Problem 1. The reaction 2 H2O2(aq) 2 H2O(l) + O2(g) is first order in H2O2 and h

Question

Problem 1. The reaction 2 H2O2(aq) 2 H2O(l) + O2(g) is first order in H2O2 and 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 1.21 g/mL). The gaseous oxygen is collected over water at 20.0 °C as it forms. What volume of O2 forms in 80.0 seconds at a barometric pressure of 742.5 mmHg? (the vapor pressure of water at this temperature is 17.5 mmHg).

A) What would be the pH of strong acid solution containing 0.0157 M HNO3?

B)Determine the [OH–] and pH for 2.200 g of LiOH in 250.0 mL of aqueous solution.

C)Determine the concentration of Ca(OH)2 in an aqueous solution that has a pH of 10.00.

Explanation / Answer

First calculate the mass of H2O2 as follows:

=150 [ml] * 1.21 [g/ml] * 30/100 = 54.45 g.

And the molecular weight of H2O2 = 34 g/mol.

Now calculate the number of mole of H2O2 in this amount.

54.45 /34 ~ 1.60 mol.

The concentration of H2O2 = 1.60 / (150/1000) = 10.67 M

Now calculate the concentration of H2O2 after 80.0 se from first order equation as follows:

ln[H2O2] = -kt + ln[H2O2] = -0.00752 * 80.0 + ln(10.67) = -0.6016 + 2.367 =1.765

[H2O2] =5.84 mol
Therefore [H2O2] after 80.0 seconds will be 5.84 mol.

difference of H2O2 = 10.67-5.84= 4.83 mol.

The amount of oxygen = ½ x 4.83 mol = 2.415 mol.

Now use the ideal gas equation we know PV=nRT or V = nRT/P = 2.415 * 0.082 * 293/742.5

= 0.078 litres or 78 mL

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