2H2O2(aq)→2H2O(l)+O2(g)

Hydrogen peroxide, H2O2, decomposes according to the equation above. This reaction is thermodynamically favorable at room temperature.

(a) A particulate representation of the reactants is shown below in the box on the left. In the box below on the right, draw the particulate representation of all the molecules that would be produced from these four reactant molecules.

b) Shown below is a potential energy diagram for the uncatalyzed decomposition of H2O2(aq).

(i) According to the diagram, is the decomposition reaction exothermic or endothermic? Justify your answer.

(ii) Manganese dioxide, MnO2(s), is an insoluble substance that acts as a catalyst for the decomposition reaction. On the diagram above, draw a curve to represent the reaction as it occurs in the presence of MnO2(s).

A student investigates the decomposition reaction in the laboratory. The student prepares two small beakers, adding 20.0mL of 9.77MH2O2(aq) to each one. Each beaker is placed on an electronic balance. The student adds 0.10g of MnO2(s) to the second beaker and records the mass of each beaker and its contents at 10-second intervals for one minute. The beakers and the data are shown below.

Time (seconds)

Mass of Beaker 1 + H2O2(aq) (grams)

Mass of Beaker 2 + H2O2(aq)+MnO2(s) (grams)

43.09

43.19

43.09

43.06

43.09

42.94

43.09

42.83

43.09

42.73

43.09

42.65

43.09

42.58

(i) The number of moles of O2(g) that was produced

(ii) The mass of H2O2(aq) that decomposed

(d) The student continues the experiment for an additional minute. For beaker 2, will the mass of H2O2(aq) consumed during the second minute be greater than, less than, or equal to the mass of H2O2(aq) consumed during the first minute? Explain your answer referring to the data in the table.

(e) Based on the data, the student claims that the catalyzed reaction has zeroth-order kinetics. Do you agree with the student’s claim? Justify your answer.

(f) A second student did the experiment using larger volumes of 9.77MH2O2(aq) . The student noticed that the reaction in beaker 2 proceeded extremely rapidly, causing some of the liquid to splash out of the beaker onto the lab table. The student claims that as a result of the loss of the liquid from the beaker, the calculated number of moles of O2(g) produced is greater than the actual number of moles of O2(g) produced during the first 60 seconds. Do you agree with the student? Justify your answer.

(g) The hydrogen peroxide used in this experiment can be prepared by the reaction of solid ammonium peroxydisulfate, (NH4)2S2O8, with water. The products are hydrogen peroxide (H2O2) and ammonium bisulfate (NH4HSO4). Write the balanced equation for the reaction.

The Correct Answer and Explanation is :

Answer:

(a) To show the particulate representation, you’d need to show the molecules of hydrogen peroxide (H2O2) breaking down into water (H2O) and oxygen gas (O2). If we start with four H2O2 molecules:

The 4 H2O2 molecules will break down into 4 molecules of H2O and 2 molecules of O2 (since the reaction is 2H2O2 → 2H2O + O2).
The resulting particulate representation would have four H2O molecules (each composed of two hydrogen atoms and one oxygen atom) and two O2 molecules (each composed of two oxygen atoms).

(b)

(i) Exothermic or Endothermic?

From the diagram, the reaction is exothermic because the products have lower energy than the reactants. This indicates that the reaction releases energy, which is characteristic of an exothermic reaction.

(ii) Effect of MnO2 Catalyst

The catalyst MnO2 would lower the activation energy of the reaction, so the curve representing the reaction with MnO2 should be lower than the uncatalyzed reaction curve, meaning it will reach the same products with less energy input. The catalyst speeds up the reaction without being consumed.

(c)
We need to calculate:
(i) The number of moles of O2 produced during the 60-second period.
The mass difference in Beaker 2:

Initial mass = 43.19 g
Final mass = 42.58 g
Mass loss = 43.19 g – 42.58 g = 0.61 g

The mass of O2 produced can be calculated as the mass of H2O2 that decomposed because the mass loss is the result of the oxygen gas escaping.

From the balanced equation 2H2O2 → 2H2O + O2, 34 g of H2O2 produces 32 g of O2 (because the molar mass of O2 is 32 g/mol and the molar mass of H2O2 is 34 g/mol).

The moles of O2 produced:
[
\text{moles of O2} = \frac{0.61 \text{ g O2}}{32 \text{ g/mol O2}} = 0.019 g/mol
]

(ii) The mass of H2O2 that decomposed:

We use the same proportion:
[
\text{mass of H2O2 decomposed} = \frac{0.61 \text{ g O2}}{32} \times 34 = 0.65 \text{ g H2O2}
]

(d) In the second minute, the reaction will likely proceed at a slower rate because the concentration of H2O2 in the beaker will be reduced. This will cause the reaction to slow down due to the decreased availability of reactant, meaning less H2O2 will decompose in the second minute.

(e) The student claims zeroth-order kinetics. For zeroth-order reactions, the rate does not depend on the concentration of the reactant. If the reaction rate was constant despite the decreasing concentration of H2O2 (i.e., the rate stayed the same over time), then the claim could be correct. However, based on the data, the mass loss decreases over time, indicating that the reaction is slowing down, which is not typical of zeroth-order kinetics. Therefore, I would disagree with the student’s claim.

(f) I disagree with the second student’s claim. The loss of liquid from the beaker does not affect the calculation of moles of O2 produced because the experiment only measures the mass loss of the entire contents, which accounts for the total mass, whether it stays in the beaker or splashes out. Therefore, the number of moles calculated should still be accurate.

(g) The balanced equation for the preparation of hydrogen peroxide from ammonium peroxydisulfate and water is:
[
(NH4)2S2O8 (s) + H2O (l) → H2O2 (aq) + NH4HSO4 (aq)
]

This is a redox reaction where ammonium peroxydisulfate decomposes to form hydrogen peroxide and ammonium bisulfate.

Now, I will generate the image of the reaction for part (a) based on the description.

Here is the diagram showing the decomposition of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2), as requested. Let me know if you need further explanations or additional images!

Answer:

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