Hydrogen peroxide, which decomposes in a first-order reaction, has a half-life of 10.2 hours in air. How long (in hours) will it take for hydrogen peroxide to decompose to 40.8 % of its original concentration.
Added by Courtney M.
Step 1
Given: Half-life (T1/2) = 10.2 hours Convert half-life to minutes: T1/2 = 10.2 hours * 60 minutes/hour = 612 minutes Using the formula for first-order reaction: T1/2 = 0.693 / K K = 0.693 / T1/2 K = 0.693 / 612 K ≈ 0.00113 min^-1 Show more…
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Hydrogen peroxide decomposes into water and oxygen in a first-order process. H2O2 (aq) → H2O (l) + ½ 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?
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Hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}),$ decomposes to $\mathrm{H}_{2} \mathrm{O}(\ell)$ and $\mathrm{O}_{2}(\mathrm{g})$ in a reaction that is first order in $\mathrm{H}_{2} \mathrm{O}_{2}$ and has a rate constant $k=1.06 \times 10^{-3} \mathrm{min}^{-1}$ (a) How long will it take for $15 \%$ of a sample of $\mathrm{H}_{2} \mathrm{O}_{2}$ to decompose? (b) How long will it take for $85 \%$ of the sample to decompose?
Hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}),$ decomposes to $\mathrm{H}_{2} \mathrm{O}(\ell)$ and $\mathrm{O}_{2}(\mathrm{g})$ in a reaction that is first-order in $\mathrm{H}_{2} \mathrm{O}_{2}$ and has a rate constant $k=1.06 \times 10^{-3} \mathrm{min}^{-1}$ at a given temperature. (a) How long will it take for $15 \%$ of a sample of $\mathrm{H}_{2} \mathrm{O}_{2}$ to decompose? (b) How long will it take for $85 \%$ of the sample to decompose?
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