Question

For this question, consider the reaction A-->B with a rate constant k=7.17x10^-4 m/s.

A.)In one experiment, the half life for A is determined to be 15.00 minutes. What was the initial concentration of A?

B.) In a different experiment, it takes 15.00 minutes for the concentration of A to decrease by 28%. What was the initial concentration of A in this trial?

ate dete ate law for this rance of pros Name: TA Name: Student ID: Section number: Aneta 10For this question, consider the reaction A B , with rate constant k = 7.17 x 10" M/s. -ans A) (6 points) in one experiment, the half life for A is determined to be 15.00 minutes. What was the initial concentration of A? [A] = B) (8 points) in a different experiment, it takes 15.00 minutes for the concentration of A to decrease by 28%. What was the initial concentration of A in this trial? [A=

Answer 1

lo Here the given neactian A B The mate constant on the neac i's tion Jiven as K 7.17 X10 M/S mol Ls- 7-17 x 10 Fnom the unit nate Unstant can tell that we The meac ti on is a 2eno onaer YOcaction meacti cn Fon 2e no ond en the ate law can be 9en ted integna TA] =Kt + CA = initial cmc entoation f A TAJ. at time Lone of A t CAJ = Agan = halt -Ri fe [AD. 2K

A Given Phalt lite t = 15 minutes -15 X60 900 N fnm e we have [AJ. mul-Ls 2 x 7.17 X10 -4 CA). mol-L-1 g4 =900 x 27-17 x10 - 129 mol/ 29 M Thus Hhhe initial mcen taation A in this tind is [A) 29 Aws B. given time, t= min The comcen tmation of A de ure as es 8

Thus, CAJ = [A]. - A), x 28 100 [A).-0.28 L4], 0 28 L41, 0-72 CA). Phave Now from ennation 1 De 0.72 LA.= - kt CA A 0.72 [A. kt 0.28 L4). =7-17 x10ml-L-1

7.17 X10-4 LA]. = M 2-30 M 0.28 nitial ume of Ain +his tmial 2-30 M Au CA).
For the sake of simplicity I have putted [A]o for initial concentration......