Question

CH 14

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 83.2 seconds at a barometric pressure of 725.2 mmHg . (The vapor pressure of water at this temperature is 17.5 mmHg)

Answer 1

The balanced chemical equation is

2 H₂O₂ (aq) -----> 2 H₂O (l) + O₂ (g)

The reaction is first order with rate constant, k = 0.00752 s^-1 at 20⁰C.

Write the rate law as

-d[H₂O₂]/dt = k[H₂O₂]

===> -d[H₂O₂]/[H₂O₂] = k.dt

The integrated rate law is

-ln [H₂O₂]│₀^t = k.[t]│₀^t

===> ln [H₂O₂]_i/[H₂O₂]_f = k.t ….(1)

Find the initial concentration of H₂O₂. We have 150.0 mL 30.0% H₂O₂ having density 1.11 g/mL.

Therefore, mass of solution = (150 mL)*(1.11 g/mL) = 166.5 g.

100 g solution contains 30.0 g H₂O₂; therefore, 166.5 g solution will contain (30.0 g)*(166.5 g/100 g) = 49.95 g H₂O₂.

Molar mass of H₂O₂ = 34.0147 g/mol.

Therefore, moles of H₂O₂ = (49.95 g)*(1 mol/34.0147 g) = 1.4685 mol.

Molar concentration of H₂O₂ = moles of H₂O₂/volume of solution in L = (1.4685 mol)/(0.150 L) = 9.79 mol/L = 9.79 M

Use equation (1) to calculate the final concentration as

ln (9.79)/[H₂O₂]_f = (0.00752 s^-1).(83.2 s) = 0.625664

===> (9.79)/[H₂O₂]_f = e^(0.625664) = 1.8695

===> [H₂O₂]_f = 9.79/1.8695 = 5.23669 ≈ 5.24

The final concentration of H₂O₂ = 5.24 mol/L.

Therefore, change in concentration of H₂O₂ = (9.79 – 5.24) mol/L = 4.55 mol/L.

Change in number of moles of H₂O₂ = moles of H₂O₂ reacted = (molar concentration)*(volume in L) = (4.55 mol/L)*(0.15 L) = 0.6825 mol.

As per the balanced stoichiometric equation, moles of O₂ produced = (0.6825 mol H₂O₂)*(1 mole O₂/2 mole H₂O₂) = 0.34125 mol.

Barometric pressure = 725.2 mmHg while vapor pressure of water at 20⁰C = 17.5 mmHg

Therefore, pressure of dry O₂ = (725.2 – 17.5) mmHg = 707.7 mmHg = (707.7 mmHg)*(1 atm/760 mmHg) = 0.931 atm.

Temperature = 20⁰C = 293 K

Use the ideal gas law, PV = nRT and plug in values:

(0.931 atm)*V = (0.34125 mol)*(0.082 L-atm/mol.K)*(293 K)

===> V = 8.806 ≈ 8.81 L

The volume of O₂ formed = 8.81 L (ans).