The following rate data were obtained at `303 K` for the following reaction:
`2A + B rarr C + D`

What is the rate law? What is the order with respect to each reactant and the overall order? Also calculate the rate constant and write its units.

Explanation

Let the order of the reaction with respect to A be x and with respect to B be y .
Therefore , rate of the reaction is given by ,
Rate = `k[A]^(x) [B]^(y)`
According to the question ,
`6.0 xx 10^(-3) = k [0.1]^(x) [0.1]^(y) " " (i)`
`7.2 xx 10^(-2) = k [0.3]^(x) [0.2]^(y) " " (ii)`
`2.88 xx 10^(-1) = k [0.3]^(x) [0.4]^(y) " " (iii) `
`2.40 xx 10^(-2) = k [0.4]^(x) [0.1]^(y) " " (iv)`
Dividing equation (iv) by (i) , we obtain
`(2.40 xx 10^(-2))/(6.0 xx 10^(-3)) = (k [0.4]^(x)[0.1]^(y))/(k [0.1]^(x) [0.1]^(y))`
`implies 4 = ([0.4]^(x))/([0.1]^(x))`
`implies 4 = ((0.4)/(0.1))^(x)`
`implies (4)^(1) = 4^(x)`
`implies x = 1 `
Dividing equation (iii) by (ii) , we obtain
`(2.88 xx 10^(-1))/(7.2 xx 10^(-2)) = (k [0.3]^(x) [ 0.4]^(y))/(k[0.3]^(x) [ 0.2]^(y))`
`implies 4 = ((0.4)/(0.2))^(y)`
`implies 4 = 2^(y)`
`implies 2^(2) = 2^(y)`
`implies y = 2 `
Therefore , the rate law is
Rate = k `[A][B]^(2)`
`implies k = ("Rate")/([A][B]^(2))`
From experiment l , we obtain
`k = (6.0 xx 10^(-3) "mol" L^(-1) "min"^(-1))/((0.1 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))`
`6.0 L^(2) "mol"^(-2) "min"^(-1)`
From experiment lll , we obtain
`k = (2.88 xx 10^(-1) "mol" L^(-1) "min"^(-1))/((0.3 "mol" L^(-1)) (0.4 "mol" L^(-1))^(2))`
`6.0 L^(2) mol^(-2) "min"^(-1)`
From experiment IV , we obtain
`k = (2 . 40 xx 10^(-2) "mol" L^(-1) "min"^(-1))/((0.4 "mol" L^(-1)) (0.1 "mol" L^(-1))^(2))`
`= 6.0 L^(2) mol^(-2) "min"^(-1)`
Therefore , rate constant , K = `6.0 L^(2) mol^(-2) "min"^(-1)`