For a certain reaction, the rate = \(k[A]^2[B]\), when the initial concentration of A is tripled keeping the concentration of B constant, the initial rate would
Answer & explanation
Correct answer: option 4
The correct answer is option 4. increase by a factor of nine.
The given rate law is:
rate = \(k[A]^2[B]\)
According to the rate law, the rate of the reaction is directly proportional to the square of the concentration of A and the concentration of B.
If the initial concentration of A is tripled while keeping the concentration of B constant, the new concentration of A would be 3 times the initial concentration ([A]_new = 3[A]_initial). Plugging this new concentration into the rate law, we have:
\(rate_{new} = k[(3[A]_{initial})]^2[B] = k[9[A]_{initial}^2][B] = 9k[A]_{initial}^2[B]\)
As we can see, the new rate \((rate_{new})\) is 9 times the initial rate \((rate_{initial})\). Tripling the concentration of A results in a nine-fold increase in the rate of the reaction, assuming the concentration of B remains constant.
Therefore, when the initial concentration of A is tripled, keeping the concentration of B constant, the initial rate would increase by a factor of 9.