Final answer:
The rate constant (k) for the reaction is 5.9×10^-2 (M·s)^-1. The rate law of the reaction is rate = k[AB]. The half-life of the reaction is 11.7 seconds when the initial concentration is 0.51 M. After 80 seconds, the concentrations of A and B would be 0.019 M each.
Step-by-step explanation:
The rate constant (k) for the reaction can be determined from the slope of the graph of 1/[AB] versus time. In this case, the slope is given as 5.9×10^-2 (M·s)^-1. Therefore, the value of the rate constant (k) is 5.9×10^-2 (M·s)^-1.
The rate law for the reaction can be determined based on the stoichiometry of the reaction. Since AB is converted into A and B, the rate law can be written as rate = k[AB].
The half-life (t1/2) of a reaction can be calculated using the equation t1/2 = 0.693/k, where k is the rate constant. Substituting the given initial concentration of 0.51 M into the equation gives t1/2 = 0.693/5.9×10^-2 (M·s)^-1 = 11.7 s.
If the initial concentration of AB is 0.220 M and the reaction mixture initially contains no products, the concentrations of A and B after 80 seconds can be calculated using the rate law. Assuming the reaction is first-order with respect to AB, the concentration of AB after 80 seconds is [AB] = [AB]0*e^(-kt), where [AB]0 is the initial concentration and t is the time.
Substituting the given values into the equation, the concentration of AB after 80 seconds is 0.220*e^(-5.9×10^-2*(80)) = 0.038 M. Since AB is converted into equal amounts of A and B, the concentrations of A and B after 80 seconds would be 0.038/2 M.