The concentration of S₂O₈^2- remaining at 800 s is ________ M.

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the integrated rate law for a first-order reaction: ln([A]t/[A]0) = -kt where [A]t is the concentration of reactant at time t, [A]0 is the initial concentration of reactant, k is the rate constant, and t is time. We are given that the initial concentration of S₂O₈^2- is 0.100 M. From the data in the table, we can see that the rate constant for the reaction is 0.0050 s^-1. We can use this information to solve for the concentration of S₂O₈^2- at 800 s: ln([A]t/[A]0) = -kt ln([S₂O₈^2-]800/0.100) = -(0.0050 s^-1)(800 s) Solving for [S₂O₈^2-]800 gives: [S₂O₈^2-]800 = 0.041 M Therefore, the answer is E) 0.041.

The Concentration of S2O82- Remaining at 800 S Is ________

The Concentration of S₂O₈^2- Remaining at 800 S Is ________