Question

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 80.9 seconds at a barometric pressure of 771.4 mmHg ? (The vapor pressure of water at this temperature is 17.5 mmHg)

Answer 1

First we will find concentration of H2O2 initially present in vessel.

volume = 150.0 ml

density = 1.11 g/ml

mass of H2O2 solution = density*volume = 1.11*150.0 = 166.5 g

H2O2 present is 30.0 % by mass

So, mass of H2O2 present in solution = 30% of 166.5 g = 0.30*166.5 = 49.95 g

Moles of H2O2 = mass/molar mass = 49.95/34 = 1.47 mol

concentration of H2O2 = moles/volume = 1.47/0.150 = 9.80 M

rate of reaction = k[H2O2]

where k = rate constant = 0.00752 s^-1

As the unit of rate constant is s-1, this means that the reaction is first order.

So, rate = 0.00752(9.80) = 0.074 M s^-1

from this rate, we can determine the concentration of oxygen formed.

rate = d[O2]/dt

dt = change in time = 80.9 s

d[O2] = change in O2 concentration

So, d[O2] = rate*dt = 0.074*80.9 = 5.99 M

As initial conc of O2 was zero. So, concentration of oxygen formed = 5.99 M

moles of O2 formed = n = 5.99*0.150 = 0.90 mol

R = gas constant = 8.314 J/K.mol

Temperature = T = 20.0 oC = 273+20 = 293 K

Now we need to calculate pressure = 771.4-17.5 = 753.9 mm Hg = 10⁵ Pa

According to ideal gas equation,

PV = nRT

V = nRT/P = 0.90*8.314*293)/10⁵ = 0.022 L = 220 ml

volume of oxygen formed = 220 ml