The reaction 2H_2O_2(aq) rightarrow 2H_2 O(l) + O_2(g) is first order in H_2O_2

Question

The reaction 2H_2O_2(aq) rightarrow 2H_2 O(l) + O_2(g) is first order in H_2O_2 and under certain conditions has a rate constant of 0.00752 s^-1 at 20.0 degree C. A reaction vessel initially contains 150.0 mL of 30.0% H_2O_2 by mass solution (the density of the solution is 1.11 g/mL). The gaseous oxygen is collected over water at 20.0 degree C as it forms. What volume of O_2 will form in 76.7 seconds at a barometric pressure of 743.8 mmHg . (The vapor pressure of water at this temperature is 17.5 mmHg) Express your answer with the appropriate units.

Explanation / Answer

Mass of aqueous solution = volume*density =150*1.11 g/ml =166.5 gms

Percentage of H2O2 (30%), Amount of H2O2 =166.5*30/100=49.95 gms

Molecular weight of H2O2= 2+32=34, Moles= mass/Molecular weight

Moles of H2O2, CAO= 49.95/34=1.469

For first order NA/NAO= e-kt,

NA= moles of H2O2 at any time ,t

k = rate constant

NA= 1.469* e-0.00752*76.7 =0.82514

Moles of H2O2 reacted = 1.469-0.82514=0.64386 moles

Moles of oxygen produced= 0.64386/2 =0.32193 moles

Partial pressure of oxygen= 743.8 ( Total pressure)- 17.5 (saturation pressure of water vapor= 726.3 mm Hg= 726.3/760 atm =0.9556

Temperature T= 20 deg.c =20+273.15= 293.15K

From PV= nRT, R =0.08206L.atm/mole.K

V= nRT/P= 0.32193*0.08206*293.15/0.9556 =8.1 L

=8100ml

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