The following rate data were obtained at 303K for the following reaction :
`2A+BtoC+D`
`{:("Experiment",,,[A]//"mol "L^(-1),,,[B]//"mol "L^(-1),,,"Initial rate of formation of "D//"mol "L^(-1)min^(-1)),(I,,,0*1,,,0*1,,,6*0xx10^(-3)),(II,,,0*3,,,0*2,,,7*2xx10^(-2)),(III,,,0*3,,,0*4,,,2*88xx10^(-1)),(IV,,,0*4,,,0*1,,,2*4xx10^(-2)):}`
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

From experiments I and IV, it may be noted that [B] is same but [A] has been made four times, the rate of reaction has also become four times. This means that w.r.t. A,
Rate `prop[A]" "…(i)`
From experiments II and III, it may be noted that [A] is kept same and [B] has been doubled, the rate of reaction has become four times. This means that w.r.t. B,
Rate `prop[B]^(2)" "...(ii)`
Combining (i) and (ii), we get the rate law for the given reaction as : Rate `=k[A][B]^(2)`
Thus, order w.r.t. A=1, order w.r.t. B=2 and overall order of the reaction `=1+2=3`
The rate constant and its units can be calculated from the data of each experiment using the expression.
`k=("Rate")/([A][B]^(2))=(molL^(-1)min^(-1))/((molL^(-1))(molL^(-1))^(2))=mol^(-2)L^(2)min^(-1)`
`{:("Expt.",,,k"("mol^(-2)L^(2)"min"^(-1)")"),(I,,,(6*0xx10^(-3))/(0*1xx(0*1)^(2))=6*0),(II,,,(7*2xx10^(-2))/(0*3xx(0*2)^(2))=6*0):}`
`{:("Expt.",,,k"("mol^(-2)L^(2)"min"^(-1)")"),(III,,,(2*88xx10^(-1))/(0*3xx(0*4)^(2))=6*0),(IV,,,(2*4xx10^(2))/(0*4xx(0*1)^(2))=6*0):}`
`:." Rate constant",k=6*0mol^(-2)L^(2)min^(-1)`
Alternatively, suppose order w.r.t. A is `alpha` and w.r.t. B is `beta`. Then the rate law will be Rate `=k[A]^(alpha)[B]^(beta)` Our aim is to find `alpha, beta` and k.
Substituting the values of expts I to IV, we get
`("Rate")_("expt"1)=6*0xx10^(-3)=k(0*1)^(alpha)(0*1)^(beta)" "...(i)`
`("Rate")_("expt"2)=7*2xx10^(-2)=k(0*3)^(alpha)(0*2)^(beta)" "...(ii)`
`("Rate")_("expt"3)=2*88xx10^(-1)=k(0*3)^(alpha)(0*4)^(beta)" "...(iii)`
`("Rate")_("expt"4)=2*4xx10^(-2)=k(0*4)^(alpha)(0*1)^(beta)" "...(iv)`
`:." "(("Rate")_("expt 1"))/(("Rate")_("expt 4"))=(6*0xx10^(-3))/(2*4xx10^(-2))=(k(0*1)^(alpha)(0*1)^(beta))/(k(0*4)^(alpha)(0*1)^(beta))" or "(1)/(4)=((0*1)^(alpha))/((0*4)^(alpha))=((1)/(4))^(alpha)" or "alpha=1`
`(("Rate")_("expt"1))/(("Rate")_("expt"3))=(7*2xx10^(-2))/(2*88xx10^(-1))=(k(0*3)^(alpha)(0*2)^(beta))/(k(0*3)^(alpha)(0*4)^(beta))" or "(1)/(4)=((0*2)^(beta))/((0*4)^(beta))=((1)/(2))^(beta)`
or `((1)/(2))^(2)=((1)/(2))^(beta)" or "beta=2`
`:." Rate law expression is : Rate"=k[A][B]^(2)." ""k can be calculated as in the first method."`