iN A LAB, the orders of two reactants, A and B, were determined experimnetally to be 2 and 3 respectively.
a. write the rate law for this reaction.
c. what would happen to the rate of the reaction if A were Doubled?
d. what would happen to the rate of the reaction if B were doubled?
e. What would happen to the rate of the reaction if both A and B were doubled?
a. rate = k(A)^2(B)^3
c. If A is doubled 2^2 = 4 so rate is 4x.
d. If B is doubled 2^3 = 8 so rate is 8x.
e. You get the idea
To write the rate law for the reaction, we need to understand the concept of the rate equation. The rate equation expresses how the rate of a reaction is affected by the concentration of the reactants. It can be determined experimentally by observing how the rate of reaction changes when the concentrations of the reactants are altered.
a. To write the rate law for this reaction, we need to write an equation that relates the rate of the reaction to the concentrations of the reactants. Given that the orders of the reactants A and B are 2 and 3, respectively, the rate law can be expressed as follows:
rate = k[A]^2[B]^3
Here, [A] and [B] represent the concentrations of reactants A and B, respectively, and k is the rate constant.
c. If the concentration of A is doubled, we can determine the effect on the rate of the reaction using the rate law equation. Assuming all other factors remain constant, if [A] is doubled, the rate of the reaction would increase by a factor of (2^2) = 4. Therefore, the rate of the reaction would become four times faster.
d. Similarly, if the concentration of B is doubled, the rate of the reaction would increase by a factor of (2^3) = 8, based on the rate law equation. Thus, the rate of the reaction would become eight times faster.
e. If both A and B concentrations are doubled, we can apply the rate law equation again. Doubling both [A] and [B] would result in a rate increase of (2^2) * (2^3) = 4 * 8 = 32. Therefore, the rate of the reaction would become 32 times faster.