Problem: The reaction 2H2O2(aq)→2H2O(l)+O2(g) is first order in H2O2 and under certain conditions has a rate constant of 0.00752 s−1 at 20.0 °C. A reaction vessel initially contains 150.0 mL of 30.0% H2O2  by mass solution (the density of the solution is 1.11 g/mL). The gaseous oxygen is collected over water at 20.0°C as it forms.What volume of O2 will form in 41.3 seconds at a barometric pressure of 721.1 mmHg. (The vapor pressure of water at this temperature is 17.5 mmHg)

FREE Expert Solution

Step 1. Initial moles of H₂O₂

$$\begin{gathered} \mathrm{g} \mathrm{total} \mathrm{gas}= 150 \cancel{\mathrm{mL}} \left(\frac{1.11 \mathrm{g}}{1 \cancel{\mathrm{mL}}}\right) \\ \mathrm{g} \mathrm{total} \mathrm{gas} = 166.5 \mathrm{g} \\ \mathrm{moles} {\mathrm{H}}_2{\mathrm{O}}_2 = 30\% (166.5 \cancel{\mathrm{g}})\left(\frac{1 \mathrm{mol}}{34\cancel{ \mathrm{g}}}\right)\hspace{0.17em} \\ \mathrm{moles} {\mathrm{H}}_2{\mathrm{O}}_2 =1.47 \mathrm{mol} \end{gathered}$$

Step 2. Moles of left after 41.3 seconds 

$$\begin{gathered} 1\mathrm{st} \mathrm{order} \mathrm{reaction}: \\ \ln {\left[\mathrm{A}\right]}_{\mathrm{t}} = -\mathrm{kt} +\ln \hspace{0.17em}{\left[\mathrm{A}\right]}_0 \\ \ln {\left[\mathrm{A}\right]}_{\mathrm{t}} = -\left(0.00752 \cancel{{\mathrm{s}}^{-1}}\right)\left(41.3 \cancel{\mathrm{s}}\right)+\ln \left(1.47 \mathrm{mol}\right) \\ \ln {\left[\mathrm{A}\right]}_{\mathrm{t}} =0.0747 \mathrm{mol} \\ {\left[\mathrm{A}\right]}_{\mathrm{t}} = 1.078 \mathrm{mol} \end{gathered}$$

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