please anwer me fast ...

please anwer me fast ... has t calculate the rate of A and B have been reduced to ned the first of At

Dear Student,

The initial rate of the reaction is
Rate = k [A]²
=  (0.5 mol ^{- 1}Ls ^{- 1})(0.6 mol L ^{- 1})²
= 0.18 mol ^{- 1}Ls ^{- 1}
When [A] is reduced from 0.1 mol L ^{- 1}to 0.25 mol ^{- 1}, the concentration of A reacted = (0.6 - 0.25) mol L ^{- 1} = 0.35 mol L ^{- 1}
Therefore, concentration of B reacted= 1/2 x 0.35 mol L^-1 = 0.175 mol L ^{- 1}
Then, concentration of B available, [B] = (0.5 - 0.175) mol L ^{- 1}
= 0.325 mol L ^{- 1}
After [A] is reduced to 0.25 mol L ^{- 1}, the rate of the reaction is given by,
Rate = k [A]²
= (0.5 mol^{- 1} L  s ^{- 1}) (0.25 mol L ^{- 1})²
= 3.125 mol^{- 1} L s ^{- 1}
After [B] is reduced to 0.25 mol L - 1, the rate of the reaction is given by,
Rate = k [B]²
= (0.5 mol ^-1L  s ^{- 1}) (0.25 mol L - 1)²
= 3.125 mol^-1 L s ^{- 1}​​​​​​
Regards,