The following results have been obtained during the kinetic studies of the reaction :
`2A+B to C+D`

Determine the rate law and rate constant for the reaction.

Explanation

Rate law may be expresswed as
Rate `=k[A]^(x)[B]^(y)`
`[Rate]_(1)=6.0xx10^(-3) =k(0.1)^(x) (0.1)^(y) -(i)`
`[Rate]_(2)=7.2 xx10^(-2) =k(0.3) ^(x) (0.2)^(y)-(ii)`
`[Rate]_(3)=2.88xx10^(-1)=k(0.3)^(x)(0.4)^(y)-(iii)`
`[Rate]_(4)=2.40xx10^(-2) =k(0.4)^(x)(0.1)^(y)-(iv)`
`((Rate)_(1))/((Rate)_(4))=(6.0xx10^(-3))/(2.40xx10^(-2))=(k[0.1]^(x) [0.1]^(y))/(k[0.4]^(x)[0.1]^(y))or 1/4 =((0.1)^(y))/((0.4)^(x))=((1)/(4))^(x) x=1`
`((Rate)_(2))/((Rate)_(3))=(7.2xx10^(-2))/(2.88xx10^(-1))=(k[0.3]^(x)[0.2]^(y))/(k[0.3]^(x)[0.4]^(y))or 1/4 =((0.2)^(y))/((0.4)^(y))=[(1)/(2)]^(y)y=2`
`therefore` Rate law expression is given by Rate `=k [A][B]^(2)`
Rate constant k, can be determined by placing the values of A, B and rate of formation of D. By taking the values for experiment II.
Rate `k[A][B]^(2)`
`k=(Rate)/([A][B]^(2))=(7.2xx10^(-2)mol L^(-1)"min"^(-1))/((0.3 mol L^(-1) )(0.2 molL^(-1))^(2))=6.0 mol^(-2) L^(2)"min"^(-1)`
`therefore k=6.0 mol^(-2)L^(2) min ^(-1).`