` Calculate the order of reaction w.rt. A and B.

Explanation

Let the order of the reaction with respect to A be x and with respect to B be y .
Therefore ,
`r_(0) = k[A]^(x) [B]^(y)`
`5.07 xx 10^(-5) = k [0.20]^(x) [0.30]^(y) " " (i)`
`5.07 xx 10^(-5) = k[0.20]^(x) [0.10]^(y) " " (ii)`
`1.43 xx 10^(-4) = k [0.40]^(x) [0.05]^(y) " " (iii)`
Dividing equation (i) by (ii) , we obtain
`(5.07 xx 10^(-5))/(5.07 xx 10^(-5)) = (k[0.20]^(x) [0.30]^(y))/(k[0.20]^(x) [0.10]^(y))`
`implies 1 = ([0.30]^(y))/([0.10]^(y))`
`implies ((0.30)/(0.10))^(0) = ((0.30)/(0.10))^(y)`
`implies y = 0`
Dividing equation (iii) by (ii) , we obtain
`(1.43 xx 10^(-4))/(5.07 xx 10^(-5)) = (k[0.40]^(x)[0.50]^(y))/(k [0.20]^(x)[0.30]^(y))`
`implies (1.43 xx 10^(-4))/(5.07 xx 10^(-5)) = ([0.40]^(x))/([0.20]^(x)) " " {:[("Since" y = 0),( [0.05]^(y) = [0.30]^(y) = 1)]:}`
` implies 2.821 = 2^(x)`
`implies "log" 2.821 = x "log" 2 " " ` (Taking log on both sides )
`implies x = ("log" 2 .821)/("log" 2)`
= `1.496`
= `1.5` (approximately )
Hence, the order of the reaction with respect to A is `1.5` and with respect to B is zero .