In a reaction , A+Brarr Product , rate is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B ) are doubled , rate law for the reaction can be written as -

In a reaction , A+Brarr Product , rate is doubled, and rate increases

1.	In a reaction , A+Brarr Product , rate is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B ) are doubled , rate law for the reaction can be written as -
Answer» rate = k [A] [B] rate `=k [A]^(2)[B]` rate = `k [A][B]^(2)` rate `=k[A]^(2)[B]^(2)` Solution :`A+Brarr` Product `R=k[A]^(alpha)[B]^(BETA)""...[1]` `2r=k[A]^(alpha)[2B]^(beta)""...[2]` `8R=k[2A]^(alpha)[2B]^(beta)""...[3]` Now , by dividing EQUATION (1) by equation (2) we get `(r)/(2r)=(k[A]^(alpha)[B]^(beta))/(k[A]^(alpha)[2B]^(beta))` or, `(1)/(2)=(1)/(2^(beta))` or, `2^(1)=2^(beta) " or, "beta=1` Again by dividing equation (2) by equation (3) we get `(2r)/(8r)=(k[A]^(alpha).[2B]^(beta))/(k[2A]^(alpha).[2B]^(beta))` or, `(1)/(2^(2))=(1)/(2^(alpha)) " or, " 2^(2)=2^(alpha)` or,`alpha=2` So, rate equation will be : `r=k[A]^(2).[B]`