For a hypothetical reaction `A + B rarr C`, suggest the rate law and order form the following data:
`|{:("Experiment",[A] (mol L^(-1)),[B] (mol L^(-1)),"Refer of reaction" (mol L^(-1) s^(-1)),),(I,0.25,0.25,3.0 xx 10^(-3),),(II,0.50,0.25,6.0 xx 10^(-3),),(III,0.50,0.50,1.20 xx 10^(-2),):}|`

Explanation

Let the rate of the reaction is
`r = k[A]^(x)[B]^(y)`
`r_(1) = 3.0 xx 10^(-3) = k[0.25]^(x)[0.25]^(y)` …(i)
`r_(2) = 6.0 xx 10^(-3) = k[0.5]^(x)[0.25]^(y)` …(ii)
`r_(3) = 1.20 xx 10^(-2) = k[0.5]^(x)[0.5]^(y)` …(iii)
Divide equation (ii) by (i)
`(6.0 xx 10^(-3))/(3.0 xx 10^(-3)) = (k[0.5]^(x)[0.25]^(y))/(k[0.25]^(x)[0.25]^(y))`
`(2)^(1)= (2)^(x) rArr x = 1`
Divide Eq. (iii) by Eq. (ii)
`(1.20 xx 10^(-2))/(6.0 xx 10^(-3)) = (k[0.5]^(x)[0.5]^(y))/(k[0.5]^(x)[0.25]^(y))`
`:. (2)^(1) = (2)^(y) rArr y = 1`
`:. r = k[A]^(1)[B]^(1)`
Total order `= 1+1 = 2`