For the reaction, A+Brarr"product" If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate increases 8 times . The differential rate equation of the reaction will be

For the reaction, A+Brarr"product" If concentration of A is doubled,

1.	For the reaction, A+Brarr"product" If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate increases 8 times . The differential rate equation of the reaction will be
Answer» `(DC)/(dt)-kC_(A)xxC_(B)` `(dC)/(dt)="k "C_(A)^(2)xxC_(B)^(3)` `(dC)/(dt)=kC_(A)^(2)xxC_(B)` `(dC)/(dt)=kC_(A)^(2)xxC_(B)^(2)` Solution :Let the ORDER with repect to A and B are x and y, respectively . Hence , Rate , `r=[A]^xx[B]^Y` On doubling the concentration of A, rate increases 4 times , `4r=[2A]^xx[B]^Y` From Eqs. (i) and (ii), `(1)/(4)=((1)/(2))^(x)` `:.""x=2` `:.` Order with respect to A is 2. If concentration of A and B both are doubled, `8R=[2A]^x[2B]^y` From Eqs. (i) and (ii), we get `(1)/(8)=(1)/(2)^(x).(1)/(2)^(y)""[becausex=2]` `(1)/(8)=(1)/(4xx2^(y))rArr2^(y)=2` `:.""y=1` Hence, the DIFFERENTIAL rate equation is `rprop[A]^(2)[B]^(1)""or(dc)/(dt)=kC_(A)^(2)xxC_(B)` `[Where, C_(A)andC_(B)=concentrations " of " A andB]`