rate=k[A2][B2]^2.
by how much would the rate change if:
i) [A2] is doubled?
ii) [B2] is halved?
*does the rate change equal with k?
If A2 is doubled the result doubles
if B2 is divided by 2, the result is divided by 2^2 or four
k doesn't change.
To determine how the rate would change if the concentration of reactants is altered, you can use the rate equation provided:
rate = k[A2][B2]^2
Let's consider each scenario separately:
i) If [A2] is doubled:
To assess the impact of doubling [A2], plug the new concentration into the rate equation and compare it to the original rate. Assuming all other variables are constant, you would have:
new rate = k(2[A2])([B2]^2) = 4k[A2][B2]^2
The new rate would be four times greater than the original rate. Therefore, doubling [A2] would result in a quadrupling of the rate.
ii) If [B2] is halved:
Similar to the previous scenario, substitute the new concentration into the rate equation and compare it to the original rate:
new rate = k[A2]((1/2)[B2])^2 = k[A2]((1/4)[B2]^2)
The new rate would be one-fourth of the original rate. Therefore, halving [B2] would result in a quarter of the initial rate.
Regarding your second question, the rate constant (k) remains constant under normal conditions. Therefore, the rate change is not directly equal to the rate constant. The rate constant represents the proportionality constant between the concentrations of reactants and the rate of reaction.