What is the concentration (M)of S₂O₈^2- remaining at 1200 s?

Question

Quizplus · Accepted Answer

Since the concentration of the reactant S₂O₈^2- decreases over time, the correct answer must be positive. The reaction is first order with a rate constant of 0.015 s^-1. Using the first-order integrated rate law, we can calculate the concentration at 1200 s: ln([S₂O₈^2-]t/[S₂O₈^2-]0) = -kt ln([S₂O₈^2-]1200/[S₂O₈^2-]0) = -(0.015 s^-1)(1200 s) ln([S₂O₈^2-]1200/[S₂O₈^2-]0) = -18 [S₂O₈^2-]1200/[S₂O₈^2-]0 = e^(-18) [S₂O₈^2-]1200 = e^(-18)[S₂O₈^2-]0 [S₂O₈^2-]1200 = 0.038[S₂O₈^2-]0 Therefore, the concentration at 1200 s is 0.038 times the initial concentration, or 0.038 M. The answer is B.

What Is the Concentration (M)of S2O82- Remaining at 1200 S

What Is the Concentration (M)of S₂O₈^2- Remaining at 1200 S