Hydrogen peroxide decomposes into water and oxygen in a first-order process. H₂O₂(aq) → H₂O(l) + 1/2 O₂(g)
At 20.0°C,the half-life for the reaction is 3.92 × 10⁴ seconds.If the initial concentration of hydrogen peroxide is 0.52 M,what is the concentration after 7.00 days?

Question

Quizplus · Accepted Answer

The half-life of a first-order reaction is given by the equation:t_1/2 = (ln 2) / kwhere t_1/2 is the half-life, and k is the rate constant.We can rearrange this equation to solve for k:k = (ln 2) / t_1/2Substituting the given values into this equation, we get:k = (ln 2) / (3.92 × 10⁴ s) = 1.77 × 10^-5 s^-1The integrated rate law for a first-order reaction is given by the equation:ln[A]_t = ln[A]₀ - ktwhere [A]_t is the concentration of the reactant at time t, [A]₀ is the initial concentration of the reactant, k is the rate constant, and t is the time.Substituting the given values into this equation, we get:ln[H₂O₂]_t = ln(0.52 M) - (1.77 × 10^-5 s^-1)(7.00 days)(86400 s/day)= ln(0.52 M) - 10.2= -9.68Solving for [H₂O₂]_t, we get:[H₂O₂]_t = e^-9.68 = 1.2 × 10^-5 MTherefore, the concentration of hydrogen peroxide after 7.00 days is 1.2 × 10^-5 M.

Hydrogen Peroxide Decomposes into Water and Oxygen in a First-Order