Final answer:
The initial rate of the reaction when concentration [A] is halved and [B] is tripled would become 0.3645 m/s.
Step-by-step explanation:
The reaction rate is given as rate=[A][B]^2. Given the initial rate of 0.0810 m/s, if the concentration of [A] is halved and [B] is tripled, the new rate can be calculated using the rate law. Halving [A] would multiply the rate by 1/2, and tripling [B] would multiply the rate by 3^2 since the concentration of [B] is squared in the rate law. So, the new rate is 0.0810 m/s × (1/2) × (3^2) = 0.0810 m/s × 1/2 × 9 = 0.3645 m/s.
The given reaction is a second-order reaction with a rate law of rate = k[A][B]^2. The initial rate of the reaction is 0.0810 m/s.
If [A] is halved and [B] is tripled, the new concentrations will be halved and tripled, respectively. Plugging these new concentrations into the rate law, the new initial rate can be calculated using the proportionality of the rate to the concentrations:
new initial rate = (0.0810 m/s) * (1/2) * (3^2) = 0.60825 m/s.