Consider the decomposition reaction: AB(g)-->A(g)+B(g). The rate law for the reaction is: rate=k[AB]^2, where k=0.00846 M^-1 min^-1. After 5.25 hours, what is the percent of AB that decomposed if the initial concentration is 2.00 M?
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Second-Order_Reactions
So if the half-life is given by Thalf=1/(k(AB))=1/.0169=59.2 min=5.32 halflives
and 5.25 hours =5.25*60min=315min=315/59.2 halflives=5.32 halflives
amount remaining/original =(1/2)^5.32=.0250
percent remain = above x 100
To calculate the percent of AB that decomposed after a certain amount of time, we need to find the concentration of AB at that time and compare it to the initial concentration.
First, let's convert the given time of 5.25 hours to minutes. There are 60 minutes in an hour, so 5.25 hours is equal to 5.25 * 60 = 315 minutes.
Next, we need to calculate the concentration of AB at 315 minutes using the given rate law and initial concentration. The rate law is given as rate = k[AB]^2, where [AB] represents the concentration of AB. Rearranging the equation, we have:
rate = k[AB]^2
k[AB]^2 = rate
[AB]^2 = rate / k
[AB] = √(rate / k)
Substituting the given values into the equation, we have:
[AB] = √(0.00846 M^-1 min^-1 / 2.00 M)
[AB] = √(0.00423 min^-1)
[AB] ≈ 0.065 M
Now we can calculate the percent of AB that decomposed. The initial concentration is 2.00 M, and at 315 minutes the concentration has decreased to approximately 0.065 M. The percent decomposition is given by:
Percent decomposition = [(initial concentration - final concentration) / initial concentration] * 100
Percent decomposition = [(2.00 M - 0.065 M) / 2.00 M] * 100
Calculating this expression:
Percent decomposition ≈ 96.75%
Therefore, approximately 96.75% of AB has decomposed after 5.25 hours.